In a game, three groups of students (A, B and C) in a class have averaged scores of -4, 0.5 and 20 respectively. The class average is 2. Group A and B account for what proportion of the entire student population?
a) 1/3 and 4/9
b) 4/9 and 2/9
c) 5/9 and 1/9
d) 2/9 and 5/9
e) 1/3 and 2/3
OA A....Please explain
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voodoo_child
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neelgandham
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In a game, three groups of students (A, B and C) in a class have averaged scores of -4, 0.5 and 20 respectively. The class average is 2. Group A and B account for what proportion of the entire student population?
Let the number of people in groups A,B,C be A,B,C. So, -4*A + 0.5*B + 20*C = 2*A + 2*B + 2*C
6A + 1.5B = 18C
4A + B = 12C
a) 3/9 and 4/9. So, C = 2/9. 4*(3/9) + 4/9 != 24/9
b) 4/9 and 2/9. So, C = 3/9. 4*(4/9) + 2/9 != 36/9
c) 5/9 and 1/9. So, C = 3/9. 20/9 + 1/9 != 36/9
d) 2/9 and 5/9. So, C = 2/9. 8/9 + 5/9 != 24/9
e) 1/3 and 2/3. So, C = 0. Can't happen!
I don't get it. Am I doing it wrong ?
Let the number of people in groups A,B,C be A,B,C. So, -4*A + 0.5*B + 20*C = 2*A + 2*B + 2*C
6A + 1.5B = 18C
4A + B = 12C
a) 3/9 and 4/9. So, C = 2/9. 4*(3/9) + 4/9 != 24/9
b) 4/9 and 2/9. So, C = 3/9. 4*(4/9) + 2/9 != 36/9
c) 5/9 and 1/9. So, C = 3/9. 20/9 + 1/9 != 36/9
d) 2/9 and 5/9. So, C = 2/9. 8/9 + 5/9 != 24/9
e) 1/3 and 2/3. So, C = 0. Can't happen!
I don't get it. Am I doing it wrong ?
Anil Gandham
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Ian Stewart
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No, you're not doing anything wrong. The question makes no sense.neelgandham wrote: I don't get it. Am I doing it wrong ?
With the information given, there is not one unique answer. For example, if the proportion of students in group B is roughly 0, then (using weighted average principles), we find 1/4 of the students are in group C and 3/4 are in group A. Or, if the proportion in group A is roughly 0, you can find that 1/13 of students are in group C, and 12/13 are in group B. So really all you can say from the information given is that somewhere between 1/13 and 1/4 of all students are in group C. And when the proportion who are in group C is close to 1/4, as would need to be true for any of the answer choices to be right, almost all of the remaining students need to be in group A, which is why none of the answer choices is even close to being possible.
Where is the question from?
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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voodoo_child
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Ian,
Thanks for your response. This question is from 700 series document that I downloaded from GMATClub. However, because of an issue with my computer while traveling, I have lost all of my data. I do remember the solution.
The OE said that after solving the individual equations (i.e. Group A, B and C), the numbers for Group C were 2 and 38. Group B : -1 0 1 2. I don't remember the #s in Group A. But I do remember that there were totally 9 numbers in Group A, B and C, i.e. 4 in B, 2 in C and 3 in A. I don't remember the numbers in A. I think that they were all -ve. (actually, they have to be -ve because the overall mean is so low. Also, Group A's mean is also -ve).
I am not sure whether that would help. My apologies in late reply because I was really busy with official work this week.
thanks
Thanks for your response. This question is from 700 series document that I downloaded from GMATClub. However, because of an issue with my computer while traveling, I have lost all of my data. I do remember the solution.
The OE said that after solving the individual equations (i.e. Group A, B and C), the numbers for Group C were 2 and 38. Group B : -1 0 1 2. I don't remember the #s in Group A. But I do remember that there were totally 9 numbers in Group A, B and C, i.e. 4 in B, 2 in C and 3 in A. I don't remember the numbers in A. I think that they were all -ve. (actually, they have to be -ve because the overall mean is so low. Also, Group A's mean is also -ve).
I am not sure whether that would help. My apologies in late reply because I was really busy with official work this week.
thanks
