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Tough problem

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by hariharakarthi » Wed Nov 18, 2009 6:25 pm
In a group of 80 college students, how many own a car?

(1) Of the students who do not own a car, 14 are male.

(2) Of the students who own a car, 42% are female.
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Source: — Data Sufficiency |

by gmater29 » Wed Nov 18, 2009 8:27 pm
Is the OA C ?
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by kpitchfo » Wed Nov 18, 2009 8:30 pm
Is it possible to solve using both statements? I want to say the answer is no, thereby making it E.
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by thephoenix » Wed Nov 18, 2009 8:38 pm
With bth also we are not getting any value , so E
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by sunnyjohn » Thu Nov 19, 2009 12:10 am
IMO:E
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by hariharakarthi » Thu Nov 19, 2009 6:46 pm
Ans is B.

Stmt 2 Alone is suff to solve the problem.

Let C be the no of cars.
then, 42% C = No of Females
42/100 * C= Should be an integer
21/50 * C = should be an integer. The fractional part of the equation should be an int.
Hence, it should be multiple of 50 ... 50,100,150...
Total no of students given as 80. So only possible value of C is 50.

Take way, read the problem carefully and relate the term given in the problem.
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by Willy » Fri Nov 20, 2009 8:33 pm
lets not force or cook answers here...its E
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by Testluv » Sat Nov 21, 2009 12:11 am
hariharakarthi's reasoning is correct.

Sometimes algebra word problems are actually testing number property concepts such as factors, multiples, remainders, etc.

It must be true that any number of people is a positive integer. (We can't have a negative or fractional number of people!)

If we let the number of car-owining students be "C" we know that 42% of C is the number of women who own cars:

0.42 * C = number of women who own cars

But, we know that the number of people who own cars (C) must be a positive integer, and we know that the number of women who own cars must also be a positive integer. Therefore:

42% of a positive integer is a positive integer

.42 * pos integer = pos integer

42/100 * pos integer = pos integer

21/50 * pos integer = pos integer

21 and 50 do not share any prime factors; we cannot simplify the fraction any further. Accordingly, in order for the equation to hold true, the positive integer on the left (which was C) must be able to divide 50--it must be a positive multiple of 50.

If the positive integer on the left hand side were not a positive multiple of 50, then the left hand side of the equation would be a noninteger, and would no longer be equal to the right hand side of the equation. But, obviously and by definition, the two sides of an equation must equal each other. So, we know that C is a positive multiple of 50. The question stem necessitates that C<=80. Therefore, C = 50.

The second statement is independently sufficient while the first statement is not.

Choose B.
Kaplan Teacher in Toronto
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