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Re: GRE Prep Club Test - A child bus ticket costs $0.40 and an a [#permalink]
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we could solve it by finding the prefect combination that satisfy the number 11 from 7 and 4 multiplies.
7*1 = 7
59-7 = 52
52 is dividable by 4 which gives 13 however 13 is too big.
7*2 = 14
59-14 = 45
45 is not dividable by 4
7*3 = 21
59-21 = 38
38 is not dividable by 4
7*4 = 28
59-28 = 31
31 is not dividable by 4
7*5 = 35
59-35 = 24
42 is dividable by 4 which gives 6
this could be a valid solution
then child ticket is 6
adult ticket is 5
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Re: GRE Prep Club Test - A child bus ticket costs $0.40 and an a [#permalink]
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Wouldn't it be shorter to try a = 6 (smallest possibility for Quantity B to be greater than A)?

5.90$ - 6 * 0.7$ = 1.7$

1.7$ / 0.4$ = 4.25 --> 4.25 children is smaller than 5 (11-6 = 5)

Therefore the only solution is that the number of children is greater than the number of adults (since in this case we also can't have 5.5 children)
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Re: GRE Prep Club Test - A child bus ticket costs $0.40 and an a [#permalink]
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Re: GRE Prep Club Test - A child bus ticket costs $0.40 and an a [#permalink]
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