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If a, b, and c are consecutive positive integers where a < b

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Carcass
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If a, b, and c are consecutive positive integers where a < b [#permalink] New post   22 Jun 2020, 09:34
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If a, b, and c are consecutive positive integers where a < b < c, and c - a = 16/b, then a + b + c =

A. 30
B. 24
C. 18
D. 12
E. 6
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B
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Carcass
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Re: If a, b, and c are consecutive positive integers where a < b [#permalink] New post   22 Jun 2020, 09:34
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Re: If a, b, and c are consecutive positive integers where a < b [#permalink] New post   22 Jun 2020, 09:58
Carcass wrote:
If a, b, and c are consecutive positive integers where a < b < c, and c - a = 16/b, then a + b + c =

A. 30
B. 24
C. 18
D. 12
E. 6


If a, b, and c are consecutive positive integers where a < b < c, then we know that:
a + 1 = b
And, more importantly, a + 2 = c
If a + 2 = c, then c - a = 2

GIVEN: c - a = 16/b
Substitute to get: 2 = 16/b
Solve: b = 8

So, a, b and c equal 7, 8, and 9 respectively
This means a + b + c = 7 + 8+ 9 = 24

Answer: B

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Re: If a, b, and c are consecutive positive integers where a < b [#permalink] New post   12 Dec 2021, 04:35
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Given that a,b,c are +ve consecutive integers and a<b<c
with that knowledge we can assume a,b,c to be: x, x+1, x+2 respectively
so c - a = x+2 - x = 2
so, 2 = 16/b => b = 8 or x+1 = 8

now, x + (x+1) + (x+2) = 7 + 8 + 9 = 24

ans = B
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Re: If a, b, and c are consecutive positive integers where a < b [#permalink] New post   26 Oct 2022, 10:06
Given that a, b, and c are consecutive positive integers where a < b < c, and c - a = 16/b and we need to find the value of a + b + c

As a, b and c are consecutive positive integers
=> b = a + 1
=> c = b + 1 = a + 2

=> c - a = a + 2 - a = 2 = \(\frac{16}{b}\) (given)
=> b = \(\frac{16}{2}\) = 8

=> a = b - 1 = 7
=> c = b + 1 = 9

=> a + b + c = 7 + 8 + 9 = 24

So, Answer will be B
Hope it helps!

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