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3 people split a $100 stereo but pay different amounts. If A pays $5

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Bunuel
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3 people split a $100 stereo but pay different amounts. If A pays $5 [#permalink] New post   12 Jul 2021, 08:15
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3 people split a $100 stereo but pay different amounts. If A pays $5 less than B, C pays more than $35, and all 3 people pay integer amounts, what is the most A could pay?

(A) 29
(B) 29.5
(C) 30
(D) 33
(E) 34
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rajatchopra1994
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Re: 3 people split a $100 stereo but pay different amounts. If A pays $5 [#permalink] New post   12 Jul 2021, 08:25
Explanation:

If a pays 30 dollars, then B pays 35
C = 100-35-30
C = 35

Which doesn't not satisfy question. So A should pay less than 30. Only integer values less than 30 is 29.

IMO-A

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Re: 3 people split a $100 stereo but pay different amounts. If A pays $5 [#permalink] New post   12 Jul 2021, 09:34
Bunuel wrote:
3 people split a $100 stereo but pay different amounts. If A pays $5 less than B, C pays more than $35, and all 3 people pay integer amounts, what is the most A could pay?

(A) 29
(B) 29.5
(C) 30
(D) 33
(E) 34

B - A= 5
C > 35
A + B + C = 100
A + B has to be maximum for A to be most.
Hence,
A + B < 65

This means A + B = 64.99999....

Thus, A ~= 29.9999
Since all are integers, A = 29

Answer A.
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Re: 3 people split a $100 stereo but pay different amounts. If A pays $5 [#permalink] New post   12 Jul 2021, 11:54
Bunuel wrote:
3 people split a $100 stereo but pay different amounts. If A pays $5 less than B, C pays more than $35, and all 3 people pay integer amounts, what is the most A could pay?

(A) 29
(B) 29.5
(C) 30
(D) 33
(E) 34


A = B - 5(To maximize A, we need to maximize B)
C > 35
So,
A + B + C = 100
B - 5 + B + C = 100
2B + C = 105
\(B = \frac{105 - C}{2}\) (Note: All pay integer amounts and to maximize B, we need to minimize C)
So, the smallest value of C which is greater than 35 and makes B an integer = 37
So, B = 34
Therefore, A = 29, IMO (A)!
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Re: 3 people split a $100 stereo but pay different amounts. If A pays $5 [#permalink]
12 Jul 2021, 11:54
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