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There are 125 members in the winter sports club. Of those,

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There are 125 members in the winter sports club. Of those,

by swerve » Fri Aug 23, 2019 11:11 am

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There are 125 members in the winter sports club. Of those, 20 do not ski or snowboard. If 1/3 of the people who ski also snowboard, and if the number of people who only snowboard is equal to 1/2 of the number of people who ski and snowboard, how many people only snowboard?

A. 15
B. 20
C. 30
D. 45
E. 60

The OA is A

Source: Veritas Prep
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hi

by Scott@TargetTestPrep » Tue Aug 27, 2019 5:21 pm
swerve wrote:There are 125 members in the winter sports club. Of those, 20 do not ski or snowboard. If 1/3 of the people who ski also snowboard, and if the number of people who only snowboard is equal to 1/2 of the number of people who ski and snowboard, how many people only snowboard?

A. 15
B. 20
C. 30
D. 45
E. 60

The OA is A

Source: Veritas Prep
We can let k = the number of people who ski, n = the number of people who snowboard, and b = the number of people who do both. From the information given in the problem, we can create the equations:

k + n - b + 20 = 125,

(1/3)k = b,

and

n - b = (1/2)b

n - b = (1/2)(k/3k)

Simplifying the third equation, we have:

n - b = (1/6)k

Substituting â…™k for n - b in the first equation, we have:

k + (1/6)k + 20 = 125

7/6 k = 105

k = 105 x 6/7

k = 90

Since the number of people who only snowboard is n - b, which is (1/6)k , the number of people who only snowboard is(1/6)(90) = 15.

Answer: A

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