All students enrolled in econometrics this semester are enro

Question

All econometrics students are enrolled in either the morning course, the evening course, or the weekend course, each of which has at least 20 students. In the morning course, the ratio of men to women is 6:11. In the evening course, the ratio of men to women is 10:7, while in the weekend course, it is 5:4. If there are 180 male econometrics students, what is the smallest possible number of female econometrics students?

A. 126
B. 146
C. 156
D. 162
E. 168

Show Answer

Official Answer

D

A. 126
B. 146
C. 156
D. 162
E. 168

paragw · Answer

To minimize women, we want to maximize the Evening class because it has the lowest ratio of women to men (7 women for every 10 men). We want to minimize the Morning class because it has the highest ratio of women to men (11 women for every 6 men).

Set Morning groups = 5 (giving 5 × 6 = 30 men)
This leaves 150 men for the other two classes (180 - 30 = 150).
To maximize Evening groups, we set Evening groups = 13 (giving 13 × 10 = 130 men).
This leaves 20 men for the Weekend class, which means Weekend groups = 4 (since 4 × 5 = 20 men).

Check total men: 30 + 130 + 20 = 180 men. (Matches perfectly!)

Morning women: 5 groups × 11 = 55
Evening women: 13 groups × 7 = 91
Weekend women: 4 groups × 4 = 16

Total Women = 55 + 91 + 16 = 162
Correct Answer: D (162)

kevincan · Answer

Fantastic! What made him set the morning groups to 5 ?

yashsharma1 · Answer

https://gmatclub.com/forum/download/file.php?id=88878 Remaining men = 180-47 = 133. Since we want to minimize the number of women, we need to maximize the sets of evening class because it has the lowest Women to Men ratio (Only 7 increases in women for a 10 increase in men). However, the Evening class (EC) can only increase the number of men in a multiple of 10 and hence to achieve 133, some combination of Morning class (MC) and Weekend class (WC) will be needed to reduce 133 to a multiple of 10. Hence, the combination of MC+WC should give us either 3 or 13 or 23 or 33... men. Lowest such combination possible is 3MC + 1WC which will give 3*6+5*1 = 23 men. Remaining men (133-23 = 110) can be covered in 11 sets of EC. Thus total sets MC = 2+3 = 5 (No of women = 5*11 = 55) EC = 2+11 = 13 (No of women = 13*7 = 91) WC = 3+1 = 4 (No of women = 4*4 = 12) Total women = 55+91+16 = 162 GMAT-Club-Forum-b7wu9ukn.png

Akansha2000 · Answer

First, let's establish the ratio multipliers for each of the three courses to represent the number of men and women.

Morning course: Men:Women = 6:11.
Let the multiplier be x. Men = 6x, Women = 11x.
Evening course: Men:Women = 10:7.
Let the multiplier be y. Men = 10y, Women = 7y.
Weekend course: Men:Women = 5:4.
Let the multiplier be z. Men = 5z, Women = 4z.

The problem states that there are at least 20 students enrolled in each course. Since we are dealing with people, x, y, and z must be integers. This gives us our minimum limits for each multiplier:

Morning total 17x>= 20 implies x>=2
Evening total 17y>= 20 implies y>=2
Weekend total 9z>= 20 implies z>=3

We are given exactly 180 men in total, giving us the primary equation:
6x + 10y + 5z = 180
Our objective is to minimize the total number of women, which is expressed as:
\text{Total Women} = 11x + 7y + 4z
The Strategy
To keep the number of women as low as possible, we should maximize enrollment in the class that has the lowest ratio of women to men.
The Evening course yields the fewest women per man, so we want to make y as large as possible. Conversely, the Morning course yields the most women per man, so we want to keep x as small as possible.
Finding the Minimums
Look closely at the total men equation:
6x + 10y + 5z = 180
Notice that 10y, 5z, and 180 are all multiples of 5. For the math to balance out, the remaining term, 6x, must also be a multiple of 5.
Since x is an integer, x itself must be a multiple of 5. The smallest possible value that satisfies both x >=2 and being a multiple of 5 is x = 5.
Substitute x = 5 into our equation:
6(5) + 10y + 5z = 180
30 + 10y + 5z = 180
10y + 5z = 150
Divide the whole equation by 5 to simplify:
2y + z = 30
Now, we want to maximize y. Let's solve for z:
z = 30 - 2y
We established earlier that the weekend course constraint requires z>=3:
30 - 2y >= 3
2y <=27
y <=13.5
Since y must be an integer, the absolute maximum value for y is 13.
If y = 13, then z = 30 - 2(13) = 4. This satisfies our z >=3 requirement perfectly.
Calculating the Final Answer
We now have our optimal, minimized distribution:

x = 5 (Morning)
y = 13 (Evening)
z = 4 (Weekend)

Plug these back into the women equation:

Morning women: 11(5) = 55
Evening women: 7(13) = 91
Weekend women: 4(4) = 16

Total Women = 55 + 91 + 16 = {162}

Edskore · Answer

This is a classic Min-Max question dressed up in ratio form. The key move is figuring out what "minimize women" actually requires.

To minimize women across all three courses, you want as many students as possible in the course with the lowest women-to-men ratio. Evening course is 7:10 (women:men), which is the smallest, so you want Evening as large as possible. Morning course is 11:6 — huge proportion of women — so you want Morning as small as possible.

Set up variables:
Let Morning multiplier = x, Evening = y, Weekend = z.
Men: 6x + 10y + 5z = 180
Minimum students per course: Morning needs 17x >= 20, so x >= 2. Evening: 17y >= 20, so y >= 2. Weekend: 9z >= 20, so z >= 3.

Now here's the subtle constraint that many people miss. Notice 10y and 5z are both multiples of 5, and 180 is a multiple of 5. So 6x must also be a multiple of 5. That means x must be a multiple of 5. The smallest valid x >= 2 that's a multiple of 5 is x = 5.

With x = 5: 30 + 10y + 5z = 180, so 10y + 5z = 150, or 2y + z = 30.
Minimize women = 11(5) + 7y + 4z = 55 + 7y + 4z.
To minimize this, maximize y (since z coefficient is smaller, but we're constrained by 2y + z = 30 and z >= 3).
Max y: z = 30 - 2y >= 3, so y <= 13.5, meaning y = 13. Then z = 4.

Women = 55 + 7(13) + 4(4) = 55 + 91 + 16 = 162.

Answer: D.

The trap on this question is not setting up the integrality constraint properly. If you forget that x must be a multiple of 5, you might try x = 2 and get a non-integer for y. Then you might panic and pick a wrong answer. The divisibility check on 6x always catches people.

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kevincan · Answer

To minimize the number of women , we want to maximize the number of men in the evening course and minimize the number of students in the morning course , subject to the following constraints: at least 20 students per course and 180 male students in total .

the number of males in the morning course needs to be a multiple of 5 : take 30 men and 55 women

Thus the number of males in the weekend course needs to be a multiple of 10: 20 men and 16 women (so that there are at least 20 students).

The 130 remaining men come from the evening course : 130 men and 91 women

162 women in all

Sunsh · Answer

MC(>=20) | EC(>=20) | WC(>=20) | Total Male = 180.... 
Ma : Fe = 6 : 11 | Ma : Fe = 10 : 7 | Ma : Fe = 5 : 4 | 6x+10y+5z = 180....(1)

MC = 17k EC = 17k WC = 9k

For Min(Fe)... MC should be as less as possible, as share of Female is larger there..
 EC should be as large as possible, 
 WC should be less as diff between Male and Female is less..

Also   6x + 10y + 5z = 180....(RHS ends with 0, 10y will always end with 0, So 6x+5z also should end with 0,.. 5 multiples end with 5 or 0, to match that multiple of 6 that ends with 0 is only acceptable)  
So 6x to be min, x min = 5...

MC (105) - Ma = 30, Fe = 55 | Now left with 150 male, so we cannot take EC(Male)= 150,
 | Lets take EC (Male) = 140, then WC(Male) = 10...and WC(Total) = 18<20... Not Possible... (5z/9 = 10, z = 18)
 | So EC(Male) = 130 , WC(Male) = 20, WC(Total) = 36>20.. Valid..
 EC(Female) = 91, WC(Female) = 16
Total Female = (55+91+16) = 162

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