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

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26 Jul 2019, 07:59
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72% (01:49) correct 28% (02:20) wrong based on 263 sessions

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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

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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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26 Jul 2019, 08:12
3
A+B=C
—> A = C - B ....... (1)
A+D=B
—> B = C - B + D
—> 2B = C + D ....... (2)

2C=3D
—> C = 3D/2

From (2),
2B = 3D/2 + D = 5D/2
—> B = 5D/4

So, B should be a multiple of 5
—> Least value = 5

IMO Option C

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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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26 Jul 2019, 08:13
2
Given 2C=3D
Now, substituting with the other equations: 2A+2B=-3A+3B => 5A=B
As, 'A, B, C, and D are positive integers' , the least value of A could be 1.
Thus least value of B is 5. .... Hence Ans C.
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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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26 Jul 2019, 08:15
1
A+B=C----1
A+D=B----2
2C=3D----3

Subtracting eq 2 from 1 we get B-D=C-B ===> 2B=C+D===> 4B=2C+2D====> 4B=5D====> B has to be 5.

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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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26 Jul 2019, 08:16
1
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

Given: A+B=C, A+D=B, 2C=3D and A, B, C, and D are positive integers

Asked: what is the least possible value of B?

A=C-B=B-D => 2B=C+D => B=(C+D)/2

2C = 3D = k
C= k/2 is a positive integer
D=k/3 is a positive integer
B= (k/2+k/3)/2 = 5k/12 is a positive integer
For B to be an integer k=12m
B=5m
Least possible value of B = 5

IMO C
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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: 26 Jul 2019, 22:16
1
A, B, C and D are positive integers.

A + B = C -> 2A + 2B = 2C -> (a)
A + D = B -> 3A + 3D = 3B -> 3B – 3A = 3D -> (b)
2C = 3D -> (c)
Putting (a) and (b) in (c)

2A + 2B = 3B – 3A => 5A = B => A = B/5 -> (d)

From (d), B has to be a multiple of 5. So eliminate options A, B, and D.

When B = 5
A=1, C= 6, D = 4. This certainly satisfies all the equations

Therefore, least possible value of B is 5. Answer C

Originally posted by Sayon on 26 Jul 2019, 08:17.
Last edited by Sayon on 26 Jul 2019, 22:16, edited 1 time in total.
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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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26 Jul 2019, 08:18
1
A. 1
B. 4
C. 5
D. 6
E. 10

By substituting variables we get
D = 4A
C = 6A
B = 5A

writing all variables in terms of B --> A = B/5
C= 6B/5
D = 4B/5

So when B=1,4 etc A,C,D will not be integers - so the minimum possible value for B is 5
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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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26 Jul 2019, 08:20
1
C

If we solve the three equations, we basically get B=5A, D=4A, and C=6A.

Now, since all parameters are positive integers, minimum possible value of A is 1, so least possible value of B is 5.
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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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26 Jul 2019, 08:20
1
Given all are positive integers, A, B, C, and D are positive integers, what is the least possible value of B?
A+B=C
A+D=B,
2C=3D

we can solve it using the options.
1 can never be a option because A+D=B(remember A&D are positive integers so min value B can have is 2)

Lets take B as 5 and A=1
C=6.D=4 and it accepts 2C=3D. We have a answer in C. A,D&E options goes away(min value of B)

when you take B=4, cases that can happen are A=1,D=3 or A=2,D=2 or A=3,D=1
case B=4,A=1,D=3=>C=5, 2C!=3D
case B=4,A=2,D=2=>C=6,2C!=3D
case B=4,A=3,D=1=>C=7,2C!=3D

So option B goes wrong.
therefore option C is the answer.
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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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26 Jul 2019, 08:20
1
Here we have A+B=C (1), A+D=B (2) and 2C=3D (3)

From (1) and (3) we have that A+B=1.5D (4).

Then, from (2) and (4) we have that A + A+D = 1.5D, so 2A = 1/2D, finally A=0.25D (5)

From (4) and (5) we have that B = 5/4D, so the least value of B (since they are + integers) must be 5.

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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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26 Jul 2019, 08:22
1
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?

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

solving
we get
A+B=C
A+D=B
2C=3D
we get
5A=B
and C=6A and D= 4A
so least value of B=5
IMO C
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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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26 Jul 2019, 08:25
1
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?

Let's solve an equation
2A+2B=2C
3A+3D=3B
3D=3B-3A
since 3D=2C
2A+2B=3B-3A
B=5A
Since A and B are positive integers we can assume the least value for A=1 and thus the least value of B=5

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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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26 Jul 2019, 08:30
1
A+B=3/2 D
A+B= 3/2(B-A)
2A+2B=3B-3A
2A+3A=B
5A=B
Taking A as min
B should be 5 (C)
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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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26 Jul 2019, 08:40
1
Quote:
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

A+B=C -> A-C=B-----------------1
A+D=B................2
2C=3D...................3

Multiplying 1 by 2 and 2 by 3 and solving both as we have

we get, 5A=B
And minimum possible value of A=1
hence C
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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: 26 Jul 2019, 08:42
1
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

A+B = C
A + D = B
Therefore, A + D = C - A, 2A + D = C
Now 2C = 3D, Substituting in above equation we get 2A = 1/2*D, D = 4A.
Therefore A + 4A = B, B = 5A.
Least value of A can be 1 as it is a positive number so least value of B = 5.

Originally posted by ruchik on 26 Jul 2019, 08:41.
Last edited by ruchik on 26 Jul 2019, 08:42, edited 1 time in total.
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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: 26 Jul 2019, 11:58
1
Given:

A+B=C -- 1

A+D=B -- 2

2C=3D -- 3

From the third equation we get:

$$D=\frac{2}{3}C$$ and we know from equation 1 that C =A+B, which means:

D= $$\frac{2}{3}(A+B)$$

Substituting value of D in equation 2 ==>

A+$$\frac{2}{3}(A+B)$$=B

$$\frac{5}{3}A + \frac{2}{3}B = B$$

$$\frac{5}{3}A=\frac{B}{3}$$

5A=B

Least possible value for A is 1 as it is a positive integer.

Thus least value of B=5

Originally posted by Ajiteshmathur on 26 Jul 2019, 08:41.
Last edited by Ajiteshmathur on 26 Jul 2019, 11:58, edited 1 time in total.
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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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26 Jul 2019, 08:44
1
Here,
2C=3D
So, C=(3D/2)

A+B=C
A+D=B
So, A+D=C-A
2A=C-D

2A=3D/2 -D
2A=D/2
4A=D

So, if A+D=B
then, A+4A=B
5A=B
We know that all A , B, C, D are positve intergers so A can be least as 1.
Putting A=1 we get B=5*1=5
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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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26 Jul 2019, 08:45
1
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?

So C= A+B , D= B-A
Let’s substitute them into the constraint 2C = 3D
2(A+B) = 3(B-A)
2A+2B = 3B-3A —> 5A =3B-2B
A= B/5
Now least +ve value =5 ,since A has to be an integer
.: A =5/5 =Integer
.: B =5
Smash that C

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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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26 Jul 2019, 08:46
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= B-D
b-D+B =C
2B+D =C
2B+2/3C =C
B=2/3C

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

IMO D.
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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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26 Jul 2019, 08:46
1

for B =5, we ned up with integer values for A,C and D which are A=1, B=5, C=6 and D = 4
only B=5, answer choice C is min possible value for B
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ   [#permalink] 26 Jul 2019, 08:46

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