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# If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ

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Manager
Joined: 10 Nov 2014
Posts: 63
Location: United States
Concentration: Marketing, Strategy
If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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Updated on: 27 Jul 2019, 10:20
2
If A+B=CA+B=C, A+D=BA+D=B, 2C=3D2C=3D and A, B, C, and D are positive integers, what is the least possible value of B?

Soln:

We are Given
i) A+B=C therefore A=C-B ( We want to find B so we do this )
ii) A+D =B
iii) 2C =3D

Replacing A in (ii) A+D = B we get C-B+D=B , simplifying it further we get C+D=2B

2C=3D , we replace in above equation C+D=2B , we get, 5D=4B

B= 5/4 D

We need least possible value of B. Also note it is positive integer. D has to be 4 to get integer value of B.

Therefore least possible value of B=5

And choice C

Originally posted by komals06 on 27 Jul 2019, 06:32.
Last edited by komals06 on 27 Jul 2019, 10:20, edited 1 time in total.
Manager
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Posts: 103
Location: India
Concentration: Finance, Strategy
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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27 Jul 2019, 07:43
1
since all are positive integers,
the least possible value of C which can give us a integer value of D is c=3, and D=2 but when we plug in these values in the first two equations we get A=1/2.
The next possible options are c=6,d=4,A=1 and B=5
So Option C is correct
Director
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Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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27 Jul 2019, 07:44
2
If A+B=C, A+D=B, 2C=3D and A, B, C, and D are positive integers, what is the least possible value of B?

A. 1
B. 4
C. 5
D. 6
E. 10

2C = 3D
thus minimum value of D = 3/2C
hence D = 3 and C =2
but C can not be 2 as a+b>2
let us take another smallest value of C = 6, D= 4
now a+b =C
a+d = b
1+4 = 5
and
A+B=C
1+5 = 6
hence B = 5
Intern
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Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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27 Jul 2019, 09:11
3
given abcd positive integers
we have
c/d=3/2=> c,d can be (3,2) (6,4) etc
b=c-a
b=a+d
=>b=(c+d)/2

3,2 not possible so option least number is option d
Intern
Joined: 19 May 2019
Posts: 25
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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28 Jul 2019, 04:29
A + B = C -> (1)
A + D = B -> Rearrange A - B = - D (2)
2C = 3D

Multiply (1) by 2 and (2) by 3:

2A + 2B = 2C = 3D Because 2C = 3D
3A - 3B = -3D

Sum both equations:

B = 5A or A = B/5

We are told that A, B, C and D are positive integers so A minimum value is 1.

If A = 1 then based on the last equality found B = 5 x 1 = 5

Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ   [#permalink] 28 Jul 2019, 04:29

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