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There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village? 1. There are 40 more pupils in the second school than there are in the first. 2. There are three times as many pupils in the second school as there are in the first. Source: GMAT Club Tests - hardest GMAT questions Statement (1) by itself is insufficient. Denote X and Y the number of pupils in school 1 and school 2 respectively. By definition of average, we have to calculate \frac{12.2X + 13.1Y}{X + Y} . S1 is not sufficient because one of the unknowns cannot be found.
Statement (2) by itself is sufficient. S2 says that Y = 3X , so X cancels out of the fraction and we can calculate the result. Just a little confused : according to statement 1. can't we make the equation Y=X+40 ? Thanks
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shrive555 wrote: Quote: There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village?
1. There are 40 more pupils in the second school than there are in the first.
2. There are three times as many pupils in the second school as there are in the first.
Statement (1) by itself is insufficient. Denote X and Y the number of pupils in school 1 and school 2 respectively. By definition of average, we have to calculate \frac{12.2X + 13.1Y}{X + Y} . S1 is not sufficient because one of the unknowns cannot be found.
Statement (2) by itself is sufficient. S2 says that Y = 3X , so X cancels out of the fraction and we can calculate the result. Just a little confused : according to statement 1. can't we make the equation Y=X+40 ? Thanks Hi shrive, you are right. We can make the equation Y=X+40, but by using this equation, can we find the required avg?? I think No. Check this out.
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Thanks Hussain15: well i assumed " zero " on the right hand side of the equation, which is wrong.
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we cant find answer by stmnt 1 but we can solve it by statement 2. because the common term will cancel out.
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It just says the answer is "B", but I don't see what B is. I got 12.875 as the average age. Is this correct?
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B.
Average * #items = SUM
12.2 * x = 12.2x School 1
13.1 * 3x = 39.3x School 2
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? * 4x = 51.5x Total
? = 51.5x/4x doesn't matter outcome. Xs cancel and yes we get the average.
Kudus for me please
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Zanini wrote: It just says the answer is "B", but I don't see what B is. I got 12.875 as the average age. Is this correct? yes.. i solve this question using the statement 2 and got the same result. its 12.875
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hi wayxi,
I dont understand. Can you explain in more detail how you came your equation below. please explain using statement 1 and 2
12.2 * x = 12.2x School 1
13.1 * 3x = 39.3x School 2
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? * 4x = 51.5x Total
Thanks
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B.
S1 : assume x for first school so for second school = x + 40 the number of pupils
x*avg1 + (x+40)*avg2 / 2x + 40 is not sufficient
s2 : x, 3x : x*avg1 + 3x*avg2 / 4x --> x get cancelled and hence new avg can be determined
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Maryann,
The formula for average goes:
Average * # Items = SUM
i.e. Average of 6 different values is 62. Then sum of the six different values are:
62 * 6 = 372
Statement 1:
x : number of students in School 1
x + 40 : number of students in School 2
SCHOOL 1: 12.1 * x = 12.1x
SCHOOL 2: 13.1 * (x+40) = 13.1x + 524
We want to the get the average of BOTH schools so we need to divide the Total SUM by Total Items.
Adding the SUM of the two schools we get 25.2x + 524.
Adding the # Items we get 2x+40
AVERAGE: ( 25.2x + 524 ) / ( 2x+40 ) . We can't work this out so its, INSUFFICENT
Statement 2:
x: number of students to School 1
3x: number of students in School 2
SCHOOL 1: 12.1 * x = 12.1x
SCHOOL 2: 13.1 * 3x = 39.3x
Adding the SUM of the schools : 12.1x + 39.3x = 51.4x
Adding #Items: x +3x = 4x
AVERAGE: (51.4x) / (4x )
The Xs cancel so we get a value. It doesn't matter what the value is as long as we're able to solve this. Its 12.85 btw. B is SUFFICENT
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@Wayxi
Sorry, I need to brush up on math fundamentals, but is the reason we cannot solve statement 1 because by trying to eliminate the variable in the denominator, we end up creating a quadratic equation with two potential answers? If not, whats the reasoning?
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What would the difficulty level of this question be?
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shrive555 wrote: There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village? 1. There are 40 more pupils in the second school than there are in the first. 2. There are three times as many pupils in the second school as there are in the first. Source: GMAT Club Tests - hardest GMAT questions Statement (1) by itself is insufficient. Denote X and Y the number of pupils in school 1 and school 2 respectively. By definition of average, we have to calculate \frac{12.2X + 13.1Y}{X + Y} . S1 is not sufficient because one of the unknowns cannot be found.
Statement (2) by itself is sufficient. S2 says that Y = 3X , so X cancels out of the fraction and we can calculate the result. Just a little confused : according to statement 1. can't we make the equation Y=X+40 ? Thanks This is a weighted average question. Say x and y are the number of pupils in the first and the second schools in the village, respectively: average=\frac{12.2x+13.1y}{x+y}. (1) There are 40 more pupils in the second school than there are in the first --> x+40=y. Not sufficient to find the average. (2) There are three times as many pupils in the second school as there are in the first --> 3x=y --> average=\frac{12.2x+13.1*3x}{x+3x}=\frac{12.2+13.1*3}{4}\approx{12.9}. Sufficient. Answer: B. General rule for weighted average questions of two groups. If you know ANY 2 of the following 3 you can find the third one:1. The weighted average - 12.9 in our case; 2. The individual averages of the groups - 12.2 and 13.1 in our case; 3. The ratio of the groups - 3x=y ( \frac{x}{y}=\frac{1}{3}) in our case (notice that if you know \frac{x}{y} you also know \frac{x}{x+y}=\frac{x}{total}).
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Bunuel wrote: General rule for weighted average questions of two groups. If you know ANY 2 of the following 3 you can find the third one:
1. The weighted average - 12.9 in our case;
2. The individual averages of the groups - 12.2 and 13.1 in our case;
3. The ratio of the groups - 3x=y (\frac{x}{y}=\frac{1}{3}) in our case (notice that if you know \frac{x}{y} you also know \frac{x}{x+y}=\frac{x}{total}). I didnt understand the general rule - if #1 and #3 are given, how can we get individual averages of both groups ? Can someone clarify ?
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glores1970 wrote: Bunuel wrote: General rule for weighted average questions of two groups. If you know ANY 2 of the following 3 you can find the third one:
1. The weighted average - 12.9 in our case;
2. The individual averages of the groups - 12.2 and 13.1 in our case;
3. The ratio of the groups - 3x=y (\frac{x}{y}=\frac{1}{3}) in our case (notice that if you know \frac{x}{y} you also know \frac{x}{x+y}=\frac{x}{total}). I didnt understand the general rule - if #1 and #3 are given, how can we get individual averages of both groups ? Can someone clarify ? Can somebody explain how one can arrive at the individual averages knowing just #1 and #3 ? I dont seem to get it.
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glores1970 interesting question. even i cannot figure how to get the individual avgs. I guess we would need to know atleast one of the individual avgs. Other wise we end up with one equation with 2 unknowns!
@Bunuel could you elaborate on this point?
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Hi,
Only B is sufficient to answer the question, cannot be solved by A .
If we go by A, we end up in an unknown variable that doesnot cancel out.
If we go by B, we clearly can get the answer as Sum of age of pupils (second school) can be expressed in form of Sum of Age of Pupils (First School)
and same applies to No Of students as well.
Thus, B can solved the problem alone.
Thanks,
Rishukvt
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B for me because only average is needed, not the exact no of students.
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