GMAT math / data sufficiency question - please help! I can't figure it out.?

Question

On a certain sight seeing tour, the number of women to the number of children was 5 to 2. What was the number of men on the sight seeing tour?

(1) On the tour, the ratio of the number of children to the number of men was 5 to 11.

(2) The number of women on the sight seeing tour was less than 30.

The answer here is "C" - the answer can be found using both statements 1 and 2. But I have no idea how they came to an answer using this information. Can someone help?

Thanks!

Answer 1

I misunderstood the question at first. You need BOTH 1 and 2 to get the answer.

Since we know the ratio of women to children is 5:2, and the ratio of children to men is 5:11, we can combine the ratios by multiplying the first by 5 and the second by 2:

5(5:2) = 25:10
2(5:11) = 10:22

So the ratio of women to children to men is 25:10:22. This ratio cannot be reduced, because there's no number that is a common factor of all 3.

Then, since we also know that there were fewer than 30 women on the tour, the only possible answer for the number of men is 22.

(In other words, the next smallest possibility that has the ratio 25:10:22 would be 50 women, 20 children, and 44 men, but that violates #2.)

Answer 2

The ratio of women to children is 5 to 2. The ratio of men to children is 11 to 5. Lets say there are 10 children (10 because 10 = 5*2). Then there are 10*5/2 = 25 women and 10*11/5 = 22 men. The numbers 10,25 and 22 dont have any common divisors (other than 1) so no smaller numbers will do. Any mutiples of these numbers will work, except that there are fewer than 30 women. So 10 kids, 25 women and 22 men is the answer.

Answer 3

I have a clumsy way of doing it - start with 2 children. Then there's five women. So there's two and a half times as many women as children. But if we had 5 children, we'd have 11 men. So there's 11/5 as many men as children. So if we start with a number of children that is divisible by both 5 and 2, let's see where that gets us.

Start with 10 children. That would make 25 women, and 11/5 times 10 men. So that would be 110/5 men, which divides evenly, for 22 men. The numer of women falls within the limit, so this solution works. Note that it would work for all multiples of 10 children, such as 20, 30 etc., but those would mean too many women. Like I said, it's a bit clumsy, because I can't precisely say why it was right to try 10 children, but it works.

Good luck.

Answer 4

Let

W = Number of women
C = Number of children
M = Number of men

Given

W/C = 5/2
C/M = 5/11
W < 30

Since we are dealing with people, W, C, and M must be integers.

W/M = (W/C)(C/M) = (5/2)(5/11) = 25/22
which cannot be reduced

So possible combinations of W and M are
(W,M) = (25,22), (50,44) etc.
But W <30
So only the first pair (25,22) is possible.

So the number of men is 22.
And we needed both statements (1) and (2) to get the answer.

Answer 5

w/c = 5/2, so w = (5/2)c
c/m = 5/11, so m = (11/5)c
w < 30.

and of course, all variables are integers.
since w = (5/2)c, c must be even.
since m = (11/5)c, c must be a multiple of 5. so c is a multiple of 10. the possibilities are:

c = 10, w = 25, m = 22
c = 20, w = 50, m = 44, and so on.

since w < 30, only the 1st solution works.

Answer 6

C/M=5/11 so the C-M pairs are (5,11), (10,22), (15,33) etc. Basically anything (5k,11k). W/C = 5/2 means W-C pairs are (5,2), (10,4), etc. Where is the overlap? When C=10,20,30,40, etc. If C=10 then there are 25 women but if C=20 or larger then the count for number of women will be too high. So C=10, W=25 and M=22

Answer 7

(a million) on my own is inadequate. for sure the width could be as great as one likes, and the dimensions is even greater effective. So the section could be as great as one likes; the section can not be desperate. (2) on my own is inadequate. All all of us understand is that the dimensions and width upload to eight meters. as an occasion, the floor could desire to degree 6 meter by utilising 2 meters (and have section 12 sq. meters), or degree 3 meters by utilising 5 meters (and have section 15 sq. meters). So the section can not be desperate. (a million) and (2) at the same time are sufficient. If w is the width in meters, we've 2w + 2(w+2) = sixteen and so devoid of even fixing the equation we see that the width w could be desperate. simply by fact the dimensions is two meters greater effective than the width, the dimensions is additionally desperate. provided that section equals length situations width, the section could be desperate. the respond is C. Lord bless you at present!

Answer 8

what is "c"? because i came up with 22 using the same process they others did