source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39397 | (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 sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives $4 plus one-half of what remains. Next, Bob receives $4 plus one-third of what remains. Finally, Chloe receives the remaining $32. How much money M did Bob receive? | [
"20",
"22",
"24",
"26"
] | B | Notice that we need not consider Ann's portion in the solution. We can just let K = the money REMAINING after Ann has received her portion and go from there.
Our equation will use the fact that, once we remove Bob's portion, we have $32 for Chloe.
So, we getK - Bob's $ = 32
Bob received 4 dollars plus one-third of what remained
Once Bob receives $4, the amount remaining is K-4 dollars. So, Bob gets a 1/3 of that as well.
1/3 of K-4 is (K-4)/3
So ALTOGETHER, Bob receives4 + (K-4)/3
So, our equation becomes:K -[4 + (K-4)/3 ]= 32
Simplify to get: K - 4 - (K-4)/3 = 32
Multiply both sides by 3 to get: 3K - 12 - K + 4 = 96
Simplify: 2K - 8 = 96
Solve: K = 52
Plug this K-value intoK - Bob's $ = 32to get: 52 - Bob's $ = 32
So, Bob's $ M= 20
Answer:
B |
AQUA-RAT | AQUA-RAT-39398 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is : | [
"Rs. 700",
"Rs. 690",
"Rs. 650",
"Rs. 698"
] | D | Explanation:
Simple Interest (SI) for 1 year = 854-815 = 39
Simple Interest (SI) for 3 years = 39 × 3 = 117
Principal = 815 - 117 = Rs.698
Answer: Option D |
AQUA-RAT | AQUA-RAT-39399 | 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 train ,140 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is | [
"270 m",
"245 m",
"235 m",
"220 m"
] | C | Explanation:
Assume the length of the bridge = x meter
Total distance covered = 140+x meter
total time taken = 30s
speed = Total distance covered /total time taken = (140+x)/30 m/s
=> 45 × (10/36) = (140+x)/30
=> 45 × 10 × 30 /36 = 140+x
=> 45 × 10 × 10 / 12 = 140+x
=> 15 × 10 × 10 / 4 = 140+x
=> 15 × 25 = 140+x = 375
=> x = 375-140 =235
Answer: Option C |
AQUA-RAT | AQUA-RAT-39400 | Lastly, the area of $$\triangle ABC$$ gives us the following identity:
$24 = \frac{1}{2} \cdot (3y) \cdot [ 2x\cdot \sin(A) ]$
which implies the area of $$\triangle ADE$$ is
$\frac{1}{2}\cdot x\cdot y \cdot \sin(A) = 4$
It should be fairly straightforward to deduce that the area of $$\triangle DEF$$ is
$24 - (4+4+9) = 7$
This analytic solution is really the same as the geometric solution I posted earlier ( http://www.hpmuseum.org/forum/thread-795...l#pid71463 )
Graph 3D | QPI | SolveSys
04-08-2017, 03:37 AM (This post was last modified: 04-08-2017 03:37 AM by Han.)
Post: #156
Han Senior Member Posts: 1,811 Joined: Dec 2013
RE: Little explorations with the HP calculators
Here is a proof (using linear algebra) that reduces the general case to the special case provided by Gerson.
Define the barycentric coordinates $$\mathbf{\lambda} = (\lambda_1, \lambda_2, \dotsm, \lambda_n )$$ of a point $$\mathbf{P}\in \mathbf{R}^m$$ relative to a set of points $$\{ \mathbf{Q}_1, \mathbf{Q}_2, \dotsm, \mathbf{Q}_n \}$$ as the solution to the equation
$\mathbf{P} =\sum_{k=1}^n \lambda_k \cdot \mathbf{Q}_k$
(This is actually a system of equations in the coordinates of the points.) We can even require that $$\sum \lambda_k = 1$$ by embedding these points in a hyperplane $$x_{m+1} = 1$$ of $$\mathbb{R}^{m+1}$$ though this is not necessary for this problem.
The following is multiple choice question (with options) to answer.
Points A, B, and, C have xy-coordinates (2,0), (8,12), and (14,0), respectively. Points X, Y, and Z have xy-coordinates (6,0), (8,4), and (10,0), respectively. What fraction C of the area of triangle ABC is the area of triangle XYZ? | [
" 1/9",
" 1/8",
" 1/6",
" 1/5"
] | A | If you notice, both triangles ABC and XYZ have a side on X axis. we can take these sides as bases for each triangle, therefore
Area of ABC is 1/2*12*12 (Height of ABC is the y coordinate of the third point (8,12))
similarly Area of XYZ is 1/2*4*4
dividing area of XYZ with that of ABC gives C= 1/9.A |
AQUA-RAT | AQUA-RAT-39401 | permutations around a round table with labelled seats
Two men, Adam and Charles, and two women, Beth and Diana, sit at a table where there are seven places for them to sit down. Two people are sitting next to each other if they occupy consecutive chairs. A non-trivial rotation defines a different seating arrangement, meaning that if all four people rotate their positions by moving k chairs to the right, it is the same way for them to be seated if and only if k divides 7.
Determine the number of ways that these four people can be seated so that every man is next to a woman and every woman is next to a man.
Answer is 224 . Here is the link with (source with explanation).
But my answer is 252. My calculations are given below.
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man any of the 3 far away seats(3)
seat remaining woman in adjacent seat(2)
so 7*2*2*3*2= 168
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the woman(1)
seat remaining woman in adjacent seat(2)
so 7*2*2*1*2= 56
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the man(1)
seat remaining woman in adjacent seat(1)
so 7*2*2*1*1= 28
So total = 168+ 56+28 = 252
Can someone help me to figure out whether any of this answer is right? If my answer is wrong, please help to understand why it has gone wrong and how correct answer can be reached.
The explanation given in the site derives the answer as 224 but it is in a different approach than mine. Is that correct?
The correct answer is $224$. Your calculations are almost fine
The following is multiple choice question (with options) to answer.
In how many different number of ways 6 men and 2 women can sit on a shopa which can accommodate persons? | [
"1250",
"1640",
"1240",
"1680"
] | D | 8p4 = 8 x 7 x 6 × 5 = 1680
D) |
AQUA-RAT | AQUA-RAT-39402 | 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.
The length of a train and that of a platform are equal. If with a speed of 90 k/hr, the train crosses the platform in one minute, then the length of the train (in meters) is? | [
"228",
"267",
"200",
"750"
] | D | Speed = [90 * 5/18] m/sec = 25 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 25 è x = 25 * 60 / 2 = 750 .Answer: D |
AQUA-RAT | AQUA-RAT-39403 | # 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 L = 775 × 778 × 781, what is the remainder when L is divided by 14? | [
"6",
"7",
"8",
"9"
] | A | L 775/14 leaves a remainder 5
778/14 leaves a remainder 8
781/14 leaves a remainder 11
5*8*11 =440
So the remainder will be the remainder of 440/14 which is 6
Ans A |
AQUA-RAT | AQUA-RAT-39404 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
Two persons A and B can complete a piece of work in 20 days and 40 days respectively. If they work together, what part of the work will be completed in 5 days? | [
"1/8",
"1/3",
"3/6",
"3/8"
] | D | A's one day's work = 1/20
B's one day's work = 1/40
(A + B)'s one day's work = 1/20 + 1/40 = 3/40
The part of the work completed in 5 days = 5 (3/40) = 3/8.
Answer:D |
AQUA-RAT | AQUA-RAT-39405 | # 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.
The division of a whole number N by 14 gives a quotient of 18 and a remainder of 2. Find N. | [
"797",
"254",
"597",
"997"
] | B | According to the division process of whole numbers, N can be written, using multiplication, as follows
N = quotient *divisor + remainder = 18* 14 + 2 = 254 correc answer B |
AQUA-RAT | AQUA-RAT-39406 | pressure, vacuum, air
Title: How to understand air pressure against a hand pump? Imagine this little though experiment with me for a moment.
A piston sort-of contraption - like a bicycle pump, but without any nozzle to push air out of. One foot tall, one inch width/depth. Lets say the pump is completely bottomed out. As such, there is no air at the bottom. It's also completely air tight, and if I pull on it, no air can slip in to displace it. ASCII explanation:
======
||
====== ||
|| ||
|| ||
|||| ||||
|||| ==> | |
|||| | |
|||| | |
---- ----
In a vacuum, I would be able to pull the pump out, fairly easily. No forces acting other than friction and gravity.
However, on earth, air pressure is acting on it strongly. Being that the top is 1 inch by 1 inch - it should be 14-15 pounds of force, correct? This seems too easy. I can lift that amount quite easily. Am I missing something? I have enough force to create a vacuum so easily?
My guess is that it gets more difficult very quickly, with pulling out the piston further. But a vacuum cant be even more of a vacuum, so that's not what's going on. The exposed bit of the pump has more surface area, does that cause more pressure on the piston? Or is it simply a linear 15 lbs the whole way through? You are correct. With just a 1 in cross section you would be able to pull as much vacuum as you wish. (most pumps are larger than this.
As a counter example, just think of a mercury barometer. It takes about 26 to 32 inches of mercury to draw the vacuum and hence we can measure the ambient pressure. If we made this barometer 100 inches tall it would not take any more mercury, we would just get 70 inches of vacuum.
The following is multiple choice question (with options) to answer.
If 2/3 of the air in a tank is removed with each stroke of a vacuum pump, how many strokes does it take before less than 1% of the original amount of air in the tank remains? | [
" 2",
" 3",
" 4",
" 5"
] | D | My approach is to find more or less good number and plug in. Lets take 90 as the total volume of an air, and the question asks after how many stokes there will be less than 0,9 air in a tank, if each stroke takes 2/3 of an air.
So after he1 strokethe volume will be 30 (90-60=30), and so on
2 stroke- 30-20=10
3 stroke- 10-6,7=3,3
4 stroke- 3,3-2,2=1,1
5 stroke- 1,1 - minus something that gets us less than 0,9 for sure. so the answer is: after 5 strokes there will be less than 1% of air left in a tank. OptionD. |
AQUA-RAT | AQUA-RAT-39407 | 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 jumps into a river from a overbridge ,at the same time a hat drops in the river which flows with the stream of the water. That man travels 10 minute upstream and returns back where he was asked to catch the hat. He catches hat at a distance of 1000 yard down at a second bridge . Find the speed of the river. | [
"50 yards per minute",
"40 yards per minute",
"60 yards per minute",
"55 yards per minute"
] | A | Suppose x is the distance travelled by the man in 10 min upstream.
Then
x/(v-u) = 10 where v is man's speed and u is river speed)
total time taken to swim by the man upstream and down stream is equal to the time taken by the hat to travel 1000 yards with the speed of river
1000/u = 10 + (x+1000)/(u+v)
now substitute x = 10(v-u) and simplify
we will get
100v = 2uv
means
2u = 100 and u = 50 yards per minute
ANSWER:A |
AQUA-RAT | AQUA-RAT-39408 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
Two cars cover the same distance at the speed of 48 and 64 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car. | [
"87",
"138",
"192",
"83"
] | C | Explanation:
48(x + 1) = 64x
X = 3
48 * 4 = 192 km
Answer: Option C |
AQUA-RAT | AQUA-RAT-39409 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
### GMAT/MBA Expert
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
total dinning bill of 5 people was $139.00 and 10% tip divided the bill evenly ?what is the bill amount each person shared . | [
"32.84",
"22.84",
"30.58",
"24.84"
] | C | dinner bill of 5 person = 139 + 10% tip
so,
10% of 139 = (139*10)/100 = 13.9
So, the actual total amount = 139+13.9 = $ 152.9
so per head bill = 152.9/5 = $ 30.58
ANSWER:C |
AQUA-RAT | AQUA-RAT-39410 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
The distance between Delhi and Mathura is 170 kms. A starts from Delhi with a speed of 20 kmph at 7 a.m. for Mathura and B starts from Mathura with a speed of 25 kmph at 8 p.m. from Delhi. When will they meet? | [
"10.50 a.m.",
"11.20 a.m.",
"10.30 a.m.",
"11.40 a.m."
] | B | D = 170 – 20 = 150
RS = 20 + 25 = 45
T = 150/45 = 3 1/3 hours
8 a.m. + 3 hrs 20 min = 11.20 a.m.
ANSWER:B |
AQUA-RAT | AQUA-RAT-39411 | 321
a) f(x)=abs(x), g(x)=-x
b) f(x)=cos(x), g(x)=x+2pi
6. ### csprof2000
287
Good point.
I guess then that my argument only works for functions which don't have an inverse where Domain(inv f) = Range(f). This, naturally, precludes functions such as abs(x) and cos(x)...
So I guess more though will have to be put into functions which are not bijections.
7. ### mnb96
638
Thanks a lot. You all made very good observations that helped me a lot.
BTW, it seems that if the function f admits an inverse there are not many choices, while if the function f is not invertible, many solutions exist but the problem is non-trivial, and it is difficult to say what kind of functions f and g would have to be, in order to satisfy f(x)=f(g(x)).
At the moment I am trying to solve the following (similar) problem:
$$f = (f \circ g) g'$$
where g is invertible, and g' denotes its derivative
If you find it interesting, suggestions are always welcome.
Thanks!
8. ### csprof2000
287
Well, similar suggestions - cases - are possible.
Assume f(x) = c, c constant. Then c = cg', g' = 1, and g = x + k, k constant.
Assume f(x) = x. Then x = gg', gdg = xdx, and (g^2)/2 = (x^2)/2 + k, k constant.
Assume f(x) = x^n. Then x^n = g^n g', g^n dg = x^n dx, [1/(n+1)]g^(n+1) = [1/(n+1)]x^(n+1) + k, k constant
The following is multiple choice question (with options) to answer.
If f(x)= 5x^3- 2x+8 and g(y)= 6y+4, then g(f(x)) = | [
"30x^3-12x+52",
"11x^2-12x+44",
"8x^3-8x+32",
"30x^3+4x+4"
] | A | g(f(x)) = 6(f(x)) +4 = 6(5x^3- 2x+8) + 4 = 30x^3-12x+52 =>A |
AQUA-RAT | AQUA-RAT-39412 | Difference in angular distance travelled by the minute hand and the hour hand in one minute is thus $6-\dfrac{1}{2} = \dfrac{11}{2}$ degrees. So, on a full rotation $(360^\circ)$, any similar event between them will be repeated every $\dfrac{360}{(11/2} = 65 \dfrac{5}{11}$ minutes.
In twenty four hours, i.e. $24 \times 60 = 1440$ minutes, the number of times angle between the minute hand and hour hand will be same, can be calculated by dividing 1440 minutes with this value, thus,
$n=\dfrac{1440}{720/11}=22$
$90^\circ$ angle (event of perpendicular disposition) will be obtained $= 22\times 2 = \textbf{44 times}$ (as this situation arises twice on a complete rotation).
S Malik
()
how hour hand covers (2*360)
as if minute hand cover 360 than hour hour should also?
is it?
The following is multiple choice question (with options) to answer.
A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7.30 a.m. They shall first cross each other at: | [
"5.30 a.m",
"6.42 a.m",
"7.55 a.m",
"7.42 a.m"
] | D | Sol.
Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two
Relative speed of A and B = (6-1) = 5 rounds per hour.
Time taken to complete one round at this speed = 1/5 hr = 12 min.
Answer D |
AQUA-RAT | AQUA-RAT-39413 | \begin{align*} a_1&=1+b_0\\ a_2&=1+b_0+b_1\\ a_3&=1+b_0+b_1+b_2\\ a_4&=1+b_0+b_1+b_2+b_3 \end{align*}
(We don’t need $b_4$ to reconstruct $A$; it’s there to make the sum a known constant.)
Thus, there is a bijection between $4$-element subsets $A$ of $\{1,\dots,10\}$ satisfying the condition that no two members are consecutive, and ordered $5$-tuples $\langle b_0,b_1,b_2,b_3,b_4\rangle$ of integers satisfying the equation $$b_0+b_1+b_2+b_3+b_4=9\tag{1}$$ together with the conditions $b_1,b_2,b_3\ge 2$ (to ensure that no two members of $A$ are consecutive), $b_0\ge 0$ (to ensure that $1$ is the smallest possible value of $a_1$), and $b_4\ge 0$ (to ensure that $10$ is the largest possible value of $a_4$.
The following is multiple choice question (with options) to answer.
If 5a + 7b = t, where a and b are positive integers, what is the largest possible value of t for which exactly one pair of integers (a, b) makes the equation true? | [
"35",
"48",
"69",
"70"
] | D | 5*a1 + 7*b1 = t
5*a2 + 7*b2 = t
5*(a1 - a2) = 7*(b2 - b1)
since we are dealing with integers we can assume that a1 - a2 = 7*q and b2 - b1 = 5*q where q is integer, so whenever we get a pair for (a;b) we can find another one by simply adding 7 to a and subtracting 5 from b or vice versa, subtracting 7 from a and adding 5 to b.
Lets check how it works for our numbers, starting from the largest:
E)74 = 5*12 + 7*2 (a1 = 12, b1 = 2), subtract 7 fromaand add 5 tobrespectively, so a2 = 5 and b2 = 7, second pair - bad
D)70 = 5*7 + 7*5 (a1 = 7, b1 = 5), if we add 7 toawe will have to subtract 5 from b but b can't be 0, so - no pair, if we subtract 7 froma, we'll get a = 0 which also isn't allowed - no pair, thus this is the only pair for (a;b) that works,good!, thus
D is the answer |
AQUA-RAT | AQUA-RAT-39414 | => 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.
A software programmer does 20% of the work in 80% of the time, and 80% of the work in the remaining 20% of the time. If the code he writes is X lines long and he was given one month (30 days) to accomplish the task, then, assuming that the programmer works at a constant rate in each of the two stages, How many lines of code were written in the last two weeks, in terms of X? | [
"13x /15",
"15x/15",
"7x/15",
"2x/30"
] | A | 30 days- X lines
80% of 30
24 days; 0.2X lines
Remaining 6 days:
0.8X lines
Need to find out for last 14 days;
We know the number of lines for last 6 days: 0.8X
How about 8 days before that;
24 days- 0.2X
1 day - 0.2X/24 lines
8 days - (0.2X/24)*8 lines = (1/15)X lines
Total number of lines in last 14 days
(1/15)x + 0.8X = (1/15)x + (8/10)X = (52/60)X = (13/15)X lines of codes
ANSWER:A |
AQUA-RAT | AQUA-RAT-39415 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
In a division sum, the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 12 to the thrice of the remainder. The dividend is | [
"74",
"78",
"86",
"126"
] | D | Divisor = (6 * 3) + 12 = 30
5 * Quotient = 30
Quotient = 6.
Dividend = (Divisor * Quotient) + Remainder
Dividend = (20 * 6) + 6 = 126.
D) |
AQUA-RAT | AQUA-RAT-39416 | python, beginner, python-3.x
Title: Flooring cost calculator I've made a small program that calculates the cost of flooring for a given area based on the price per sqft and total sqft. I have only been programming for a few days now so I am sure that there are some things I could be doing better.
from decimal import *
import time
print("This program will help you find the cost of floor covering\n")
def program():
exit_program = ["q", "Q"]
global move_on
move_on = ["y", "Y"]
global more_info
more_info = ["n", "N"]
cost_of_tile = Decimal(input("What is the cost of the tile per square foot?"))
global answer
answer = input("Do you know how big the room is in sqft? Y/N")
if answer in move_on:
global sqft
sqft = Decimal(input("Please enter the room's size in SQUARE FEET"))
print("The cost to cover your floor will be $%s" % (sqft * round(cost_of_tile, 2)))
labor()
answer = input("Would you like to run the calculator again for another room? Y/N")
should_i_stay()
elif answer in more_info:
answer = input("Do you know the length and the width of the room? Y/N")
if answer in move_on:
width = Decimal(input("What is the width of the room?"))
length = Decimal(input("What is the length of the room?"))
sqft = width * length
print("Your room is %s sqft" % sqft)
print("The cost to cover your room will be $%s" % (sqft * round(cost_of_tile, 2)))
labor()
elif answer in more_info:
print("Sorry we need more information ")
answer = input("Would you like to run the calculator again? Y/N")
should_i_stay()
elif answer in exit_program:
print("Exiting")
exit()
The following is multiple choice question (with options) to answer.
The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs. 1000 per sq. metre. | [
"Rs. 15000",
"Rs. 15550",
"Rs. 15600",
"Rs. 20625"
] | D | Solution
Area of the floor = (5.5 × 3.75) m2 = 20.625 m2
∴ Cost of paving = Rs. (1000 × 20.625) = 20625. Answer D |
AQUA-RAT | AQUA-RAT-39417 | 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.
What must be added to each term of the ratio 7:11, So as to make it equal to 3:4? | [
"8",
"7.5",
"6.5",
"5"
] | D | Solution: Let x be added to each term.
According to question,
(7+x)/(11+x) = 3/4;
Or, 33+3x = 28+4x;
Or, x = 5.
Answer: Option D |
AQUA-RAT | AQUA-RAT-39418 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ? | [
"270",
"250",
"245",
"320"
] | A | Let the smaller number be x. Then larger number = (x + 1365).
x + 1365 = 6x + 15
5x = 1350
x = 270
Smaller number = 270.
ANSWER A |
AQUA-RAT | AQUA-RAT-39419 | homework-and-exercises, statics
My way to solve this:
since ladder is not moving
Torque on the ladder
$$ -\frac{mgL}{2}\sin\theta-\mu N_{f}L\cos\theta+N_{f}L\sin\theta=0$$
Horizontal forces
$$ N_{w}-f_{f}=0$$
Vertical forces
$$ N_{f}+f_{w}-mg=0$$
where $N_{f}$ and $N_{w}$ is normal forcé due to floor and Wall, $f_{w}$ and $f_{f}$ friction due to Wall and floor.
Now I solve for $N_{f}$ the first equation, to get:
$$N_{f}=\frac{mg\tan\theta}{2(\tan\theta-\mu)}$$
Sincé the angle is the situation where the ladder is about to start sliding, friction forces will have their máximum value, so:
$$f_{f}=\mu N_{f}$$
from 2º equation:
$$N_{w}=f_{f}$$
and for same reason like befor
$$f_{w}=\mu N_{w}$$
substituting all this values in the 3º equation I get:
$$\frac{\tan\theta}{2(\tan\theta-\mu)}(1+\mu^{2})=1$$
and solving this for $\tan\theta$:
$$\tan\theta=\frac{2\mu}{1+\mu^{2}}$$
but aparently this is wrong. Could someone check this and point where is the mistake. Ty You can take the last two equations to solve for the normal forces and use them in the torque equation
The following is multiple choice question (with options) to answer.
A ladder 14 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 9 feet from the base of the wall. If the top of the ladder slips down 1 feet, how many feet will the bottom of the ladder slip? | [
"4",
"10",
"18",
"19"
] | B | 14^2-9^2=115
it means that the height is equal to 10.72.
since the top of the ladder slips down 1 feet, then the height of the wall =10.72-1=9.72
the bottom =sqrt(14^2-9.72^2)=sqrt(196-94.47)=10.07~=10
ans is B |
AQUA-RAT | AQUA-RAT-39420 | 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.
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The following is multiple choice question (with options) to answer.
What day of the week was 1 January 1901 | [
"Monday",
"Tuesday",
"Saturday",
"Friday"
] | B | Explanation:
1 Jan 1901 = (1900 years + 1st Jan 1901)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
1st Jan 1901 = 1 odd day
Total number of odd days = (0 + 1 + 1) = 2
2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.
Answer: Option B |
AQUA-RAT | AQUA-RAT-39421 | x 0.6
=
y 2.6
Could somebody check these please?
Thanks!
they don't work, show us the work that you did c00ky and we can show you where you went wrong
note: rather than using latex, you can show us matrixes like this:
|2 3 5|
|3 4 5|
11. I mixed the 3 and 4 the wrong way around earlier on.
I now have x = -2 and y = 2 which seems to work!
Woohoo !
The following is multiple choice question (with options) to answer.
((-1.9)(0.6) – (2.6)(1.2))/2.0 = ? | [
"-0.71",
"-2.13",
"1.07",
"1.71"
] | B | Dove straight into calculation
((-1.9)(0.6) – (2.6)(1.2))/2.0 = -2.13
Answer B |
AQUA-RAT | AQUA-RAT-39422 | # 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.
Jayant opened a shop investing Rs. 30,000. Madhu joined him 2 months later, investing Rs. 45,000. They earned a profit of Rs. 52,000 after completion of one year. What will be Madhu's share of profit? | [
"Rs. 27,000",
"Rs. 26,000",
"Rs. 30,000",
"Rs. 36,000"
] | B | 30,000 *12=45,000*8
1:1
Madhu's share=1/2*52,000
i.e. Rs. 26,000
ANSWER:B |
AQUA-RAT | AQUA-RAT-39423 | # 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.
lexus car agency has 189 cars.he sold some cars at 9% profit and rest at 36% profit thus he gains 17% on the sale of all his cars.the no. of cars sold at 36% profit is? | [
"56",
"37",
"38",
"39"
] | A | ratio of 36% profit cars to 9% profit cars = 8:19
so no. of cars sold at 36% profit = 189*8/27= 56 cars
ANSWER:A |
AQUA-RAT | AQUA-RAT-39424 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
Tom, working alone, can paint a room in 10 hours. Peter and John, working independently, can paint the same room in 5 hours and 2 hours, respectively. Tom starts painting the room and works on his own for two hour. He is then joined by Peter and they work together for two hour. Finally, John joins them and the three of them work together to finish the room, each one working at his respective rate. What fraction of the whole job was done by Peter? | [
"1/3",
"1/2",
"1/4",
"1/5"
] | B | Let the time when all three were working together be t hours. Then:
Tom worked for t+4 hour and has done 1/10*(t+4) part of the job;
Peter worked for t+2 hour and has done 1/5*(t+2) part of the job;
John worked for t hours and has done 1/2*t part of the job:
1/10*(t+4)+1/5*(t+2)+1/2*t=1 --> multiply by 10 --> (t+4)+(2t+2)+5t=10 --> t=1/2;
Hence Peter has done 1/5*(1/2+2)=1/5 *5/2=1/2
Answer: B |
AQUA-RAT | AQUA-RAT-39425 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
A group consists of 5 men, 5 women and 3 children. In how many ways can 2 men , 2 women and 1 child selected from the given group? | [
"300 ways",
"400 ways",
"500 ways",
"360 ways"
] | A | Two men, two women and one child can be selected in 5C₂ * 5C2 * 3C₁ ways
5*4/2*1*5*4/2*1*3
= 300 ways
Answer : A |
AQUA-RAT | AQUA-RAT-39426 | # 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 butler is promised Rs. 100 and a cloak as his wages for a year. After 7 months he leaves this service, and receives the cloak and Rs.20 as his due. How much is the cloak worth? | [
"22",
"98",
"92",
"72"
] | C | Explanation:
Let be the price of cloak is = x
According to the Question he should get 7/12th of 100 and 7/12th of cloak.
712(100)+712(x)=20+x712(100)+712(x)=20+x
⇒ x = 92.
Answer: C |
AQUA-RAT | AQUA-RAT-39427 | physical-chemistry, thermodynamics, phase, gas-phase-chemistry, liquids
Title: Liquification of ideal solution of two liquids from vapor phase A and B form an ideal solution. In a cylinder piston arrangement, $\pu{2.0 mol}$ of vapor of liquid A and $\pu{3 mol}$ of vapor of liquid B are taken at $\pu{300 torr}$ and $T~\pu{K}$. At what pressure $30\%$ of the total amount of substance of vapor will it liquefy? Given $\ce{Pa^{.}}=\pu{300 torr}$, $\ce{Pb^{.}}=\pu{600 torr}$, ($\sqrt{48.25}=6.95$). Kindly provide a conceptual and brief solution with an explanation.
The answer comes out to be $\pu{445 torr}$. This question is from a preparatory-test for The indian jee Advanced exam, And i think it can be a discussion. As i put my input into it already, i dont think this should be a homework-question, But instead, it raises a thought and creates new perspectives
$L$ is the amount of substance of liquid
$V$ is the amount of substance of vapor
$x$ is the mole fraction A in the liquid
$y$ is the mole fraction of A in the vapor
The following is multiple choice question (with options) to answer.
A barrel holds 4 litres of paint and 4 litres of turpentine. How many quarts of turpentinemust be added to the container to create a mixture that is 3 parts petrol to 5 parts turpentine by volume? | [
"4/3",
"5/3",
"7/3",
"8/3"
] | D | An alternate to the alligation method is the more direct/algebraic method:
Let x be the amount of turpentine to be added.
New total amount of turpentine = 4+x
Total amount of paint = 4
New total = 4+4+x=8+x
Final ratio required (for turpentine) = 5/(5+3)=5/8
Thus, (4+x)/(8+x) = 5/8 --> solving for x you get x = 8/3.
D is thus the correct answer. |
AQUA-RAT | AQUA-RAT-39428 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A car covers a certain distance at aspeed of 50 kmph in 5 hours. To cover the same distance in 2 hrs, it must travel at a speed of? | [
"125 km/hr",
"678 km/hr",
"782 km/hr",
"789 km/hr"
] | A | Distance = (50 x 5) =250 km.
Speed = Distance/Time
Speed = 250/2 = 125 kmph
Answer : A |
AQUA-RAT | AQUA-RAT-39429 | python, python-3.x, calculator
Title: Taxi fare calculator, trip recorder, and fuel calculator I solved this practice problem in preparation for school exams.
The owner of a taxi company wants a system that calculates how much money his taxis take in one day.
Write and test a program for the owner.
Your program must include appropriate prompts for the entry of data.
Error messages and other output need to be set out clearly and understandably.
All variables, constants and other identifiers must have meaningful names.
You will need to complete these three tasks. Each task must be fully tested.
TASK 1 – calculate the money owed for one trip
The cost of a trip in a taxi is calculated based on the numbers of KMs traveled and the type of taxi that you are traveling in. The taxi company has three different types of taxi available:
a saloon car, that seats up to 4,
a people carrier, that seats up 8,
a mini-van, that seats up to 12.
For a trip in a saloon car the base amount is RM2.50 and then a charge of RM1.00 per KM. For a trip in a people carrier the base amount is RM4.00 and a charge of RM1.25 per KM. For a trip in a mini- van the base amount is RM5.00 and a charge of RM1.50 per KM. The minimum trip length is 3KM and the maximum trip length is 50KM. Once the trip is complete a 6% service tax is added.
TASK 2 – record what happens in a day
The owner of the taxi company wants to keep a record of the journeys done by each one of his taxis in a day. For each one of his three taxis record the length of each trip and the number of people carried. Your program should store a maximum of 24 trips or 350km worth of trips, whichever comes first. Your program should be able to output a list of the jobs done by each one of the three taxis.
TASK 3 – calculate the money taken for all the taxis at the end of the day.
At the end of the day use the data stored for each taxi to calculate the total amount of money taken and the total number of people carried by each one of the three taxis. Using the average price of RM2.79 per litre use the information in the table below to calculate the fuel cost for each taxi:
The following is multiple choice question (with options) to answer.
Jim’s Taxi Service charges an initial fee of $2.25 at the beginning of a trip and an additional charge of $0.25 for each 2/5 of a mile traveled. What is the total charge for a trip of 3.6 miles? | [
"$3.15",
"$4.5",
"$4.80",
"$5.05"
] | B | Let the fixed charge of Jim’s Taxi Service = 2.25 $
and charge per 2/5 mile(.4 mile) = .25$
Total charge for a trip of 3.6 miles = 2.25 + (3.6/.4) * .25
= 2.25 + 9*.25
= 4.5 $
Answer B |
AQUA-RAT | AQUA-RAT-39430 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
If it is 7:27 in the evening on a certain day, what time in the morning was it exactly 2,880,717 minutes earlier? (Assume standard time in one location.) | [
"7:22",
"7:24",
"7:27",
"7:30"
] | D | 7:27minus 2,880,717in any way must end with 0, the only answer choice which ends with 0 is D.
Answer: D. |
AQUA-RAT | AQUA-RAT-39431 | 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 covering a distance of 90 km, A takes 2 hours more than B. If A doubles his speed, then he would take 1 hour less than B. A's speed is: | [
"5 km/h",
"8 km/h",
"10 km/h",
"15 km/h"
] | D | Let A's speed be X km/hr.
Then, 90/x - 90/2x = 3
6x = 90
x = 15 km/hr.
Answer : D. |
AQUA-RAT | AQUA-RAT-39432 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
What will come in place of the x in the following number series? 1, 4, 10, 22, 46, x | [
"84",
"92",
"94",
"98"
] | C | 1
1 × 2 + 2 = 4
4 × 2 + 2 = 10
10 × 2 + 2 = 22
22 × 2 + 2 = 46
46 × 2 + 2 = 94 C |
AQUA-RAT | AQUA-RAT-39433 | newtonian-mechanics, rotational-dynamics, estimation, stability
Title: Human Lean Equation For a medical experiment I am doing, I need an equation to find the angle at which someone will lean before falling. I am not mathematically inclined in terms of advanced stuff, I am more so of a trigonometry person.
I assume factors that will be needed are BMI, including height and weight, but that is really all I have. For example, I am a 5'11" (180.3cm) tall female. I weigh 165 pounds (74.8kg). I want to find out, computationally (because I could measure myself), how far I could lean before I fall.
Any ideas of how I could go upon this? Even though your body is not a simple, homogeneous, rigid object, we can calculate a little... With a lot of assumptions.
I assume, your feet stand next to each other, not in a step-forward-position. And I assume, the question is about leaning forward/backward, not sideways.
And I assume that the arms have to be aligned at the side of the body.
First step, we need to locate your center of gravity (CG). Even if you feel light/heavy hearted, we assume you to be symmetric. We also assume you to be front-back-symmetric, except for your feet.
Now we need the hight-coordinate of CG. You can find it via plancking across a bar. Let's assume your result is that CG is 99cm from floor up.
Next up, we need your shoe size. I guess you at an american women's size of 11.5, which equates to 27cm foot-length.
If you ever took dance classes, you may remember that you were instructed to "put your weight over your heel", or over the center of your feet. Let's assume that this little shift can be done without "leaning".
As you are "more of a trigonometry person", here comes the fun part:
We think of you (having "weight over heel") as a L-shape, with your feet being 27cm, and the stem of the "L" being 99cm (height of CG). Now you gradually lean forward, and when CG passes over your toes, you fall.
The following is multiple choice question (with options) to answer.
In a certificate by mistake a candidate gave his height as 20% more than actual height. In the interview panel, he clarified that his height was 5 feet 5 nches. Find the % correction made by the candidate from his stated height to his actual height? | [
"13",
"20",
"30",
"43"
] | A | His height was = 5 feet 5 inch = 5 + 60 = 65 inch. Required % correction =65*(1.20-1) = 13
A |
AQUA-RAT | AQUA-RAT-39434 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A person crosses a 600 metre long street in 5 minutes. What is his speed in km per hour? | [
"8.2",
"4.2",
"6.1",
"7.2"
] | D | Distance =600 metre =0.6 km
Time = 5 min = 1/12 hr
speed = dist/time = 0.6/(1/2) = 7.2 km/hr
ANSWER D |
AQUA-RAT | AQUA-RAT-39435 | Sum the two parts gives $\pi/3$.
• Can you explain to why the bound of $\rho$ is between $0 \: \text{and} \:1$ in this case..? – misheekoh Apr 5 '17 at 4:41
• @misheekoh because you are integrating over the (half) unit sphere's volume, to cover this volume the radius has to change from $0$ to $1$. – Taozi Apr 5 '17 at 4:47
• @misheekoh I added to the answer a direct evaluation of the surface flux integration. – Taozi Apr 5 '17 at 5:03
• Thank you. Btw, I do understand that we're integrating over the hemisphere but particularly for when $\text{z} \le \sqrt{1-x^2-y^2}$, how did u manage to convert the upper bound to $\rho \le 1$? – misheekoh Apr 5 '17 at 5:06
• @misheekoh Square both sides of this inequality to have $z^2 \le 1 - x^2 -y^2$, rewrite as $x^2 +y^2+z^2 \le 1$, since $x^2 +y^2+z^ = \rho^2$, this is just $\rho^2 \le 1$, or $\rho\le 1$. – Taozi Apr 5 '17 at 5:09
The following is multiple choice question (with options) to answer.
The volume of the sphere QQ is (dfrac{37}{64}%)less than the volume of sphere PP and the volume of sphere RR is (dfrac{19}{27}%) less than that of sphere QQ. By what is the surface areaof sphere RR less than the surfacearea of sphere PP? | [
"50%",
"60%",
"75%",
"80%"
] | C | Let the volume of sphere PP be 64 parts.
Therefore volume of sphere QQ
=64−3764%=64−3764% of 6464
=64−37=27=64−37=27 parts.
The volume of RR
=27−1927×27=27−1927×27
=27−19=8=27−19=8 parts.
Volume ratio:
=P:Q:R=64:27:8=P:Q:R=64:27:8
Radius ratio:
=P:Q:R=4:3:2=P:Q:R=4:3:2
The surface area will be 16:9:516:9:5
Surface area of RR is less than the surface area of sphere PP
16k−4k=12k16k−4k=12k
Now,
=12k16k×100=12k16k×100
=75%=75%
Thus surface area of sphere RR is less than the surface area of sphere P by 75%
C |
AQUA-RAT | AQUA-RAT-39436 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is: | [
"4.04%",
"4.14%",
"4.16%",
"4.20%"
] | A | 100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
Percentage error = 404 x 100 % = 4.04%
100 x 100
A) |
AQUA-RAT | AQUA-RAT-39437 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
Pipe A fills a tank of capacity 700 liters at the rate of 40 liters a minute. Another pipe B fills the same tank at the rate of 30 liters a minute. A pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute. If pipe A is kept open for a minute and then closed and pipe B is open for a minute and then closed and then pipe C is open for a minute and then closed and the cycle is repeated, when will the tank be full? | [
"42 minutes",
"14 minutes",
"39 minutes",
"40 minutes 20 seconds"
] | A | In one cycle they fill 40+30-20 = 50 liters
700 = 50*n => n = 14
here n = number of cycles.
total time = 14*3 = 42 as in one cycle there are 3 minutes.
thus 42 minutes
ANSWER:A |
AQUA-RAT | AQUA-RAT-39438 | # Proof for divisibility by $7$
One very classic story about divisibility is something like this.
A number is divisible by $2^n$ if the last $n$-digit of the number is divisible by $2^n$. A number is divisible by 3 (resp., by 9) if the sum of its digit is divisible by 3 (resp., by 9). A number $\overline{a_1a_2\ldots a_n}$ is divisible by 7 if $\overline{a_1a_2\ldots a_{n-1}} - 2\times a_n$ is divisible by 7 too.
The first two statements are very well known and quite easy to prove. However I could not find the way on proving the third statement.
PS: $\overline{a_1a_2\ldots a_n}$ means the digits of the number itself, not to be confused with multiplication of number.
The following is multiple choice question (with options) to answer.
If x is divisible by 2 and 7, which of the following must divide evenly into x? I. 60 II. 32 III. 14 | [
"I,II only",
"I,III only",
"I only",
"III only"
] | D | If x is divisible by 2,7 means it will be = or > 14, 28,42 etc...
That are not divisible by 60, 32.
So, the answer is D |
AQUA-RAT | AQUA-RAT-39439 | # 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.
Ramu bought an old car for Rs. 42000. He spent Rs. 13000 on repairs and sold it for Rs. 66900. What is his profit percent? | [
"17%",
"19%",
"18%",
"21.6%"
] | D | Total CP = Rs. 42000 + Rs. 13000
= Rs. 55000 and SP
= Rs. 66900
Profit(%) = (66900 - 55000)/55000 * 100
= 21.6%
Answer: D |
AQUA-RAT | AQUA-RAT-39440 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 36 hours and 16 hours respectively. The ratio of their speeds is? | [
"4:5",
"4:3",
"4:6",
"4:9"
] | C | Let us name the trains A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √16 : √36 = 4:6
Answer:C |
AQUA-RAT | AQUA-RAT-39441 | (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.
Betty picked a number of pears, each of which she sold, canned, or poached. If she canned 30 percent more than she poached, and she poached 10 percent less than she sold. If there were a total of 614 pears, then how many pears did she sell? | [
"190",
"200",
"210",
"220"
] | B | Let x be the amount of pears that she sold.
Then she poached 0.9x.
Then she canned 1.3*0.9x=1.17x.
The total amount of pears is x+0.9x+1.17x=3.07x
The fraction of pears that she sold is 1/3.07=100/307.
The number of pears she sold is (100/307)*614=200.
The answer is B. |
AQUA-RAT | AQUA-RAT-39442 | the largest and smallest integers in the group. This method should take three decimal arguments, and return the smallest of the three. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. It is similar to modulus – ‘%’ operator of C/C++ language. Note that this is not the modulo * operation (the result can be negative). Print its smallest integer divisor greater than 1. Find the least common multiple and greatest common divisor of the given three numbers : Find the least common multiple and the greatest common divisor of numbers 12, 24, 48, 192 and 288. LCM(Least Common Multiple) is the smallest positive number which is divisble by both the numbers. The macro below does the. During the reconstruction of the railway tracks 40 meter long pieces of rails were replaced by 15 meter long pieces. (For example: 7/3 = 3 and 10/2 = 5). Dynamically Calculate Prime Numbers. The smallest divisor (other than 1) of a composite number is a/an a) odd number b) even number c) prime number d) composite number. Thus the smallest number we can represent would be: 1E(-(MIN_EXP-digits-1)*4), eg, for digits=5, MIN_EXP=-32767, that would be 1e-131092. For example divisors of 6 are 1, 2, 3 and 6, so divisor_sum should return 12. gcd(a, b) = 5 3 · 7 2 = 6125. On a number line, we get hundredths by simply dividing each interval of one-tenth into 10 new parts. Example: The smallest weird number is 70. Write a Java program that computes the absolute value of the product of 2 numbers given by the user. Here dividend is 205 and divisor is 2 therefore remainder is 1. The last divisor will be the HCF of given numbers. In particular, we have added an entirely new Ch. The SMALL function will automatically ignore TRUE and FALSE values, so the result will be the nth smallest value from the set of actual numbers in the array. In this tutorial we will write couple of different Java programs to find out the GCD of two numbers. The first perfect number is 6, because
The following is multiple choice question (with options) to answer.
The second smallest prime number is? | [
"67",
"71",
"73",
"21"
] | A | The second smallest prime number is 67.
A) |
AQUA-RAT | AQUA-RAT-39443 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
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The following is multiple choice question (with options) to answer.
If the length of an edge of cube M is four times the length of an edge of cube N, what is the ratio of the volume of cube N to the volume of cube M? | [
"1/32",
"1/16",
"1/8",
"1/64"
] | D | The length of cube N = 1;
The length of cube M = 4;
The ratio of the volume of cube N to the volume of cube M = 1^3/4^3 = 1/64
Answer : D |
AQUA-RAT | AQUA-RAT-39444 | For some reason I find it easier to think in terms of letters of a word being rearranged, and your problem is equivalent to asking how many permutations there are of the word YYYYBBBBB.
The formula for counting permutations of words with repeated letters (whose reasoning has been described by Noldorin) gives us the correct answer of 9!/(4!5!) = 126.
The following is multiple choice question (with options) to answer.
The number of permutations of the letters of the word 'REMEMBER' is? | [
"7!/(2!)2 3!",
"8!/(2!)2 8!",
"8!/(8!)2 3!",
"8!/(2!)2 3!"
] | D | n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern 'REMEMBER' consists of 8 letters of which there are 2R's, 3E's, 2M's and 1B.
Number of arrangements = 8!/(2!)2 3!
Answer: D |
AQUA-RAT | AQUA-RAT-39445 | 80&4352803871250866057941284067804134&18,17,27,27,29,30\\ 81&11395788481298729007027429355459050&18,17,27,29,29,30\\ 82&29834561572649264127826347038980028&18,17,29,29,29,30\\ 83&78107896236653244353987095675034922&18,17,29,29,29,32\\ 84&204489127137313012297066889013936544&18,17,29,29,31,32\\ 85&535359485175296566394760822973220276&18,17,29,31,31,32\\ 86&1401589328388601697072447570729328594&18,17,31,31,31,32\\ 87&3669408499990537951967408951471847094&18,17,31,31,31,34\\ \end{array}
The following is multiple choice question (with options) to answer.
The number 311311311311311311311 is: | [
"divisible by 3 but not by 11",
"divisible by 11 but not by3",
"divisible by both 3 and 11",
"neither divisible by 3 nor by 11."
] | D | Sum of digits= 35 and so it is not divisible by 3.
(Sum of digits at odd places)- (Sum of digits at even places)= 19-16=3, not divisible by 11.
So, the given number is neither divided by 3 nor by 11.
ANSWER:D |
AQUA-RAT | AQUA-RAT-39446 | ### Show Tags
18 Feb 2015, 07:51
1
Using smart numbers:
Jennifer Purchases watches for $5: B=5 Markup of 100%; now selling for$10: x=10
She sells 5 watches: N=5
Her total profit will be (N*x)-(B*N)
(5*10)-(5*5)= 25
T=25
Now using the variables we must look at the answer choices that will solve for the % markup of the watches, which we chose to be 100.
B=5
T=25
N=5
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
### Show Tags
18 Feb 2015, 21:51
Cost price per watch = b
Selling price per watch = $$\frac{t}{n}$$
Let x = percent of the markup from her buy price to her sell price
$$x = 100 *\frac{t}{n} * \frac{1}{b} = \frac{100t}{nb}$$
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Posts: 50544
Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
### Show Tags
22 Feb 2015, 11:17
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?
A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
The following is multiple choice question (with options) to answer.
During a two-week period, the price of an ounce of silver increased by 5 percent by the end of the first week and then decreased by 10 percent of this new price by the end of the second week. If the price of silver was x dollars per ounce at the beginning of the two-week period, what was the price, in dollars per ounce, by the end of the period? | [
"0.945x",
"0.985x",
"0.995x",
"1.05x"
] | A | The price at the end is 0.9(1.05x)=0.945x
The answer is A. |
AQUA-RAT | AQUA-RAT-39447 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Simple interest on a certain sum is 9 over 25 of the sum. Find the rate per cent and time, if both are equal. | [
"8% and 8 years",
"6% and 6 years",
"10% and 10 years",
"6 % and 6 years"
] | D | 9â„25 P = P×R×R/100
⇒ R2 = 900â„25 ⇒ R^2 = 36 R= 6
Also, time = 6 years
Answer D |
AQUA-RAT | AQUA-RAT-39448 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
if A=2^65 AND B=2^64+2^63.....+2^0 | [
"A is larger than B by 2^64 times",
"B is larger than A by 1",
"A is larger than B by 1",
"A is equal to B"
] | C | lets say A : 2^4=16 , B: 2^3+2^2+2^1+2^0 = 15 so A is larger than B by 1 always
ANSWER:C |
AQUA-RAT | AQUA-RAT-39449 | If stock price is $$X = (0,30)$$; we use the 7 puts (A) for profit = 7(40 - X); 9 (B) puts are used giving a loss = 9(30 - X). Total earning = 7(40-X) -9(30-X) + 2$= 2X + 12$ > 0
- 3 years, 2 months ago
Assume that the $40 put is still priced at$10.
What would be the price of the $30 put option, where there will be no arbitrage opportunity? Staff - 3 years, 2 months ago Log in to reply Let the price at which the 30$ put options is priced be $$M$$.
Now suppose that an arbitrage opportunity does exist. It is easily proved that the arbitrage opportunity(AO) must consist of buying 40$(A) put options and selling 30$ (B) put options. Let x (A) puts be bought and y (B) puts be sold for the AO.
So we spend $$10x$$ for the (A) puts and gain $$My$$ for the (B) puts. Total gain =$$My - 10x$$
If the stock price is above 40; both puts remain unused. Therefore net earnings = $$My - 10x.$$ Since earning is greater than 0 in an AO, $$My - 10x \geq 0 \implies My > 10x \implies (30-M)My > 10(30-M)x$$. Also, since $$M < 10$$, we have $$y > x$$
If the stock price is $$= P = (0,30)$$; both puts are used. Earning on (A) puts = $$x(40 - P)$$. Loss on B puts$$= y(30 - P).$$ Total earning = $$x(40 - P) - y(30-P) + My - 10x = 30x - (30 - M)y + (y-x)P$$
Since $$y - x > 0$$. Total earning is minimum when $$P = 0$$.
The following is multiple choice question (with options) to answer.
X, Y and Z together invested Rs.106000, X invested Rs.4000 more than Y, and Rs.10000 more than Z. what is the share of X out of annual profit of Rs.26500 ? | [
"Rs.10000",
"Rs.10800",
"Rs.10500",
"Rs.11000"
] | A | Let Z share be ‘a’ then
Y = a + Rs.6000 [as X investment is Rs.10000 more than Z, and Rs.4000 more than Y, then Y share more than Z by Rs.10000 – Rs.4000 = Rs.6000]
X = a + Rs.10000.
X + Y + Z = 3a + 16000 = Rs.106000.
3a = 106000 – 16000 = Rs.90000.
a = 90000 / 3 = Rs.30000.
therefore Z investment = Rs.30000.
Y = Rs.30000 + Rs.6000 = Rs.36000.
X = Rs.30000 + Rs.10000 = Rs.40000.
Ratio of shares X : Y : Z = 20 : 18 : 15 = [53 units].
Investment Ratio = Profit sharing Ratio.
X = 26500 × 20 / 53 = Rs.10000
Option A |
AQUA-RAT | AQUA-RAT-39450 | Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.
So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.
We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.
Regards,
Saquib
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The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
A trade analyst announced 10% reduction in the unit price of bike. As a result, the sales volume went up by 25%. What was the net effect on the sales revenue? | [
"no change",
"decreases by 12.5%",
"increases by 12.5%",
"increases by 10%"
] | C | Explanation :
Reduction of price = (100 -10)% = 90% = 0.9
Increase of sale = (100+25)% = 125% = 1.25
Total effect = 0.90X 1.25 = 112.5%, Increases by 12.5%
Answer : C |
AQUA-RAT | AQUA-RAT-39451 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
10, 50, 13, 45, 18, 38, ?.
What a number should replace the question mark? | [
"34",
"27",
"25",
"46"
] | C | C
25
There are two alternate sequences: start at 10 and add 3, 5, 7;
start at 50 and deduct 5, 7, 9. |
AQUA-RAT | AQUA-RAT-39452 | 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.
Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 4 seconds, the speed of each train (in km/hr) is | [
"42",
"36",
"108",
"120"
] | C | Explanation:
Distance covered = 120+120 = 240 m
Time = 4 s
Let the speed of each train = v. Then relative speed = v+v = 2v
2v = distance/time = 240/4 = 60 m/s
Speed of each train = v = 60/2 = 30 m/s
= 30×36/10 km/hr = 108 km/hr
Answer: Option C |
AQUA-RAT | AQUA-RAT-39453 | Examveda
# In an institute, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 90, find the total number of students getting 50% concessions if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?
A. 360
B. 280
C. 320
D. 330
E. 350
### Solution(By Examveda Team)
Let us assume there are 100 students in the institute.
Then, number of boys = 60
And, number of girls = 40
Further, 15% of boys get fee waiver = 9 boys
7.5% of girls get fee waiver = 3 girls
Total = 12 students who gets fee waiver
But, here given 90 students are getting fee waiver. So we compare
12 = 90
So, 1 = $$\frac{{90}}{{12}}$$ = 7.5
Now number of students who are not getting fee waiver = 51 boys and 37 girls
50% concession = 25.5 boys and 18.5 girls (i.e. total 44)
Hence, required students = 44 × 7.5 = 330
1. 60%*15%+40%*7.5%=12%
12%=90
1=750
750-90=660
50%= 330
2. let total students = x
then
(15/100*60/100*x)+(7.5/100*40/100*x)=90
900x+300x=90,0000
x=750
number of students who are not getting fee waiver=750-90=660
50% of those not getting a fee waiver are eligible=660/2=330
required students=330
Related Questions on Percentage
The following is multiple choice question (with options) to answer.
In a college, 50% of total 400 Arts students are Locals. 25% of students from 100 Science students are locals and 85% of total 120 Commerce students are locals. What is the total percentage of locals from Arts, Science and Commerce. | [
"322",
"340",
"333",
"327"
] | D | locals from Arts = 50% of 400 = 200
locals from Science = 25% of 100 = 25
locals from Commerce = 85% of 120 = 102
total locals = 200+25+102 = 327
D |
AQUA-RAT | AQUA-RAT-39454 | # 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.
Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women? | [
"1/3",
"2/5",
"1/2",
"3/5"
] | D | probability = Desired Outcomes/Total Outcomes
Event = Selection of 3 employees to assign office from 5 such that 2 are men and 1 is woman
Total Outcomes = Selection of any 3 employees from 5 employees = 5 C 3 = 10
Desired Outcomes = Selection of 2 Men from 3 Men AND selection of 1 woman from 2 Women. = (3 C 2) * (2 C 1) = 3*2 = 6
Probability = 6/10 --------> 3/5
ANSWER:D |
AQUA-RAT | AQUA-RAT-39455 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
An outlet pipe can empty 2/3 rd of a cistern in 12 minutes. In 8 minutes, what part of the cistern will be emptied? | [
"4/9",
"2/3",
"3/4",
"5/6"
] | A | 2/3 ---- 12
? ----- 8 ==> 4/9
ANSWER A |
AQUA-RAT | AQUA-RAT-39456 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
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The following is multiple choice question (with options) to answer.
A couple who own an appliance store discover that if they advertise a sales discount of 10% on every item in the store, at the end of one month the number of total items sold increases 20%. Their gross income from sales for one month increases by what percent? | [
"2%",
"4%",
"5%",
"8%"
] | D | Let List price of an item = 100
Discount on each item = 10%
Discounted price of an item = .9*100 = 90
If they advertise a sales discount of 10% on every item in the store, at the end of one month the number of total items sold increases 20%
Originally if 10 items were sold in a month , with the new discount 12 items will be sold
Original revenue = No of items* Price of an item = 10 * 100 = 1000
New revenue = 12 * 90 = 1080
Increase in gross income = 1080 - 1000 = 80
% increase in gross revenue = 80/1000 * 100% = 8%
Answer D |
AQUA-RAT | AQUA-RAT-39457 | # 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.
Mohit sold an article for Rs. 18000. Had he offered a discount of 22% on the selling price, he would have earned a profit of 8%. What is the cost price of the article? | [
"13000",
"27767",
"16688",
"26678"
] | A | Let the CP be Rs. x.
Had he offered 22% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 - 22/100(18000) = Rs. 14040
=> 1.08x = 14040
=> x = 13000
\Answer: A |
AQUA-RAT | AQUA-RAT-39458 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
_________________
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A total of 22 men and 26 women were at a party, and the average [#permalink]
### Show Tags
04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
_________________
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Abhishek....
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Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
The average age of three persons is 52 years. Their ages are in the proportion of 1:5:7. What is the age in years of the oldest one among them. | [
"9 years",
"12 years",
"60 years",
"70 years"
] | D | Explanation:
Let the age of three persons be x, 5x, 7x
= 13x/3 = 52
x= 12 years
Answer: Option D |
AQUA-RAT | AQUA-RAT-39459 | 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 apples 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 Rs.20.50. Find the total cost of 4 kg of apples, 3 kg of rice and 5 kg of flour? | [
"777.4",
"877.4",
"577.4",
"477.4"
] | B | Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.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 = Rs.877.40
Answer: B |
AQUA-RAT | AQUA-RAT-39460 | 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.
If A is thrice as fast as B and together can do a work in 21 days. In how many days A alone can do the work? | [
"36",
"42",
"28",
"54"
] | C | A’s one day’s work= 1/X
B’s one day’s work= 1/3x
A + B’s one day’s work= 1/x + 1/3x = 1/21
= 3+1/3x = 4/3x = 1/21
x = 21*4/3 = 28
ANSWER:C |
AQUA-RAT | AQUA-RAT-39461 | "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 n is divided by 5 the remainder is 3. What is the remainder when (n + 3)^2 is divided by 5? | [
"0",
"1",
"2",
"3"
] | B | n = 5x+3, for some integer x
(n+3)^2=(5x+6)^2=5y+36, for some integer y
When we divide this by 5, the remainder is 1.
The answer is B. |
AQUA-RAT | AQUA-RAT-39462 | # 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.
A team scored a total of 126 points. If each player on the team scored at least 14 points, then what is the greatest possible number of players on the team? | [
"A)6",
"B)7",
"C)8",
"D)9"
] | D | 126/14=9 plus remainder.
The answer is D. |
AQUA-RAT | AQUA-RAT-39463 | ### 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.
The sale price sarees listed for Rs.400 after successive discount is 20% and 15% is? | [
"297",
"272",
"342",
"762"
] | B | 400*(80/100)*(85/100)
= 272
Answer: B |
AQUA-RAT | AQUA-RAT-39464 | 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 $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
The following is multiple choice question (with options) to answer.
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women? | [
"11760",
"266",
"5040",
"3000"
] | A | Req no of ways = 8C5 * 10C6
=11760
ANSWER A |
AQUA-RAT | AQUA-RAT-39465 | 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 $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
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.
The following is multiple choice question (with options) to answer.
P is three times as fast as Q and working together, they can complete a work in 14 days. In how many days can Q alone complete the work? | [
"18 1/2 days",
"17 1/2 days",
"19 1/2 days",
"14 1/4 days"
] | A | P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 14 days, which means Q can do the work in 48 days.
Hence, P can do the work in 18 1/2 days.
Answer:A |
AQUA-RAT | AQUA-RAT-39466 | # 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.
If an investor puts $800 in a savings account that earns 10 percent annual interest compounded semiannually, how much money will be in the account after one year? | [
"$878",
"$880",
"$882",
"$884"
] | C | 1.05*1.05*800=$882
The answer is C. |
AQUA-RAT | AQUA-RAT-39467 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
There are 3 bags, in 1st there are 9 Mangoes, in 2nd 8 apples & in 3rd 6 bananas. There are how many ways you can buy one fruit if all the mangoes are identical, all the apples are identical, & also all the Bananas are identical? | [
"648",
"432",
"23",
"2 to the power(23)-1"
] | C | Since bags are different so it should be 9+8+6=23
ANSWER:C |
AQUA-RAT | AQUA-RAT-39468 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
How many three digits number can be formed by using the digits 1,2,3,4,5? | [
"152",
"125",
"225",
"325"
] | B | Here the repetition of digits allowed. The unit place of the number can be filled in by any of the given numbers in 5 ways. Simillarly each one of the ten's and the hundred's place can be filled in 5 ways. Therefor the total number of required numbers =5*5*5 =125 Answer : B |
AQUA-RAT | AQUA-RAT-39469 | # 2010 AMC 10B Problems/Problem 25
## Problem
Let $a > 0$, and let $P(x)$ be a polynomial with integer coefficients such that
$P(1) = P(3) = P(5) = P(7) = a$, and
$P(2) = P(4) = P(6) = P(8) = -a$.
What is the smallest possible value of $a$?
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$
## Solution
We observe that because $P(1) = P(3) = P(5) = P(7) = a$, if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$, $R(x)$ has roots when $P(x) = a$; namely, when $x=1,3,5,7$.
Thus since $R(x)$ has roots when $x=1,3,5,7$, we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$.
Then, plugging in values of $2,4,6,8,$ we get
The following is multiple choice question (with options) to answer.
Can you please walk me through how to best approach this problem? Thanks
If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# = 6, what is the value of #5#? | [
"5",
"0",
"-8",
"-3"
] | C | #p# = ap^3 + bp - 1
#-5# = 6
putting p = -5 in above equation
-125a -(5b +1) = 6 or
#-5# = (125a+5b+1) = -6
therefore 125a+5b = -7 .....(1
now putting p = 5
#5# = 125 a+5b - 1
using equation 1(125a+5b = -7)
#5# = -7-1 = -8
hence C |
AQUA-RAT | AQUA-RAT-39470 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is? | [
"81.6",
"81.7",
"81.0",
"281.6"
] | A | SI = 40 + 40
CI = 40 + 40 + 1.6 = 81.6.Answer: A |
AQUA-RAT | AQUA-RAT-39471 | of a circle and therefore its area is a fraction of the area of the whole circle. com Area and Perimeter of Circle and Semi-Circle Perimeter. 1) A circle has a radius of 5 cm. Area of rectangle = 4 × 8 = 32 m² A circle of radius 8 cm is cut into 6 parts of equal size, as shown in the diagram. A button is made in the shape of a circle, with two congruent rectangles cut out, as shown in the diagram. The area of a partial circle is the area of the circular segment equal to the area of the circular sector minus the area of the triangular portion formed from the. So cropping is quick, highly secured and consumes less bandwidth. The dimensions are m = 80. To find the perimeter of a rectangle, add the lengths of the rectangle's four sides. Cut a rectangle (more or less about 8 X 16 inches) of your 'swimsuit' stretch fabric. Tie the ribbon to zipper pull. The perimeter of a plane 2-D figure is the length of the outer periphery of the figure. **Many of our plants are listed as Not a Good Fit for Kid Traffic/Dog Traffic/Around Pools. Geometry calculator solving for arc length given central angle and circle radius. – Project-Based Learning. But we could get a better result if we divided the circle into 25 sectors (23 with an angle of 15° and 2 with an angle of 7. Semi-circular arcs are drawn with the sides of the rectangle as diameters. 99) Find great deals on the latest styles of Half circle area rugs. To find the total area, find the area of each part and add them together. The figure can be any regular or irregular polygon; it does not have to be a regular geometric figure. Angles Games. 68 cm 2 Example #2: Find the surface area of a cylinder with a radius of 4 cm, and a height of 3 cm. A filled rounded rectangle with b=0 (or a=0) is called a stadium. 14 for pie , and do not round your answer. cropping is much Faster, since we are not uploading your images to our server. The rectangle has a dimensions 5 length and 4 width, the semicircle has a radius of 2 using 3. Hence the diameter of a circle with area equal to 40 square cm is found to be 7. The round cut-out
The following is multiple choice question (with options) to answer.
A 25 cm wide path is to be made around a circular garden having a diameter of 4 meters. Approximate area of the path is square meters is? | [
"3.34 sq m",
"3.38 sq m",
"3.24 sq m",
"3.36 sq m"
] | A | Area of the path = Area of the outer circle - Area of the inner circle
= ∏{4/2 + 25/100}2 - ∏[4/2]2
= ∏[2.252 - 22] = ∏(0.25)(4.25) { (a2 - b2 = (a - b)(a + b) }
= (3.14)(1/4)(17/4)
= 53.38/16
= 3.34 sq m
Answer:A |
AQUA-RAT | AQUA-RAT-39472 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
If the gross salary for first year is Rs. 45000 and an employee has two options to choose from for increment .Either he can opt for 4% yearly increase or Rs. 25000/- yearly increase on the gross salary. If an employee choose the first option, after how many years his salary will be more than the option two ? | [
"15 years",
"16 years",
"17 years",
"18 years"
] | D | if initial annual gross salary = Rs. 450000/-
check when
450000+25000*n = 450*(1+4/100)^n
After 18 years
ANSWER:D |
AQUA-RAT | AQUA-RAT-39473 | $7|61$ gives $61=7\cdot 8 +5$
He would have 5 cows left over. So 61 can't be an answer
5. Hello, swimalot!
A cowboy was asked how many cows were on the ranch.
He replied that he was unsure, but he knew that when he counted them
by twos, threes, fours, fives, or sixes, he always had one left over.
When counted by sevens, he had none left over.
What is the smallest number of cows on the ranch?
Let $N$ = number of cows on the ranch.
The LCM of 2, 3, 4, 5, 6 is $60$
. . Hence: . $N \:=\:60a + 1$ ... for some integer $a.$
Since $N$ is divisible by 7: . $60a + 1 \:=\:7b$ ... for some integer $b.$
Solve for $b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}$
Since $b$ is an integer, $4a + 1$ must be divisible by 7.
The first time this happens is: $a = 5$
Therefore: . $N \;=\;60(5)+1 \;=\;\boxed{301}$
6. Originally Posted by TheEmptySet
we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.
here is our list,
56,63,70,77,84,91
If you check all the other conditions you will see that they hold.
I hope this helps.
Hello Tessy
91 doesn't work for ... four
91=88+3
7. You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html
8. I will do the Chinese Remainder Theorem for you:
The following is multiple choice question (with options) to answer.
A ranch has both horses and ponies. Exactly 5/7 of the ponies have horseshoes, and exactly 2/3 of the ponies with horseshoes are from Iceland. If there are 3 more horses than ponies, what is the minimum possible combined number of horses and ponies on the ranch? | [
"33",
"37",
"41",
"45"
] | D | 5/7*P are ponies with horseshoes, so P is a multiple of 7.
2/3*5/7*P = 10/21*P are Icelandic ponies with horseshoes, so P is a multiple of 21.
The minimum value of P is 21. Then H = P+3 = 24.
The minimum number of horses and ponies is 45.
The answer is D. |
AQUA-RAT | AQUA-RAT-39474 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
In 1986 the book value of a certain car was 2/3 of the original price, and in 1988 its book value was 1/2 of the original purchase price. By what percent did the book value for this car decrease from 1986 to 1988? | [
"16 2/3 %",
"25 %",
"33 1/3 %",
"50%"
] | B | 1986; 2/3
1988; 1/2
% decrease = change/original*100
2/3−1/2 / 2/3∗100
1/6∗3/2∗100=14∗100=25
Answer: "B" |
AQUA-RAT | AQUA-RAT-39475 | thanks in advance.
2. Originally Posted by Simo
Dear all,
I need to know if is it possible to find the mean of a normal distribution knowing the ratio of the area under the "bell" curve on the left and on the right of a given point.
To be more precise I need to solve the following equation
$\frac{1}{\sigma \sqrt{2\pi }}\int_{x=\bar{X}}^{+\infty }\exp \left( \frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}\right) dx=\frac{k}{\sigma \sqrt{2\pi }}\int_{x=-\infty}^{\bar{X} }\exp \left( \frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}\right) dx$
where I know the ratio $k$, the variance $\sigma ^{2}$ and the point $\bar{X}$ is fixed. i need to find the mean of the distribution $\mu$ that achieve the ratio of $k$ between the two portion of area.
thanks in advance.
hey mate, just so Im reading your your question correctly, you have an integral equation of the form
int(f(x)) = k*inf(g(x)) and you want to solve for k?
Regards,
David
3. David,
my problem is in the form
$\int_{x=\bar{X}}^{+\infty }f(\mu,x) dx=k\int_{x=-\infty}^{\bar{X} }f(\mu,x) dx$
and I want to solve it for $\mu$. I already know $k$ and $\bar{X}$.
4. Hey Simo,
The following is multiple choice question (with options) to answer.
A Bell Curve (Normal Distribution) has a mean of − 1 and a standard deviation of 1/8 . How many integer values Q are within three standard deviations of the mean? | [
"0",
"1",
"3",
"6"
] | B | Got the question correctly -- the second item -- [highlight]bthe list of elements in the set is required.[/highlight] is not required.
With the new information, there is only one integer value (-1) that is between (-1.375, -0.625) i.e., falls within the three 3 SD range.B |
AQUA-RAT | AQUA-RAT-39476 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 kms in 4 hours, then the speed of the first train is: | [
"70 km/hr",
"75 km/hr",
"80 km/hr",
"87.5 km/hr."
] | D | Let the speed of two trains be 7x and 8x km/hr.
Then, 8x = 400/4 = 100
Let the speed of two trains be 7x and 8x km/hr.
x = 100/8 = 12.5
Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.
Answer : D. |
AQUA-RAT | AQUA-RAT-39477 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
Rahul's Mathematics test had 75 problems, 10 arithmetic, 30 algebra, 35 geometry problems.
Although he answered 70% of arithmetic, 40% of arithmetic and 60% of geometry problems correctly,
still he got less than 60% problems right. How many more questions he would have to answer more to get
passed | [
"5",
"6",
"7",
"8"
] | A | Explanation:
Number of questions attempted correctly = (70% of 10 + 40% of 30 + 60% of 35)
= 7 + 12 + 21 = 40.
Questions to be answered correctly for 60% = 60% of total quations
= 60 % of 75 = 45.
He would have to answer 45 - 40 = 5
Answer: Option A |
AQUA-RAT | AQUA-RAT-39478 | homework-and-exercises, newtonian-mechanics, rotational-kinematics
Now your doubt is that why (2) holds it might happen that $\alpha$ can be positive or negative and thus $\omega_i$ can be greater than or less than $\omega_f$.
In the above problem it is given that wheel covers 90 revolutions in 15 sec. So on averge it angular velocity is $\frac{90}{15}=6$ revolutions/sec.
But final velcity is $10$ revolutions/sec.
This means that at some point of time the velocity is less than $6$ revolutions/sec only then we get an average of 6 revolution/sec.
But we are given that angular acceleration is constant. So velocity can either decrease or increase in that time interval. It can't be the case that at some interval velocity will decrease and then increase as for that angular acceleration has to change sign so it won't remains constant giving us contradiction.
As at some interval angular velocity has to take value less than $6$ revolutions/sec.
This suggests that over the whole interval velocity will increase continuously thus initial angular velocity will be less that final angular velocity.
So, the take home message is that in the equation $(2)$ the information of angular acceleration is inherent, though it might not be so much evident from the form of equation but we can see this from the derivation.
Hope that helps!
The following is multiple choice question (with options) to answer.
A wheel that has 6 cogs is meshed with a larger wheel of 9 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is: | [
"49",
"4",
"12",
"14"
] | D | Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
9 : 6 :: 21 : x 9 * x = 6 x 21
x = (6 x 21)/9
x = 14.
Answer is D. |
AQUA-RAT | AQUA-RAT-39479 | (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 soda machine sells both bottles and cans, and no other items. Bottles cost $1.50 each, while cans cost $0.75 each. If on one day, the soda machine sold 250 total beverages and yielded $315, how many more bottles than cans were sold? | [
"60",
"80",
"90",
"115"
] | C | b+c=250_______(1)
1.5b+0.75c=315
2b+c=420_______(2)
(2)-(1)
b=170
c=80
b-c=170-80=90
ANSWER:C |
AQUA-RAT | AQUA-RAT-39480 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
The probability that a visitor at the mall buys a pack of candy is 10%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy? | [
"0.009",
"0.015",
"0.027",
"0.036"
] | C | One case is: candy - candy - no candy
The probability is 1/10*1/10*9/10 = 9/1000
There are 3 such cases, so we should multiply this probability by 3.
P(exactly 2 buy candy) = 9/1000 * 3 = 27/1000 = 0.027
The answer is C. |
AQUA-RAT | AQUA-RAT-39481 | # Sort of a challenge
#### topsquark
##### Well-known member
MHB Math Helper
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
-Dan
#### Bacterius
##### Well-known member
MHB Math Helper
Well the Earth's radius is $6371 ~ \text{km}$, so the original rope has a length of $2 \pi \times 6371 ~ \text{km}$. If we add $6 ~ \text{m} = 6 \times 10^{-3} ~ \text{km}$ to this rope, its new radius is:
$$\frac{2 \pi \times 6371 + 6 \times 10^{-3}}{2 \pi} \approx 6371.0009 ~ \text{km}$$
So the rope now "floats" about $0.0009 ~ \text{km} = 90 ~ \text{cm}$ above the ground. To be exact, $95.5 ~ \text{cm}$ (this is $\frac{6}{2 \pi} ~ \text{m}$).
Wait, what? My mind is blown
#### MarkFL
Staff member
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
The following is multiple choice question (with options) to answer.
A rope of which a calf is tied is increased from 12 m to 23 m, how much additional grassy ground shall it graze? | [
"1222",
"1289",
"1210",
"1129"
] | C | π (232 – 122)
= 1210
Answer:C |
AQUA-RAT | AQUA-RAT-39482 | ### Show Tags
19 May 2015, 12:37
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
### Show Tags
20 May 2015, 03:32
BrainLab wrote:
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
Dear BrainLab
Perfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation
$$(\frac{5}{4})^4*X = 6250$$
will take lesser time to solve (especially if you know that $$5^4 = 625$$) than $$(1.25)^4*X = 6250$$
Hope this was useful!
Japinder
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] 18 Jan 2016, 23:36
The following is multiple choice question (with options) to answer.
A number increased by 40% gives 1680. The number is? | [
"1680",
"1600",
"1200",
"1500"
] | C | Formula = TOTAL=100% ,INCREASE = "+" DECREASE= "-"
A number means = 100 %
That same number increased by 40 % = 140 %
140 % -------> 1680 (140 × 12= 1680)
100 % -------> 1200 (100 × 12 = 1200)
Option 'C' |
AQUA-RAT | AQUA-RAT-39483 | # 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 dishonest dealer professes to sell his goods at Cost Price but still gets 40% profit by using a false weight. What weight does he substitute for a kilogram? | [
"714 1/7 grams",
"714 2/7 grams",
"714 8/3 grams",
"714 1/3 grams"
] | B | If the cost price is Rs.100, then to get a profit of 40%, the selling price should be Rs.140.
If 140kg are to be sold, and the dealer gives only 100kg, to get a profit of 40%.
How many grams he has to give instead of one kilogram(1000 gm).
140 gm ------ 100 gm
1000 gm ------ ?
(1000 * 100)/140 = 714 2/7 grams.
Answer:B |
AQUA-RAT | AQUA-RAT-39484 | Another answer that use almost the same idea: the sum or subtraction of two even or odd number is an even number. How many odd number we have?
Replacing 100 with $n$ and using Brian M. Scott's solution, we want a partition of $\{1, 2, ..., n+1\}$ into two sets with equal sums.
The sum is $\frac{(n+1)(n+2)}{2}$, and if $n=4k$, this is $(4k+1)(2k+1)$ which is odd and therefore impossible.
If $n = 4k+1$, this is $(2k+1)(4k+3)$ which is also odd, and therefore impossible.
If $n = 4k+2$, this is $(4k+3)(2k+2)$, so it is not ruled out, and each sum must be $(4k+3)(k+1)$.
if $n = 4k+3$, this is $(2k+2)(4k+5)$ which is also not ruled out, and each sum must be $(k+1)(4k+5)$.
Now I'll try to find a solution for the not impossible cases. (I am working these out as I enter them.)
The following is multiple choice question (with options) to answer.
Set A contains all the even numbers between 2 and 50 inclusive. Set B contains all the even numbers between 62 and 110 inclusive. What is the difference between the sum of elements of set B and the sum of the elements of set A? | [
"1500",
"3100",
"4400",
"6200"
] | A | Each term in set B is 60 more than the corresponding term in set A.
The difference of the sums = 25 * 60 = 1500.
The answer is A. |
AQUA-RAT | AQUA-RAT-39485 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
Each shelf of a bookcase contained 12 books. If the librarian took out 21 books and rearranged the remaining books so that all shelves but the last one contained 8 books and that last shelf contained 3 books, how many shelves does the bookcase have? | [
"4",
"5",
"6",
"7"
] | A | Let x be the number of shelves.
12x - 21 = 8(x-1) + 3
4x = 16
x= 4
The answer is A. |
AQUA-RAT | AQUA-RAT-39486 | # computing probability that at least one letter matches with envelope
Four letters to different insurers are prepared along with accompanying envelopes. The letters are put into the envelopes randomly. Calculate the probability that at least one letter ends up in its accompanying envelope.
### Attempt
Since this is tedioues, we can do
$$P(at \; least \; one ) = 1 - P( no \; match )$$
We notice that sample space is $4!$ since for letter 1 it has 4 choices but letter 2 has 3 choices and so on. Now, we wanna count in how many ways we get no match.
Let start with first one, we only have $3$ choices for this since it can go to either 2,3,4 envelope.
Now, as for the second one, we have to possibilities. If the first letter went to the second envelope, then the second letter now will have 3 different choices, but if the first letter didnt go to second letter, then the second letter will have 2 choices. Assume the former. Then we have 3 choices for this stage.
Now, for the third one (envelope 2 is taken already and assume letter 2 went to letter 1) then it would have 1 choice only and
last one must go to envelope 3.
Thus, we have $3 \times 3 \times 1 \times 1 = 9$ choices in total
Thus,
$$P(at \; least \; 1 \; letter) = 1 - \frac{9}{24}$$
IS this correct? I still feel as is something wrong because I assumed the letter 2 went to 1 and letter 1 went to envelope 2. Can we do that?
The following is multiple choice question (with options) to answer.
Lanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address? | [
"1/24",
"1/8",
"1/4",
"1/3"
] | D | Total Lanya# of ways - 4! = 24.
Desired:
A-Mapped to the only correct address----------x 1 way only
B-Mapped to other two incorrect addresses - x 2 ways
C-Mapped to other two incorrect addresses - x 2 ways
D-Mapped to other two incorrect addresses - x 2 ways
Therefore, 1*2*2*2/24 = 1/3.? |
AQUA-RAT | AQUA-RAT-39487 | 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.
10 years ago, the average age of a family of 4 members was 24 years. Two children having been born (with age diference of 2 years), the present average age of the family is the same. The present age of the youngest child is ? | [
"1",
"7",
"3",
"9"
] | C | Total age of 4 members, 10 years ago = (24 x 4) years = 96 years.
Total age of 4 members now = [96 + (10 x 4)] years = 136 years.
Total age of 6 members now = (24 x 6) years = 144 years.
Sum of the ages of 2 children = (144 - 136) years = 8 years.
Let the age of the younger child be years.
Then, age of the elder child = years.
So,
Age of younger child = 3 years.
Answer: C |
AQUA-RAT | AQUA-RAT-39488 | 1. ## Having Trouble
Hi all,
I am just new to this site and it looks like such an awesome resource.
Can anyone help me with this???
If x + y = 1 and x³ + y³ = 19, find the value of x² + y²
2. Originally Posted by Joel
Hi all,
I am just new to this site and it looks like such an awesome resource.
Can anyone help me with this???
If x + y = 1 and x³ + y³ = 19, find the value of x² + y²
$\left\{\begin{array}{lclcr}
x&+&y&=&1\\
x^3&+&y^3&=&19
\end{array}\right.$
$\Rightarrow x^3+(1-x)^3=19,$
and after expanding and simplifying,
$3x^2-3x-18=0$
$\Rightarrow3(x-3)(x+2)=0.$
You should be able to finish.
3. Hello, Joel!
Welcome aboard!
I have a back-door approach . . .
If . $\begin{array}{cccc}x + y &=& 1 & {\color{blue}[1]} \\ x^3 + y^3 &=& 19 & {\color{blue}[2]} \end{array}$ . find the value of $x^2+y^2.$
Cube [1]: . $(x+y)^3 \:=\:1^3\quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:1$
$\text{We have: }\;\underbrace{(x^3 + y^3)}_{\text{This is 19}} + \;3xy\underbrace{(x+y)}_{\text{This is 1}} \:=\:1 \quad\Rightarrow\quad 19 \;+ \;3xy \:=\:1 \quad\Rightarrow\quad xy \:=\:-6$
The following is multiple choice question (with options) to answer.
If x+y=25 and x2y3 + y2x3=25, what is the value of xy? | [
"1",
"2",
"3",
"4"
] | A | xy=1
As x+y=25
x2y3+y2x3=25
x2y2(y+x)=25
Substituting x+y
x2y2=1
xy=1
ANSWER:A |
AQUA-RAT | AQUA-RAT-39489 | 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 train moves fast a telegraph post and a bridge 264 m long in 8 sec and 20 sec respectively. What is the speed of the train? | [
"79.4",
"79.0",
"79.5",
"79.2"
] | D | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 264)/20 = y
y = 22
Speed = 22 m/sec = 22 * 18/5 = 79.2 km/hr.
Answer:D |
AQUA-RAT | AQUA-RAT-39490 | +0
# At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
0
2788
3
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Feb 25, 2015
#3
+99377
+5
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$
so
$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$
$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$
Obviously there is not a neg number of people so there must be 12 people.
Feb 25, 2015
#1
0
can you even solve that
Feb 25, 2015
#2
+98196
+5
We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
The following is multiple choice question (with options) to answer.
If 12 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes? | [
" 10•9•8•7•6•5•4•3•2•1",
" 10•10",
" 10•9",
" 66"
] | D | We got #12 people who shake each other's hands once ==> a pair of 2
12!/10!2! = 12*11 / 2*1 = 66.
Hence answer D. |
AQUA-RAT | AQUA-RAT-39491 | 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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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.
A pupil's marks were wrongly entered as 79 instead of 45. Due to that the average marks for the class got increased by half. The number of pupils in the class is : | [
"30",
"80",
"20",
"68"
] | D | Let there be x pupils in the class.
Total increase in marks = (X*1/2) = X/2.
X/2 = (79 - 45)
=> X/2 = 34
=> X = 68.
ANSWER:D |
AQUA-RAT | AQUA-RAT-39492 | 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 can do a piece of work in 40 days; B can do the same in 30 days. A started alone but left the work after 10 days, then B worked at it for 10 days. C finished the remaining work in 10 days. C alone can do the whole work in? | [
"days",
"days",
"days",
"1/2 days"
] | A | Explanation:
10/40 + 10/30 + 10/x = 1
x = 24 days
Answer: A |
AQUA-RAT | AQUA-RAT-39493 | java, xml, swing
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/12 00:00:00</date>
<riddle>ipso lorem 7</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/19 00:00:00</date>
<riddle>ipso lorem 8</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/26 00:00:00</date>
<riddle>ipso lorem 9</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/02 00:00:00</date>
<riddle>ipso lorem 10</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/09 00:00:00</date>
<riddle>ipso lorem 11</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/16 00:00:00</date>
<riddle>ipso lorem 12</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/23 00:00:00</date>
<riddle>ipso lorem 13</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/30 00:00:00</date>
<riddle>ipso lorem 14</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
</weeks>
The following is multiple choice question (with options) to answer.
What will be the day of the week 15th August, 2010? | [
"Sunday",
"Monday",
"Tuesday",
"Thursday"
] | A | 15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days.
Jan. Feb. March April May June July Aug.
(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days
227 days = (32 weeks + 3 days) 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days.
Given day is Sunday.
Answer: Option A |
AQUA-RAT | AQUA-RAT-39494 | of$5$, and$-2=5\pmod{7}$. – joriki Jun 10 '11 at 0:20 Regarding divisibility by 7 I created an algorithm that must be applied repetitively to each period of a large number. If the last sum results in a multiple of 7 then the tested number is also a multiple of 7, otherwise it is not. The sum of each period must be added to the sum of the next period. N = abc algorithm: – ( (x) + a + b + c + (y) ) mod 7 (x) must be mentally inserted before the hundreds and (y) must be mentally inserted after the ones in such a way that 7 divides both (x)a and c(y). In this rule no digits are inserted before or after the tens. Numerical examples: 1) N = 462; – ( (1) + 4 + 6 + 2 + (1) ) mod 7 ≡ Ø 2) N = 863; – ( (2) + 8 + 6 + 3 + (5) ) mod 7 ≡ 4 3) N = 1.554; – ( 1 + (4) ) mod 7 ≡ 2; – ( 2 + (3) + 5 + 5 + 4 + (2) ) mod 7 ≡ Ø 4) N = 68.318; – ( 6 + 8 + (4) ) mod 7 ≡ 3; – ( 3 + (6) + 3 + 1 + 8 +(4) ) mod 7 ≡ 3 5) N = 852.655; – ( (2) + 8 + 5 + 2 + (1) ) mod 7 ≡ 3; – ( 3 + (5) + 6 + 5 + 5 + (6) ) mod 7 ≡ 5 To determine the remainder take the digit that forms with the last sum a multiple of 7. Remainders of the second, fourth and fifth examples: 2) Last sum is 4 → 4(2); 2 is the remainder 4) Last sum is 3 → 3(5); 5 is the remainder 5) Last sum is 5 → 5(6); 6 is the remainder The application of this rule must be performed in a dynamic way. It is not necessary to write the inserted numbers; all you need is to add the written numbers and the mentally inserted numbers at sight, determine the difference of each sum and its next multiple of 7
The following is multiple choice question (with options) to answer.
What must be added to each of the numbers 7, 11 and 19, so that the resulting numbers may be in continued proportion? | [
"3",
"5",
"4",
"-3"
] | D | Let X be the required number, then
(7 + X) : (11 + X) :: (11 +X) : (19 + X)
(7 + X) (19 + X) = (11 + X)2
X2 + 26X + 133 = X2 + 22X + 121
4X = - 12 or X = - 3
ANSWER:D |
AQUA-RAT | AQUA-RAT-39495 | 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 X. If the board is to be split up into 2 three-person subcommittees, what percent E of all the possible subcommittees that include Michael also include Anthony? | [
" 20%",
" 30%",
" 40%",
" 50%"
] | C | The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability E= 1/5+4/5*1/4 = 2/5.C |
AQUA-RAT | AQUA-RAT-39496 | 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.
Find the simple interest on Rs.520 for 7 months at 8 paisa per month? | [
"277",
"270",
"291",
"266"
] | C | I = (520*7*8)/100 = 291
Answer: C |
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