DavidG@VeritasPrep wrote: ↑Fri Sep 08, 2017 11:23 am
Jerylseah97 wrote:At a stadium, there were an equal number of men and women. 79 Men and women left the stadium. In the end, there were 6 times as many men as women who remained behind. How many women were there at first?
I'm going to go out on a limb and posit that this is not a GMAT question. Folks, please try to post the answer choices. The point of many problems is that we can logically deduce the correct without having to grind through any formal algebra.
We can let x = the number of men (and also, the number of women) originally at the stadium, y = the number of men who left, and thus 79 - y = the number of women who left. We can create the equation:
(x - y) / (x - (79 - y)) = 6/1
(x - y) / (x - 79 + y) = 6
x - y = 6x - 474 + 6y
474 = 5x + 7y
x = (474 - 7y) / 5
Let’s find the smallest positive integer value for y such that x will be an integer. In that case, y will be 2.
If y = 2, x = (474 - 14)/5 = 92 and 79 - y = 77. That is, there are 92 men and 92 women originally at the stadium, and 2 men and 77 women left. Let see if this works: (92 - 2)/(92 - 77) = 90/15 = 6. We see that it works.
If y = 7, x = (474 - 49)/5 = 85 and 79 - 7 = 72. That is, there are 85 men and 85 women originally at the stadium, and 7 men and 72 women left. Let see if this works: (85 - 7)/(85 - 72) = 78/13 = 6. We see that it also works.
If y = 12, x = (474 - 84)/5 = 78 and 79 - 12 = 67. That is, there are 78 men and 78 women originally at the stadium, and 12 men and 67 women left. Let see if this works: (78 - 12)/(78 - 67) = 66/11 = 6. We see that it also works.
We see that this problem has more than one solution. The number of women originally at the stadium could be 92, 85, 78, etc.
