Every student at Darcy School is in at least one of three clubs: horse

MANHATTAN GMAT OFFICIAL SOLUTION:

Although the setup of this problem seems to call for a Venn Diagram with three circles, this question is actually solely about ratios. We need to figure out the ratio of students who are in only one club to the ratio of students in exactly two clubs to the ratio of students in all three clubs.

One : Two : Three Clubs = ?

We can use the unknown multiplier to help us represent the two ratios. The ratio of the number of students in exactly two clubs to the number of students in exactly one club is 4:3.



And the ratio of the number of students in exactly two clubs to the number of students in at least two clubs is 5:7. Make sure to use a different variable for this ratio. Also, note that this ratio is students in 2 clubs to students in at least two clubs. That means that there are really only 2 students in three clubs for every 5 people in two clubs.



Now we have two different ways of expressing the number of students in exactly two clubs. We know that the number of students in two clubs is a multiple of both 4 and 5. Therefore, the number of students in two clubs must be a multiple of 20. Multiply the first ratio by 5 and the second ratio by 4.



We can now combine the two ratios into one ratio, which will be 15z : 20z : 8z.

The number of students in the school must be a multiple of 15 + 20 + 8 = 43. The only answer that is a multiple of 43 is 86.

The correct answer is E.


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Look at the explanation in image attached

Answer: option E

First ratio is 4/3, second ratio is 5/7. The 4 and 5 represent exactly two clubs. By simplifying, that is multiplying the first by 5 and the second by 4, we get the ratio of the 3 as 15:20:28 which adds up to 63.

So the answer is A?

You are right till the calculation of 15:20:28

but please understand that 20 (the number of students in exactly two clubs) is subset of 28 (Number of students in at least two clubs) only. SO you don't have to add them.

So total will be only sum of 15 and 28 = 43

I hope this helps!

This is a good question to test the concept of fractions and ratios.

It took me a while to solve this but once you know the trick, it is quite easy.

From the question, we know that \(\frac{Exactly 2 Clubs}{Exactly 1 Club}= \frac{4}{3}\) and \(\frac{Exactly 2 Clubs}{At least 2 Clubs}\) \(= \frac{5}{7}\)

The number of students in exactly 2 clubs is our link between 2 information here, so we have to make them equal by multiplying \(\frac{4}{3}\) by \(\frac{5}{5}\) and \(\frac{5}{7}\) by \(\frac{4}{4}\)

You will get the ratios of Exactly 1 Club : Exactly 2 Clubs : At least 2 Clubs (or Exactly 2 clubs + Exactly 3 clubs) \(= 15 : 20 : 28\)

You now know that the ratio of students in exactly 2 clubs is \(20\), therefore the ratio of the students in exactly 3 clubs is \(8\).

You will get the final ratio of Exactly 1 Club : Exactly 2 Clubs : Exactly 3 clubs \(= 15 : 20 : 8\)

So, total student must be multiple of \(15+20+8 = 43\) and the only answer that is multiple of \(43\) is \(86\)

I used venn diagram to get to the correct answer choice...
suppose A=all 3
B, C, D - only in 2
E, F, G - in 1 only

we are told:
(B+C+D)/(F+G+E)=4/3
F+G+E = 3(B+C+D)/4

and we know that:
(B+C+D)/(A+B+C+D)=5/7
or that A=2(B+C+D)/5

now..
all together is A+B+C+D+E+F+G
2(B+C+D)/5 + B+C+D+3(B+C+D)/4
we have 2 fractions..we can add those two to get 23(B+C+D)/20
so B+C+D must be a factor of 20.
suppose B+C+D=20
then A=40/5 = 8
and E+G+F=15
or total = 43.
since we do not have 43 in our answer choices, we need to find an answer that is a multiple of 43.
so E.

Answer is Clearly E

see attached image .

Let number of students in exactly one club = x
Let number of students in exactly two clubs = y
Let number of students in exactly three clubs = d
so number of students in atleast two clubs = y + d
Given : \(y/x = 4/3\) => \(x = (3/4)y\)
\(y/(y+d) = 5/7\)
=> \(2y = 5d\) => \(d = 2y/5\)
=> \(y + d\) = \(y + 2y/5\) = \(7y/5\)

Total : x + y + d = \((3/4)y + 7y/5 = (43/20)y\)

=> y multiple of 20, if y = 20, total = 43 (no options)
=> y = 40, total = 86 => (E)

Hi,

no big calculations needed, since we can compare the two ratios because both have a common comparator, namely the number of students in exactly two classes. We we have 4:3, 5:7, and 4,5 respectively represent the people in exactly two classes for both, expanding to 20 we get.

28:20:15

Total Students=Students in exactly 1 class + students in at least 2 -> 28+15=43 -> M43 is the answer -> (E)

a= one class
b= two classes
c= at least two classes
d= three classes

So, enrollment is equal to

a+b+d

From the problem statement:

b/a=4/3 so b=(4/3)a

b/c=b/(b+d)=5/7, so

5(b+d)=7b, so d=b(2/5)=

(2/5)(4/3)a= a(8/15)

So a+b+d = a+(4/3)a +(8/15)a

which equals 43a, meaning the total has to be a multiple of 43.

The only answer that fits is 86.

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