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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, 09:58
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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, 18:27
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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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Archit3110
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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, 10:07
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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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink] New post  24 May 2020, 01:02
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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CrackverbalGMAT
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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink] New post  29 Oct 2021, 22:44
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a, b, and c are consecutive positive integers where a < b < c

So,b = a + 1

c = a + 2

a + b + c = a + a+1 + a+2

=3a + 3

=3(a+1)

=3b...(1) [As a + 1 =b]

c - a = 16/b

=>a+2 - a =16/b

=>b=8

So, in equation 1,

a +b +c = 3b =3 * 8 =24
(option b)
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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink] New post  26 Oct 2022, 11:07
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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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Re: If a, b, and c are consecutive positive integers where a < b < c, and [#permalink]
26 Oct 2022, 11:07
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