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