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If a, b, and c are consecutive positive integers where a < b < c, and
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Bunuel
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If a, b, and c are consecutive positive integers where a < b < c, and [#permalink] New post  22 May 2020, 08:58
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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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BrentGMATPrepNow
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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink] New post  22 May 2020, 17:27
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Bunuel 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
PS20355


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

Cheers,
Brent
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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink] New post  22 May 2020, 09:07
Bunuel 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


PS20355

c - a = 16/b
x+2-x= 16/x+1
x= 7
so a+b+c ; 7+8+9 ; 24
OPTION B
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NitishJain
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Joined: 11 Feb 2019
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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink] New post  24 May 2020, 00:02
B.

a,b, c are consecutive integers and a<b<c so the difference c-a = 2;
We have given c-a = 16/b
OR b = 16/(c-a). ==> b = 8

==> a= 7; b= 8; c =9
a+b+c = 7+8+9 = 24
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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink]
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