source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37797 | +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.
A number is doubled and 5 is added. If the resultant is trebled, it becomes 129. What is that number? | [
"12",
"29",
"27",
"19"
] | D | Explanation:
Let the number be x.
Therefore,
3(2x + 5) = 129
6x + 15 = 129
6x =114
x = 19
ANSWER:D |
AQUA-RAT | AQUA-RAT-37798 | 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 running at the speed of 60 km/hr crosses a pole in 6 seconds. Find the length of the train? | [
"100 meter",
"170 meter",
"156 meter",
"168 meter"
] | A | Speed = 60*(5/18) m/sec
= 50/3 m/sec
Length of Train (Distance)
= Speed * Time
(50/3) * 6
= 100 meter
Answer:A |
AQUA-RAT | AQUA-RAT-37799 | Here is the situation: My friend and I are at an impasse. I believe I'm correct, but he's so damn stubborn he won't believe me. Also, I'm not the most articulate at explaining things. Hopefully some of you guys can help me explain this to him in a way he'll understand. Here is the problem:
A DVD is either at his parent's house or his own. The probability that it's at his house is 30%. If the DVD is at his own house, there is a 90% chance it's on the porch, and a 10% chance it's in the living room. What is the % chance the DVD is on the porch?
My friend says you take 90% of 30% which is 27 and that is the % chance it's on the porch. Is this correct? I don't believe so.
I believe that regardless of where the DVD is, the chance of it being anywhere in his house is still 30% overall. Location inside his house won't change those odd because the porch and the living room are both part of the house. If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.
Now, if you rephrase the question and ask, "The DVD Is either at my parents, my porch, or my living room. What is the % chance it's on my porch?", the answer is 33%. If there are three places it could be, then there is 33.333% chance it's on the porch. Even if it's a 90% chance it's at his house, if there are only three places it can be, it remains the same.
I think the correct way of answering the question is: There is a 30% chance the DVD is at my house. If it is at my house, there is a 90% chance it's on my porch. They are two separate odds and you can't take a percentage of the overall odds since the locations are inside the house.
Is this correct or am I wrong? And regardless, please give me your explanation.
The following is multiple choice question (with options) to answer.
EASY,INC recently conducted a survey and found that 60,000 of its customers live in rural areas. if the number of it customers who live in urban areas is 200 percents grater than the numbers of customers who live in rural areas, how many of EASY,INC customers live in urban areas ? | [
"200,000",
"180,000",
"360,000",
"480,000"
] | B | Let's say that we have a starting value of X...
100% of X = X
100% MORE than X = X + X = 2X
200% MORE than X = X + 2X = 3X
300% MORE than X = X + 3X = 4X
Etc.
Here, we're told that 60,000 customers live in rural areas and the number who live in urban areas is 200% GREATER than the number who live in rural areas. That means that the number who are in urban areas is 3 times the number in rural areas...
3(60,000) = 180,000
Final Answer:B |
AQUA-RAT | AQUA-RAT-37800 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
Evaluate: |7 - 8(3 - 12)| - |5 - 11| = ? | [
"40",
"50",
"73",
"70"
] | C | According to order of operations, inner brackets first. Hence
|7 - 8(3 - 12)| - |5 - 11| = |7 - 8*(-9)| - |5 - 11|
According to order of operations, multiplication within absolute value signs (which may be considered as brackets when it comes to order of operations) next. Hence
= |7 + 72| - |5 - 11|
= |79| - |-6|
= 79 - 6 = 73
correct answer C)73 |
AQUA-RAT | AQUA-RAT-37801 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? | [
"34",
"40",
"68",
"88"
] | D | EXPLANATION
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Answer D |
AQUA-RAT | AQUA-RAT-37802 | The largest it can be is actually also the smallest it can be. In fact, if the numbers 1 through $n$ are written and the same process followed, the end result will be $(n+1)! - 1$ no matter what order you combine numbers.
Let's take a smaller set, just $\{a, b, c\}$, to see why. If you group $a$ and $b$ first, you'll end up with $$(ab+a+b)c+(ab+a+b)+c=a+b+c+ab+ac+bc+abc$$ If you group $b$ and $c$ first, you get $$(bc+b+c)a+(bc+b+c)+a=a+b+c+ab+ac+bc+abc$$ And just for completeness, grouping $a$ and $c$ first gives $$(ac+a+c)b+(ac+a+c)+b=a+b+c+ab+ac+bc+abc$$ At the end of the $n$ numbers, you will always end up with the sum of the individual numbers, plus the sum of the products of the numbers taken 2 at a time, plus the sum of the products taken 3 at a time, all the way up to the product of all the numbers. If you added 1 to the final sum, you could factor the final result into $(a+1)(b+1)(c+1)...$ giving $(n+1)!$ in this case.
Similarly, a starting list of the number $k$ written $n$ times will result in $(k+1)^n-1$.
The following is multiple choice question (with options) to answer.
In a set of three numbers,the difference between the largest and the second largest numbers is added to the smallest number.The average of the largest,second largest and the new number formed exceeds the average of the original three numbers by 16.The largest number exceeds the second largest number by how much ? | [
"5",
"10",
"30",
"15"
] | D | Let the numbers in increasing order be A<B<C...
since the new number(N) increase the average of new set of numbers by 5 after replacing the smallest number,A,
the new number,N, exceeds the smallest,A, by 3*5=15 orN=15+A..(1)
and what is the new number,N,----C-B (diff in the largest and second largest)+A...(2)...
from (1)(2) C-B=15
Answer:D |
AQUA-RAT | AQUA-RAT-37803 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
Murari and Madan Started a business with investments of Rs. 13000 and Rs. 52000. After 5 months Arun joins with a capital of Rs. 52000. At the end of a year they got a profit of Rs. 14250. Find the share of Arun. | [
"Rs. 3650",
"Rs. 5250",
"Rs. 6750",
"Rs. 2250"
] | B | Explanation:
Rs. 5250
Answer: Option B |
AQUA-RAT | AQUA-RAT-37804 | # Homework Help: Derive a formula for motion with constant acceleration and constant deceleration
1. Jul 11, 2012
### LoA
1. The problem statement, all variables and given/known data
A subway train travels over a distance $s$ in $t$ seconds. it starts from rest and ends at rest. In the first part of its journey it moves with constant acceleration $f$ and in the second part with constant deceleration $r \,$.
Show that $s \, = \, \frac {[\frac {fr} {f \, + \, r}] \, t^2} {2}$
2. Relevant equations
I know that $s \,= \, (\frac {1}{2})(-r )t^2\, +\, v_it$, where $v_i \,=\, ft$ but I'm not sure where to go from there. In particular, I can't figure out how to connect the seemingly separate equations for distance generated by the different accelerations into one function of time for the entire interval.
3. The attempt at a solution
By assuming that total acceleration is a sum of the given accelerations, I've gotten something that looks awfully close to the desired result, but am still not quite there:
$s \,=\,\frac{1}{2} \,(f\,+\,r)\,t^2 \, + \,v_i\,t$
I feel like I'm missing something.
Last edited: Jul 11, 2012
2. Jul 11, 2012
### !)("/#
You have to integrate the aceleration two times to get the movement ecuation.
You know that:
$x (t) = x_0 + v_0 (t-t_0) + \frac{1}{2} a (t-t_0)^2$
$v (t) = v_0+ a (t-t_0)$
Part one we start from time, position and initial velocity cero and also positive aceleration f:
$x (t) =\frac{1}{2} f t^2$
$v (t) = a t$
The following is multiple choice question (with options) to answer.
The speed of a subway train is represented by the equation z=s^2+2s for all situations where 0≤s≤7, where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving. In kilometers per hour, how much faster is the subway train moving after 7 seconds than it was moving after 5 seconds? | [
"4",
"28",
"15",
"48"
] | B | Given: z=s^2+2s for 0≤s≤7
z(5) = 5^2 + 2*5 = 35
z(7) = 7^2 + 2*7 = 63
Therefore z(7) - z(3) = 63 - 35 = 28 km/hr
Option B |
AQUA-RAT | AQUA-RAT-37805 | Alternatively, the lcm of 54 and 72 can be found using the prime factorization of 54 and 72:
• The prime factorization of 54 is: 2 x 3 x 3 x 3
• The prime factorization of 72 is: 2 x 2 x 2 x 3 x 3
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(54,54) = 216
In any case, the easiest way to compute the lcm of two numbers like 54 and 72 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 54,72. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 54 and 72
What is the least common multiple of 54 and 72 used for? Answer: It is helpful for adding and subtracting fractions like 1/54 and 1/72. Just multiply the dividends and divisors by 4 and 3, respectively, such that the divisors have the value of 216, the lcm of 54 and 72.
$\frac{1}{54} + \frac{1}{72} = \frac{4}{216} + \frac{3}{216} = \frac{7}{216}$. $\hspace{30px}\frac{1}{54} – \frac{1}{72} = \frac{4}{216} – \frac{3}{216} = \frac{1}{216}$.
## Properties of LCM of 54 and 72
The most important properties of the lcm(54,72) are:
• Commutative property: lcm(54,72) = lcm(72,54)
• Associative property: lcm(54,72,n) = lcm(lcm(72,54),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The following is multiple choice question (with options) to answer.
Two numbers N and 12 have LCM = 42 and GCF = 6. Find N. | [
"35",
"56",
"21",
"87"
] | C | The product of two integers is equal to the product of their LCM and GCF. Hence.
12* N = 42 * 6
N = 42*6 / 12 = 21 correct answer C |
AQUA-RAT | AQUA-RAT-37806 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
My brother is 4 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age of when my brother was born, then, what was the age of my father and mother respectively when my brother was born? | [
"32 yrs, 23 yrs",
"32 yrs, 29 yrs",
"32 yrs, 22 yrs",
"35 yrs, 33 yrs"
] | C | Sol.
Clearly, my brother was born 4 years before I was born and 4 years after my sister was born.
So, father's age when brother was born = (28 + 4) = 32 years.
mother's age when was born = (26 - 4) years = 22 years.
Answer C |
AQUA-RAT | AQUA-RAT-37807 | c#, formatting
/// <summary>
/// Calculates and returns the general discount
/// based on the <see cref="AccountStatus"/>
/// </summary>
/// <param name="accountStatus"></param>
/// <returns></returns>
private static decimal GetDiscountPercentage(AccountStatus accountStatus)
{
switch(accountStatus)
{
case AccountStatus.SimpleCustomer:
return 0.10m;
case AccountStatus.ValuableCustomer:
return 0.30m;
case AccountStatus.MostValuableCustomer:
return 0.50m;
default:
return 0.00m;
}
}
/// <summary>
/// Applying the discounts (if any) on the price and returns the final price (after discounts).
/// </summary>
/// <param name="price"></param>
/// <param name="accountStatus"></param>
/// <param name="timeOfHavingAccountInYears"></param>
/// <returns></returns>
public static decimal ApplyDiscount(decimal price , AccountStatus accountStatus , int timeOfHavingAccountInYears)
{
decimal loyaltyDiscountPercentage = GetLoyaltyDiscountPercentage(timeOfHavingAccountInYears);
decimal discountPercentage = GetDiscountPercentage(accountStatus);
decimal priceAfterDiscount = price * (1.00m - discountPercentage);
decimal finalPrice = priceAfterDiscount - ( loyaltyDiscountPercentage * priceAfterDiscount );
return finalPrice;
}
}
since all methods don't need several instance, and the nature of the class is unchangeable, making the class static would be more appropriate.
You can then reuse it :
var finalPrice = DiscountManager.ApplyDiscount(price, accountStatus, timeOfHavingAccountInYears);
The following is multiple choice question (with options) to answer.
A single discount equivalent to the discount series of 20%, 30% and 5% is? | [
"31.9",
"31.7",
"46.8",
"31.5"
] | C | 100*(80/100)*(70/100)*(95/100) = 53.2
100 - 53.2 = 46.8
Answer:C |
AQUA-RAT | AQUA-RAT-37808 | 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 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform. | [
"310 meter",
"335 meter",
"345 meter",
"350 meter"
] | D | Explanation:
Speed = Distance/time = 300/18 = 50/3 m/sec
Let the length of the platform be x meters
then
Distance=Speed∗Timex+300=50/3 X 39
=>3(x+300)=1950=>x=350 meters
ANSWER IS D |
AQUA-RAT | AQUA-RAT-37809 | This implies:
$\displaystyle y+3=x+2\:\therefore\:x-y=1\:\therefore\:x+y=2x-1$
Now solve:
$\displaystyle\frac{x^2-2x-3}{x+2}=\frac{2}{3}$
To find x.
#### MarkFL
Staff member
Here's another method:
$\displaystyle\frac{x^2-2x-3}{x+2}=\frac{2}{3}$
implies:
(1) $\displaystyle 3x^2-8x-13=0$
$\displaystyle\frac{y^2-4}{y+2}=\frac{2}{3}$
implies:
(2) $\displaystyle 3x^2-2y-18=0$
Adding (1) and (2) we get:
(3) $\displaystyle 3x^2+3y^2-8x-2y-31=0$
If we multiply (1) by 2 we have:
$\displaystyle 6x^2-16x-26=0$
which we may write as:
$\displaystyle -2x-8(x-1)+6x(x-1)-34=0$
Using $\displaystyle y=x-1$ this becomes:
(4) $\displaystyle-2x-8y+6xy-34=0$
Adding (3) and (4) we obtain:
$\displaystyle 3x^2+6xy+3y^2-10x-10y-65=0$
(5) $\displaystyle 3(x+y)^2-10(x+y)-65=0$
Now we have a quadratic in $\displaystyle x+y$.
The following is multiple choice question (with options) to answer.
3x + y = 21 , and x + 3y = 1. Find the value of 2x + 2y | [
"20",
"18",
"16",
"11"
] | D | Add these two equations
4x + 4y = 22
Divide by 2 ( to Get 2x + 2y)
Answer will be D. 11 |
AQUA-RAT | AQUA-RAT-37810 | The remainder is $5$ because the remainder of $17$ is $5$.
It's no different with polynomials.
There remainder is $-9$ because the remainder of $5x + 1$ is $-9$.
\ \ \begin{align}p\ &\equiv\,\ 5\,\ x\,\ +\ 1\!\!\pmod{\!(x + 2)f}\quad\ \ [\,f = x + 1\rm \ \ in\ OP\,]\\ \Rightarrow\ \ p\ &\equiv\,\ 5\ \,\color{#c00}x\,\ +\ 1\!\!\pmod{\color{#0a0}{x+2}}\\ &\equiv\, 5(\color{#C00}{-2})\!+ 1\quad {\rm by}\quad\, \ \color{#0a0}{x + 2}\equiv 0\,\Rightarrow\, \color{#c00}{x\equiv -2} \end{align}
Remark Sometimes authors write such congruences as $\!\!\pmod{\color{#c00}{x\equiv -2}}\,$ to make such evaluations clearer. The 2nd line uses the fact that congruences always persist mod factors of the modulus by $$\,p\equiv q\!\!\pmod{\!mn}\,\Rightarrow\, \underbrace{m\mid mn\mid p-q\,\Rightarrow\, m\mid p-q}_{\large\text{\it{transitivity} of divisibility}}\,\Rightarrow\,p\equiv q\!\!\pmod{\!m}$$
The following is multiple choice question (with options) to answer.
What is the dividend. divisor 17, the quotient is 9 and the remainder is 5? | [
"140",
"143",
"144",
"158"
] | D | D = d * Q + R
D = 17 * 9 + 5
D = 153 + 5
D = 158 |
AQUA-RAT | AQUA-RAT-37811 | doesn't rule out anything and thus it could be either coin with equal probability. to bet you even money that it is the two headed coin. One coin has been specially made and has a head. He selects a coin at random and flips it twice. One coin is chosen at random and flipped, coming up heads. Perform the following experiment. on each side. If the same coin is tossed twice, find the probability that it is the two-headed coin. There are three coins One is two headed coin, another is biased coin that comes up tails 25% of the times and the other is unbiased coin One of the three coins is chosen at random and tossed , it shows head What is the probability that - Math - Probability. I've read that if you flip a coin 10 times and it comes up heads every time then it's still a 50/50 chance to be a heads or tails on the 11th flip. I'm not a mathematician so please bear with me. The hypotheses areH1-the coin is two headed, and H2 the coin is fair. Condtion that the face observed is already heads. Find the probability that heads appears twice. A coin is chosen at random and tossed 2 times. Remember that P(A given B) = P(A and B)/P(B) So let's say that A is the event that he chose the 2-headed coin, and B is an event denoted by H(N), which indicates that the coin was tossed N times, and came up heads each time, so the answer in our first case is P(A given H(1)), and the answer our last case is P(A given H(3)). What is the conditional prob-ability that both are boys given that at least one of them is a boy? 2. There are three coins. This is the currently selected item. (The fair coin lands on heads 50% of the time). All k times the coin landed up heads. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of time. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. What is the conditional probability that it is the fair coin. The rest are fair. onditional probability is a tool for
The following is multiple choice question (with options) to answer.
John tossed a fair coin 3 times. What is the probability that the coin landed heads up exactly twice? | [
"0.125",
"0.175",
"0.275",
"0.375"
] | D | The probability is exactly same as the cion landed tail up once. That is 3*(1/2)^3 = 3/8 =0.375.
The answer is, therefore, (D). |
AQUA-RAT | AQUA-RAT-37812 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B's contribution in the capital? | [
"Rs. 7500",
"Rs. 8000",
"Rs. 8500",
"Rs. 9000"
] | D | A invests 3500 Rs for 12 months (Why 12 - because profit is given of 1 year) in the ratio of 2.
B invests X Rs for 7 months (Why 7 - because B joins A after 5 months and profit is given of 1 year) in the ratio of 3.
3500*12+X*7 = 2/3.
42000+7X = 2/3.
7X = 3/2*42000.
X = 9000.
ANSWER D |
AQUA-RAT | AQUA-RAT-37813 | (D)
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Posts: 64
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink]
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05 Oct 2019, 22:16
Bunuel wrote:
If 20 typists can type 48 letters in 20 minutes, then how many letters will 30 typists working at the same rate complete in 1 hour?
A. 63
B. 72
C. 144
D. 216
E. 400
This i how i solved...
20 Typist can type 48 letters / 20 Min therofre in 1 Min 20 Typist can type = 48/20 = 2.4 Letters per Min
In 60 Min 20 Typist can type = 2.4 *60 =144 Words per min...
We are also told now there are 30 Typist so if 20 typist can type 144 words so 30 Typist can type X words per min.....
Solve for X
20/144 = 30 / x .....
x = 72*3 = 216 Words per min
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink] 05 Oct 2019, 22:16
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The following is multiple choice question (with options) to answer.
20 people can write 50 book in 15 days working 8 hour a day.then in how many day 200 can be written by 60 people? | [
"200",
"100",
"50",
"70"
] | B | work per day epr hour per person= 50/(15*8*20) // eq-1
people= 60; let suppose day=p; per day work for 8 hours
acc. to condition
work per day epr hour per person= 200/(p*8*60) // eq-2
eq-1==eq-2;
p= 100
ANSWER:B |
AQUA-RAT | AQUA-RAT-37814 | 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.
Sandy can do a job in 21 days and Molly can do the same job in 42 days. If they work together, in how many days will they complete the job? | [
"8",
"10",
"12",
"14"
] | D | Sandy can do 1/21 of the job each day.
Molly can do 1/42 of the job each day.
The combined rate is 1/21 + 1/42 = 1/14 of the job each day.
The job will take 14 days.
The answer is D. |
AQUA-RAT | AQUA-RAT-37815 | 10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420
(B) 840
(C) 1260
(D) 2520
(E) 5040
Q9)
20% of total applied at X = 15
100% of total applied at X = 75
Only applied at X = 60
25% of total applied at Y = 15
100% of total applied at Y = 60
Only applied at Y = 45
Only applied at X + Only applied at Y = 60 + 45 = 105
ANS = D
Q10)
1 x 2 x 3 x 4 x 5 x 6 x 7 = 2^4 x 3^2 x 5 x 7
Option A = 420 = 2^2 x 3 x 5 x 7 (factored this out)
Option B = 2 x 420 = 2^3 x 3 x 5 x 7 (just add a 2 to A)
Option C = 3 x 420 = 2^2 x 3^2 x 5 x 7 (just add a 3 to A)
Option D = 2 x 1260 = 2^3 x 3^2 x 5 x 7 (just add a 2 to C)
Option E = 2 x 2520 = 2^4 x 3^2 x 5 x 7 (just add a 2 to D)
Actually after doing up to A you can quickly just figure out how many more 2's and 3's and determine it's E.
for q10 we need to find the lowest number - so should be 420
its divisible by all the integers from 1-7 inclusive
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Re: Good set of PS 2 [#permalink] 17 Oct 2009, 03:38
1
KUDOS
Bunuel wrote:
Please find below new set of PS problems:
The following is multiple choice question (with options) to answer.
The least number which when increased by 4 each divisible by each one of 22, 32, 36 and 54 is : | [
"427",
"859",
"860",
"4320"
] | C | Solution
Required number = (L.C.M. of 24, 32, 36, 54) - 4 = 864 - 4 = 860. Answer C |
AQUA-RAT | AQUA-RAT-37816 | 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.
A can give B 60 meters start and C 120 meters start in a kilometer race. How much start can B give C in a kilometer race? | [
"63.83",
"111.67",
"111.64",
"111.11"
] | A | A runs 1000 m while B runs 940 m and C runs 880 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 880)/940=936.17 m.
B can give C = 1000 - 936.17= 63.83 m.
Answer:A |
AQUA-RAT | AQUA-RAT-37817 | # How to calculate the area covered by any spherical rectangle?
Is there any analytic or generalized formula to calculate area covered by any rectangle having length $l$ & width $b$ each as a great circle arc on a sphere with a radius $R$?
Note: Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.
• What do you mean by "rectangle"? Do you just want a quadrilateral, or some stronger condition? (Note that you cannot have four right angles in a quadrilateral on the sphere.) – Nick Matteo Mar 25 '15 at 13:16
• Yes, a quadrilateral having equal opposite sides (but not parallel) each as a great circle arc on a sphere. – Harish Chandra Rajpoot Mar 25 '15 at 13:21
• What do you mean by length and width? As the sides are not parallel, we don't have the usual idea of width and length. Do you simply mean the lengths of the sides? – robjohn Apr 4 '15 at 14:12
• Yes, length & width are simply the sides of the rectangle as great circles arcs on the spherical surface. – Harish Chandra Rajpoot May 28 '15 at 23:56
Assume we are working on a sphere of radius $1$, or consider the lengths in radians and the areas in steradians.
Extend the sides of length $l$ until they meet. This results in a triangle with sides $$w,\quad\frac\pi2-\frac l2,\quad\frac\pi2-\frac l2$$
The following is multiple choice question (with options) to answer.
The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 900 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units? | [
"120",
"99",
"88",
"72"
] | A | Given that the area of the square = 900 sq.units
=> Side of square = √900 = 30 units
The radius of the circle = side of the square = 30 units Length of the rectangle = 2/5 * 30 = 12 units
Given that breadth = 10 units
Area of the rectangle = lb = 12 * 10 = 120 sq.units
Answer: A |
AQUA-RAT | AQUA-RAT-37818 | A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
As one of our solutions is 'it cannot be determined' we cannot use the answers in this question.
Therefore, we'll go for a direct calculation, a Precise approach.
Let's write down what we know, going from the start of the question:
Men + Women = 150
Men = 75 --> Women = 150 - 75 = 75
Women officers = 2/15 * Women = 2/15 * 75 = 10
male officers = 16 - 10 = 6
male non-officers = 75 - 6 = 69.
Bunuel are we assuming that the club contains only men and women (and not, say, children)?
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
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01 Jan 2018, 03:24
Bunuel wrote:
The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club?
A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
Total members 150
Men 75
So women = 150-75 = 75
16 members are officers
2/15 of women are officers
So women officers = 75 * 2/15 = 10
So men officers = 16-10 = 6
So men who are not officers = 75-6 = 69
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Of the employees in a company, 50 % are female who have a master degree. If 20 % of the female employees do not have a master degree, what percent of the employees in the company are female? | [
"60%",
"62.5%",
"63.25%",
"63.5%"
] | B | Let E be total no. of employees and F be total no of female employees.
Question asked what is F/E X 100%?
No. of female employees with Masters = 0.5 x E (From the question)
No. of female employees without masters = 0.2 x F (From the question)
Therefore No. of female employees with masters = F - 0.2 F = 0.8 F
The 2 expressions equal each other therefore 0.8F = 0.5E; F/E = 0.5/0.8 = 62.5%
Ans: B |
AQUA-RAT | AQUA-RAT-37819 | => 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.
Rs.1540 were divided among John, Paul & Tom in such a way that John had Rs.80 more than Paul and Tom had Rs 60 more than John . How much was Paul’s share? | [
"Rs.575",
"Rs.580",
"Rs.585",
"Rs.590"
] | B | Let Paul gets Rs x. Then We can say John gets Rs (x + 80 ) and Tom gets Rs ( x + 140) .
x + 80 + x + x + 140 = 1540
3x = 1320
x = 440 .
Paul’s share = Rs ( 440 + 140 ) = Rs.580
B |
AQUA-RAT | AQUA-RAT-37820 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B | [
"10",
"8",
"6",
"5"
] | A | Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, B's age = 2x = 10 years
Answer:A |
AQUA-RAT | AQUA-RAT-37821 | (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 collection of books went on sale, and 2/3 of them were sold for $4 each. If none of the 36 remaining books were sold, what was the total amount received for the books that were sold? | [
"$288",
"$135",
"$90",
"$60"
] | A | Since 2/3 of the books in the collection were sold, 1/3 were not sold. The 36 unsold books represent 1/3 of the total number of books in the collection, and 2/3 of the total number of books equals 2(36) or 72. The total proceeds of the sale was 72($4) or $288. The best answer is therefore A.
Answer: A. |
AQUA-RAT | AQUA-RAT-37822 | Question
# The number of positive numbers less than $$1000$$ and divisible by $$5$$ ( no digits being replaced) is
A
150
B
154
C
166
D
None of these
Solution
## The correct option is B $$154$$Here, the available digits are $$0,1,2,3,4,5,6,7,8,9.$$The numbers can be of one, two or three digits and in each of them unit's place must have $$0$$ or $$5$$ as they must be divisible by $$5$$.The number of numbers of one digit $$=1$$$$(\because$$ is the only number$$)$$.The number of numbers of two digits divisible by $$5=$$ number of all the numbers of two digits divisible by $$5-$$ number of numbers of two digits divisible by $$5$$ and having $$0$$ in ten's place $$\displaystyle =^{ 2 }{ { P }_{ 1 } }\times ^{ 9 }{ { P }_{ 1 } }-1$$,$$(\because$$unit's place can be filled by either $$0$$ or $$5$$ in first category and only by $$5$$ in the second category$$)$$$$=2\times 9-1=17$$.The number of numbers of numbers of three digits divisible by $$5=$$ number of all the numbers of three digits divisible by $$5=$$ number of numbers of three digits divisible by $$5=$$ number of numbers of three digits divisible by $$5$$ and having $$0$$ in hundred's place $$\displaystyle =^{ 2 }{ { P }_{ 1 } }\times ^{ 9 }{ { P }_{ 2 } }\times ^{ 8 }{ { P }_{ 1 } }\times 1=2\times 9\times 8-8=136.$$$$\therefore$$ required number of numbers $$=1+17+136=154$$ Mathematics
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The following is multiple choice question (with options) to answer.
The least number of four digits which is divisible by 15, 25, 40 and 75 is: | [
"9000",
"9400",
"9600",
"9800"
] | C | Greatest number of 4 digits is 9999. L.C.M of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399. Required number = 9999 - 399 = 9600
ANSWER:C |
AQUA-RAT | AQUA-RAT-37823 | 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.
Calculate the effect changes in dimension of a rectangle will have on its area, if length is increased by 40% and its breadth is decreased by 25%? | [
"5% increase",
"7% increase",
"6% increase",
"4% increase"
] | A | let L and B be 100 each
100 * 100 = 10000
L increase by 40% = 140
B decrease by 25% = 75
140 * 75 = 10500
5% increase
ANSWER:A |
AQUA-RAT | AQUA-RAT-37824 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B complete a work in 30 days. A alone can do it in 40 days. If both together can do the work in how many days? | [
"1.0875 days",
"0.1873 days",
"0.0673 days",
"0.0583 days"
] | D | 1/30 + 1/40 = 0.0583 days
ANSWER:D |
AQUA-RAT | AQUA-RAT-37825 | # 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 shopkeeper purchased 70 kg of potatoes for Rs. 420 and sold the whole lot at the rate of Rs. 6.50 per kg. What will be his gain percent? | [
"12 4/3 %",
"18 1/4 %",
"8 1/3 %",
"6 1/3 %"
] | C | C
C.P. of 1 kg = 420/7 = Rs. 6
S.P. of 1 kg = Rs. 6.50
Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 % |
AQUA-RAT | AQUA-RAT-37826 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are 56 lights which are functional and each is controlled by a separate On/Off switch. Two children X and Y start playing with the switches. X starts by pressing every third switch till he reaches the end. Y, thereafter, presses every fifth switch till he too reaches the end. If all switches were in Off position at the beggining, How many lights are switched On by the end of this operation? | [
"18",
"19",
"17",
"15"
] | C | Editing my solution:
Number of switches = 56
Number of switches turned on by X: 3, 6, ... 54 = 18
Number of switches turned on by Y: 5, 10, ....55 = 11
Few switches are turned on by X and later turned off by Y: LCM(3,5) = 15x = 15, 30,....90 = 6.
Subtract the above 6 switches from both X and Y as they are turned off.
Number of switches that are turned on = (18- 6) + (11 - 6) = 17
Answer: C |
AQUA-RAT | AQUA-RAT-37827 | ### 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.
Sanoop bought 8 t-shirts at an average price (arithmetic mean) of Rs.526. If Sanoop returned 2 t-shirts to the retailer, and the average price of the remaining t-shirts was Rs.505, then what is the average price, of the three returned t-shirts? | [
"560",
"589",
"562",
"563"
] | B | Total price of 8 t-shirts= 8*526=4208
Total price of 6 t-shirts=5*505=3030
Total price of 2 t-shirts=4208-3030=1178
Average price of 3 t-shirts=1178/2=589
Correct option Answer:B |
AQUA-RAT | AQUA-RAT-37828 | # Probability a 9-digit number has the digits 2,4, and 6 next to each other.
The integers $1,2,3,....,9$ are arraned (at random) in a row, resulting in a $9$-digit integer (without replacement). What is the probability that:
The result is even? $\frac49$ or $\frac{4(8!)}{9!}$
The result is divisible by $5$? The number must end in $0$ or $5$. Edit: As Andre pointed out we have no $0$. So, $\frac19$. or $\frac{(8!)}{9!}$
The digits $2, 4,$ and $6$ are next to each other (in any order)? The above two I have confidence in, it is this last one I'm a little confused on.
$9\choose 3$ ways to position $2,4,6$ in the 9-digit number and $3!$ ways they can be arranged. This doesn't appear to be right as after you divide by $9!$ you get $.13$% which seems unreasonably low. So, I thought to multiple by $6!$ to account for the number of ways the other $6$ numbers can be arranged. This gives you:
$9\choose 3$$3!6!/9!$
which equals $1$, and obviously isn't right. Any suggestions as to where I went wrong would be great.
-
You're right! Silly me. – Vincent Jun 16 '14 at 22:47
For $2,4,6$ next to each other, imagine choosing the $3$ spots that will be reserved for our three favoured guests. This can be done in $\binom{9}{3}$ equally likely ways.
The three spots can be in a row in $7$ ways, for the leftmost of the spots can be in any of the positions $1$ to $7$. Thus the required probability is $\frac{7}{\binom{9}{3}}$. This simplifies nicely.
The following is multiple choice question (with options) to answer.
Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 15 phone numbers by using the readable digits followed by 15 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly? | [
"1/9",
"10/243",
"1/18",
"10/271"
] | C | If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).
If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).
P = 15/(9*9*3+9*3) = 1/18.
Answer: C. |
AQUA-RAT | AQUA-RAT-37829 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
If six boys sit in a row, then what is the probability that three particular boys are always together is? | [
"1/5",
"1/4",
"1/3",
"1/7"
] | A | Six boys can be arranged in a row in 6! ways.
Treat the three boys to sit together as one unit then there 4 boys and they can be arranged in 4! ways.
Again 3 boys can be arranged among them selves in 3! ways.
Favorable outcomes = 3!4!
Required probability = 3!4!/6! = 1/5
A |
AQUA-RAT | AQUA-RAT-37830 | Substitute 0.085 for interest rate, $3000 for the amount deposited and$6000 for the amount after t years in the formula, “A=Pert” as,
$6000=$3000e0.085t
Divide both sides by $3000 as,$6000$3000=$3000e0.085t$3000e0.085t=2 Take natural logarithm on both sides as, ln(e0.085t)=ln2 Now, apply the inverse property of lnex as, 0.085t=ln2 Divide both sides by 0.085 as, 0.085t0.085=ln20.085t0.6930.0858.15 years So, the deposited amount will be doubled in approximately 8.15 years when the interest rate is 8.5%. Now, as the amount will become quadruple of the deposited amount after t years, then the amount after t years is 4P, that is, A=4P=4($3000)=\$12000
Substitute 0
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The following is multiple choice question (with options) to answer.
A sum of Rs. 2665 is lent into two parts so that the interest on the first part for 8 years at 3% per annum may be equal to the interest on the second part for 3 years at 5% per annum. Find the second sum? | [
"1686",
"1640",
"6829",
"2780"
] | B | (x*8*3)/100 = ((2665 - x)*3*5)/100
24x/100 = 39975/100 - 15x/100
39x = 39975 => x = 1025
Second sum = 2665 – 1025 = 1640
Answer: B |
AQUA-RAT | AQUA-RAT-37831 | Therefore, x = 10 cm ------vertical edge of the rectangle.
(30-x) = 20 cm -----------horizontal edge rectangle.
Therefore, the rectangle must be 10cm by 20cm. -----answer.
(max V2 = pi(20^2)(10) = 4000pi cu.cm. also, same as max V1)
The net of the generated cylinder?
The same as in V1.
3. Thanx for that mate
4. Hello, mathsB!
I have no idea what a "net" is . . .
Consider a rectangle of perimeter 60Cm.
Form a cylinder by revolving this rectangle about one of its edges.
What dimensions of the rectangle will result in a cylinder of maximum volume?
(Include a net of the final shape)
We have a rectangle with dimensions $L \times W$.
Code:
* - - - - - *
| |
W | |
| |
* - - - - - *
L
The perimeter is 60 cm: . $2L + 2W \:=\:60\quad\Rightarrow\quad W \:=\:30 - L$ .[1]
Revolve the rectangle about a vertical side.
The cylinder has radius $L$ and height $W$.
. . It volume is: . $V \:=\:\pi r^2h \:=\:\pi L^2W$. . [2]
Substitute [1] into [2]: . $V \;=\;\pi L^2(30-L)$
Now maximize this function.
. . $\text{(I got: }L = 20,\;W = 10)$
5. Originally Posted by Soroban
I have no idea what a "net" is . . .
...
Hello,
only for the record: I've attached a "net" of a cube and a "net" of perpendicular circular cylinder. Maybe you now can provide me with the correct English expression
The following is multiple choice question (with options) to answer.
How many meters of carpet 50cm, wide will be required to cover the floor of a room 30m * 20m? | [
"1000",
"1050",
"1100",
"1200"
] | D | 50/100 * x = 30 * 20 => x = 1200
ANSWER:D |
AQUA-RAT | AQUA-RAT-37832 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
All of the students of the Music High School are in the band, the orchestra, or both. 90 percent of the students are in only one group. There are 91 students in the band. If 60 percent of the students are in the band only, how many students are in the orchestra only? | [
"30",
"33",
"36",
"39"
] | D | 60% of students are in the band only, so 30% of students are in the orchestra only.
Since 10% of students are in both, then 70% of students are in the band which is 91 students.
The number of students in the orchestra is (30/70)*91 = 39.
The answer is D. |
AQUA-RAT | AQUA-RAT-37833 | ### Show Tags
05 Sep 2017, 11:01
2
In traveling from city A to city B, John drove for 1 hour at 50 mph and for 3 hoursat 60 mph. What was his average speed for the whole trip?
(A) 50
(B) 53.5
(C) 55
(D) 56
(E) 57.5
The total distance traveled from City A to City B is 50*1 + 60*3 = 230
Since John drove for 4 hours to travel this distance,
the average speed of the whole trip is $$\frac{Distance}{Time} = \frac{230}{4}$$ = 57.5(Option E)
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Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink]
### Show Tags
17 Jan 2019, 03:38
How does this work using the harmonic mean?
$$\frac{1/50+3/60}{4}$$=$$\frac{7/100}{4}$$=57.1
Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink] 17 Jan 2019, 03:38
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The following is multiple choice question (with options) to answer.
Yesterday it took Robert 3 hours to drive from City A to City B. Today it took Robert 2.5 hours to drive back from City В to City A along the same route. If he had saved 30 minutes in both trips, the speed for the round trip would be 80 miles per hour. What is the distance between city A and city B? | [
" 90",
" 180",
" 150",
" 240"
] | B | 2d/80 = 4.5 ( because time = 3 + 2.5 - 1 hrs)
=> d = 180
Answer - B |
AQUA-RAT | AQUA-RAT-37834 | So total 6 ways we can assign project to three groups.
So total number of ways we can do the required is 15 Ways of distributing people in 3 groups X 6 ways of distributing 3 projects among the three formed groups.
= 90
Hope this helps
Probus
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Posts: 54
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
05 Mar 2019, 19:03
Suppose the six students are 1,2,3,4,5,6
Now student 1 HAS to be paired with someone else. He/she can be paired with 2,3,4,5, or 6. So that's 5 possibilities.
Once that has occurred (without loss of generality suppose 1 pairs with 2. 3 can then pair with 4,5, or 6. That's 3 possibilities. Only one pair is left. 5x3=15 groups of 3.
For each group there are 3x2x1 different arrangements of the topics. 15x6=90.
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Joined: 22 Oct 2017
Posts: 19
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
06 Mar 2019, 03:35
arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540
Can someone please explain to me why the red part of 15*3!, because I really can't figure out the reasoning behind the permutation of the assignments,
guaranteed kudos for accurate responses
Thanks
Director
Joined: 04 Aug 2010
Posts: 546
Schools: Dartmouth College
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
06 Mar 2019, 03:57
arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540
The following is multiple choice question (with options) to answer.
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements E are possible? | [
" 30",
" 60",
" 90",
" 180"
] | C | 90 is the number of ways you can assign 3 teams formed out of 12 people to 3 different tasks.
But now you can order the 3 tasks in 3! ways. T1 T2 T3 or T2 T1 T3.... etc etc.
I was confused between 90 and 540 but since question used the wordarrangementsdecided to go with complete arrangements E including the order of tasks.
could you explain the highlighted step... i'm getting 90 = 15 * 3!
suppose the students are numbered 1,2,3,4,5,6 and tasks are X,Y and Z
one of the 15 possible ways of forming teams is 12, 34, 56. these teams can be assigned to 3 tasks in 3! = 6 ways
X-- Y-- Z
12-- 34-- 56
12-- 56-- 34
34-- 12-- 56
34-- 56-- 12
56-- 12-- 34
56-- 34-- 12
so the answer should be 15*6 = 90
But now you can fruther decide which task you want to perform first X Y or Z..C |
AQUA-RAT | AQUA-RAT-37835 | ## Solution
Consider $f(2)$. The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$. Since $100\le (x+1)^2-x^2=2x+1$, this first happens at $x\ge \lfloor 99/2\rfloor = 50$. The perfect squares from here go: $2500, 2601, 2704, 2809\dots$. Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$. After the ones and tens make $100$, the hundreds place will go up by $2$, thus reaching our goal. Since $10^2=100$, the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$. The missing number is one less than the number of hundreds $(k=2)$ of $3600$, or $35$.
Now consider f(4). Instead of the difference between two squares needing to be $100$, the difference must now be $10000$. This first happens at $x\ge 5000$. After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$. This will take place at $\left(5000+100\right)^2=5100^2= 26010000$. Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.
The following is multiple choice question (with options) to answer.
The difference between the place values of two fives in the numerical 86510529 is | [
"0",
"4905",
"49050",
"490500"
] | D | Required difference = (500000-500)
=490500.
Answer is D |
AQUA-RAT | AQUA-RAT-37836 | newtonian-mechanics, kinematics, free-fall
Title: Free fall equations I've been looking for this on the internet for a long time, but nowhere explains it clearly, and I need to explicitly calculate this and not other things:
If I have the height at which an object is, and it falls, how do I calculate at what speed and at what height it will be at x time? If the object starts falling from rest at height $h$, then its velocity at time $t$ later is
$$\vec v=-gt\hat z$$
and its vertical position is
$$z=h-\frac12gt^2$$
where $g$ is the magnitude of the constant and uniform downward gravitational acceleration. This neglects air drag, and assumes that the gravitational field is the same everywhere.
These equations come from integrating the kinematical equation for free-fall,
$$a_z=\frac{d^2z}{dt^2}=-g,$$
and applying the initial conditions that $z=h$ and $\vec v=0$ at $t=0$.
The following is multiple choice question (with options) to answer.
The velocity of a falling object in a vacuum is directly proportional to the amount of time the object has been falling. If after 5 seconds an object is falling at a speed of 90 miles per hour, how fast will it be falling after 18 seconds? | [
"18 miles per hour",
"90 miles per hour",
"216 miles per hour",
"324 miles per hour"
] | D | Since Velocity is proportional to Time
Velocity =k*Timewhere k is the constant of proportionality
Time= 5 seconds
Velocity = 5k = 90 miles per Hour
i.e. k = 18
i.e. The relation between Velocity and Time becomes
Velocity =18*Time
Time= 18 seconds
Velocity = 18*18 = 324 miles per hour
Answer: Option D |
AQUA-RAT | AQUA-RAT-37837 | materials, density
densitytotal = masstotal / volume =
volume = masstotal / densitytotal = 0.2 kg / 2300 kg/m3 = 0.00008696 m3
masssand = 2700 kg/m3 * 0.0000869 m3 = 0.2348 kg
massplastic = 1400 kg/m3 * 0.0000869 m3 = 0.1217 kg
But I don't think this step was the right choice since the mass of the sand came out to be larger than the mass of the sample. Was I correct in trying to figure out the value of the mass of sand and plastic or was that the wrong approach to determine the material's composition? If it's the first case, how should I find out the mass values? Your mistake is that you have calculated the mass of a sample with the same volume as ours consisting entirely of sand, and a sample with the same volume as ours consisting entirely of sand.
volume = mass_total / density_total = 0.2 kg / 2300 kg/m^3 = 0.00008696 m^3
gives the initial volume of the sample before the salt is removed, however by multiplying this by the densities of sand and plastic, we calculate the mass of a 0.0000869 m^3 sample of sand and a 0.0000869 m^3 sample of plastic.
First we need to find the new volume and density of the sample after the salt has been removed. This can be done by calculating the volume of salt removed:
volume_salt=0.03kg_salt/2100 kg/m^3=0.00001429m^3
By subtracting this volume from the initial volume you already calculated:
0.00008696m^3-0.00001429m^3=0.00007267m^3
dividing the new mass by this gives:
0.170kg/0.00007267m^3=2,339kg/m^3
The following is multiple choice question (with options) to answer.
A cement mixture is composed of 3 elements. By weight, 1/2 of the mixture is sand, 1/5 of the mixture is water, and the remaining 15 pounds of the mixture is gravel. What is the weight of the entire mixture in pounds? | [
"30",
"40",
"50",
"60"
] | C | Let the total weight be x.
Sand content= (1/2)x
Water content= (1/5)x
Gravel=x-(1/2)x-(1/5)x=(3/10)x=15
x=50
Then answer will be C=50 |
AQUA-RAT | AQUA-RAT-37838 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in upstream is 30 kmph and the speed of the boat downstream is 70 kmph. Find the speed of the boat in still water and the speed of the stream? | [
"10 kmph",
"20 kmph",
"12 kmph",
"97 kmph"
] | B | Speed of the boat in still water
= (30+70)/2
= 50 kmph. Speed of the stream
= (70-30)/2
= 20 kmph.
Answer: B |
AQUA-RAT | AQUA-RAT-37839 | # How to determine the most probable value of the perimeter of a quadrilateral?
The problem is as follows:
The diagonals of a quadrilateral measure $$8\,cm$$ and $$10\,cm$$. Using this information find the most probable perimeter of this quadrilateral indicated as a range.
The alternatives given in my book are:
$$\begin{array}{ll} 1.&\textrm{Between 14 cm and 16 cm}\\ 2.&\textrm{Between 18 cm and 36 cm}\\ 3.&\textrm{Between 38 cm and 40 cm}\\ 4.&\textrm{Between 42 cm and 50 cm}\\ 5.&\textrm{Between 12 cm and 14 cm}\\ \end{array}$$
In this problem I don't know what sort of strategy can be used?. Does it exist a relationship between the diagonals of a quadrilateral?.
So far what it come to my mind was to use the triangle inequality which states that the sum of the two sides of a triangle is greater than the third side.
Given a quadrilateral ABCD then this means the diagonals are from $$AC$$ and from $$BD$$.
$$AC
$$BD
$$AC
$$BD
Given that the diagonals are AC and BD then its sum must be less than: By summing the expressions given:
$$2AC+2BD<2AB+2BC+2CD+2AD$$
$$AC+BD
This must be the upper bound in the perimeter of the quadrilateral. But how about the lower bound?. Well, for this part I'm using the reverse triangle inequality which states that any side is greater than the difference of the other two sides provided one is greater than the other.
In this case I'm assuming $$AB>BC$$ and $$BC>CD$$ and $$CD>AD$$ and $$AB>AD$$
Then this would mean:
$$AC>AB-BC$$
BD>BC-CD
$$AC>CD-AD$$
$$BD>AB-AD$$
The thing here is that you may not get the perimeter directly.
Suming these:
The following is multiple choice question (with options) to answer.
What is the ratio between perimeters of two squares one having 3 times the diagonal then the other? | [
"3: 4",
"3: 9",
"3: 2",
"3: 1"
] | D | d = 3d d = d
a√2 = 3d a√2 = d
a = 3d/√2 a = d/√2 => 3: 1
Answer: D |
AQUA-RAT | AQUA-RAT-37840 | 3,5,-5... The first term in the sequence of #'s shown above is 3. Each even # term is 2 more than the previous term and each odd # term, after the first, is -1 times the previous term. For example, the second term is 3+2, and the
8. ### Math *URGENT
Please give the answers and solutions for each. 1.If the second term is 2 and the seventh term of a geometric sequence is 64, find the 12th term. 2. Which term if the geometric sequence 18,54,162,486,... is 3,188,646? 3. Determine
9. ### maths
The fifth term of an arithmetic sequence is 23 and the 12th term is 72. What is the value of the 10th term. Which term has a value of 268.
10. ### Sj
Mathematics......Under number patterns-Geometric series If a question goes.determine the expression for the nth term of the following sequence if the a) 4th term is 24 and the 7th term is 192 in a geometric sequence.what formula
More Similar Questions
The following is multiple choice question (with options) to answer.
14, 23, 32, 41, 50, 59, …
In the sequence above, each term is 9 more than the previous term. What is the 41st term Z of the sequence? | [
"360",
"365",
"369",
"374"
] | D | First term , a= 14
Common difference , d = 9
nth term , tn = a + (n-1)d
41st term Z, t41 = a+ 40*d = 14 + 40*9 = 374
Answer D |
AQUA-RAT | AQUA-RAT-37841 | \over 4\pi} + {3 \over 8}\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A}} \end{array}\right.\tag{6}$$
The following is multiple choice question (with options) to answer.
Evaluate: 45678 * 8*4*8 = ? | [
"730848",
"705265",
"730846",
"730596"
] | A | According to order of operations, 8?4?8 (division and multiplication) is done first from left to right
8/4 = 2* 8 = 16
Hence
45678 * 8*4*8 = 45678 * 16 = 730848
correct answer A |
AQUA-RAT | AQUA-RAT-37842 | $$C^2_w$$ # of selections of 2 women out of $$w$$ employees;
$$C^2_{10}$$ total # of selections of 2 representatives out of 10 employees.
Q is $$\frac{C^2_w}{C^2_{10}}>\frac{1}{2}$$ --> $$\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}$$ --> --> $$w(w-1)>45$$ --> $$w>7$$?
(1) More than 1/2 of the 10 employees are women --> $$w>5$$, not sufficient.
(2) The probability that both representatives selected will be men is less than 1/10 --> $$C^2_{(10-w)}$$ # of selections of 2 men out of $$10-w=m$$ employees --> $$\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}$$ --> $$\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$, not sufficient
(1)+(2) $$w>5$$ and $$w>6$$: $$w$$ can be 7, answer NO or more than 7, answer YES. Not sufficient.
Hope it's clear.
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Joined: 20 Feb 2012
Posts: 40
Re: If 2 different representatives are to be selected at random [#permalink]
### Show Tags
27 Feb 2012, 09:11
10
37
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?
The following is multiple choice question (with options) to answer.
Twenty percent of the skiers that tried out for the ski patrol were selected. Of those that were not selected, 15 skiers were CPR certified and 25 were not. How many skiers tried out for the ski patrol? | [
"40",
"50",
"60",
"70"
] | B | Not Selected = 80%
If total is x.
Then 0.8x=15+25=40
x=50
Hence answer is B. |
AQUA-RAT | AQUA-RAT-37843 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$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$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Raman mixed 24 kg of butter at Rs. 150 per kg with 36 kg butter at the rate of Rs. 125 per kg. At what price per kg should he sell the mixture to make a profit of 40% in the transaction? | [
"Rs.129",
"Rs.198",
"Rs.189",
"Rs.128"
] | C | CP per kg of mixture = [24(150) + 36(125)]/(24 + 36) = Rs. 135
SP = CP[(100 + profit%)/100]
= 135 * [(100 + 40)/100]
= Rs.189.
Answer:C |
AQUA-RAT | AQUA-RAT-37844 | Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x
Solving, xt will get cancelled and you will get:
11000 = 11x
x is 1000
sum of both investments is 2x = 2000 which is Option D
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Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
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02 Jul 2017, 07:29
1
1
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000
Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.
The following is multiple choice question (with options) to answer.
A part of certain sum of money is invested at 10% per annum and the rest at 16% per annum, if the interest earned in each case for the same period is equal, then ratio of the sums invested is? | [
"8:5",
"4:9",
"4:3",
"4:1"
] | A | 16:10
= 8:5
Answer: A |
AQUA-RAT | AQUA-RAT-37845 | Therefore, x = 10 cm ------vertical edge of the rectangle.
(30-x) = 20 cm -----------horizontal edge rectangle.
Therefore, the rectangle must be 10cm by 20cm. -----answer.
(max V2 = pi(20^2)(10) = 4000pi cu.cm. also, same as max V1)
The net of the generated cylinder?
The same as in V1.
3. Thanx for that mate
4. Hello, mathsB!
I have no idea what a "net" is . . .
Consider a rectangle of perimeter 60Cm.
Form a cylinder by revolving this rectangle about one of its edges.
What dimensions of the rectangle will result in a cylinder of maximum volume?
(Include a net of the final shape)
We have a rectangle with dimensions $L \times W$.
Code:
* - - - - - *
| |
W | |
| |
* - - - - - *
L
The perimeter is 60 cm: . $2L + 2W \:=\:60\quad\Rightarrow\quad W \:=\:30 - L$ .[1]
Revolve the rectangle about a vertical side.
The cylinder has radius $L$ and height $W$.
. . It volume is: . $V \:=\:\pi r^2h \:=\:\pi L^2W$. . [2]
Substitute [1] into [2]: . $V \;=\;\pi L^2(30-L)$
Now maximize this function.
. . $\text{(I got: }L = 20,\;W = 10)$
5. Originally Posted by Soroban
I have no idea what a "net" is . . .
...
Hello,
only for the record: I've attached a "net" of a cube and a "net" of perpendicular circular cylinder. Maybe you now can provide me with the correct English expression
The following is multiple choice question (with options) to answer.
The area of the maximum size of the circle described from the 10 inch square? | [
"75.5 sq inch",
"76.5 sq inch",
"77.5 sq inch",
"78.5 sq inch"
] | D | 10 inch square means sides of sq are 10inch
then circle is within square so diameter would be 10 inch maximum and radius is 5inch
so area of the circle is = pi*r*r=3.14*5*5=78.5 sq inch
ANSWER:D |
AQUA-RAT | AQUA-RAT-37846 | 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.
What approximate value will come in place of the question mark(?) in the below question?
(47% of 1442 - 36% of 1412) + 63 = ? | [
"237",
"278",
"278",
"252"
] | D | Explanation:
Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252
Answer: D) 252 |
AQUA-RAT | AQUA-RAT-37847 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Three numbers are in the ratio 4 : 5 : 6 and their average is 30 . The largest number is: | [
"33",
"77",
"36",
"18"
] | C | Explanation:
Let the numbers be 4x, 5x and 6x. Therefore,
(4x+5x+6x)/3 = 30
15x = 90
x = 6
Largest number = 6x = 36.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37848 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is | [
"45 kg",
"75kg",
"25kg",
"35 kg"
] | A | Explanation:
Weight of the teacher = (35.4 x 25 - 35 x 24) kg = 45 kg.
Answer: A |
AQUA-RAT | AQUA-RAT-37849 | # Math Help - Distance (Word Problem)
1. ## Distance (Word Problem)
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
2. Originally Posted by Beastkun
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
1. Let d denote the covered distance and t the elapsed time. Then you know:
$\dfrac dt = 50$
and
$\dfrac{d}{t-3}=60$
2. Solve for d.
Spoiler:
You should come out with d = 900
3. I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
4. Originally Posted by Beastkun
I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
Substitute d = 50t from the first equation into the second equation and solve for t and hence d.
5. What prove it is doing is simultaneous equations.
The following is multiple choice question (with options) to answer.
Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home? | [
"(x + y) / t",
"2(x + t) / xy",
"2xyt / (x + y)",
"2(x + y + t) / xy"
] | C | Say the distance to school is 10 miles, x=5 miles per hour and y=1 miles per hour, then:
Time Bob spent biking would be 5/5=1 hour, and time he s pent walking would be 5/1=5 hours, so t=1+5=6 hours.
Now, plug x=5, y=1, and t=6 into the answer choices to see which one yields the distance of 10 miles. Only answer choice C fits.
Answer: C. |
AQUA-RAT | AQUA-RAT-37850 | 1 = 1
4 = 2^2
6 = 2 * 3
9 = 3^2
13 = 13
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
33 = 3 * 11
36 = 2^2 * 3^2
37 = 37
46 = 2 * 23
49 = 7^2
52 = 2^2 * 13
54 = 2 * 3^3
61 = 61
64 = 2^6
69 = 3 * 23
73 = 73
78 = 2 * 3 * 13
81 = 3^4
88 = 2^3 * 11
94 = 2 * 47
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
109 = 109
117 = 3^2 * 13
118 = 2 * 59
121 = 11^2
132 = 2^2 * 3 * 11
141 = 3 * 47
142 = 2 * 71
144 = 2^4 * 3^2
148 = 2^2 * 37
150 = 2 * 3 * 5^2
157 = 157
166 = 2 * 83
169 = 13^2
177 = 3 * 59
181 = 181
184 = 2^3 * 23
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
208 = 2^4 * 13
213 = 3 * 71
214 = 2 * 107
216 = 2^3 * 3^3
222 = 2 * 3 * 37
225 = 3^2 * 5^2
229 = 229
241 = 241
244 = 2^2 * 61
249 = 3 * 83
253 = 11 * 23
256 = 2^8
262 = 2 * 131
276 = 2^2 * 3 * 23
277 = 277
286 = 2 * 11 * 13
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
297 = 3^3 * 11
312 = 2^3 * 3 * 13
313 = 313
321 = 3 * 107
324 = 2^2 * 3^4
325 = 5^2 * 13
333 = 3^2 * 37
334 = 2 * 167
337 = 337
349 = 349
352 = 2^5 * 11
358 = 2 * 179
361 = 19^2
366 = 2 * 3 * 61
373 = 373
376 = 2^3 * 47
382 = 2 * 191
384 = 2^7 * 3
388 = 2^2 * 97
393 = 3 * 131
The following is multiple choice question (with options) to answer.
Which of the following numbers is between 1⁄4 and 1⁄5? | [
".45",
".35",
".29",
".22"
] | D | 1/4 = .25
1/5 = .2
The only number between these two is 0.22.
The answer is D. |
AQUA-RAT | AQUA-RAT-37851 | 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$$.Physics
The following is multiple choice question (with options) to answer.
A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is? | [
"16 kmph",
"88 kmph",
"54 kmph",
"18 kmph"
] | C | S = 240/16 * 18/5
= 54 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-37852 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$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$$
$$...................................$$
The following is multiple choice question (with options) to answer.
A survey of employers found that during 1993 employment costs rose 3.5 percent, where employment costs consist of salary costs and fringe-benefit costs. If salary costs rose 3 percent and fringe-benefit costs rose 8.5 percent during 1993, then fringe-benefit costs represented what percent of employment costs at the beginning of 1993 ? | [
" 16.5%",
" 9%",
" 35%",
" 55%"
] | B | The amount by which employment costs rose is equal to 0.035(salary costs + fringe benefit costs);
On the other hand the amount by which employment costs rose is equal to 0.03*salary costs + 0.085*fringe benefit costs;
So, 35(S+F) = 30S+85F --> S = 10F --> F/S = 1/10 --> F/(S+F) = 1/(1+10) = 1/11 = 0.09.
Answer: B. |
AQUA-RAT | AQUA-RAT-37853 | = 0 \\ a^7+ b^7 + c^7 = -7mk^2$$ Then your identity reads $$\left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\ \left( -m \right) \left( -mk^2 \right) = (mk)^2$$
The following is multiple choice question (with options) to answer.
The value of C+5C(7C)(5CC)when C= 7 is: | [
"420754",
"420777",
"420170",
"420175"
] | D | Solution:
C+5C(7C)(5CC)
Put the value of C= 7 in the above expression we get,
7+ 35(7 x 7)(5x7 x 7)
= 7+ 35(49)(5 × 49)
= 7 + 420175
=420175
Answer : D |
AQUA-RAT | AQUA-RAT-37854 | Thus the hands have been coincident 118 times after the starting position.
Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.
Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)
To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.
Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.
That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.
In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.
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? | [
"7.42 am",
"7.45 am",
"8.30 am",
"9.45 am"
] | A | 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 A |
AQUA-RAT | AQUA-RAT-37855 | $\frac {0.367}{0.978} = 0.375.$
13. ## Re: Probability with a "known"
Another way to do this is with combinations (since you don't care about the order that you pick the fruit). You will get the same answer either way. With combinations, you can just divide the number of outcomes. So, if we figure out the number of outcomes with (at least two red AND at least one red or green) divided by the number of outcomes with at least one apple (red or green), that will give us the same probability.
Just as ebaines did, we want the number of outcomes with exactly 0 red and exactly 1 red, then take the compliment (total number of outcomes minus outcomes that give us only 0 or 1 red). For exactly 0 red, that is $\binom{24}{5}$ outcomes.
For exactly 1 red, that is $\binom{24}{4}\binom{8}{1}$ outcomes (note: since we are doing combinations, we don't care which of the apples is red). So, there are $\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}$ outcomes with at least 2 red.
For at least one apple (red or green), we want the number of outcomes with 0 apples and take the compliment: $\binom{32}{5} - \binom{16}{5}$.
So, the probability that you get at least 2 red apples given that you pick at least one apple (red or green):
$\dfrac{\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}}{\binom{32}{5} - \binom{16}{5}} = \dfrac{73,864}{197,008} = \dfrac{1319}{3518} \approx 0.375$
14. ## Re: Probability with a "known"
The following is multiple choice question (with options) to answer.
A basket of 1430 apples is divided equally among a group of apple lovers. If 45 people join the group, each apple lover would receive 9 apples less. How many apples did each person get before 45 people joined the feast? | [
"20.",
"21",
"22",
"23"
] | C | if 1430 is divisible by anyone of the answer choices.
A.1430/20 = 143/2
B 1430/21 = 1430/21
C 1430/22 = 65
If 1430 apple was divided among 65 people, each would have received 22.
After addition of 45 people the answer should be 13. 1430/110 = 13.
C is the answer. |
AQUA-RAT | AQUA-RAT-37856 | java, performance, algorithm, programming-challenge
{8,5,3} satisfy both conditions
s[8] = 'a', s[5] = 'b' and s[3] = 'c'
(5 + 1)² = (8 + 1)(3 + 1)
We find 2 triples that match the 2 conditions so the answer is 2.
The code
gist link =
https://gist.github.com/JulienRouse/cafbce417bbc2a4f6303df10df20d445
Code here:
/**
* Consider a string s of length n with alphabet {a,b,c}.
* We look for the number of different triplet (i,j,k) where 0<=i,j,k<n, satisfying!
* - s[i] = "a", s[j] = "b", s[k] = "c"
* - (j + 1)² = (i + 1)(k + 1)
* @param s A string we look the triplet in
* @return Number of different triplets such as enonced above.
*/
public static int geometricTrickv2(String s){
Set<Integer> indexA = new HashSet<>(s.length());
Set<Integer> indexC = new HashSet<>(s.length());
List<Integer> indexB = new ArrayList<>();
for(int i=0;i<s.length();i++){
if(s.charAt(i)=='a')
indexA.add(i);
if(s.charAt(i)=='b')
indexB.add(i);
if(s.charAt(i)=='c')
indexC.add(i);
}
The following is multiple choice question (with options) to answer.
In a coded language
If
tomato = 3
apple = 2
math = 1
then
mathematics = ? | [
"1",
"2",
"3",
"4"
] | D | mathematics=4
because no. of vowels in mathematics=4
ANSWER:D |
AQUA-RAT | AQUA-RAT-37857 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Posts: 140
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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.
Mike needs 30% to pass. If he scored 212 marks and falls short by 22 marks, what was the maximum marks he could have got? | [
"343",
"777",
"780",
"867"
] | C | If Mike had scored 22 marks more, he could have scored 30%
Therefore, Mike required 212 + 22 = 234 marks
Let the maximum marks be m.
Then 30 % of m = 234
(30/100) × m = 234
m = (234 × 100)/30
m = 23400/30
m = 780
Answer:C |
AQUA-RAT | AQUA-RAT-37858 | 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.
Latha travels from A to B a distance of 1000 miles in 10 hours. He returns to A in 5 hours. Find his average speed? | [
"100mph",
"125.98mph",
"156.15mph",
"133.3mph"
] | D | Speed from A to B = 1000/10 = 100 mph
Speed from B to A = 1000/5 = 200 mph
Average speed = 2*100*200 / 300 = 133.3mph
Answer is D |
AQUA-RAT | AQUA-RAT-37859 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
A man bought 20 shares of Rs. 50 at 5 discount, the rate of dividend being 4 %. The rate of interest obtained is | [
"5%",
"5.28%",
"5.30%",
"5.50%"
] | B | Shares = 20 shares
50 at 5 discount
rate divident 4%
req ===> 5.28%
ANSWER B |
AQUA-RAT | AQUA-RAT-37860 | 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.
Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is
45 kmph. For how many minutes does the bus stop per hour? | [
"12",
"11",
"10",
"9"
] | C | SPEED OF THE BUS EXCLUDING STOPPAGES = 54 KMPH
SPEED OF THE BUS INCLUDING STOPPAGES = 45 KMPH
LOSS IN SPEED INCLUDING STOPPAGES = 54-45 = 9 KMPH
IN HOUR BUS COVERS 9 KM LESS DUE TO STOPPAGES
HENCE DISTANCE/SPEED = 9/54 = 1/6 HR = 60/6 MIN = 10 MIN
ANSWER C |
AQUA-RAT | AQUA-RAT-37861 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular garden is to be twice as long as it is wide. If 180 yards of fencing, including the gate, will completely enclose the garden, what will be the length of the garden, in yards? | [
"60",
"70",
"80",
"90"
] | A | ALTERNATE APPROACH
Backsolving ( Using answer options to reach the correct answer ) can work wonders here if one is fast in calculations.
Given Perimeter is 180
So, 2 ( L + B ) = 180
Or, L + B = 90
Now use the answer options ( Given Length ; Breath will be half the length)
(A) 60
L = 60 ; B = 30
L + b = 90
(B) 70
L = 70 ; B = 35
L + b = 105
(C) 80
L = 80; B = 40
L + b = 120
(D) 90
L = 90; B = 45
L + b = 135
(E) 100
L = 100; B = 50
L + b = 150
Thus you see no, need of any calculations, U can reach the correct option only by checking options ; correct answer will be (A) |
AQUA-RAT | AQUA-RAT-37862 | $$(2)-(1')\Rightarrow f\, =\, -2e\, +\, 765 \: \: \: ---(3)\\ (3)\rightarrow (1)\Rightarrow 22e = 21(-2e+765)+319 \\ e=256 \\ f=253$$
The following is multiple choice question (with options) to answer.
653×1001=? | [
"649011",
"540041",
"654321",
"653001"
] | D | 653×(1000+1)
653000+1
653001
answer D |
AQUA-RAT | AQUA-RAT-37863 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
Three candidates contested an election and received 3834 votes, 7926 votes, and 16240 votes respectively. What percentage of the total votes did the winning candidate receive? | [
"55%",
"56%",
"57%",
"58%"
] | D | The total number of votes polled = (3834 + 7926 + 16240) = 28000
The winner's percentage = 16240/28000 * 100 = 58%
The answer is D. |
AQUA-RAT | AQUA-RAT-37864 | ### Show Tags
22 May 2018, 02:57
belagerfeld wrote:
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
This type of sums can also be solved using Fermet's Theorem approach.
Refer photo attached below:
Attachments
WhatsApp Image 2018-05-22 at 3.18.54 PM.jpeg [ 90.53 KiB | Viewed 2464 times ]
Re: What is the units digit of 2222^(333)*3333^(222) ? &nbs [#permalink] 22 May 2018, 02:57
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
If n = (33)^43 + (43)^32 what is the units digit of n? | [
"0",
"2",
"4",
"6"
] | B | First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^32 is the same as that of 3^32. So, we need to find the units digit of 3^43 + 3^33.
Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}:
3^1=3 (the units digit is 3)
3^2=9 (the units digit is 9)
3^3=27 (the units digit is 7)
3^4=81 (the units digit is 1)
3^5=243 (the units digit is 3 again!)
...
Thus:
The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3).
The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).
Therefore the units digit of (33)^43 + (43)^32 is 7 + 5 = 2.
Answer: B. |
AQUA-RAT | AQUA-RAT-37865 | Since 2m+1 is greater than P1, x1 is positive. An even number minus an odd plus an odd equals an even, and half an even number is an integer. Hence, x1 is a solution to S.
In the second case, where 2m+1 is less than P1, let’s try N = 2m+1 as a solution. This yields: $x_1 = (\frac{2S}{2^{m+1}} – 2^{m+1} + 1)/2 \\ S = 2^m\cdot P_1 \cdot S_2 \implies \\ x_1 = (\frac{2 \cdot 2^m \cdot P_1 \cdot S_2}{2^{m+1}} – 2^{m+1} + 1)/2 \\ = (P_1 \cdot S_2 – 2^{m+1} + 1)/2$
Since P1 is greater than 2m+1, x1 is again positive. P1 and S2 are both odd, hence P1S2 is odd. An odd number minus an even plus an odd is an even number, and half of an even number is an integer. Hence, x1 is a solution to S.
This proves that S has a solution R for all values that are not powers of 2; it also illustrates why 23•16 requires 23 numbers (23 < 32) while 23•8 requires only 16 (16 < 23). As we would expect, 592 (37•16) requires 32 numbers, not 37 (3+4+...+34); 1184 (37•32) requires 37 (14+15+...+50). All of this proves the original hypothesis: A positive integer S is equal to the sum of consecutive positive integers iff it has odd prime factors.
## 1 Comment
1. paulhartzer
It occurred to me later that this provides a direct proof all odd numbers have N=2 solutions: All odd numbers are of the form 2^0•P_1•S_2. Since 2^(0+1) = 2 will always be lower than P_1, N=2^(0+1)=2 will always provide a solution.
The following is multiple choice question (with options) to answer.
If x is a whole number, then x2(x2 - 1) is always divisible by | [
"12",
"24",
"12-x",
"multiple of 12"
] | A | Sol.
Putting x = 2, we get 22 (22 - 1) = 12.
So, x2(x2 - 1) is always divisible by 12.
Answer A |
AQUA-RAT | AQUA-RAT-37866 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of first 12 even numbers is? | [
"10",
"11",
"12",
"13"
] | D | Sum of 12 even numbers = 12 * 13 = 156
Average = 156/13 = 13
ANSWER:D |
AQUA-RAT | AQUA-RAT-37867 | #### Jomo
##### Elite Member
It has already been pointed out that $$\displaystyle log((3x- 1)(3x+ 1))= log(9x^2- 1)= log(16)$$.
Now I would NOT write that as $$\displaystyle log(9x^2- 1)- log(16)= log\left(\frac{9x^2- 1}{16}\right)= 0$$.
Instead I would just use the fact that the logarithm is "one to one" so that $$\displaystyle 9x^2- 1= 16$$.
$$\displaystyle 9x^2= 17$$
$$\displaystyle x^2= \frac{17}{9}$$
$$\displaystyle x= \pm\frac{\sqrt{17}}{3}$$.
Dr Halls,
I truly respect you as a mathematician. I just do not know what to say sometimes about your ability to do basic math! Logs do have a restricted domain! I am sure that you know that better than I do.
I am just messing with you. I am always amused at some things that brilliant people say and actually post.
Have a great day!
Steve
#### HallsofIvy
##### Elite Member
I certainly won't say that I know mathematics better than you do, but I do often amaze myself with how careless I can be!
#### eddy2017
##### Senior Member
I am printing the whole thread and stuying it now that i am learning how to do logs.This is amazing. I don't think there is another site like this online. No, none like this!
The following is multiple choice question (with options) to answer.
log 3600 is equal to | [
"2log6 +1",
"6log2 + 1",
"2log6 +2",
"6log2 + 2"
] | C | log 3600=log 36 + log 100=2log 6 + 2.
log 36=log 6^2=2log 6
log 100=2.
-->2log 6+2
ANSWER:C |
AQUA-RAT | AQUA-RAT-37868 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m? | [
"2828",
"2277",
"2797",
"3900"
] | D | Area = (l + b – d) d
(80 + 60 – 10)10 => 1300 m2
1300 * 3
= Rs.3900
D |
AQUA-RAT | AQUA-RAT-37869 | # 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 goods at the cost price but uses a weight of 600 grams per kg, what is his percent? | [
"75%",
"67%",
"29%",
"55%"
] | B | 600 --- 400
100 --- ? => 67%
Answer: B |
AQUA-RAT | AQUA-RAT-37870 | 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 of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction? | [
"8",
"10",
"12",
"11"
] | C | Speed = 120/10 = 12 m/sec
speed of the second train = 120/5 = 8 m/sec
relative speed = 12+8 = 20 m/sec
Req time = 120+120/20 = 12 sec
ANSWER C |
AQUA-RAT | AQUA-RAT-37871 | # 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.
Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received Rs.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 Rs.605 as interest. What was the value of his total savings before investing in these two bonds? | [
"Rs.5500",
"Rs.11000",
"Rs.22000",
"Rs.2750"
] | D | Solution:
Explanatory Answer
Shawn received an extra amount of (Rs.605 – Rs.550) Rs.55 on his compound interest paying bond as the interest that he received in the first year also earned interest in the second year.
The extra interest earned on the compound interest bond = Rs.55
The interest for the first year =550/2 = Rs.275
Therefore, the rate of interest =55/275 * 100 = 20% p.a.
20% interest means that Shawn received 20% of the amount he invested in the bonds as interest.
If 20% of his investment in one of the bonds = Rs.275, then his total investment in each of the bonds = 275/20 * 100 = 1375.
As he invested equal sums in both the bonds, his total savings before investing = 2*1375 = Rs.2750.
Answer D |
AQUA-RAT | AQUA-RAT-37872 | 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 is 360 meter long is running at a speed of 36 km/hour. In what time will it pass a bridge of 140 meter length? | [
"11 seconds",
"38 seconds",
"50 seconds",
"88 seconds"
] | C | Speed = 36 Km/hr = 36*(5/18) m/sec = 10 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
= 500 * (1/10) = 50 seconds
Answer: C |
AQUA-RAT | AQUA-RAT-37873 | Correct. This is a much easier problem, isn’t it?
In particular, I would explain what you’ve done in this way:
Since one particular team can be singled out as “the one with the four friends”, making the teams no longer indistinguishable, we need only to count ways to fill out the rest of this one team, and do not need to divide by 2 in the end. There are 8 other students, from which we choose 2 in addition to the four friends, so there are 8C2 = 28 ways.
We could instead choose the 6 for the other team, putting the remaining 2 on the team with the friends; since 8C6 = 8C2, this gives the same result.
We have either two slots to fill from the 8 remaining students, or six slots (on the other team):
## Problem 3: every school on each team
Here is his answer to the third part:
Meanwhile, for the third part, I am solving it this way:
Since we want at least one student from each school, Team 1 will be (3C1 × 4) × 8C2 and Team 2 will take the remaining students. Hence, we have 336 ways. But this includes the combinations with the second team that might not have all students from all 4 schools.
So we subtract by (3C1 × 4) × 4C1 = 48. Thus my final answer is 336 – 48 = 288 ways.
What do you think about my solutions and answers?
### A subtractive method
My response continued:
This problem is much more complicated, and this is a good attempt, though flawed. You’ve done it a different way than I did, and got a different answer. So let’s see who’s right! We won’t want to check by listing this time, so we’ll just compare our answers.
You’ve used a subtractive method, which can be good. (I used addition.) You focus on one team (which suggests we may have to divide by 2 at the end, since the teams are indistinguishable). But you haven’t explained the details of your (3C1 × 4) × 8C2. I think you are saying that for each of the 4 schools, you are picking 1 student, and then picking 2 more from the remaining 8. There are a couple problems here.
The following is multiple choice question (with options) to answer.
A project has three test cases. Three teams are formed to study the three different test cases. James is assigned to all three teams. Except for James, each researcher is assigned to exactly one team. If each team has exactly 11 members, then what is the exact number of researchers required? | [
"33",
"34",
"36",
"31"
] | D | 11*3-2( because james is included in all the three teams)
=33-2 =31
D |
AQUA-RAT | AQUA-RAT-37874 | # How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.
The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.
I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.
I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.
Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\to 720/6 = 120$.
So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.
I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.
The following is multiple choice question (with options) to answer.
There are 15 players in a chess group, and each player plays each of the others once. Given that each game is played by two players, how many total games will be played? | [
"10",
"30",
"105",
"60"
] | C | 15 players are there.
two players play one game with one another.
so 15C2=15x14/2=105
SO OPTION C is correct |
AQUA-RAT | AQUA-RAT-37875 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Find the odd man out. 10, 20, 30, 40, 50, 60, 72 | [
"15",
"72",
"51",
"90"
] | B | Explanation :
All except 72 are multiples of 10
Answer : Option B |
AQUA-RAT | AQUA-RAT-37876 | 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, each 100 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is? | [
"28 km/hr",
"17 km/hr",
"60 km/hr",
"16 km/hr"
] | C | Let the speed of the slower train be x m/sec.
Then, speed of the train
= 2x m/sec.
Relative speed
= ( x + 2x) = 3x m/sec.
(100 + 100)/8
= 3x => x = 25/3.
So, speed of the faster train = 50/3
= 50/3 * 18/5
= 60 km/hr.
Answer: C |
AQUA-RAT | AQUA-RAT-37877 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
Two vessels contains equal number of mixtures milk and water in the ratio 7:2 and 8:1. Both the mixtures are now mixed thoroughly. Find the ratio of milk to water in the new mixture so obtained? | [
"1:3",
"9:13",
"5:1",
"11:3"
] | C | The ratio of milk and water in the new vessel is = (7/9 + 8/9) : (2/9 + 1/9)
= 15/9 : 3/9 = 5:1
Answer is C |
AQUA-RAT | AQUA-RAT-37878 | 2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
The following is multiple choice question (with options) to answer.
Evaluate:434 - 12*3*2 = ? | [
"426",
"526",
"626",
"726"
] | A | According to order of operations, 12?3?2 (division and multiplication) is done first from left to right
12**2 = 4* 2 = 8
Hence
434 - 12*3*2 = 434 - 8 = 426
correct answer A |
AQUA-RAT | AQUA-RAT-37879 | $r = 0.962407319393878$ Compare to the value of the correlation coefficient $r$ found in part c), the small difference is due to the rounding errors.
Problem 2
The table below is the total world wheat production (x) and export (y) in million metric tons from 1962 to 2020 (61 years).(Data from The world bank at http://data.worldbank.org/).
a) Use any software applications such as Google Sheets, Excell, LibreOffice .. to make a scatter plot of y versus x.
b) Use any software to find the sums of squares $SS_x$, $SS_y$ and cross products $SS_{xy}$
c) Calculate the correlation using the formula and the sums calculated in part b).
d) Calculate the correlation $r$ using excel .
Solution Problem 2
a)
b)
There are 61 pairs of data values $(x_i,y_i)$, hence $m = 61$.
Excel was also used to do all the calculations of sums (red):
$\sum x_i = 31061.991$ , $\sum y_i = 6267.028$,
$\sum x_i y_i = 3556496.296$ , $\sum x_i^2 = 17328675.78$ , $\sum y_i^2= 742779.5112$
The following is multiple choice question (with options) to answer.
When 15% is lost in grinding wheat, a country can export 30 lakh tons of wheat. On the other hand, if 10% is lost in grinding, it can export 40 lakh tons of wheat. The production of wheat in the country is : | [
"20 lakh tons",
"80 lakh tons",
"200 lakh tons",
"800 lakh tons"
] | C | Solution
Let the total production be x lakh tons.
Then, 15% of X - 10% of X = (40 - 30) lakh tons
⇔ 5% of x = 10 lakh tons ⇔ x = ( 10x100/5 )= 200 lakh tons.
Answer C |
AQUA-RAT | AQUA-RAT-37880 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
I chose a number and divide it by 4. Then I subtracted 24 from the result and got 10. What was the number I chose? | [
"170",
"700",
"800",
"900"
] | A | Solution:
Let x be the number I chose, then
x/4−24=10
x/5=34
x=170
Answer: A |
AQUA-RAT | AQUA-RAT-37881 | Probability in getting two numbers exactly the same after rolling three dices?
What is the probability of getting two numbers the same having three dices? I had this on my exam, it sounded super easy but isn't so.
Solution: I know that there are $$6^3$$ combinations for all three dices. Then I just wrote the following:
And the combinations that I want are: $$first-second:(1,1,x),(2,2,x),(3,3,x),(4,4,x),(5,5,x),(6,6,x)$$
$$second-third:(x,1,1),(x,2,2),(x,3,3),(x,4,4),(x,5,5),(x,6,6)$$ $$first-third:(1,x,1),(2,x,2),(3,x,3),(4,x,4),(5,x,5),(6,x,6)$$ wich gives me the combination of $$6+6+6=18$$ and final result $$18/216$$, which is incorrect. What am I doing wrong?
• Exactly two the same, or at least two the same? – tilper Sep 7 '16 at 14:27
• Exactly the same – eugene_sunic Sep 7 '16 at 14:28
• Refer to Lovsovs' answer-hint. What you've done so far doesn't exclude, for example, (1,1,1). – tilper Sep 7 '16 at 14:29
• Still unclear. Do you mean exactly 2 the same? – drhab Sep 7 '16 at 14:29
• Yes exactly to the same... – eugene_sunic Sep 7 '16 at 14:29
Hint: For each of your eighteen cases, what can $x$ be?
The following is multiple choice question (with options) to answer.
Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is | [
"5/17",
"5/12",
"5/11",
"5/18"
] | B | Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9)
= 5/12
Answer:B |
AQUA-RAT | AQUA-RAT-37882 | A. $28.50 B.$27.00
C. $19.00 D.$18.50
E. $18.00 We can compute a weighted average to solve. Let’s assume that 2 units of Q and 1 unit of P were produced last year. So the total revenue is 2 x 17 + 1 x 20 =$54, and thus the average revenue per unit is thus 54/3 = $18. Answer: E _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Retired Moderator Joined: 21 Aug 2013 Posts: 1386 Location: India Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 02 Jul 2017, 20:49 1 Ratio of quantity of P and Q sold, P:Q = 1:2. Thus, average revenue per unit = (20*P + 17*Q)/(P+Q) = (20*1 + 17*2)/(2 + 1) = 54/3 =$ 18
Intern
Joined: 28 Aug 2016
Posts: 11
Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink]
### Show Tags
18 Apr 2019, 14:30
1
1
AbdurRakib wrote:
A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is $20.00 and the revenue from the sale of each unit of Q is$17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?
The following is multiple choice question (with options) to answer.
If a certain toy store's revenue in November was 2/5 of its revenue in December and its revenue in January was 1/2 of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January? | [
"1/4",
"1/2",
"10/3",
"2"
] | C | N = 2D/5
J = N/2 = D/5
The average of November and January is (N+J)/2 = 3D/5 / 2 = 3D/10
D is 10/3 times the average of November and January.
The answer is C. |
AQUA-RAT | AQUA-RAT-37883 | 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.
A’s speed is 17/15 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat? | [
"1/16",
"2/15",
"2/17",
"1/8"
] | C | Let x be the fraction of the distance that B runs.
Let v be the speed at which B runs.
The time should be the same for both runners.
Time = D / (17v/15) = xD/v
(15/17)*D/v = x*D/v
x = 15/17
B should have a head start of 2/17 of the full distance.
The answer is C. |
AQUA-RAT | AQUA-RAT-37884 | 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 moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.50 km and 1.0 km respectively. The time taken by the slower train to cross the faster train in seconds is? | [
"48",
"9",
"7",
"60"
] | D | :
Relative speed = 60 + 90 = 150 km/hr.
= 150 * 5/18 = 125/3 m/sec.
Distance covered = 1.50 + 1.0 = 2.5 km = 2500 m.
Required time = 2500 * 3/125
= 60 sec.
Answer: D |
AQUA-RAT | AQUA-RAT-37885 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
A sum of Rs.4800 is invested at a compound interest for three years, the rate of interest being 10% p.a., 20% p.a. and 25% p.a. for the 1st, 2nd and the 3rd years respectively. Find the interest received at the end of the three years. | [
"2888",
"3120",
"2877",
"2999"
] | B | Let A be the amount received at the end of the three years.
A = 4800[1 + 10/100][1 + 20/100][1 + 25/100]
A = (4800 * 11 * 6 * 5)/(10 * 5 * 4)
A = Rs.7920
So the interest = 7920 - 4800 = Rs.3120
Answer: B |
AQUA-RAT | AQUA-RAT-37886 | Author Message
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Joined: 26 Apr 2010
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
A corporation triples its annual bonus to 50 of its employees. What T percent of the employees’ new bonus is the increase? | [
"50%",
"662⁄3%",
"100%",
"200%"
] | B | T of the employees’ new bonus is the increase Hence B. |
AQUA-RAT | AQUA-RAT-37887 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
In order to obtain an income of Rs. 3800 from 70% stock at Rs. 280, one must make an investment of | [
"Rs.16000",
"Rs.15000",
"Rs.15250",
"Rs.15200"
] | D | Explanation :
Market Value = Rs. 280
Required Income = Rs. 3800.
Here face value is not given. Take face value as Rs.100 if it is not given in the question
To obtain Rs.70 (ie,70% of the face value 100), investment = Rs.280
To obtain Rs.3800, investment = 280/70×3800=Rs.15200
Answer : Option D |
AQUA-RAT | AQUA-RAT-37888 | ruby, e-commerce
Here is a revised version of the code:
def get_discounts
discount = 0
@specials.each do |special|
if item = @items[special.sale_item]
multiples = (item[:quantity] / special.quantity).floor
discount += multiples * special.discount
# Suspicious side-effect...
# item[:quantity] -= multiples * special.quantity
end
end
return discount
end
The following is multiple choice question (with options) to answer.
the percentage discount offered for"Buy 6 get 1 free is"? | [
"14.29%",
"15.29%",
"17.29%",
"19.29%"
] | A | Total Items = 7
Discounted = 1
Percentage = 1/7*100
=14.29%
ANSWER:A |
AQUA-RAT | AQUA-RAT-37889 | Circular motion question
1. Nov 10, 2005
donjt81
This is the question...
A small wheel of radius 1.4cm drives a large wheel of radius 15cm by having their circumferences pressed together. If the small wheel turns at 407 rad/s, how fast does the larger one turn? Answer in rad/s
This is what I was thinking...
radius of smaller wheel = .014m
radius of larger wheel = .15m
circumference of smaller wheel = 2*pi*r = 2*3.14*.014 = .08792
angular velocity of smaller wheel (given) = 407 rad/s
angular velocity = circumference/time
time = circumference/angular velocity
=.08792/407 = .000216s
circumference of larger wheel = 2*pi*r = 2*3.14*.15 = .942
angular velocity = circumference/time
=.942/.000216 = 4361.11 rad/s
Does this approach look right?
2. Nov 10, 2005
BerryBoy
I disagree... By intuition, you can predict that the larger wheel is going to turn more slowly.. Try another approach..
Hint: Consider the fact that the speeds of circumferences are equal.
Eq: speed = w x r
w = angular velocity
r = radius
Does this help?
Sam
Last edited: Nov 10, 2005
3. Nov 10, 2005
donjt81
You are right... the larger wheel should go slower.
but since the speed of smaller wheel is 407 rad/s wont the larger wheel speed be the same?
so is the answer to the problem 407 rad/s for the larger wheel? but that doesnt make sense because the larger wheel is supposed to go slower...
I am confused...
4. Nov 10, 2005
BerryBoy
OK, so if the speed at the circumfrence is:
v = w x r (as I stated above).
If the wheels are in contact this speed is equal on both wheels (not the angular velocity). Therefore:
wsmall x rsmall = wlarge x rlarge
I can't give you anymore hints without doing it now.
The following is multiple choice question (with options) to answer.
The radius of the wheel of a bus is 100 cms and the speed of the bus is 66 km/h, then the r.p.m. (revolutions per minutes)of the wheel is | [
"175",
"250",
"300",
"330"
] | A | Radius of the wheel of bus = 100 cm. Then,
circumference of wheel = 2Ï€r = 200Ï€ = 628.57 cm
Distance covered by bus in 1 minute
= 66â„60 × 1000 × 100 cms
Distance covered by one revolution of wheel
= circumference of wheel
= 440 cm
∴ Revolutions per minute = 6600000/60×628.57 = 175
Answer A |
AQUA-RAT | AQUA-RAT-37890 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
The ratio of boys to girls in a school is 9 to 11. After 36 more girls join, the ratio becomes 9 to 13. How many boys are there in the school? | [
"102",
"115",
"132",
"162"
] | D | Boys : Girls
9 : 11
let the boys be 9x and the girls be 11x. Now, 36 more girls are added -------> 9:13(new ratio).
so,
9x/(11x+36) =9/13.
solving,
We get : x=18
so No. of boys in the school are :9x= 9*18= 162
hence (D) |
AQUA-RAT | AQUA-RAT-37891 | 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.
In how many ways letters of the word REPEAT be arranged? | [
"360",
"120",
"240",
"180"
] | A | REPEAT has 6 letters, out of which E repeated 2 times. Hence total ways = 6! / (2!) = 360ways
A |
AQUA-RAT | AQUA-RAT-37892 | We are now left with $5$ more elements $\{3, 4, 5, 6, 7\}$. For the left subtree's two leaves, we have $5 * 4$ ways. i.e. first choosing one of the 5 nodes, then choosing one of the remaining 4 nodes.
Now $3$ nodes left. Out of these $3$, one will be the least among them, and that will definitely become the parent of the two remaining leaves(in the right subtree). Now with $2$ nodes left, we can arrange them in $2$ ways.
This gives $(5 * 4) * 2 = 40$ ways.
We can have the same number of ways, if we fixed $2$ at the right subtree instead of left. So total ways:
$= 40 * 2$
$= \mathbf{80}$
+1
+1
Yes, answer will be same @Warlock lord
+1
Great explanation
1 has to be root. Now,
case 1: 2 and 3 are on 2nd level.
so, 4,5,6 and 7 can occupy any of 4 positions in the 3rd level as all are less than 2 and 3. So, 4! arrangements. And 2 and 3 can be arranged in 2 ways in 2nd level (mirror image). Hence 2*24=48
case 2: 2 and 4 are on 2nd level
Now, 3 can only be below 2 and not 4 as it is MIN heap. So, 2 cases are possible 3XXX or X3XX, which gives 3!+3!=12 arrangements. And 2 and 4 can be arranged in 2 ways in 2nd level (mirror image). Hence 2*(6+6)=24
case 3: 2 and 5 are on 2nd level
Now, 3 and 4 can only be below 2 and not 5 as it is MIN heap. So, 2 arrangements are possible for 3 and 4 and similarly 2 for 6 and 7. And 2 and 5 can be arranged in 2 ways in 2nd level (mirror image). Hence, 2*(2+2)=8
Hence, total 48+24+8=80
+2
this was my exact approach in d exam
0
I hope, Hence, 2*(2+2)=8 LINE SHOULD BE 2*(2*2)=8
The following is multiple choice question (with options) to answer.
The least number, which when divided by 12, 15, 20 and 63 leaves in each case a remainder of 8 is: | [
"448",
"488",
"542",
"1268"
] | D | Required number = (L.C.M. of 12, 15, 20, 63) + 8
= 1260+ 8
= 1268.
answer :D |
AQUA-RAT | AQUA-RAT-37893 | 0.00698 acre1 acre = 43560 square feet304 sq ft * 1 acre/43560 sq ft = 0.00698 acre
### Number of sq ft per acre?
1 sq mi=5280ft x 5280ft = 27878400 sq ft/sq mi/ 27,878,400 sq ft/sq mi divided by 640acres/sq mi = 43,560 sq ft /acre 1 acre = 43,560 sq ft
### How many square feet are in 1.8 acre?
1 acre = 43,560 sq ft → 1.8 acre = 1.8 × 43,560 sq ft = 78,408 sq ft
### What does 0.2 acres equals sq feet?
1 acre = 43,560 sq ft .2 acre * 43560 sq ft = 8712 sq ft
### What percentage of acres in 160 feet x 180 feet?
160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.
### How many ft are in a acre?
There are 43559.66 sq ft in an acre.
### What is a fifth of an acre?
an acre is 43,560 sq ft 1/5th of that is 8,712 sq ft
### How many acres is 11026 sq ft?
43,560 sq ft = 1 acre11,026 sq ft = 0.2531 acre (rounded)
### Figure how many acres in 60000 sq ft?
(60,000 sq ft) / (43,560 sq ft / acre) = 1.3774 acre (rounded)
### What is the convertion of 435 acre to square feet?
1 acre = 43,560 sq ft → 435 acre = 435 x 43,560 sq ft = 18,948,600 sq ft
### How many sq ft for one acre 14 guntas in India?
The following is multiple choice question (with options) to answer.
What will be the cost of building a fence around a square plot with area equal to 25 sq ft, if the price per foot of building the fence is Rs. 58? | [
"1160",
"2287",
"2977",
"2668"
] | A | Let the side of the square plot be a ft.
a2 = 25=> a = 5
Length of the fence = Perimeter of the plot = 4a = 20 ft.
Cost of building the fence = 20 * 58 = Rs. 1160.
Answer:A |
AQUA-RAT | AQUA-RAT-37894 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a 1000 m race, A beats B by 100 m and B beats C by 100 m. In the same race, by how many meters does A beat C? | [
"A)190m",
"B)829m",
"C)822m",
"D)929m"
] | A | By the time A covers 1000 m, B covers (1000 - 100) = 900 m.
By the time B covers 1000 m, C covers (1000 - 100) = 900 m.
So, the ratio of speeds of A and C =
1000/900 * 1000/900 = 1000/810 So, by the time A covers 1000 m, C covers 810 m.
So in 1000 m race A beats C by 1000 - 810 = 190 m.
Answer:A |
AQUA-RAT | AQUA-RAT-37895 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
If 5/w + 5/x = 5/y and wx = y, then the average (arithmetic mean) of w and x is | [
"1/2",
"1",
"2",
"4"
] | A | Given: 5/w + 5/x = 5/ywx=y
Find: (w+x)/2 = ?
5(1/w + 1/x) = 5(1/y) - divide both sides by 5
(1/w + 1/x) = 1/y
(x+w)/wx = 1/wx - sub'd in y=wx
x+w - 1 = 0
x+w = 1
Therefore (w+x)/2 = 1/2
Ans: A |
AQUA-RAT | AQUA-RAT-37896 | 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 car averages 70 mph for the first 4 hours of a trip and averages 60 mph for each additional hour. The average speed for the entire trip was 65 mph. How many hours long is the trip? | [
"8",
"9",
"10",
"11"
] | A | Let the time for which car averages 60 mph = t
65*(t+4)= 70*4 + 60 t
=>5 t= 20
=> t = 4
Total duration of the trip = 4+4=8
Answer A |
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