IMO D

10% of 60 women failed = 6 women failed
14% of 100 people failed = 14 people failed

Thus, 14 - 6 = 8 men failed
total number of men = 40

thus failed percentage = 8/40 = 20%

Got same answer as shovan85. I used the overlapping sets strategy of MGMAT to solve this.

I now know what I got wrong. My answer was 8%, which was actually the percentage of failed men out of all classmates.

I guess that the misunderstanding of the exact question is one of the greatest hurdles on the GMAT... at least for me.

Thanks for helping me hone my comprehension skills.

ikaplan wrote:While I was practicing today, I ran across the following problem.
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In a class of 100 people, there are 60 women and 40 men. 10 percent of the women failed the final exam. If a total of 14 percent of the class failed the exam, what percent of the men failed the exam?

(A) 6
(B) 8
(C) 18
(D) 20
(E) 112

My answer was incorrect.

I look forward to your replies.

Thanks!

If we understand the concept of weighted averages, and keep an eye on the choices, we can solve this with 0 calculations.

We know that there are more women than men in the class. Accordingly, the women carry more weight in the class average and the overall average will be closer to the women's than the men's.

On a number line:

10 ------ 14 ----------- men's average

If the men's average were 18, then the women and men would be equally weighted. Accordingly, the men's average must be more than 18. Only (D) fits the bill - done!

Even if we were strategically guessing at this question, we can eliminate A, B and E as nonsensical (the overall average has to be between the women's and men's, eliminating A and B, and you can't have more than 100%, eliminating E).

The best trained GMAT takers know when you don't need to do any math!

Stuart Kovinsky wrote:

ikaplan wrote:While I was practicing today, I ran across the following problem.
------------------------------------------------------
In a class of 100 people, there are 60 women and 40 men. 10 percent of the women failed the final exam. If a total of 14 percent of the class failed the exam, what percent of the men failed the exam?

(A) 6
(B) 8
(C) 18
(D) 20
(E) 112

My answer was incorrect.

I look forward to your replies.

Thanks!

If we understand the concept of weighted averages, and keep an eye on the choices, we can solve this with 0 calculations.

We know that there are more women than men in the class. Accordingly, the women carry more weight in the class average and the overall average will be closer to the women's than the men's.

On a number line:

10 ------ 14 ----------- men's average

If the men's average were 18, then the women and men would be equally weighted. Accordingly, the men's average must be more than 18. Only (D) fits the bill - done!

Even if we were strategically guessing at this question, we can eliminate A, B and E as nonsensical (the overall average has to be between the women's and men's, eliminating A and B, and you can't have more than 100%, eliminating E).

The best trained GMAT takers know when you don't need to do any math!

Can you please explain it further , I am not able to understand it, How to do it without any calculation.......

goyalsau wrote:
Can you please explain it further , I am not able to understand it, How to do it without any calculation.......

Well, I can try, but if you have specific questions those are much easier to answer than general ones.

The weighted average formula is:

Overall average = (% weight group 1)(average of group 1) + (% weight group 2)(average of group 2) + ...

Accordingly, the more heavily weighted a group, the bigger a factor it will be in the overall average, pulling the overall average towards itself.

For example, let's look at a simple 2 group system:

A math class wrote a test scored out of 100. The 10 females in the class averaged 90 and the 5 males in the class averaged 60. What was the overall class average?

While we could solve this just using the regular average formula, it's much quicker to do so using weighted averages. Applying the formula:

Overall average = (10/15)(90) + (5/15)(60) = (2/3)(90) + (1/3)(60) = 60 + 20 = 80

(10/15 and 5/15 represent the proportion of the class that's female and male, respectively.)

Now let's examine our overall average, 80: it's closer to the female average (90) than the male average (60), because the females carry more weight. This trend will always apply to weighted average scenarios.

So, back to the original question:

we know that 60% of the people are women and only 40% are men. Consequently, the overall average will be closer to the women's average than the men's. Since the women's average is 4 away from the overall average (14-10=4), the men's average must be MORE than 4 away; hence, the men's average must be greater than 18, leaving (D) as the only possibly answer.

Stuart Kovinsky wrote:

we know that 60% of the people are women and only 40% are men. Consequently, the overall average will be closer to the women's average than the men's. Since the women's average is 4 away from the overall average (14-10=4), the men's average must be MORE than 4 away; hence, the men's average must be greater than 18, leaving (D) as the only possibly answer.

Thanks Stuart,
I need to ask one more thing,

Women are more in the class to the average must have been closer to them, or We can say they are the leaders in deciding the average of the class because their strength is more in class.

Now we know that 10 % of Women failed, and over all 14 % students failed,

So the rest 4 % of the contribution must been from the difference group or I must say that there contribution must be 4% greater than that of average of class to counter the deficiency created by the women in the class.

I think this is what you are trying to say,
Now my question is we are not considering the ratio of women and men that is 60 : 40 While applying this fact, Can we do this is any question or this is just a coincidence that the ratio is same, and we have to consider the ratio as well.

If we have to consider it , then How we use this ratio Without doing calculation...

goyalsau wrote:

Stuart Kovinsky wrote:

we know that 60% of the people are women and only 40% are men. Consequently, the overall average will be closer to the women's average than the men's. Since the women's average is 4 away from the overall average (14-10=4), the men's average must be MORE than 4 away; hence, the men's average must be greater than 18, leaving (D) as the only possibly answer.

Thanks Stuart,
I need to ask one more thing,

Women are more in the class to the average must have been closer to them, or We can say they are the leaders in deciding the average of the class because their strength is more in class.

Now we know that 10 % of Women failed, and over all 14 % students failed,

So the rest 4 % of the contribution must been from the difference group or I must say that there contribution must be 4% greater than that of average of class to counter the deficiency created by the women in the class.

I think this is what you are trying to say,
Now my question is we are not considering the ratio of women and men that is 60 : 40 While applying this fact, Can we do this is any question or this is just a coincidence that the ratio is same, and we have to consider the ratio as well.

If we have to consider it , then How we use this ratio Without doing calculation...

First, 10% of women failed and 14% of the class failed, so it's actually the men who are more deficient!

Second, the 14% overall failure rate doesn't mean that the men account for 4% of the failures; it means that the men's failure rate must be higher than 14% to achieve that average failure rate.

We could use the ratios to solve if we actually wanted to do so. There's an inverse ratio between % weight and distance from the average. We can see this relationship on the number line:

women -------- x --------- average ----------------- y ---------------- men

If the ratio of women:men is 6:4, the ratio of x:y will be 4:6. So, if x=4, then y=6 and men = 14+6 = 20.

To your last question: if we understand these concepts, we don't need to actually do any of the math to arrive at (D) as the correct choice. So, while we certainly can solve algebraically, we can avoid doing so by applying the basic principles and keeping an eye on the choices.