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

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New post 26 Jul 2019, 08:46
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A + B = C ............... (1)
A + D = B ................(2)
2C = 3D ..................(3)

B (min) = ?
All are integers.

B = A + D {As per equation 2}
B = C - B + D {As per equation 1}
2B = C + D
2B = 3/2D + D {As per equation 3}
2B = 5/2D
D=4/5B

Now, D is an integer, Looking at the options,
B(min) has to be 5.
Ans should be (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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New post 26 Jul 2019, 08:47
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A+B=C
A+D=B,
-> 2A + D = C

C=3/2 x D
-> A = (C-D)/2 = (3D/2 - D)/2 = D/4
Minimum value of D is 4 so A=1
B = A+D = 1+4 = 5

C is correct.
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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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New post 26 Jul 2019, 08:48
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\(A+B=C\) , \(A+D=B\), \(2C=3D\)

Rearranging

\(B=C-A\)
\(B=D+A\)

Adding the two equations,

\(2B = C+D\)

Multiplying by 2,

\(4B = 2C + 2D\)

Substituting \(3D\) for \(2C\) (as we're given \(2C=3D\))

\(4B = 3D + 2D = 5D\)

\(B = \frac{5}{4}D\)

Since we have \(D\) in the numerator, for \(B\) to be minimum possible integer \(D\) has to be the minimum possible positive multiple of 4 (That is \(D=4\))

When \(D=4\), \(B=5\)

So, the minimum value for \(B\) is \(5\)

Answer is (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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New post 26 Jul 2019, 08:56
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Given: (1)A+B=C (2)A+D=B (3)2C=3D and A, B, C, and D are positive integers

Easy elimination of B=1 as sum of two positive integers will be a minimum 2
D must be a multiple of 2 and C multiple of 3 (from 3rd equation)

Method: (1) + (2) gives 2A+D=C which when substituted in (3) gives 4A=D
4A=D substituted in 2A+D=C gives C=6A
4A=D substituted in (2) gives 5A=B

Thus we have 5A=B; 6A=C; 4A=D. To have the least value for B, A must have the least possible +ve integer which is 1.
So A=1,B=5,C=6 and D=4.

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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New post 26 Jul 2019, 08:58
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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 \(B\)?


Let us write down the first equation:
\(A + B = C\)
We also know that \(A + D = B\). We can also write it down as \(A = B - D\).
Let us replace \(A\) with this new equation. In this case:\(A + B = B + B - D = C\)
\(2B - D = C\)
We also know that \(2C = 3D\). So, \(C= \frac{3}{2}D\).
By replacing \(D\) in the previous equation we get \(2B - D = C = \frac{3}{2}D\)
\(2B = \frac{5}{2}D\)
\(B = \frac{5}{2}D : 2 = \frac{5}{4} D\)
Putting it into the equation \(A + D = B\), we get \(A + D = \frac{5}{4} D\) => \(A = \frac{1}{4} D\)
Since \(A\) is the smallest of all the integers, and it is positive, it can be at least 1. Thus, \(D = 4A = 4\). In this case \(B = \frac{5}{4}D = 5\).

Answer: 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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New post 26 Jul 2019, 09:02
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The least possible value of B ?

\(2C=3D\) --> Equation (1): \(C=\frac{3}{2}D\)
\(A+B=C\) --> Equation (2): \(A+B=\frac{3}{2}D\)
\(A+D=B\) --> Equation (3): \(A-B=-D\)

Equation (2) - Equation (3) :
\(2B=\frac{5}{2}D\) --> \(B=\frac{5}{4}D\) .....Equation (3)

If \(B\) and D each have to be positive integer, then we deduce that the least possible value of D is 4 from Equation (4).
Consequently, the least possible value of B is \(B=\frac{5}{4}(4) = 5\).

QUICK VERIFICATION:
\(D=4;\) \(C=\frac{3}{2}D = 6;\)
\(B=5;\) \(A-B=-D\) --> \(A-5=-4\) --> \(A=1\)
A,B,C,D are all positive integers and satisfy all known equations.

Answer is (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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New post 26 Jul 2019, 09:04
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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?


All variables are positive integers. We can try to determine their relations with eachother

A+B=C... So, A<C & B<C........(1)
A+D=B.... So, A<B & D<B...... (2)
combining (1) and (2)..... A<B<C & D<B ........(3)... we dont know if D is less than A or greater than A

A+B=C
2A+2B=2C... multiplying both sides by 2
2A+2B=3D..... because given that 2C=3D
2A+2A+2D=3D.... because A+D=B is given
4A=D...... So, A<D...... (4)

Combining (3) and (4).... A<D<B<C


A is the smallest integer of the 4, so A = 1 is the least possible value that A can have
We can use A = 1 to find least possible value of B

4A=D.... from (4)
so D = 4

B = A+D... Given

So, B = 5

Any other value of A will result in a higher value of B.

ANSWER: C-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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New post 26 Jul 2019, 09:12
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IMO C

Solving 2 equation derived from solution we reach at result
A= 4d : Putting in equation 2, we get , b =5d
Minimum value which satisfy is B=5, putting A=1
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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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New post 26 Jul 2019, 09:13
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Given:

A+B=C --->(Eq-1)
A+D=B --->(Eq-2)
2C=3D --->(Eq-3)

A, B, C & D are positive integers.

Question
Least possible value of B.

Method:

To make finding the minimum value of B simpler, lets try to isolate it in an equation with just one other variable:

- Note that Eq-3 only has two variables C and D and Eq-1 tells us that the sum of A & B is C, so it looks like a good place to start:
2C=3D ---> C=\(\frac{3D}{2}\) --->(Eq-4).

- Before we substitute Eq-4 into Eq-1, we can remove the D by considering Eq-2:
A+D=B ---> D=B-A --->(Eq-5)

- We can now substitute Eqs- 4 & 5, to give us another equation with only A & B as the variables:
A+B=C ---> A+B=\(\frac{3}{2}\)*(B-A) ---> A+B=\(\frac{3B}{2}\) - \(\frac{3A}{2}\) ---> \(\frac{5A}{2}=\frac{B}{2}\) ---> \(B=5A\)

As A and B are both positive integers, to minimize the value of B we need to pick the minimum value of A, which is 1, which would mean the least possible value of B is 5. The answer is thus 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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New post 26 Jul 2019, 09:15
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IMO-C

2C = 3D
=> D = 2C/3

A + D = B
=> A - B = -D
=> A - B = -2C/3 .......(I)

A + B = C ...........(II)

Adding I & II
2A = C/3
=> A= C/6

From (I), B= 5C/6
D = 2C/3


Therefore , A ,B ,C ,D resp. are [ C/6 , 5C/6, C, 2C/3 ]
For A ,B ,C ,D to be positive integer, C (min) = 6

B= 5x6/6 = 5 (Ans)
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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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New post 26 Jul 2019, 09:18
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IMO C

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?

Given,
A+B=C —— (1)
A+D=B —— (2)
2C=3D —— (3)

From (1) - (2) => B-D=C-B => 2B=C+D => B=5D/4 [By applying (3)]
From (1) + (2) => 2A = C-D => A=D/4 [By applying (3)]
From (3) => C=3D/2

So, A=D/4, B=5D/4, C=3D/2

As A, B, C, and D are positive integers, D MUST be a factor of 4

Hence, the minimum value of D=4, that leads to B=5

So, the 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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New post 26 Jul 2019, 09:21
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A,B,C,D > 0
A + B = C ---- (1)
A + D = B = > A - B = -D ----(2)
D = \(\frac{2C}{3}\)

From 1 and 2 equation:

2B = C + D = C + \(\frac{2C}{3}\) = \(\frac{5C}{3}\)
B = \(\frac{5C}{6}\)

Thus, If we take C = 6, then B = 5

IMO the answer is 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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New post 26 Jul 2019, 09:23
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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


C=3D/2
A=C-B
A=B-D
C-B=B-D,…C+D=2B,…B=(C+D)/2
B=(C+D)/2,…([3D/2]+D)/2,…[5D/2]/2,…5D/4
C=3D/2=integer, so D must be a multiple of 2
B=5D/4=integer, and the smallest multiple of 2 that is divisible by 4 is 4
B=5

Answer (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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New post 26 Jul 2019, 09:31
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Given that: A + B = C ..........................(i)
thus, A = C - B
A + D = B ..........................(ii)
Subtitute A = C - B in equation (ii)
we have, C - B + D = B
C + D = 2B ..........................(iii)

Since 2C = 3D, thus C = (3/2)D
Subtitute C = (3/2)D for C in equation (iii)

we have, (3/2)D + D = 2B
D = (4/5)B
Since A,B,C and D are all positive integers, the least possible value of B will make D to also be an integer

Let B = 5, we have; D = 4
Therefore, C = (3/2) x 4
thus, C = 6
Since, A = C - B = 6 - 5
therefore, A = 1

Thus, A = 1, B = 5, C = 6, and D = 4

Hence answer choice C.

Thank you so much.

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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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New post 26 Jul 2019, 09:39
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A+B=C --> B=C-A
A+D=B --> B=D+A
Let's sum both equations
2B=C+D
Let's multiply by two
4B=2C+2D

We know that 2C=3D
Substitute 2C to get
4B=3D+2D
4B=5D and both D and B are positive integers, THe least value of B must be 5

IMO
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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New post 26 Jul 2019, 09:50
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A+B=C

A+D=B

2C =3D


I have always learned that if options are increasing

Then test the middle value

Let B =5

C-A=5

A+D=5


WE GET D=4 AND C=6

AMD A=1( MOST IMPORTANT)

BY THIS WE GET IT IS B=5

IT IS 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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New post 26 Jul 2019, 10:25
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ANSWER :C
GIVEN,
A+B=C,A+D=B,2C=3D
C-B+D=B
2B=C+D
4B=2C+2D
FROM GIVEN
4B=2D+3D
4B=5D
B=5D/4
Bmin=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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New post 26 Jul 2019, 10:43
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If A + B = C, A + D = B, 2C = 3D

Solving Using Proportions

A + B = C----- from question stem
A + D = B----- from question stem
Adding both
2A=C-D

C = 1.5D----- from question stem
2A=1.5D-D=0.5D

2A=(1/2)D
4A=D
A/D=1/4------------1

A + D = B
5A=B

A/B=1/5--------------2

C/D = 3/2=6/4-------3

A:B:C:D = 1:5:6:4-----4

Least value of B=5

ANSWER 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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New post 26 Jul 2019, 10:45
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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?

As answer choices given in increasing order we can put the each value instead of B and wee if the given task still holds true.

A. 1
A+1=C
A+D=1 cannot be true because A,B,C,D are integers.

B. 4
A+4=C
A+D=4
1+3=4 --> D=3, C=5 --> 2C=3D not equal
2+2=4 --> D=2, C=6 --> 2C=3D not equal
3+1=4 --> D=1, C=7 --> 2C=3D not equal

C. 5
A+5=C
A+D=5
1+4=5 --> D=4, C=5, 2C=3D IS EQUAL

D. 6
E. 10


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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New post 26 Jul 2019, 10:46
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Given :
B+A=C
B-A=D

So, B=(C+D)/2 ... substituting 2C=3D we get B=(5/4)D

So D is multiple of 4... Dmin will give Bmin.. If D=4, B=5

Hence IMO C.

Check if A and C are positive integers to make sure above values are correct.
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Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ   [#permalink] 26 Jul 2019, 10:46

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