source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35797 | # How to distribute $k$ prizes in $p$ student with each student would receive $(k-1)$ prizes at most?
A teacher decided to encourage the kids by distributing prizes to them. Each of the prizes was different from the other. The total number of prize was $k$, and total number of kids was $p$. To encourage the kids, the number of prizes was more than the number of kids. But, the teacher imposed a restriction on herself that each kid would receive $(k – 1)$ prizes at the most. How many ways she could distribute the prizes?
Answer is $p^k-p$
If I understood the problem correctly, it is no-where stated that a student can get no prize at all, and the answer seems to be using this assumption,however I am a bit confused why the answer is not $p^{k-1}$?
If the restriction is that no student can get more than $(k-1)$ prizes then what is wrong with starting with $(k-1)$ distinct prizes and distributing those into $p$ distinct groups (students)?
-
Then what happens with the last prize? That one needs to be distributed too, right? But then it makes a difference if the first k-1 prizes went to a single student or more students... – N. S. Nov 6 '11 at 23:34
Suppose there were no restrictions whatsoever and the problem simply said "How can you distribute $k$ prizes among $p$ students?".
For each prize, there are $p$ possibilities for the recipient of that prize. In total, that means there are $p \cdot p \cdot \dots \cdot p$ (with $k$ copies of $p$ in the product) possible ways to assign the prizes. This is the $p^k$ part.
Now let's impose the restriction that no student receives all the prizes (this is the same as saying no student receives more than $k-1$ prizes). Of the $p^k$ configurations we just enumerated, which ones are illegal under this new restriction? Exactly $p$ of them, since there is one illegal configuration for each student (namely, giving that student all the prizes).
Subtracting the illegal configurations from our previous enumeration, we get $p^k - p$ legal configurations.
The following is multiple choice question (with options) to answer.
In a function they are distributing noble prize. In how many ways can 3 prizes be distributed among 4 boys when No boy gets more than one prize? | [
"12",
"15",
"18",
"24"
] | D | Sol. In this case, repetitions are not allowed. So, the first prize can be given in 4 ways. The second in 3 ways and the third in 2 ways. But fundamental principle, (4 x 3 x 2) ways = 24 ways
4: Or 4p = — 4:- 4X3X2X1- 24 ways
D |
AQUA-RAT | AQUA-RAT-35798 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
A boy traveled from the village to the post-office at the rate of 12.5 kmph and walked back at the rate of 2 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village. | [
"5 km",
"7 km",
"9 km",
"10 km"
] | D | Explanation :
Solution: Average speed = 2xy/(x+y) km/hr
= (2*12.5*2)/(12.5+2) km/hr
= 50/14.5 km/hr.
Total distance = (50/14.5 * 29/5) km.
= 20 km.
Required distance = 20/2 = 10 km.
Answer : D |
AQUA-RAT | AQUA-RAT-35799 | • $P_2$ will fly $\big[1-(d+r+y)\big]$ distance away from the airport in the counter-clockwise direction to meet up with $P_3$.
• At this point, $P_2$ will donate $z$ fuel to $P_3$.
• $P_2$ and $P_3$ will then both fly back $z$ distance, arriving at a distance of $1-d-r-y-z$ from the airport with no fuel.
• After refuelling at the airport, $P_1$ will fly the distance towards $P_2$ and $P_3$ and refund each of them for that much fuel. All three planes will then head back to the airport together.
From this, we must have
• $0 \leqslant s\leqslant d/3$: $P_1$ can fly $s$ distance forward and backwards, and refund $P_2$ for $s$ distance
• $z\geqslant 0$: cannot donate negative fuel
• $2x + 1-d-r-y \leqslant d+r+y$: $P_3$ must not run out of fuel before $P_2$ can reach it again
• $1-d-r-y - z \leqslant d/4$: $P_1$ can reach $P_2$ and $P_3$, refund them both, and the three of them will have enough fuel to head back to the airport
• $2x + 2s + 1-d-r-y - z\leqslant d+r+y + z$: $P_2$ and $P_3$ must not run out of fuel before $P_1$ can reach them again
Putting these together:
The following is multiple choice question (with options) to answer.
The first flight out of Phoenix airport had a late departure. If the next three flights departed on-time, how many subsequent flights need to depart from Phoenix on-time, for the airport's on-time departure rate to be higher than 40%?
I will see what is the quickest way to solve it then I will provide the explanation | [
"6",
"2",
"9",
"10"
] | B | The following approach might be the easiest one and less error prone.
We need on-time departure rate to be higher than 4/10, so it should be at least 5/11, which means that 5 out of 11 flights must depart on time. Since for now 3 out of 4 flights departed on time then 5-3=2 subsequent flights need to depart on-time.
Answer: B. |
AQUA-RAT | AQUA-RAT-35800 | We have $30$ pairs from which to choose the second pair. However, once we choose a pair, we need to get rid of $4$ other pairs with the same man (since there are $6-2=4$ women who haven't been picked yet) and the $5$ other pairs with the same woman (since there are $7-2=5$ men who haven't picked yet), leaving us with $30-1-4-5=20$ pairs left.
We have $20$ pairs from which to choose the third pair, at which point we are done. This gives us an answer of: $$42*30*20=25200$$
Thus, both methods done correctly give us the same answer.
The following is multiple choice question (with options) to answer.
There are 5 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair. what is the number of pair that can be formed. | [
"19",
"20",
"30",
"32"
] | B | First of all you should remember that there is a difference in left and right sock.
now no. of way to select any of the sock = 5
and for second = 4
so total methods = 5*4 = 20
ANSWER:B |
AQUA-RAT | AQUA-RAT-35801 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
Due to construction, the speed limit along an 4-mile section of highway is reduced from 55 miles per hour to 35 miles per hour. Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | [
"A) 3.12",
"B) 8",
"C) 2.49",
"D) 15"
] | C | Old time in minutes to cross 4 miles stretch = 4*60/55 = 4*12/11 = 4.36
New time in minutes to cross 4 miles stretch = 4*60/35 = 4*12/7 = 6.85
Time difference = 2.49
Ans:C |
AQUA-RAT | AQUA-RAT-35802 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman in his 20th innings makes a score of 90, and thereby increases his average by 2. What is his average after the 20th innings? He had never been ’not out’. | [
"52",
"37",
"39",
"43"
] | A | Average score before 20th innings
= 90 - 2 × 20= 50
Average score after 20th innings
=> 50 + 2 = 52
ANSWER:A |
AQUA-RAT | AQUA-RAT-35803 | 4. ## Re: find length of rectangle given diagonal and area
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Have a look at this webpage.
5. ## Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
The following is multiple choice question (with options) to answer.
The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle? | [
"6m",
"5m",
"2m",
"8m"
] | D | Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l2 + b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l - b = 8 - 6
= 2m.
Answer:D |
AQUA-RAT | AQUA-RAT-35804 | ## 5 Feb 2019 Enter the compounding period and stated interest rate into the effective interest rate formula, which is: r = (1 + i/n)^n-1. Where: r = The effective
1 Apr 2019 Based on the method of calculation, interest rates are classified as nominal interest rate, effective interest rate and annual percentage yield 10 Jan 2018 The simple interest rate is the interest rate that the bank charges you for taking the loan. It is also commonly known as the flat rate, nominal rate or 10 Apr 2019 The advertised rate (also known as nominal rate) is the interest the bank charges you on the sum you borrow. Note that there are different ways to 10 Feb 2019 TaxTips.ca - The effective rate of interest depends on the frequency of compounding. (e.g. 6% compounded monthly), the stated rate is the nominal rate. Interest earned on chequing and savings accounts is usually 3 Oct 2017 In this situation, with an effective interest rate of 17.2737 percent, there is very little margin for missing out on making an amortization payment. 1) If I'm given a 7% semi-annual nominal rate, does that mean the annual nominal rate is simply 14%?. No. 7% semi-annual is 3.5% every six months. So annual 27 Nov 2016 Going further, since a nominal APR of 12% corresponds to a daily interest rate of about 0.0328%, we can calculate the effective APR if this
### 19 Apr 2013 The interest rate per annum is only the nominal interest rate. This nominal rate is equal to the effective rate when a loan is on annual-rest basis
The following is multiple choice question (with options) to answer.
The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is? | [
"8.09%",
"6.07%",
"1.09%",
"6.09%"
] | D | Explanation:
Amount of Rs. 100 for 1 year when compounded half-yearly = [100 * (1 + 3/100)2] = Rs. 106.09 Effective rate = (106.09 - 100) = 6.09%
Answer: D |
AQUA-RAT | AQUA-RAT-35805 | are equal. h = a 2 b = a √ 2 L = ( 1 + √ 2 ) a S = a 2 4 h = a 2 b = a 2 L = ( 1 + 2 ) a S = a 2 4 select element Since the triangle is right angled, Hypotenuse 2 = Base 2 + Height 2 ⇒ Hypotenuse 2 = 4 2 + 4 2 ⇒ Hypotenuse … It's equal to 10.33 ft. However, since right triangles are special triangles, we will use the hypotenuse to explain the properties of right triangles. Now, in an isosceles right triangle, the other two sides are congruent. In an isosceles right triangle, the equal sides make the right angle. Find its area. Solve the isosceles right triangle whose side is 6.5 cm. For example, if we know a and b we know c since c = a. The hypotenuse is the side of the triangle opposite the right angle. The isosceles right triangle, or the 45-45-90 right triangle, is a special right triangle. Isosceles triangle calculator Isosceles triangle calculator computes all properties of an isosceles triangle such as area, perimeter, sides and angles given a sufficient subset of these properties. The ratio of the sides opposite the two 45° angles is 1:1. For any integer ≥, any triangle can be partitioned into isosceles triangles. You can find the hypotenuse: Given two right triangle legs; Use the Pythagorean theorem to calculate the hypotenuse from right triangle sides. Right isosceles Calculate area of the isosceles right triangle which perimeter is 41 cm. Isosceles triangle is a polygon with three vertices (corners) and three edges (sides) two of which are equal. If a triangle has an angle of 90° in it, it is called a right triangle. Angles are equal and we assume the equal sides to calculate the angles. Of only the right triangle has an angle of all the three.... Hypo- ‘ under ’, and the hypotenuse ( L ): 2 is. Angle β = 14.5° and leg b = 2.586 ft are displayed well. Hypoteneuse of a right triangle hypotenuse, appears solving the
The following is multiple choice question (with options) to answer.
The perimeter of an isosceles right triangle is 8 + 8 sq rt 2. What is the length of the hypotenuse of the triangle? | [
"4",
"5.656",
"5",
"6"
] | B | side of triangle is a then perimeter = a+ a +a.sqrt2 ( right angle and pythagorus) =2a +a .sqrt 2 = 8 + 8 Sqrt2 or,
a. ( 2+ sqrt2) = 8(1 + sqrt2), a= 8*(1+sqrt2)/2+sqrt2 =8*2.414/3.414 =0.707*8 then hypotenuse = 5.656
B |
AQUA-RAT | AQUA-RAT-35806 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
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Posts: 140
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Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Shekar scored 76, 65, 82, 67 and 85 marks in Mathematics, Science, Social studies, English and Biology respectively. What are his average marks? | [
"65",
"69",
"75",
"85"
] | C | Explanation :
Average= (76+65+82+67+85)/5 = 375/5 =75
Hence average=75
Answer : C |
AQUA-RAT | AQUA-RAT-35807 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
If cost of 15 eggs be 75 rupees, then find out the cost of 5 dozen eggs. | [
"300",
"400",
"500",
"600"
] | A | Cost of 15 eggs = 75 rupees
Cost of 1 egg = 75/15 = 5 rupees,
Cost of 5 dozens, i.e. 5 x 12 = 60 x 5 = 300
Answer : A |
AQUA-RAT | AQUA-RAT-35808 | ### Show Tags
09 Oct 2015, 14:30
jimjohn wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
I found an almost identical question with the same question stem, but different statements.
Here it is: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
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Posts: 80
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
07 Feb 2016, 10:19
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8803
Location: Pune, India
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
08 Apr 2016, 21:31
MrSobe17 wrote:
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
z can take any value in that case. Think of a case in which z = 12.
$$5^2 * 3 * 2^2 = 3 * x^y$$
Here x = 10
_________________
The following is multiple choice question (with options) to answer.
If 5 < x < 9 and y = x + 3, what is the greatest possible integer value of x + y? | [
"19",
"20",
"21",
"22"
] | B | x+y = x+x+3 = 2x+3
We need to maximize this value and it needs to be an integer.
2x is an integer when the decimal of x is .0 or .5
The largest such value is 8.5
Then x+y = 8.5 + 11.5 = 20.
The answer is B. |
AQUA-RAT | AQUA-RAT-35809 | ### smslca
yeah I got it from other forum.as
Let n = [m/3] ; [] mean integer part
and k = m mod 3 ; mod mean modulus (remainder of the integer divide m/3)
That is:
m --> n , k
3 --> 1, 0
4 --> 1, 1
5 --> 1, 2
6 --> 2, 0
7 --> 2, 1
8 --> 2, 2
9 --> 3, 0
...............
Then P(n,k) = n*(6*n-1) + k*(4*n+1) - 2
The following is multiple choice question (with options) to answer.
A student was asked to find the arithmetic mean of the numbers 3,11,7,9,15,13,8,19,17,21,14 and x. He found the mean to be 12. What should be the remainder in the place of x? | [
"5",
"6",
"7",
"8"
] | C | We have (3+11+7+9+15+13+8+19+17+21+14+x) / 12 = 12
137+x = 144
x = 7
Answer is C |
AQUA-RAT | AQUA-RAT-35810 | Hint: You may suppose, w.l.o.g. that $$|x-2|<\frac12$$, in which case $$|x-1|>\frac12$$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$
$$\delta<\min(1,\varepsilon)/2$$ should do the job.
The following is multiple choice question (with options) to answer.
Find the minimum value of 5cosA + 12sinA + 12 | [
"-1",
"1",
"2",
"3"
] | A | Solution:
5cosA +12sinA + 12 = 13(5/13 cosA +12/13sinA) + 12
Now, for any values of B we can get sinB =5/13 and we can replace cosB = 12/13.
We see that our assumption is right because we satisfy the condition sin^2B + cos^2B=1 so we get 13(sinBcosA +cosBsinA) + 12 =13(sin(A+B))+12. Therefore we know that minimum value of sinx=-1 and greatest is 1 so the least value of the expression becomes when sin(A+B) =-1 then the value of the whole term becomes 13.(-1) + 12 = -13 +12 = -1
Answer A |
AQUA-RAT | AQUA-RAT-35811 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average (arithmetic mean) of 4 different integers is 70. If the largest integer is 90, what is the least possible value of the smallest integer? | [
"1",
"19",
"29",
"13"
] | D | Total of Integers=70*4=280
Lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer.
But all integers are distinct. So if the largest integer is 90, then the middle 2 will be 88 and 89
Lowest of least possible integer = 280-(90+89+88)=280-267=13
Answer: D |
AQUA-RAT | AQUA-RAT-35812 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
In a school with 620 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years 9 months, then the number of girls in the school is | [
"150",
"200",
"155",
"350"
] | C | Sol.
Let the number of grils be x.
Then, number of boys = (620 - x).
Then, (11 3/4 × 620)
⇔ 11x + 12(620 - x) ⇔ x = 7440 - 7285 ⇔ 155.
Answer C |
AQUA-RAT | AQUA-RAT-35813 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
An article costing rs.300 is sold at 10% discount on a mark-up price. What is the selling price after
discount? | [
"270",
"116",
"126",
"136"
] | A | 300*90/100=270
ANSWER:A |
AQUA-RAT | AQUA-RAT-35814 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained? | [
"9:1",
"9:8",
"9:2",
"9:7"
] | A | Explanation:
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1
Answer: A |
AQUA-RAT | AQUA-RAT-35815 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
The average age of three boys is 15 years. If their ages are in the ratio 3:5:7 the average age of the young boy is ______ years? | [
"18",
"8",
"9",
"19"
] | C | avg of three boys is 15 years given ratio of their ages are 3:5:7
let x be the age
(3x+5x+7x)/3 =15 implies x=3
then the age of the young boy is 3x3=9
ANSWER:C |
AQUA-RAT | AQUA-RAT-35816 | Starting from the two guys sitting next to each other, we have M M F M F M F M F M F
Those six guys can be rearranged in those six slots, and no arrangement will be counted twice --- the clump of two guys breaks the symmetry, so each arrangement is unique. That's a permutation of six individuals, 6P6 = 720.
Then the females be arranged in those five slots --- again, since the two guy group breaks the symmetry, any arrangement of the women will be unique. That's 5P5 = 120.
By the Fundamental Counting Principle, we multiply these to get the total number of seating arrangements.
Total number of possibilities = (720)*(120) = 86400, answer C.
Notice, this question would become considerably more difficult if we merely changed it to seven guys and five woman --- now twelves total seats, and it's still the case that no two women can sit together.
Now, we could have a clump of three guys sitting together, and then all the other guys alternating with women ---- OR, we could have two clumps of two guys next to each other, and these two clumps could be distributed at different positions around the table --- we would have to count all those arrangements, and then for each one, figure out all the permutations of males & females. A much trickier question.
Does all this make sense?
Mike
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Re: There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]
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The following is multiple choice question (with options) to answer.
The number of ways in which six men and six women can be seated in a row, so that no two girls sit together is? | [
"6! * ⁷P₆",
"5! * 7",
"6! * 7",
"4! * 4"
] | A | We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six women to be arranged. This can be done in ⁷P₆ ways.
Hence required number of ways = 6! * ⁷P₆
A |
AQUA-RAT | AQUA-RAT-35817 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
Medals are to be awarded to three teams in a 10-team competition. If one medal is gold, one medal is silver, and one medal is bronze, how many different ways are there to award the three medals to teams in the competition? | [
"10!/7!",
"10!/(3!7!)",
"10!/3!",
"7!/3!"
] | A | Choosing 3 teams out of 10 when order of the teams matters - P310=10!7!
Or: choosing which 3 teams out of 10 will get the medals - C310C103 and arranging them - 3!3!, so total - C310∗3!=10!7!
Or:
1-2-3-4-5-6-7-8-9-10 (teams);
G-S-B-N-N-N-N-N-N-N (GSB - medals, N - no medal);
Permutation of 10 letters out of which 7 N's are identical is 10!7!10!7! (so you'll get 10!7!10!7! different ways of assigning the medals to the teams).
Answer: A. |
AQUA-RAT | AQUA-RAT-35818 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a 1000 m race, A beats B by 40 m and B beats C by 100 m. In the same race, by how many meters does A beat C? | [
"145 m",
"136 m",
"268 m",
"129 m"
] | B | By the time A covers 1000 m, B covers (1000 - 40) = 960 m.
By the time B covers 1000 m, C covers (1000 - 100) = 900 m.
So, the ratio of speeds of A and C =
1000/960 * 1000/900 = 1000/864. So, by the time A covers 1000 m, C covers 864 m.
So in 1000 m race A beats C by 1000 - 864 = 136 m.
Answer:B |
AQUA-RAT | AQUA-RAT-35819 | # How many of the 9000 four digit integers have four digits that are increasing?
How to find the number of distinct four digit numbers that are increasing or decreasing?
The correct answer is $2{9 \choose 4} + {9 \choose 3} = 343$. How to get there?
-
For every collection of $4$ distinct digits, there is a unique way to arrange the digits so that they are decreasing, and a unique way to arrange them so that they are increasing. This gives $2 \binom{10}{4}$ different sequences that are increasing or decreasing, but not all correspond to a four-digit number. For increasing sequences, you cannot have $0$ as a first digit. This rules out $\binom{9}{3} = 84$ possibilities. So the total is $2(210) - 84=336 \neq 343$...?
-
The analysis that was used goes as follows:
Not using $0$: We choose $4$ non-zero digits. Once we have done that, we can arrange them in increasing order in $1$ way, and in decreasing order in $1$ way, for a total of $2\binom{9}{4}$.
Using $0$: They can only be decreasing. And we need to choose $3$ non-zero digits to go with the $0$. This can be done in $\binom{9}{3}$ ways.
I prefer Juanito's approach. Note that the sum is not $343$, so if the book got that, there is a computational error.
The following is multiple choice question (with options) to answer.
find how many 4 digits no. is less then 1239 such that all 4 digits are distinct. | [
"68",
"69",
"70",
"71"
] | B | case 1- 1000-1099- there are 56 distinct digits
case 2- 1200-1209- there are 7 distinct digits
case 3- 1230-1239(excluding 1239)- there are 6 distinct digits
hence, 56+7+6=69
ANSWER:B |
AQUA-RAT | AQUA-RAT-35820 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A fruit seller sells mangoes at the rate of Rs.14 per kg and thereby loses 15%. At what price per kg, he should have sold them to make a profit of 15%? | [
"Rs.11.81",
"Rs.12",
"Rs.18.94",
"Rs.12.31"
] | C | Solution
85 : 14 = 115 : x
x= (14×115/85)
= Rs.18.94
Hence, S.p per Kg = Rs.18.94
Answer C |
AQUA-RAT | AQUA-RAT-35821 | The answer is: 133,320.
List the $4! = 24$ permutations of the four digits
,. . and consider their sum.
. . $\begin{array}{cc}
&2468 \\ &2486 \\ &2648 \\ &2684 \\ &2846 \\ &2864 \\ &\vdots \\
&8246 \\ &8264 \\ &8426 \\ &8462 \\ &8624 \\ + & 8642 \\ \hline
\end{array}$
We find that each column has: six 2's, six 4's, six 6's, six 8's.
The total of each column is: . $6\!\cdot\!2 + 6\!\cdot\!4 + 6\!\cdot\!6 + 6\!\cdot\!8 \:=\:120$
Hence, the addition has the form:
. . $\begin{array}{cccccc}
&&& 1 & 2 & 0 \\ && 1 & 2 & 0 \\ & 1 & 2 & 0 \\ 1 & 2 & 0 \\ \hline 1 & 3 & 3 & 3 & 2 & 0 \end{array}$
Therefore, the sum is: . $133,\!320$
Edit: Plato beat me to it . . . *sigh*
.
The following is multiple choice question (with options) to answer.
How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits F equal to 28? | [
"14",
"24",
"28",
"48"
] | D | First, look for all 4 digits without repeat that add up to 28. To avoid repetition, start with the highest numbers first.
Start from the largest number possible 9874.
Then the next largest number possible is 9865.
After this, you'll realize no other solution. Clearly the solution needs to start with a 9 (cuz otherwise 8765 is the largest possible, but only equals 26). With a 9, you also need an 8 (cuz otherwise 9765 is the largest possible, but only equals 27). With 98__ only 74 and 65 work.
So you have two solutions. Each can be rearranged in 4!=24 ways. So F=24+24=48.D |
AQUA-RAT | AQUA-RAT-35822 | ### Show Tags
13 Mar 2015, 20:17
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
So far if the average of the two numbers is an INTEGER they can be written in (a+b)(a-b) form . so that narrows us down to Odd + Odd and Even+Even cases .
Special consideration need to taken for those cases in which one number is ODD and other is multiple of 4 , i.e. in this case
if the set is $${ 2,5,7,8 }$$, then possible pairs are :
7*8 = 56 = 14*4 = 28*2 none of these pairs (7,8) , (14,4), and (28*2) can be expressed in (a+b) (a-b) form .
5*8= 40 = 10*4 = (7+3) (7-3), so yes we can write $$5*8$$ as $$(7+3) * (7-3)$$
2,5,7,8
total number of cases = 4C2 = 6
favorable cases : (odd,odd) (5,7) , (Even,Even) (2,8) , and one special case as shown above (5,8) so $$3/6=1/2$$
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The following is multiple choice question (with options) to answer.
How many unique positive odd integers less than 80 are equal to the product of a positive multiple of 5 and an odd number? | [
"4",
"8",
"11",
"12"
] | B | The Question basically asks how many positive odd integers less than 80 are odd multiples of 5
So we have 5,15,25,35,45,55, 65,75
=8
B |
AQUA-RAT | AQUA-RAT-35823 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row downstream at 18 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively? | [
"13,3",
"12,6",
"15,3",
"14,4"
] | D | Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 18 --- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 28 => x = 14, y = 4.
ANSWER:D |
AQUA-RAT | AQUA-RAT-35824 | filters
a6 = 128 + (224 *w2 * w1 * w3) - (56 *w2 * w1*w1 * w3) + (448 *w2 * w1 * w4) - (112 *w2 * w1*w1 * w4) + (448* w1 * w3 * w4) + (224 *w2 * w3 * w4) - (640 *w1) - (320 *w2) - (320 *w3) - (640 *w4) + (64 *w2 * w1) + (112 *w2 * w1*w1) + (32 *w1*w1) - (224 *w2 * w1 * w3 * w4) - (168 *w2 * w1*w1 * w3 * w4) + (64 *w1 * w3) + (32 *w2 * w3) + (112 *w1*w1 * w3) + (128 *w1 * w4) + (64 *w2 * w4) + (64 *w3 * w4) + (224 *w1*w1 * w4) - (112 *w1*w1 * w3 * w4) + (32 *w4*w4) + (224 *w4*w4 * w1) - (56 *w4*w4 * w1*w1) - (168 *w4*w4 * w1 * w2 * w3) + (210 *w4*w4 * w1*w1 * w2 * w3) - (112 *w4*w4 * w1 * w2) - (84 *w4*w4 * w1*w1 * w2) + (112 *w4*w4 * w2) + (112 *w4*w4 * w3) - (112 *w4*w4 * w1 * w3) - (84 *w4*w4 * w1*w1 * w3) - (56 *w4*w4 * w2 * w3)
The following is multiple choice question (with options) to answer.
Factor: 6x4y3 – 96y3 | [
"A) 3y3(x2 + 4)(x + 2)(x -2)",
"B) 6y3(x2 + 4)(x + 2)(x -2)",
"C) 3y3(x2 + 4)(x + 2)(x -3)",
"D) 3y3(x2 + 4)(x + 3)(x -2)"
] | B | 6x4y3– 963.
= 6y3(x4 – 16).
= 6y3[(x2)2 - 42].
= 6y3(x2 + 4)(x2 - 4).
= 6y3(x2 + 4)(x2 - 22).
= 6y3(x2 + 4)(x + 2)(x -2).
Answer: (B) 6y3(x2 + 4)(x + 2)(x -2) |
AQUA-RAT | AQUA-RAT-35825 | Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.
Hence your final answer is $$3!\cdot 10!\cdot \frac{4!}{2!2!}$$
Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.
Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.
So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$
For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
The following is multiple choice question (with options) to answer.
Seven children — A, B, C, D, E, F, G and H— are going to sit in seven chairs in a row. The children CF have to sit next to each other, and the others can sit in any order in any remaining chairs. How many possible configurations are there for the children? | [
"600",
"720",
"1440",
"10080"
] | D | for such questions , we can treat both as one then total person=7..
these 7 can be arranged in 7! ways.. but within these 7, one consists of two people who can be arranged in 2 ways CF or FC.. so ans =7!*2=10080
ans D |
AQUA-RAT | AQUA-RAT-35826 | java, beginner, homework, palindrome
int Palindrome = reverse(input);
if (Palindrome == input){
return true;
}
else
return false;
}
public static void main(String[] args) {
int integer = 0;
Scanner input = new Scanner(System.in);
System.out.print("Enter a positive, multi-digit integer: ");
integer = input.nextInt();
while (integer <= 9 && integer > 0)
{
System.out.println(integer + " is a single digit. Please re-enter another integer: ");
integer = input.nextInt();
if (isPalindrome(integer) && (integer > 0 && integer > 9))
{
System.out.println(integer + " is a palindrome");
return;
}
else if (!isPalindrome(integer) && (integer > 0 && integer > 9))
{
System.out.println(integer + " is not a palindrome");
return;
}
The following is multiple choice question (with options) to answer.
A “palindromic integer” is an integer that remains the same when its digits are reversed. So, for example, 43334 and 516615 are both examples of palindromic integers. How many 6-digit palindromic integers are both even and greater than 600,000? | [
"150",
"200",
"240",
"300"
] | B | The first digit and last digit are the same so the 2 possibilities are 6 or 8.
The second and third digits can be any number from 0 to 9.
The total number of palindromic integers is 2*10*10 = 200
The answer is B. |
AQUA-RAT | AQUA-RAT-35827 | Seven draws
The first six draws are Others: . ${9\choose6}\text{ ways } \hdots\:P(\text{6 Others}) \:=\:\frac{{9\choose6}}{{15\choose6}}$
And the 7th is a 75w: . $\frac{6}{9}$
. . Hence: . $P(\text{7 draws}) \;=\;\frac{{9\choose6}}{{15\choose6}}\left(\frac{6 }{9}\right)$
Eight draws
The first seven draws are Others: . ${9\choose7}\text{ ways }\hdots\:P(\text{7 Others}) \:=\:\frac{{9\choose7}}{{15\choose7}}$
And the 8th is a 75w: . $\frac{6}{8}$
. . Hence: . $P(\text{8 draws}) \;=\;\frac{{9\choose7}}{{15\choose7}}\left(\frac{6 }{8}\right)$
Nine draws
The first eight draws are Others: . ${9\choose8}\text{ ways }\hdots\:P(\text{8 Others}) \:=\:\frac{{9\choose8}}{{15\choose8}}$
And the 9th is a 75w: . $\frac{6}{7}$
. . Hence: . $P(\text{9 draws}) \;=\;\frac{{9\choose8}}{{15\choose8}}\left(\frac{6 }{7}\right)$
The following is multiple choice question (with options) to answer.
The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 19 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12 | [
"I only",
"II only",
"III only",
"I and III only"
] | B | B
for 7 we will have mean as 7.16(approx) and median as 7 but for 3 we will have mean and median = 6.5 and for 12 we will have median and mean =8 |
AQUA-RAT | AQUA-RAT-35828 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A company bought for its 7 offices 3 computers of brand N and 3 computers of brand M. In how many ways could computers be distributed among the offices if each office can have only 1 computer. | [
"196",
"140",
"256",
"292"
] | B | This problem has to do withcombinations. Here's the general idea: if you have a set of n elements, and you are going to choose r of them (r < n), then the number of combinations of size r one could choose from this total set of n is:
# of combinations = nCr = (n!)/[(r!)((n-r)!)]
where n! is the factorial symbol, which means the product of every integer from n down to 1. BTW, nCr is readn choose r.
In this problem, let's consider first the three computers of brand M. How many ways can three computer be distributed to seven offices?
# of combinations = 7C3 = (7!)/[(3!)(4!)] = (7*6*5*4*3*2*1)/[3*2*1)(4*3*2*1)]
= (7*6*5)/(3*2*1) = (7*6*5)/(6) = 7*5 = 35
There are 35 different ways to distribute three computers to 7 offices. (The massive amount of cancelling that occurred there is very much typical of what happens in the nCr formula.)
One we have distributed those three M computers, we have to distribute 3 N computers to the remaining four offices. How many ways can two computer be distributed to four offices?
# of combinations = 4C3 = 4
For each of the 35 configurations of distributing the M computers, we have 4 ways of distributing the N computers to the remaining offices. Thus, the total number of configurations is 35*4 = 140. Answer choice =B |
AQUA-RAT | AQUA-RAT-35829 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A bookstore has a shelf that contains biographies which normally sell for $20 each and mysteries that normally sell for $12 each. During a sale, the biographies and mysteries are discounted at different rates so that a customer saves a total of $19 from the normal price by buying 5 discounted biographies and 3 discounted mysteries. If the sum of the discount rates for the two types of books is 27 percent, what is the discount rate on mysteries? | [
"10%",
"11%",
"12.5%",
"13%"
] | C | Let B be the discount on biographies and M be the discount on mysteries
So., B + M = 0.27 -----(1)
And
(20*5 + 12*3) - (20*5*(1-B) + 12*3*(1-M)) = 19
--> 100(1-(1-B)) + 36(1-(1-M) = 19
100B + 36M = 19 ------(2)
Solving 12., we get M = 0.125 = 12.5%
C |
AQUA-RAT | AQUA-RAT-35830 | I have to design a laboratory activity to answer the question, "What is the relationship between the diameter of a circle and the area of the circle?" My teacher said to use a previous activity as a templete to help me. For the
6. ### Math
Two perpendicular diamters(cutting each other at right angles) of a circle cut the circumference at four(4) points. A square is formed by joining the 4 points. If the circumference of the circle is 132cm, find the area of the
7. ### Algebra
The Circumference and area of a circle of radius r are givin by 2 [pie] r and [pie] r[2], respectively use 3.14 for the constant [pie] A. What is the circumference of a circle with a radius of 2 m? B.What is the area of a circle
8. ### Geometry
A circle with a radius of 1/2 ft is dilated by a scale factor of 8. Which statements about the new circle are true? Check all that apply. A.The length of the new radius will be 4 feet. B.The length of the new radius will be 32
9. ### Math
23.What is the approximate circumference of a circle with a radius of 6 centimeters? A.12 B.18 C.24 D.36 D 24.What is the length of the diameter if the radius is 15? A.5 B.15 C.30 D.45 C 27.What is the circumference of the circle
10. ### math
suppose you copy the circle using a size factor of 150%.what will bet he radius,diameter,circumference,and the area of the image? radius will be 1.5 times as big diameter will be 1.5 times as big but the formula for area will be
11. ### area of circle
My circle has a radius of 8ft. What is the distance of the circle? What "distance" are you talking about? Diameter? Circumference? What is the area of a circle with the radius of 32 cm????
More Similar Questions
The following is multiple choice question (with options) to answer.
If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ? | [
"Area is quadrupled",
"Area is tripled",
"Area is doubles",
"Area become half"
] | A | Explanation:
2πR1=4π
=>R1=22πR2=8π
>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π
So the area quadruples.
Option A |
AQUA-RAT | AQUA-RAT-35831 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A man rows his boat 75 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream? | [
"5 kmph",
"6 kmph",
"9 kmph",
"8 kmph"
] | B | Speed downstream = d/t = 75/(2 1/2) = 30 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (30 - 18)/2 = 6 kmph
Answer:B |
AQUA-RAT | AQUA-RAT-35832 | 2000 small packets were sold.
1 8
Q:
The angle between the minute hand and the hour hand of a clock when the time is 4.15 is
A) 0 B) 37.5 C) 27 D) 15
Explanation:
Angle between hands of a clock
When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H 'o clock
=> $\fn_jvn \small 30\left ( H -\frac{M}{5} \right )+\frac{M}{2}$ degrees
Here H = 4, M = 15 and the minute hand is behind the hour hand.
Hence the angle
$\fn_jvn \small 30\left ( H -\frac{M}{5} \right )+\frac{M}{2}$ = 30[4-(15/5)]+15/2 = 30(1)+7.5 = 37.5 degrees
The following is multiple choice question (with options) to answer.
The length of minute hand of a clock is 5.4 cm. What is the area covered by this in 10 minutes | [
"15.27",
"16.27",
"17.27",
"18.27"
] | A | area of circle is pi*r^2
but in 10 minutes area covered is (10/60)*360=60 degree
so formula is pi*r^2*(angle/360)=3.14*(5.4^2)*(60/360)=15.27 cm^2
ANSWER:A |
AQUA-RAT | AQUA-RAT-35833 | # Remainder on division with $22$
What is the remainder obtained when $$14^{16}$$ is divided with $$22$$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $$22$$, except the remainder?
How should I proceed?
• $14^{16} = (22 - 8)^{16}$ – ab123 Jul 3 '18 at 10:37
• $8^{16} = 64^8 = (66 - 2)^8$ – ab123 Jul 3 '18 at 10:41
• @ab123 Why not make an answer of that? – Arthur Jul 3 '18 at 10:41
You can use binomial expansions and see that $$14^{16} = (22 - 8)^{16}$$ implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$ Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 \implies 2^8 = 256 \text{ divided by } 22 \implies \text{remainder = 14}$
Since $$14^2 \equiv -2$$ so $$14^{16} \equiv (-2)^8 \equiv 16^2\equiv (-6)^2 \equiv 14$$
or
$$14^2 = 22k -2$$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22\cdot 11+14$$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7\cdot14^{15}$ by $11$.
The following is multiple choice question (with options) to answer.
If the remainder is 12 when the integer n is divided by 22, what is the remainder when 2n is divided by 11? | [
"0",
"2",
"3",
"6"
] | B | n = 22k+12
2n = 2(22k+12) = 4*11k + 24 = 4*11k + 2*11 + 2 = 11j+2.
The answer is B. |
AQUA-RAT | AQUA-RAT-35834 | # Math Help - Prob. Question
1. ## Prob. Question
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
2. Originally Posted by I-Think
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
If exactly three boys have their original books then the other two books are switched. Do you see how this helps?
3. Hello, I-Think!
Five boys place their books in a bag.
The books are are drawn out in random order and given back to the boys.
What is the probability that exactly 3 boys will receive their original book?
The only way I could think of to solve this problem is to
list the number of ways 3 boys could receive their original books.
You don't have to list them . . .
Select three of the five boys who will get their own books.
. . There are: . $_5C_3 \:=\:{5\choose3} \:=\:10$ ways.
The other two boys have simply switched books: . $1$ way.
Hence, there are: . $10\cdot1 \:=\:10$ ways for 3 boys to get their own books.
There are: . $5! \,=\,120$ ways to return the books.
The following is multiple choice question (with options) to answer.
A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. These book are occasionally loaned out through an inter-library program. If, by the end of the month, 70 percent of books that were loaned out are returned and there are 66 books in the special collection at that time, how many books of the special collection were loaned out during that month? | [
"20",
"30",
"35",
"40"
] | B | Total = 75 books.
65% of books that were loaned out are returned --> 100%-70%=30% of books that were loaned out are NOT returned.
Now, there are 66 books, thus 76-68=7 books are NOT returned.
{loaned out}*0.30=7 --> {loaned out}=30.
Answer: B. |
AQUA-RAT | AQUA-RAT-35835 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A is twice as good a workman as B and they took 9 days together to do the work B alone can do it in? | [
"25 days",
"88 days",
"21 days",
"27 days"
] | D | WC = 2:1
2x + x = 1/9
x = 1/27 => 27 days
Answer: D |
AQUA-RAT | AQUA-RAT-35836 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
When 2/3 of the garments in the shipment were inspected, 18 of the garments passed inspection and the remaining 2 garments failed. How many of the uninspected garments must pass inspection in order that 70 percent of the garments in the shipment pass? | [
"10",
"3",
"8",
"7"
] | B | B = 9
2/3x= 20
x=30
for 90% approval we need 21 garments approved.
already approved = 18
we need 3 more.
Answer : B |
AQUA-RAT | AQUA-RAT-35837 | 3 BIOGRAPHIES books: $\dfrac{4\cdot 3 \cdot 2}{32!} \cdot 6 =24$
4 BIOGRAPHIES books: 1
then the result is 115.
-
Suppose the biographies are of $A$, $B$, $C$, and $D$. Among the ways you counted when initially you chose two biographies, there were the biographies of $A$ and of $B$. Among the choices you counted when you chose two more books was the biography of $C$ and novel $N$. So among the choices counted in your product was choosing $A$ and $B$, then choosing $C$ and $N$. That made a contribution of $1$ to your $168$.
But among the choices you counted when you chose two biographies, there were the biographies of $A$ and $C$. And among your "two more" choices, there was the biography of $B$ and novel $N$. So among the choices counted in your product, there was the choice of $A$ and $C$, and then of $B$ and $N$. That made another contribution of $1$ to your $168$.
Both of these ways of choosing end us up with $A$, $B$, $C$, and $N$. So does choosing $B$ and $C$ on the initial choice, and $A$ and $N$ on the next. Still another contribution of $1$ to your $168$.
So your product counts the set $\{A, B, C, N\}$ three times. This it does for every combination of three biographies and one novel. It also overcounts the set $\{A, B, C, D\}$.
One could adjust for the overcount. In some problems that is a useful strategy. Here it takes some care.
But a simple way to solve the problem is to count separately the ways to choose two bios, two novels; three bios, one novel; four bios, no novels and add up.
-
The following is multiple choice question (with options) to answer.
David has d books, which is 1/2 times as many as Jeff and 3 as many as Paula. How many books do the three of them have altogether, in terms of d? | [
" 5/6*d",
" 10/3*d",
" 7/3*d",
" 7/2*d"
] | B | Although we could plug in a real value for d, the problem can be just as easily solved by setting up equations. However, let’s start by defining some variables. Since we are given that David has d books, we can use variable d to represent how many books David has.
number of books David has = d
number of books Jeff has = j
number of books Paula has = p
We are given that David has 3 times as many books as Jeff. We can now express this in an equation.
d = 3p
d/3 = p
We are also given that David has ½ as many books as Paula. We can also express this in an equation.
d = (1/2)j
2d = j
Notice that we immediately solved forj in terms of d and p in terms of d. Getting j and p in terms of d is useful when setting up our final expression. We need to determine, in terms of d, the sum of the number of books for David, Jeff, and Paula. Thus, we have:
d + d/3 + 2d
Getting a common denominator of 3, we have:
3d/3 + d/3 + 6d/3 = 10d/3 = 10/3*d
The answer is B |
AQUA-RAT | AQUA-RAT-35838 | Hint $\rm\ \ n\in\Bbb Z\:\Rightarrow\:2n\in\Bbb Z,\$ i.e. $\rm\ \dfrac{a-b}3\in\Bbb Z \ \Rightarrow\ \dfrac{2a-2b}3\, =\, 2\,\left(\dfrac{a-b}3\right)\in2\,\Bbb Z\subset \Bbb Z$
-
The following is multiple choice question (with options) to answer.
2 : 3 : : 23 : ? | [
"25",
"28",
"22",
"29"
] | D | D
29
3 is the next prime number after 2. Similarly, 29 is the next prime number after 23. |
AQUA-RAT | AQUA-RAT-35839 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
If two-third of a bucket is filled in 100 seconds then the time taken to fill the bucket completely will be . | [
"90 seconds",
"150 seconds",
"60 seconds",
"100 seconds"
] | B | 2/3 filled in 100 seconds
1/3 filled in 50 secs
then 2/3+1/3=100 + 50 seconds = 150 seconds
ANSWER: B |
AQUA-RAT | AQUA-RAT-35840 | I'll have a go and answer this the maths-lite way (though there are a number of answers with more mathematic rigor and .. dare I say it vigor posted here already).
Note that there is:
• 1 result with a face value 1
• 3 results with a face value 2,
• 5 results with a face value 3,
• 7 results with a face value 4,
• 9 results with a face value 5, and
• 11 results with a face value 6
The Average is defined to be: $$\text{Average} = \frac{\text{Sum of the Results}}{\text{Total number of Results}}$$
The Sum of the Results is: $$\begin{eqnarray} \text{Sum} &=& (1 \times 1) + (3 \times 2) + (5 \times 3) + (7 \times 4) + (9 \times 5) + (11 \times 6) \nonumber \\ &=& 1 + 6 + 15 + 28 + 45 + 66 \nonumber \\ &=& 161 \nonumber \end{eqnarray}$$
The Total number of Results is: $6 \times 6 = 36$
So the Average is: $$\text{Average} = \frac{161}{36} \approx 4.472$$
-
This is very much delayed, but consider the case with an $n$-sided die. As has already been observed, the expected value of the maximum of two $n$-sided die is
$${1 \over n^2} \sum_{k=1}^n (2k^2-k)$$
and we can write out this sum explicitly. In particular, we can expand to get
$${1 \over n^2} \left( \left( 2 \sum_{k=1}^n k^2 \right) - \sum_{k=1}^n k \right)$$ and recalling the formulas for those sums, this is
$${1 \over n^2} \left( {2n(n+1)(2n+1) \over 6} - {n(n+1) \over 2} \right)$$
or after some rearrangement
The following is multiple choice question (with options) to answer.
Find the product of the place value and face value of 3 in 5769354 | [
"900",
"9000",
"90",
"9"
] | A | Explanation:
Place value = Local value
Face value = Absolute value
The place value of 3 in 5769354 is 3 x 100 = 300
The face value of 3 in 5769354 is nothing but 3.
=> 300 x 3 = 900
Answer: Option A |
AQUA-RAT | AQUA-RAT-35841 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
Find the length of the wire required to go 15 times round a square field containing 69696 m2. | [
"15840",
"3388",
"2667",
"8766"
] | A | a2 = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
Answer: A |
AQUA-RAT | AQUA-RAT-35842 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
Anthony and Michael sit on the six-member board of directors for company T. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony? | [
"20%",
"30%",
"40%",
"50%"
] | C | Soln: Let the commitees be I and II
Assuming that Anthony and Michael T go into commitee I, there is just one more place left to be filled in that commitee and it can be taken by any of the 4 remaining people. Thus 4 ways.
Since the Anthony and Michael can also go into commitee II, we get 4 ways for that commitee also.
So in total = 8 ways
Now total number of ways of choosing 3 from 6 people is
= 6C3
= 6 * 5 * 4/3!
= 20 ways
Therefore Probability is = (8 /20) * 100 = 40% |
AQUA-RAT | AQUA-RAT-35843 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The cash difference between the selling prices of an article at a profit of 10% and 6% is Rs. 3. The ratio of the two selling prices is? | [
"52:56",
"55:53",
"52:50",
"22:56"
] | B | Let C.P. of the article be Rs. x.
Then, required ratio = 110% of x / 106% of x
= 110/106 = 55/53 = 55:53
Answer: B |
AQUA-RAT | AQUA-RAT-35844 | # Prove that two distinct number of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if $a$ is even and have $gcd=2$ if $a$ is odd
Prove that two distinct number of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if a is even and have $gcd=2$ if a is odd
My attempt:
If $a$ is even, let $a = 2^{s}k$ for some integers $k, s$
Then, $$a^{2^{n}} + 1 = 2^{2^{n}s}\times k^{2^n} + 1$$ and $$a^{2^{m}} + 1 = 2^{2^{m}s}\times k^{2^m} + 1$$ To prove that they're relatively prime, we need to show that their gcd = 1. And I was stuck here, how could I prove that gcd of two numbers is $1$?
A hint would be sufficient. Thanks.
-
You don't need to sign your posts; your name/signature is automatically added to the post. – Arturo Magidin Feb 28 '11 at 2:57
@Arturo Magidin: Thanks, I will next time. – Chan Feb 28 '11 at 2:59
## 3 Answers
Hint:
Consider the following proof when $a=2$: http://planetmath.org/encyclopedia/FermatNumbersAreCoprime.html
Try adapting it to work for all $a$.
Hint 2: Factor $a^{2^n}-1$.
The following is multiple choice question (with options) to answer.
If n = 2gh, where g and h are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ? | [
" Two",
" Three",
" Four",
" Six"
] | C | Answer - 'C' - Four.
For a number 2gh with primes gh,there will be four even divisors - 2,2*g,2*h,2*g*h |
AQUA-RAT | AQUA-RAT-35845 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
A sum of Rs.63000 is divided into three parts such that the simple interests accrued on them for three, six and nine years respectively may be equal. Find the amount deposited for 3 years. | [
"6000",
"2000",
"6500",
"4500"
] | A | Let the amounts be x, y, z in ascending order of value. As the interest rate and interest accrued are same for 3 years 6 years and 9 years i.e. 3x = 6y = 9z = k.
L.C.M. of 3,6,9 = 18 So x:y:z: = 6000 : 3000 :2000
The amount deposited for 3 years = 6000
ANSWER:A |
AQUA-RAT | AQUA-RAT-35846 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
A train travelled from station P to Q in 8 hours and came back from station Q to P is 6 hours. What would be the ratio of the speed of the train while traveling from station P to Q to that from station Q to P? | [
"3:6",
"3:1",
"3:8",
"3:4"
] | D | Since S # 1/t
S1 : S2 = 1/t1 : 1/t2
= 1/8 : 1/6
= 3 : 4
Answer: D |
AQUA-RAT | AQUA-RAT-35847 | Just to check each case: \begin{align*} f(S_T)=\begin{cases} 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + 0 + 0 = 3 & \text{if }S_T\leq 30, \\ 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + (30 - S_T) + 0 = 33- S_T & \text{if }30
I know @KeSchn already answered but hope this helps since this is how I usually do these. Of course, you can do this multiple ways but this gets to a correct answer relatively quickly.
Edit: Gordon's answer is definitely the way to go if you're comfortable with indicator functions. It does everything the graphical methods do without requiring any visualizing etc
The following is multiple choice question (with options) to answer.
Evaluate
30!/28! | [
"970",
"870",
"770",
"670"
] | B | Explanation:
=30!/28!
=30∗29∗28!/28!=30∗29=870
Option B |
AQUA-RAT | AQUA-RAT-35848 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
The cost of 10 kg of mangos is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is $20.50. Find the total cost of 4 kg of mangos, 3 kg of rice and 5 kg of flour? | [
"347.4",
"987.4",
"877.4",
"637.4"
] | C | C
$877.40
Let the costs of each kg of mangos and each kg of rice be $a and $r respectively.
10a = 24r and 6 * 20.50 = 2r
a = 12/5 r and r = 61.5
a = 147.6
Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5
= 590.4 + 184.5 + 102.5 = $877.40 |
AQUA-RAT | AQUA-RAT-35849 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Krishan and Nandan jointly started a business. Krishan invested four times as Nandan did and invested his money for double time as compared to Nandan. Nandan earned Rs. 4000. If the gain is proportional to the money invested and the time for which the money is invested then the total gain was? | [
"Rs.20000",
"Rs.24000",
"Rs.28000",
"Rs.36000"
] | D | 4:1
2:1
------
8:1
1 ----- 4000
9 ----- ? => Rs.36,000
Answer: D |
AQUA-RAT | AQUA-RAT-35850 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
This week, Florry bought the same number of music albums she bought last week, but for a third of the total price. This week, the average (arithmetic mean) price per album | [
"is twice last week's average",
"is three times last week's average",
"is a half of last week's average",
"is a third of last week's average"
] | D | let the number of music albums brought by Florry last week be '3'
let the total price of the music albums brought by Florry last week be $30 (i.e., $10 per album)
The average price per album purchased last week= $10
Florry brought same number of albums this week but for a total price of $10 ( i.e., one third the total price last week)
Therefore this weeks average price per album= $10/3 one-third of last week's average
Hence D |
AQUA-RAT | AQUA-RAT-35851 | # probability that no two spiders end up at the same vertex?
Eight spiders are located on the eight vertices of a cube.
When a bell rings, each spider moves (at random, independent of the others) to an adjacent vertex.
What is the probability that no two spiders end up at the same vertex?
How would I start this problem?
• After writing a computer program, I think the answer is $\dfrac{1}{81}$. Anyone have an idea how to prove this? – JimmyK4542 Dec 17 '14 at 22:52
• @JimmyK4542 My idea at the moment is to look at the number of functions obeying the conditions, and finding the probability that a random function from this set is injective. I'm currently needing to count how many functions are injective. There are 3^8 possible maps, given "moves to an adjacent vertex" means it cant stay where it is. – FireGarden Dec 17 '14 at 22:58
The key to this problem is to exploit symmetry as much as possible to reduce casework.
Label the vertices on the top face in CCW order $A,B,C,D$ and the vertices on the bottom face in CCW order $E,F,G,H$ where $AE$, $BF$, $CG$, $DH$ are edges connecting the top and bottom faces.
The four spiders at vertices $A,C,F,H$ will end up at one of the vertices $B,D,E,G$ (since there are no edges between the vertices $A,C,F,H$). Similarly, the four spiders at vertices $B,D,E,G$ will end up at one of the vertices $A,C,F,H$. Hence, no two spiders end up at the same vertex iff no two of the spiders at $A,C,F,H$ end up at the same vertex and no two of the spiders at $B,D,E,G$ end up at the same vertex.
There are $3^4$ ways for the four spiders at vertices $A,C,F,H$ to move. We count the ways they can move to $4$ different vertices as follows:
Case I: The spider at $A$ moves to $B$.
The following is multiple choice question (with options) to answer.
On Sunday morning, Pugsley and Wednesday are trading pet spiders. If Pugsley were to give Wednesday two of his spiders, Wednesday would then have eight times as many spiders as Pugsley does. But, if Wednesday were to give Pugsley ten of her spiders, Pugsley would now have fourteen fewer spiders than Wednesday had before they traded. How many pet spiders does Pugsley have before the trading game commences? | [
"4",
"6",
"8",
"10"
] | B | If Pugsley were to give Wednesday four of his spiders, Wednesday would then have five times as many spiders as Pugsley does:
(w + 2) = 8(p - 2)
If Wednesday were to give Pugsley four of her spiders, Pugsley would now have four fewer spiders than Wednesday had before they traded:
p + 14 = w - 10
Solving gives p = 6 and w = 30.
Answer: B. |
AQUA-RAT | AQUA-RAT-35852 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years, The ratio of the number of boys to the number of girls in the class is : | [
"2:9",
"2:3",
"2:1",
"2:2"
] | B | Explanation:
Let the ratio be k : 1. Then,
k * 16.4 + 1 * 15.4 = (k + 1) * 15.8
<=> (16.4 - 15.8) k = (15.8 - 15.4) <=> k = 0.4/0.6 = 2/3.
Required ratio = 2/3 : 1 = 2 : 3.
Answer: B |
AQUA-RAT | AQUA-RAT-35853 | That is, the elevator travels at a rate of 8 floors per 2 minutes.
How many floors does an elevator travel in 30 seconds?
Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.
So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors
Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2
(A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT!
(B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE
(C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE
(D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE
(E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE
For more information on this question type and this approach, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935
Cheers,
Brent
_________________
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Posts: 1224
Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ? | [
"30",
"20",
"10",
"50"
] | A | Suppose their paths cross after x minutes.
Then, 11 + 57x = 51 - 63x = 120x = 40
x = 1/3
Number of floors covered by David in (1/3) min. = (1/3 x 57)= 19.
So, their paths cross at (11 +19) i.e., 30th floor.
Answer is A. |
AQUA-RAT | AQUA-RAT-35854 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
the numbers 272738 and 232342,when divided by n, a 2 digit number leave a remainder 13 and 17 respectively.Find the sum of digits of n? | [
"4",
"5",
"6",
"7"
] | D | We have to find the h.c.f of both
1.272738-13
2.232325-17
H.c.f is 25
Ans is 7
ANSWER:D |
AQUA-RAT | AQUA-RAT-35855 | \begin{align} &\sum_{k = 0}^{n - 1}{1 \over {n \choose k}\pars{n - k}} =\sum_{k = 0}^{n - 1}\int_{0}^{1}t^{k}\pars{1 - t}^{n - k - 1}\,\dd t =\int_{0}^{1}\pars{1 - t}^{n - 1}\sum_{k = 0}^{n - 1}\pars{t \over 1 - t}^{k}\,\dd t \\[3mm]&=\int_{0}^{1}\pars{1 - t}^{n - 1}\, {\bracks{t/\pars{1 - t}}^{n} - 1 \over t/\pars{1 - t} - 1}\,\dd t =\int_{0}^{1}{t^{n} - \pars{1 - t}^{n} \over 2t - 1}\,\dd t \\[3mm]&=\int_{-1}^{1} {\bracks{\pars{1 + t}/2}^{n} - \bracks{\pars{1 - t}/2}^{n} \over t}\,{\dd t \over 2} ={1 \over 2^{n}}\int_{0}^{1} {\pars{1 + t}^{n} - \pars{1 - t}^{n} \over t}\,\dd t \\[3mm]&=-\,{n \over 2^{n}}\bracks{% \color{#00f}{\int_{0}^{1}\ln\pars{t}\pars{1 + t}^{n - 1}\,\dd t} +
The following is multiple choice question (with options) to answer.
Calculate the sum of first 60 natural numbers. | [
"1839",
"2830",
"1830",
"1831"
] | C | Solution
We know that(1+2+3+.....+60) = n(n+1)/2
Therefore (1+2+3+....+60) =(60×61 / 2)
= 1830.
Answer C |
AQUA-RAT | AQUA-RAT-35856 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
1.14 expressed as a per cent of 1.9 is: | [
"6%",
"10%",
"60%",
"90%"
] | C | Solution: Required percentage = (1.14*100)/1.9 = 60%.
Answer: Option C |
AQUA-RAT | AQUA-RAT-35857 | Goal: 25 KUDOZ and higher scores for everyone!
Senior Manager
Joined: 13 May 2013
Posts: 429
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
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30 Jul 2013, 16:45
1
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip & then at a average speed of 60 miles/hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan's avg speed in miles/hr for the entire trip
What we have here are equal distances for both segments.
First segment: 30 miles/hour and covered 30 miles, therefore it took one hour.
Second segment: 60 miles/hour and covered 30 miles, therefore it took 1/2 hour.
(Total distance / total time)
(60 / [1hr+ 1/2hr])
(60 / 1.5) = 40 miles avg. speed.
A. 35
B. 40
C. 45
D. 50
E. 55
(B)
When don't we simply add the distances/speeds together to get the average?
Intern
Joined: 23 Dec 2014
Posts: 48
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
03 Feb 2015, 16:58
Rate x Time = Distance
Going: 30 x 1 = 30
Returning: 30 x .5 = 30
Avg speed = Total distance/Total Time
=(30+30)/ (1+.5)
=40
Intern
Joined: 25 Jan 2016
Posts: 1
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
10 Feb 2016, 21:17
Narenn wrote:
jsphcal wrote:
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?
a. 35
b. 40
c. 45
d. 50
e. 55
The following is multiple choice question (with options) to answer.
Jane biked 21 2/3 miles in 3 hours and 20 minutes. What was her average rate of speed in miles per hour? | [
"6 1/2",
"7",
"7 1/2",
"7 3/4"
] | A | D = 21(2/3) = 65/3
T = 3(1/3) = 10/3
S = D/T = 6 1/2
Answer = A |
AQUA-RAT | AQUA-RAT-35858 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
At 15:00 there were 20 students in the computer lab. At 15:03 and every three minutes after that, 3 students entered the lab. If at 15:10 and every ten minutes after that 7 students left the lab, how many students were in the computer lab at 15:44 ? | [
"7",
"14",
"25",
"27"
] | C | Initial no of students + 3 * (1 + No of possible 3 minute intervals between 15:03 and 15:44) -
8 *(1 + No of possible 10 minute intervals between 15:10 and 15:44)
20 + 3*14 -8 * 4 = 25
C |
AQUA-RAT | AQUA-RAT-35859 | Example 2
In a large population of adults, 45% have a post secondary degree.
If people are selected at random from this population,
a) what is the probability that the third person selected is the first one that has a post secondary degree?
b) what is the probability that the first person with a post secondary degree is randomly selected on or before the 4th selection?
Solution to Example 2
a)
Let "having post secondary degree" be a "success". If a person from this population is selected at random, the probability of "having post secondary degree" is $p = 45\% = 0.45$ and "not having post secondary degree" (failure) is $1 - p = 1 - 0.45 = 0.55$
Selecting a person from a large population is a trial and these trials may be assumed to be independent. This is a geometric probability problem. Hence
$P(X = 3) = (1-0.45)^2 (0.45) = 0.1361$.
b)
On or before the 4th is selected means either the first, second, third or fourth person. The probability may be written as
$P(X \le 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$
Substitute by the formula $P(X = x) = (1 - 0.45)^{x-1} 0.45$ to write
$P(X \le 4) = (1 - 0.45)^{1-1} 0.45 + (1 - 0.45)^{2-1} 0.45 + (1 - 0.45)^{3-1} 0.45 + (1 - 0.45)^{4-1} 0.45 = 0.9085$
## Sums of the terms of a Geometric sequence
The following is multiple choice question (with options) to answer.
Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them? | [
"5/8",
"5/2",
"5/1",
"5/6"
] | D | P(at least one graduate) = 1 - P(no graduates)
= 1 - 6C3/10C3 = 1 - (6 * 5 * 4)/(10 * 9 * 8)
= 5/6
Answer: D |
AQUA-RAT | AQUA-RAT-35860 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A 250 m long train is running at a speed of 55 Km/hr. It crossed a platform of length 300 m in ? | [
"41.1 sec",
"20.2 sec",
"36 sec",
"50.4 sec"
] | C | Speed = 55 Km/hr
(to convert km/hr in to M/s)
= 55 x 5/18 M/s
Distance = 250 m + 300 m ( If questions is about train crossing a post you need to consider only the length of Train, )
= 550 m
Time = Distance / Speed
= 550 x 18 / (5 x 55)
= 36 sec
Ans is :C |
AQUA-RAT | AQUA-RAT-35861 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | [
"4",
"8",
"16",
"20"
] | B | Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 36
9x = 36
x = 4.
Required difference = (2x + x) - (2x - x) = 2x = 8.
Answer:B |
AQUA-RAT | AQUA-RAT-35862 | Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x
Solving, xt will get cancelled and you will get:
11000 = 11x
x is 1000
sum of both investments is 2x = 2000 which is Option D
Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1326
Location: Viet Nam
Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
### Show Tags
02 Jul 2017, 07:29
1
1
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000
Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.
The following is multiple choice question (with options) to answer.
If a sum of money borrowed at 5 percent/annum SI amounts to Rs.1020 in 4 yrs, then the sum of money borrowed is? | [
"Rs. 850",
"Rs. 900",
"Rs. 910",
"Rs. 1000"
] | A | Sol We have, A : Rs.1U2D
Let the pr xnpal be R5 x
Then mterest u — 12 : wzo — x
Therefore, by u xglurmula.
_umxx
’ Rx’: (
, _ma>< 1nzn—xv
\4Ve|\a\e,Jc— 5“
:> ~(:51DD—S*(or6x:51UU
or, J: : w 350
. The sum of money burrowed : Rs. 850
A) |
AQUA-RAT | AQUA-RAT-35863 | 2. World War 1 started July 28, 1914. What day of the week was it?
3. Little Boy is the name of the atomic bomb dropped in Hiroshima, Japan on August 6, 1945. What day of the week that happened?
4. Marie Skłodowska-Curie is the real name of Polish Marie Curie, the first woman who won Nobel Prize in both Chemistry and Physics with her work in radioactivity. She was born on November 7,1867. What day was that?
5. I was born September 22, 1989. What day of the week does that fallen?
### Dan
Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
### 2 Responses
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The following is multiple choice question (with options) to answer.
What was the day of the week on 15th August, 1947 ? | [
"Wednesday",
"Tuesday",
"Friday",
"Thursday"
] | C | 15th August, 1947 = (1946 years + Period from 1st Jan.,1947 to 15th Aug., 1947)
Counting of odd days :
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 × 2) + (36 × 1)]odd days
= 58 odd days
⇒ 2 odd days.
= 227 days = (32 weeks + 3 days) = 3 odd days.
Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.
Hence, the required day was ‘Friday’.
Answer C |
AQUA-RAT | AQUA-RAT-35864 | Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $$A_1, A_2, A_3, B_1, C_1$$ are selected. You count this selection three times.
The following is multiple choice question (with options) to answer.
A multiple choice test consists of 4 questions, and each question has 5 answer choices. In how many ways can the test be completed if every question is unanswered? | [
"24",
"120",
"625",
"720"
] | C | 5 choices for each of the 4 questions, thus total of 5*5*5*5 = 5^4 = 625 ways.
Answer: C. |
AQUA-RAT | AQUA-RAT-35865 | # Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
The following is multiple choice question (with options) to answer.
A shopping cart contains only apples, oranges, and pears. If there are twice as many oranges as apples, and twice as many pears as oranges, then the apples are equivalent to what fraction of the pears? | [
"1/4",
"1/3",
"1/2",
"2/3"
] | A | O = 2A
P = 2O = 4A
A = P/4
The answer is A. |
AQUA-RAT | AQUA-RAT-35866 | When we add $10$ red, we end up with $4k+10$ red. The blues remain unchanged at $7k$.
So the new proportion is $(4k+10): 7k$. We are told that the proportion $(4k+10): 7k$ is $6:7$. So $$\frac{4k+10}{7k}=\dfrac{6}{7}.$$ If we multiply through by $7k$, we get $4k+10=6k$, and therefore $k=5$. It follows that there are $35$ blues in the bag.
-
Let $x$ be the number of red cubes and $y$ be the number of blue cubes.
To start, the ratio of red cubes to blue cubes is 4:7, or for every 4 red cubes, there are 7 blue cubes. Hence, we have:
$7x = 4y$.
When 10 more red cubes are added to the bag, the ratio of red cubes to blue cubes shifts to 6:7, or:
$7(x+10) = 6y$.
Expanding, we get a system:
$7x = 4y$
$7x + 70 = 6y$
Can you solve the system of equations from here?
The following is multiple choice question (with options) to answer.
Kelly and Jody packed boxes with books. If Jody packed 10% of the total number of boxes, what was the ratio of the number of boxes Kelly packed to the number that Jody packed? | [
"1 to 6",
"1 to 4",
"2 to 5",
"9 to 1"
] | D | Explanation:
If Jody packed 10% of the boxes then Kelly packed 90%. The ratio of the number of boxes Kelly packed to the number Jody packed is thus 90%/10%=9/1
Answer: Option D |
AQUA-RAT | AQUA-RAT-35867 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A Certain car costs Rita $1500 afterwards she spent $50 over it. Now at what price must Rita sell this car in order to make a profit of 7% ? | [
"$1712",
"$1715",
"$1720",
"$1725"
] | A | Cost Price = $1500 + $50 = $1600
Profit % = 7
Selling Price = ?
Selling Price = (100 + Profit %) * Cost Price /100
= (100 + 7) * 1600 / 100
= 1712
Answer is A |
AQUA-RAT | AQUA-RAT-35868 | Math Help - Probability
1. Probability
I'm having a lot of trouble with permutations and combinations, so I was hoping someone here could help me out with a few questions. Please make sure that you explain your working; I really want to understand this topic... The first question asks;
A queue has 4 boys and 4 girls standing in line. Find how mant different arrangements are possible if;
a) The boys and girls alternate.
b) 2 particular girls wish to stand together.
c) All the boys stand together.
d) Also find the probability that 3 particular people will be in the queue together if the queue forms randomly.
EDIT: If it's any help, here are the answers...
a) 1152
b) 10080
c) 2880
d) 3/28
EDIT: I've got a new problem now;
A table has 4 boys and 4 girls sitting around it.
a) Find the number of ways of sitting possible if the boys and girls can sit anywhere around the table.
b) If the seating is arranged at random, find the probability that;
i) 2 particular girls will sit together.
ii) All the boys will sit together.
2. Hello, Flay!
We have to "talk" our way through these problems . . .
A queue has 4 boys and 4 girls standing in line.
Find how mant different arrangements are possible if;
a) The boys and girls alternate.
There are 2 possible arrangements: . $BGBGBGBG\,\text{ and }\,GBGBGBGB.$
The four boys can be placed in 4! ways.
The four girls can be placed in 4! ways.
Therefore, there are: . $2 \times 4! \times 4! \;=\;\boxed{1152}$ ways.
b) 2 particular girls wish to stand together.
Suppose the two girls are $X$ and $Y.$
Duct-tape them together.
Note that there are 2 possible orders: . $XY\,\text{ or }\,YX.$
Now we have seven "people" to arrange: . $\boxed{XY}\,,G,G,B,B,B,B$
. . and they can be arranged in ${\color{blue}7!}$ ways.
The following is multiple choice question (with options) to answer.
A group of 7 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the six other students from the group? | [
"5.76%",
"4.76%",
"15.75%",
"20.67%"
] | B | The question basically asks about the probability that Bob and Lisa sit at the ends.
The total # of sitting arrangements is 7!.
Desired arrangement is either BVWXYZL or LVWXYZB. Now, VWXYZ can be arranged in 5! ways, therefore total # of favorable arrangements is 2*5!.
P=(favorable)/(total)=(2*5!)/7!=1/21.
Answer: B. |
AQUA-RAT | AQUA-RAT-35869 | Difficulty:
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Question Stats:
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Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled? | [
"2",
"2.5",
"3",
"3.5"
] | C | Sol.
Part filled by (A + B + C) in 1 hour = (1/5 + 1/10 + 1/30) = 1/3.
∴ All the three pipes together will fill the tank in 3 hours.
Answer C |
AQUA-RAT | AQUA-RAT-35870 | Kudos [?]: 53125 [5] , given: 8043
Re: problem solving question on ratios [#permalink] 16 Dec 2010, 13:47
5
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spyguy wrote:
can someone explain in further detail the relationship between the teaching assistants to the number of students in any course must always be greater than 3:80 and how to reason through this portion? I understand how to solve for x. Once I was at this point I think was stumped on which number to select and inevitably chose to round up. My rational being .33 of a student is not possible therefore it must represent the position of an entire student. Thoughts? Help?
At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80. At this university, what is the maximum number of students possible in a course that has 5 teaching assistants?
A. 130
B. 131
C. 132
D. 133
E. 134
Given: $$\frac{assistants}{students}>\frac{3}{80}$$ --> $$assistants=5$$, so $$\frac{5}{s}>\frac{3}{80}$$ --> $$s_{max}=?$$
$$\frac{5}{s}>\frac{3}{80}$$ --> $$s<\frac{5*80}{3}\approx{133.3}$$ --> so $$s_{max}=133$$.
$$\frac{assistants}{students}>\frac{3}{80}$$ relationship means that if for example # of assistants is 3 then in order $$\frac{assistants}{students}>\frac{3}{80}$$ to be true then # of students must be less than 80 (so there must be less than 80 students per 3 assistants) on the other hand if # of students is for example 80 then the # of assistants must be more than 3 (so there must be more than 3 assistants per 80 students).
The following is multiple choice question (with options) to answer.
In a secondary school the seats for Maths, physics and social studies are in the ratio 8 : 5 : 7. there is a proposal to increase these seats by 75%, 40% and 50% respectively. What will be the new ratio? | [
"5 : 3 : 4",
"4 : 2 : 3",
"10 : 7 : 9",
"1 : 3 : 4"
] | B | Explanation :
Solution: Originally, let the number of seats for Maths, physics and social studies be 8x, 5x and 7x respectively.
Number of increased seats are (175% of 8x), (140% of 5x) and (150% of 7x).
i.e. (175*7x/100), (140*5x/100) and(150*7x/100) i.e. 14x, 7x and 21x/2.
.'. Required ratio = 14x : 7x : 21x/2 = 28x : 14x : 21x = 4 : 2 : 3
Answer : B |
AQUA-RAT | AQUA-RAT-35871 | Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
Hm, i got stuck cuz I got something a little different:
YOURS: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
MINE: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{3}{m}=\frac{9}{w}+5$$
In the above equation you also have for 2 men: $$\frac{2}{m}$$ - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
X is 3 times as fast as Y and is able to complete the work in 40 days less than Y. Find the time in which they can complete the work together? | [
"15 days",
"16 days",
"17 days",
"18 days"
] | A | If X can complete work in x days and y can complete work in 3x days.
Then 2x = 40
x= 20 days.
so X can complete work in 20 days and y can complete work in 60 days.
they can complete the work together in 1/(1/20 + 1/60) = 1/ (4/60) = 15 days
ANSWER:A |
AQUA-RAT | AQUA-RAT-35872 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
The C.P of 10 pencils is equal to the S.P of 12 pencils. Find his gain % or loss%? | [
"18 2/3%",
"58 2/3%",
"16 2/3%",
"46 2/3%"
] | C | Option C
Explanation:
10 CP = 12 SP
12 --- 2 CP loss
100 --- ? => 16 2/3% |
AQUA-RAT | AQUA-RAT-35873 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
Of the 100 athletes at a soccer club, 40 play defense and 66 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between | [
"10 to 20",
"10 to 40",
"26 to 40",
"30 to 70"
] | C | First of all notice that since only 40 athletes play defense, then the number of athletes that play both midfield and defense cannot possibly be greater than 40.Eliminate D and E.
{Total} = {defense} + {midfield} - {both} + {neither}
100 = 40 + 66 - {both} + {neither}
{both} = {neither} + 6.
Since the least value of {neither} is given to be 20, then the least value of {both} is 20+6=26.Eliminate A and B.
Answer: C. |
AQUA-RAT | AQUA-RAT-35874 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
When positive integer N is divided by positive integer J, the remainder is 10. If N/J = 205.08, what is value of J? | [
"125",
"100",
"75",
"150"
] | A | When a number is divided by another number, we can represent it as : Dividend = Quotient * Divisor + Remainder
So, Dividend/Divisor = Quotient + Remainder/Divisor
Given that N/J = 205.08
Here 205 is the quotient.
Given that Remainder = 10
So, 205.08 = 205 + 10/J
So, J = 125
Answer: A |
AQUA-RAT | AQUA-RAT-35875 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is: | [
"18",
"24",
"42",
"81"
] | B | Explanation:
Let the ten's and unit's digit be x and 8/x respectively.
Then,
(10x + 8/x) + 18 = 10 * 8/x + x
9x2 + 18x - 72 = 0
x2 + 2x - 8 = 0
(x + 4)(x - 2) = 0
x = 2
So, ten's digit = 2 and unit's digit = 4.
Hence, required number = 24.
ANSWER IS B |
AQUA-RAT | AQUA-RAT-35876 | Here is an idea which should work. Let $n >0$ be a positive integer. Define $k_0=n$, and recursively, as long as $k_j \neq 0$ define $$k_{j} =2^{k_{j+1}} m_{j+1} \, \mbox { with } m_{j+1} \mbox{ odd }$$ This process ends after $t$ steps, when $k_t=0$, or equivalently $k_{t-1}$ is odd. Now define f(n) =( \frac{m_1-1}{2}, \frac{m_2-1}{2},.., ... 2 If something is not in B \cup C, then it is in neither B nor C. Because if it was in B, then it is in B or C, which is B \cup C. So the first mistake is in the second sentence. You could show that A - (B \cup C) \subseteq A - (B \cap C). Because the former one is what is in A, but not in B nor C. The latter is what is in A, but not ... 2 That's a very creative idea. I don't know of an established set theory symbol for this (e.g. http://www.rapidtables.com/math/symbols/Set_Symbols.htm doesn't even define such an "operation") and it doesn't seem to be a standard symbol in \LaTeX. One problem I can see is that the symbol \ominus is quite widely used to denote the symmetric difference of ... 2 It's not awful, at least as an idea, but I would certainly say it's unnecessary, and I strongly dislike your particular notation choice. Basically it's a symbol that's too different and unintuitive to have to remember for too unimportant and uncommon a situation to be worthwhile. In my experience, once one gets to mathematics classes, textbooks, etc. that ... 2 In your post you only proved that B = \overline A \cup X \implies A\cap B \subseteq X \subseteq B, although
The following is multiple choice question (with options) to answer.
For any integer k greater than 1, the symbol k* denotes the product of all the fractions of the form 1/t, where t is an integer between 1 and k, inclusive. What is the value of 1*/4* ? | [
"5",
"5/4",
"4/5",
"1/4"
] | A | When dealing with 'Symbolism' questions, it often helps to 'play with' the Symbol for a few moments before you attempt to answer the question that's asked. By understanding how the Symbol 'works', you should be able to do the latter calculations faster.
Here, we're told that K* is the PRODUCT of all the fractions of the form 1/T, where T is an integer between 1 and K, inclusive.
Based on this definition....
IF....
K = 2
K* = (1/1)(1/2) = 1/2
IF....
K = 3
K* = (1/1)(1/2)(1/3) = 1/6
We're asked to find the value of 5*/4*
Now that we know how the Symbol 'works', solving this problem shouldn't be too difficult. You can actually choose to do the math in a couple of different ways....
5* = (1/1)(1/2)(1/3)(1/4)(1/5)
Don't calculate this just yet though....since we're dividing by 4*, many of those fractions will 'cancel out.'
4* = (1/1)(1/2)(1/3)(1/4)
We're looking for the value of:
(1/1)(1/2)(1/3)(1/4)(1/5) / (1/1)(1/2)(1/3)(1/4)
Since the first four fraction in the numerator and denominator cancel out, we're left with just one fraction:
4/5
A |
AQUA-RAT | AQUA-RAT-35877 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Amber works 28 days a month at d dollars per day for m months out of the year. Which of the following represents her monthly pay? | [
"m/(20d)",
"20d",
"10md/6",
"28d"
] | D | ANSWER: D The passage states that she works 28 days a month at d dollars per day, so 28 d is her monthly pay |
AQUA-RAT | AQUA-RAT-35878 | ### Show Tags
16 Jan 2019, 08:15
As we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .
Cost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88
Profit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
25 Feb 2019, 09:26
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails
(2.60 * 12)/0.13 = 240 nails (Ans)
_________________
"Please hit +1 Kudos if you like this post"
_________________
Manish
The following is multiple choice question (with options) to answer.
At a florist shop on a certain day, all corsages sold for either $20 or $30. If 8 of the corsages that sold for $30 had instead sold for $20, then the store's revenue from corsages that day would have been reduced by 10 percent. What was the store's actual revenue from corsages that day? | [
"$200",
"$400",
"$600",
"$800"
] | D | Let, No. of corsages @ $20 = x, no. of corsages @ $30 = y and revenue= r
so, 20x+30y=r.........(1)
Now,
Given the situation, 20(x+8) + 30(y-8)= r-.1r
=> 20x+160+30y-240 = .9r
=> 20x+30y = .9r+80............(2)
so, r = .9r+80
=> r = 800
The answer is D. |
AQUA-RAT | AQUA-RAT-35879 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
How many times digit 6 is used while writing numbers from 100 to 1100? | [
"648",
"320",
"252",
"225"
] | B | There are 100 numbers which begin with 600
Next, In every 10 numbers such as 100 to 110, 110 to 120, 120 to 130 6 comes at least once. Number of such intervals = End limit - First no. / interval. Our range of numbers is 100 - 1000
1000 - 100 = 900/10 = 90
Number of 10s interval in this is 90. So 90 '6s'
So far we have calculated 190.
The total now comes to 280. The nearest to which is 320. Hence B. |
AQUA-RAT | AQUA-RAT-35880 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
In the accounting branch of a commercial institute, the percentage of male and female workers is 48% and 52% respectively. In this branch, 40% of males and 20% of females are age 25 or older. If one of the workers is selected at random, what is the probability that the worker is under 25 years old? | [
"0.3",
"0.25",
"0.45",
"0.7"
] | D | Percentage of male worker = 48
Percentage of female worker = 52
Let total number of workers = 1000
number of male workers= 480
number of female workers = 520
Number of male age 25 or older = (4/10)*480 = 192
Number of female age 25 or older =(2/10)*520 = 104
Total number of workers 25 or older = 192+104 = 296
Total number of workers under 25 = 1000 - 296 = 704
Probability that a worker randomly selected is under 25 years old = 704/1000 = .704
Alternatively , since the answer options are not close , we can use estimation here .
Percentage of male worker 25 or older = (4/10)*48 = 20 % approx
Percentage of female worker 25 or older = (2/10)*52 = 10 % approx
Percentage of total worker 25 or older = 30 %
Percentage of total worker under 25 = 70 %
Probability that a worker randomly selected is under 25 years old = .7
Answer D |
AQUA-RAT | AQUA-RAT-35881 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
For a certain set of numbers, if x is in the set, then x - 3 is also in the set. If the number 2 is in the set , which of the following must also be in the set ?
a) 4
b) -4
c) -5 | [
"a only,",
"b only,",
"c only,",
"a and b only"
] | B | Notice that the question asks which of the following MUST be in the set, not COULD be in the set.
Since 2 is in the set, then so must be 2-3=-1. Similarly, since -1 is in the set, then so must be -1-3=-4. Could 4 and -1 be in the set? Certainly, but we don't know that for sure.
Answer: B.
Regarding your question: we don't know which is the source integer in the set, if it's 1, then 4 won't be in the set but if the source integer is say 7, then 4 will be in the set. So, 4 may or may not be in the set. |
AQUA-RAT | AQUA-RAT-35882 | ## Solution
Let the length of $AD$ be $x$, so that the length of $AB$ is $2x$ and $\text{[}ABCD\text{]}=2x^2$.
Because $ABCD$ is a rectangle, $\angle ADC=90^{\circ}$, and so $\angle ADE=\angle EDF=\angle FDC=30^{\circ}$. Thus $\triangle DAE$ is a $30-60-90$ right triangle; this implies that $\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}$, so $\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}$. Now drop the altitude from $E$ of $\triangle DEF$, forming two $30-60-90$ triangles.
Because the length of $AD$ is $x$, from the properties of a $30-60-90$ triangle the length of $AE$ is $\frac{x\sqrt{3}}{3}$ and the length of $DE$ is thus $\frac{2x\sqrt{3}}{3}$. Thus the altitude of $\triangle DEF$ is $\frac{x\sqrt{3}}{3}$, and its base is $2x$, so its area is $\frac{1}{2}(2x)\left(\frac{x\sqrt{3}}{3}\right)=\frac{x^2\sqrt{3}}{3}$.
To finish, $\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}$
The following is multiple choice question (with options) to answer.
Suppose that AB = AC = CD and
AD = BD. What is the measure of
6 ADC in degrees? | [
"24",
"28",
"32",
"36"
] | D | Let be the measure of \ADC. Then since ADC is isosceles, we obtain that
the measures of \CAD and \DCA are and 180 2, respectively. It follows that
\ACB has measure 2. Since ACB is isosceles, we obtain that the measures of \CBA
and \BAC are 2 and 180 4, respectively. Since ADB is isosceles, we deduce that
the measures of \CBA and \BAD are the same. This gives that 2 = (180 4) +
which implies 5 = 180. Therefore, = 36.
correct answer D |
AQUA-RAT | AQUA-RAT-35883 | is 0) So slope = (y - 0)/(0 - x) = -y/x Hi karishma Slope of line 1 is -y1/x1 slope of line 2 is -y2/x2 if we multiply both the slopes wont the product be positive?? Kindly help Kudos [?]: [0], given: 20 Intern Joined: 13 May 2014 Posts: 38 Kudos [?]: 77 [0], given: 1 Concentration: General Management, Strategy Re: GMAT Prep: XY-Coordinate Plane Ques [#permalink] ### Show Tags 18 May 2014, 08:15 adityagogia9899 wrote: VeritasPrepKarishma wrote: bsjames2 wrote: In the xy-coordinate plane, line l and line k intersect at the point (4,3). Is the product of their slopes negative? 1) The product of the x-intercepts of lines l and k is positive. 2) The product of the y-intercepts of lines l and k is negative. Please explain your answer This question has a great takeaway - something I am sure you know intuitively but you may not think of it while doing this question because it is seldom written out: The slope of a line is -(y intercept)/(x intercept) Above, sreehari uses this concept to solve the question very efficiently. If you are wondering why it is so, think what 'intercept' represents... The point of x intercept is (x, 0) (where y co-ordinate is 0) The point of y intercept is (0, y) (where x co-ordinate is 0) So slope = (y - 0)/(0 - x) = -y/x Hi karishma Slope of line 1 is -y1/x1 slope of line 2 is -y2/x2 if we multiply both the slopes wont the product be positive?? Kindly help Hi Aditya, y1 and x1 are the intercepts on y-axis and x-axis respectively by line 1 (or l) ; And we still don't know the intercepts are on negative or positive. So it's possible that y1<0 or y1 >0 and similarly , x1<0 or x>0. Similar cases hold
The following is multiple choice question (with options) to answer.
In the coordinate plane, Line A has a slope of -1 and an x-intercept of 2. Line B has a slope of 5 and a y-intercept of -10. If the two lines intersect at the point (a,b), what is the sum a+b? | [
"0",
"1",
"2",
"3"
] | C | The equation of Line A is y = -x + 2
The equation of Line B is y = 5x - 10
5x - 10 = -x + 2
x = 2
y = 0
The point of intersection is (2,0) and then a+b = 2.
The answer is C. |
AQUA-RAT | AQUA-RAT-35884 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 5% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue? | [
"2",
"5",
"100/21",
"110/20"
] | C | Last months revenue can be calculated as 220x / 1.10 = 200x
Now consider the next statement - > if the price of ticket is hiked by 5% the new price is 1.05 x.
So the new revenue when only 200 tickets are sold = 200 * 1.05x = 210x.
We need to find the percentage difference ; so we have (new-old/new*100) 210x-200x/210x*100 = 100/21
Answer: C |
AQUA-RAT | AQUA-RAT-35885 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio of 2 numbers is 1:2 and their H.C.F. and L.C.M are 10 and 20. Find the numbers. | [
"20, 10",
"10, 20",
"20, 30",
"30, 20"
] | B | Let the numbers be x and 2x
Their H.C.F. = 10
L.C.M = 20
ie) 2x^2 = 200
x^2 = 100, x = 10 t The numbers are x and 2x , ie)10, 20 respectively.
Answer is B |
AQUA-RAT | AQUA-RAT-35886 | The totient of $210$ - the number of values between $1$ and $210$ that are relatively prime to $210$ - is $(2-1)(3-1)(5-1)(7-1)=48$. Using this, we can say that there are $48\cdot5=240$ numbers not divisible by these four numbers up to $1050$. Some of these of course are out of range of the original question; we'll have to figure out what those are.
The totient of $30$ is $8$; from $991$ to $1050$ there are $16$ numbers relatively prime to $30$. We'll take these out for now, leaving $224$. But two numbers we removed - $1001$ and $1043$ - are divisible by $7$ and so weren't part of the original $240$ so we have to un-remove them, adding them back to the total. Further, $991$ and $997$ are below $1000$ so shouldn't have been removed either. This gives $224+2+2=228$ numbers relatively prime to $210$, so $1000-228=772$ numbers are divisible by $2$, $3$, $5$, or $7$.
• Note that this doesn't work when any of the numbers we're testing divisibility for are composite. – Dan Uznanski Jan 1 '18 at 14:34
• how would you change it to work for composite numbers? – oliver Apr 27 '20 at 13:51
Your solution is correct (or, at least, your result matches mine). Huzzah!
The following is multiple choice question (with options) to answer.
By how much is two-fifth of 500 greater than five - seventh of 210 ? | [
"50",
"75",
"120",
"None of these"
] | A | Solution
2/5of 500 - 5/7 of 210 = 50.
Answer A |
AQUA-RAT | AQUA-RAT-35887 | evolution, population-dynamics
Title: How many humans have been in my lineage? Is it almost the same for every human currently living? If I were to count my father, my grandfather, my great-grandfather, and so on up till, say chimps, or the most common ancestor, or whatever that suits the more accurate answer, how many humans would there have been in my direct lineage?
And would it be almost the same for every human being currently living? A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is likely to be higher. Research has shown that the great apes (chimps, gorilla, orangutan) have generation times comparatble to humans, in the range of 18-29 years (Langergraber et al. 2012).
Using 7 million years and 20 years yields an estimated 350000 ancestral generations for each living human. A more conservative estimate, using an average generation time of 28, would result in 250000 generations. However, some have argued that the human-chimp split is closer to 13 million years old, which would mean that approximately 650000 generations have passed (using a generation time of 20 years).
The exact number of ancestral generations for each human will naturally differ a bit, and some populations might have higher or lower numbers on average due to chance events or historical reasons (colonizations patterns etc). However, due to the law of large numbers my guess would be that discrepancies are likely to have averaged out. In any case, the current estimates of the human-chimp split and average historical generation times are so uncertain, so that they will swamp any other effects when trying to calculate the number of ancestoral generations.
However, this is only answering the number of ancestral generations. The number of ancestors in your full pedigree is something completely different. Since every ancestor has 2 parents, the number of ancestors will grow exponentially. Theoretically, the full pedigree of ancestors can be calculated using:
The following is multiple choice question (with options) to answer.
The present average age of a family of five members is 26 years. If the present age of the youngest member in the family is ten years, then what was the average age of the family at the time of the birth of the youngest member ? (Assume no death occurred in the family since the birth of the youngest? | [
"80 years",
"70 years.",
"20 years.",
"30 years."
] | C | Present total age of the members = 26(5) = 130 years.
Present age of the youngest member = 10 years
Present total age of the remaining four members = 130 -10 = 120 years
Their average age at the time of the birth of the youngest member = [120 - (4 * 10)] / 4 = 30 - 10 = 20 years.
Answer: C |
AQUA-RAT | AQUA-RAT-35888 | Claim: If $m$ and $n$ are integers, and $n$ is even, then $mn$ is even.
Proof: Because $n$ is even, we can write $n=2l$, where $l$ is an integer. Then $mn=m(2l)=2(ml)$ is a multiple of $2$. In other words, $mn$ is even.
Notice that each step is justified—though if we had an extremely strict teacher, we might want to point out that we got $mn=m(2l)$ by multiplying both sides of $n=2l$ by $m$, and that $m(2k)=2(ml)$ follows from the associativity and commutativity of multiplication. But you're probably safe in omitting these details unless you've been told otherwise.
Also note that I didn't assume that $m$ is odd, because this makes the equations a little more complicated. Here is a more faithful recreation of your proof:
Claim: If $m$ is an odd integer and $n$ is an even integer, then $mn$ is even.
Proof: Because $m$ is odd and $n$ is even, we can write $m=2k+1$ and $n=2l$, where $k$ and $l$ are integers. Then $mn = (2k+1)(2l) = 2\cdot(l(2k+1))$ is a multiple of $2$, hence even.
Define $2k_1k_2+k_2=k_3$ as an integer number.
You could further formalize your text, if you want (it is always handy to write down the definitions or the theorems you use, it also makes it easier for others to read and make it look more complete): By definition: $$m \text{ even }\iff \exists k \in \mathbb Z \text{ such that }2k = m$$
The following is multiple choice question (with options) to answer.
If m and n are positive integers, and m=2n and k=2m, then - | [
"2m is a factor of k.",
"m/2 is a factor of k.",
"3m is a factor of k/2.",
"k is a factor of m."
] | B | m=2n --> m is even.
k=2m --> m is factor of k, thus m/2 = integer must also be a factor of k.
Answer: B. |
AQUA-RAT | AQUA-RAT-35889 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A? | [
"240",
"288",
"266",
"216"
] | A | (3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
Answer: A |
AQUA-RAT | AQUA-RAT-35890 | But suppose we weren't done. The table would continue \begin{align*} 4 && 2 && - \\ 5 && 3 && 2 \\ 6 && 4 && 3 \\ 7 && 4 && - \\ 8 && 5 && 4 \\ 9 && 6 && 5 \end{align*} where we stop when we hit $n=9$, which is guaranteed to have a solution. Notice, there is no reason to find the value in the 3rd column when the integer part in the second column has not increased.
Note that we can actually calculate the multiples of $\log_2 13/8 = 0.700 \dots$ "by eye", so the only work in this particular problem is the integer part of the multiples of $\log_2 12/8$.
• Calculation of log to base 2 is easy in floating point. A crude approximation is to write a dot, then the mantissa portion of the floating pint representation. If you have a decimal number between 1 and 2, just subtract 1. It is exact for 1 and 2, and a bit low at the middle of the interval. For more on logs, see Doerfler's wonderful book. – richard1941 Jul 12 '17 at 0:26
$12 \lt 2^{m/n} \lt 13$
$\log_2 12 \lt \frac{m}{n} \lt \log_2 13$
$[3; 1, 1, 2, 2, \dots] \lt \frac{m}{n} \lt [3; 1, 2, 2, 1, \dots]$
The continued fractions match up to $[3; 1]$. Since $[3; 1, 1]$ and $[3; 1, 2]$ are underestimations of $\log_2 12$ and $\log_2 13$, we take $[3; 1, 1]$ (the "smaller" one lexicographically) and add $1$ to the last number (round "up"), so the answer is $[3; 1, 2]$.
The following is multiple choice question (with options) to answer.
If log 27 = 1.437, then the value of log 9 is: | [
"0.934",
"0.945",
"0.954",
"0.958"
] | D | log 27 = 1.437
log (3cube ) = 1.437
3 log 3 = 1.437
log 3 = 0.479
log 9 = log(3square ) = 2 log 3 = (2 x 0.479) = 0.958.
Answer: Option D |
AQUA-RAT | AQUA-RAT-35891 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
P can do the work in 12 days and Q can do the same work in 15 days. If they work together for 4 days, what is the fraction of work that is left? | [
"1/5",
"2/5",
"7/15",
"11/15"
] | B | 4/12 + 4/15 = 1/3 + 4/15 = 9/15 = 3/5 (completed work)
The work that is left is 1 - 3/5 = 2/5
The answer is B. |
AQUA-RAT | AQUA-RAT-35892 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A person jogged 10 times along the perimeter of a rectangular field at the rate of 12 kilometers per hour for 30 minutes. If field has a length that is twice its width, find the area of the field in square meters. | [
"10,000 square meters",
"20,000 square meters",
"30,000 square meters",
"40,000 square meters"
] | B | Let us first find the distance d jogged
distance = rate ? time = (12 km / hr) ? 30 minutes
= (12 km/hr) ? 0.5 hr = 6 km
The distance of 6 km corresponds to 10 perimeters and therefore 1 perimeter is equal to
6 km / 10 = 0.6 km = 0.6 ? 1000 meters = 600 meters
Let L and W be the length and width of the field. The length is twice the width. Hence
L = 2 W
The perimeter is 600 meters and is given by
2 (L + W) = 600
Substitute L by 2 W
2 (2 W + W) = 600
Simplify and solve for W
4 W + 2 W = 600
6 W = 600
W = 100
Find L
L = 2 W = 200
Find the area A of the rectangle
A = L * W = 200 * 100 = 20,000 square meters
correct answer B |
AQUA-RAT | AQUA-RAT-35893 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A merchant marks his goods up by 40% and then offers a discount of 10% on the marked price. What % profit does the merchant make after the discount? | [
"8%",
"14%",
"26%",
"15%"
] | C | Let the price be 100.
The price becomes 140 after a 40% markup.
Now a discount of 10% on 126.
Profit=126-100
26%
answer C |
AQUA-RAT | AQUA-RAT-35894 | the number of committees that exist where the man and women serve together is given by,
$$\begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 7 \\ 2 \end{pmatrix} = 210$$
so the total number of committees in this case amounts to,
$$1120 - 210 = 910$$
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
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The following is multiple choice question (with options) to answer.
A committee of 6 is chosen from 6 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed? | [
"275",
"700",
"1404",
"2620"
] | A | The only cases possible are :
1. 2 men and 4 women : 6C2 * 5C4 = 75
2. 3 men and 3 women: 6C3*5C3 = 200
Rest of the cases will either have 1 or 0 men (not allowed) or will have 1 or 2 or 0 women (not allowed)
Total possible combinations = 75+200 = 275. Thus A is the correct answer. |
AQUA-RAT | AQUA-RAT-35895 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B can do a piece of work in 6 days. With the help of C they finish the work in 3 days. C alone can do that piece of work in? | [
"40 days",
"16 days",
"6 days",
"5 days"
] | C | C
6 days
C = 1/3 – 1/6 = 1/6 =>6 days |
AQUA-RAT | AQUA-RAT-35896 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Sheela deposits Rs. 3400 in bank savings account. If this is 15% of her monthly income. What is her monthly income in? | [
"22666",
"20000",
"25000",
"22235"
] | A | Explanation :
15% of income = Rs. 3400
100% of income = 3400 X 100/15 = Rs. 22666
Answer : A |
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