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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, 10:50
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?

Here, minimum possible values for C & D is '2' & '3' respectively.

A+B = C
A-B = D

2A = C+D
So,
A + B = ((C+D)/2) + B = C

B = (C-D)/2
insert minimum values for C & D.

B = 1/2 ..........not possible because B is a positive integer.

If we put C = 6 & D=4 (next possible values.).
B = (6-4)/2=1


ANSWER: A
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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, 11:11
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A+B = C
A+D = B
2C = 3D C = 3/2D

A+B+D = 2B
C+D = 2B

3/2D + D = 2B

D = 4B/5

A+D = B
A+4/5B = B
A= 1/5B

A+B = C
1/5 B + B = C

SO A = 1/5 B
C = 6/5 B
D = 4/5 B
and as A B C D are positive integers least value of B is 5

So 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, 11:12
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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?

Sol :- A+B=C............(1)
A+D=B or this equation can be written as A-B=-D............(2)

Adding Equations 1&2, we get
2A=C-D or A=1/2(C-D).............(3)

Also, In the question stem, it is given that 2C=3D or C=3/2D............(4)

Putting Eqn.4 in Eqn. 3 we get,
A=1/4D
Putting this value in equation 2 we get,

1/4D+D=B ==> 5/4D=B

Since A, B, C, D are positive integers, D must at least be 4 and if D is 4, B is 5

So, The least number B could b is 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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New post 26 Jul 2019, 11:28
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.................. A+B = C ..........( i )
A+D=B => A-B = -D
................. _________
................... 2A = C - D
.................. _________
2C=3D => 2A = C - 2/3 C =1/3 C .........(ii)

USING (ii) IN (i)

A + B = 6A
=> B= 5A

AS A IS A POSITIVE INTEGER, A's MINIMUM VALUE IS 1 AND THEREFORE, B's MINIMUM VALUE 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, 11:49
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Given that \(A + B = C\), \(A + D = B\), \(2C = 3D\)
2C=3D but C and D are positive integers so C will be multiple of 3 and D will be multiple of 2.
Taking C=3, D=2 and substituting in the other 2 equations does not give us integer value for B
But if we take C=6, D=4 we get B=5 (Integer value)
So C is the correct choice imo.
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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, 12:09
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IMO correct answer is C - Explanation is given as attachment -
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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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New post 26 Jul 2019, 12:58
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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?

It seems so easy question at first, however, last day of GOT, it has to have its spice. So, expand all the equations
A+B = C;
A+D = B; A-B = -D
2B = C+D
2C = 3D ; C = \(\frac{3D}{2}\)
2B = \(\frac{3D}{2}\) + D
2B = \(\frac{5D}{2}\)
B = \(\frac{5D}{4}\)

Because we have been asked for least value of B , we will do number plugging from least value

A. 1
B = 1; D = \(\frac{4}{5}\) ; which is a decimal no, as per the question, all numbers are +ve integers. so this cannot be the value of B

B. 4
this will also result in Decimal value for D, If you notice B has to be multiple of 5 to be divisible by \(\frac{4}{5}\) , there are only two options

C. 5
Lets test this one, B = 5, D = 4; A = 1, C = 6, all are integers and we got our answer :)

No need to test other options as C is the least among other two , 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, 14:00
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A+B=C, A+D=B, 2C=3D
A+D=B--->D =B - A
2C=3D
-->2(A+B) = 3(B - A)
--->B = 5A
---->The answer can be either C or E as 5 & 10 are divisible by 5, and A, B, C, and D are positive integers that's why we can eliminate A, B, and D.
Among C & E, C yields to the smallest positive integer.
--->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, 14:08
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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

A + B = C ......1
A + D = B......2
Substituting B in 1
2A + D = C....3

Now 2C = 3 D
hence C/D= 3/2
hence substitute the ratio in the eq 3
A= 1/2
therefore B = 5/2
but as B has to be least positive integer, it will be 5.
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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New post 26 Jul 2019, 14:15
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from the given, \(a,b,c,d>0\) integers
\(b+a=c\)
\(b-a=d\)
\(2c=3d\)

so \(2(b+a) = 3(b-a)\)
\(2b+2a = 3b-3a\)
\(5a=b,\)
if the lowest value of \(a = 1\), then the lowest possible value of \(b = 5\)

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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New post 26 Jul 2019, 14:49
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From question: 2C =3D

Let C=3, and D=3...then

A+B =3
B-A = 2
----------sum
2B =5...B=2.5.........B must be integer. So discard

Let C =6 & D = 4

A+B=6
B-A=4
---------sum
2B =10...B =5

Answer : 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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New post Updated on: 27 Jul 2019, 00:00
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What we know for sure from the stem is the following:

1. \(A+B = C\)

2. \(A+D = B\)

3. \(2C = 3D\)

What we need to figure out the least possible value of integer \(B\). So let’s solve the first equation for \(B\): \(A+B=C\) or \(B=C-A\). The second equation tells us that \(B=A+D\). Now by adding these two equations we get \(2B=C+D\). From the second equation we know that \(2C=3D\). In order to create similar values and substitute them we multiply \(2B=C+D\) by \(2\) and get \(4B=2C+2D\).

Substituting \(2C\) for \(3D\) we get \(4B=2D+3D\) or \(4B=5D\).

So \(\frac{B}{D}=\frac{5}{4}\) or \(B=5x\)

Considering that \(B\) is an integer, the least possible value for \(x\) is \(1\) or \(B=5*1=5\). Hence the least possible value of \(B\) is \(5\)

Hence C
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Originally posted by JonShukhrat on 26 Jul 2019, 15:51.
Last edited by JonShukhrat on 27 Jul 2019, 00:00, 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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New post 26 Jul 2019, 16:49
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A+B=C ......(1)
A+D=B, hence D=B-A ......(2)
2C=3D ......(3)
Substitute C=A+B and D=B-A into (3)
2(B-A)=3(A+B)
2B-2A=3A+3B
B=5A
If A=1, then B=5,
C=A+B=6
D=B-A=5.
Since the condition of A,B,C, and D being positive integers are satisfied and we have one of them (A) equal to 1, then all corresponding variables have taken the smallest possible values.
Therefore, 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, 18:00
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(1) A + B = C
(2) A + D = B
(3) 2C = 3D

A,B,C,D = POS Integer.

From (3), C = 3D/2. Plug this into (1), A+B = 3D/2, Rearranging B = 3D/2 - A
Add this equation with (2), 2B = 3D/2 + D = 5D/2,
B = 5D/4.

Since D and B must be positive integer, smallest D value that makes B integer is 4.

B = 5(4)/4 = 5.

Ans 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, 20:17
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From the question stem, we can simplify to get the possible values for B.

B = A + D and 2C = 3D
Hence, A + B = C --> A + A + D = C or 2A + D = 3/2 D, --> D = 4A
So, substituting for D in A + D = B ==> B = 5A

Assuming A is the smallest positive integer, 1, smallest possible value of B = 5

Correct 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, 20:19
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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

solving the above equation

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

First Method

adding 1 and 2

2b = c + d

sub 3 in this
2b = c + 2c/3
2b = 5c/3
6b = 5c as b and c are positive integers, the above equation is possible for b = 5,10,15,20 and c = 6,12,18,24 etc ( as all are positive integers so they cannot be zero).

hence minimum value possible for b is 5,

second method

2c = 3d so the numbers possible are for c= 3,6,9,12 ... so on for D = 2, 4,6,8 .. so on respectively. ( both cannot be zero as given all are positive integers)

sub lowest corresponing value c =3 and d =2 in equation 1 and 2
B + A = C , gives B+a = 3
B - A = D.gives b-a =2

adding 2b = 5 so 5 is not integer

so substituting second lowest corresponing value c =6 and d =4 in equation 1 and 2
B + A = C , gives B+a = 6
B - A = D.gives b-a =4

adding 2b = 10 b=5 , yes we got it this is the lowest possible integer value for B

so ans is C for value 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, 21:15
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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?

2C = 3D
2(A+B) = 3(B - A)
2A + 2B = 3B - 3A
B = 5A

Minimum value of A = 1, hence minimum value of B = 5. This value is also consistent with other variables.
Substituting in equations C = 6, D = 4 and all the equations hold good.

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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New post 26 Jul 2019, 22:01
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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?

Lets put equation 1 in 2.

A + D = B
A + D = C - A
2A + D = C
4A + 2 D = 2C
we have 2C = 3D
4A + 2D = 3D
4 A = D
D is a multiple of 4

We can write 2nd equation as,
A + 4 A = B

B must be a multiple of 5. If we put this value then it satisfies both equations.

Correct 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, 22:08
The answer is E.

First, we know A B C and D have to be larger than 0.

Since none of the numbers are 0, then since A + B = C, then A and B < C. Also, since A + D = B, then A and D < B. Hence, combine it and you have A and D < B < C. So we know that B cannot be 1, eliminate answer choice a.

Also since 2C = 3D, then C=1.5D and since both are integers, then D must be an even number.

And since A + D = B, then either both A and B are odd numbers or both even numbers.

Knowing the above, then try with B=4. if B=4, and A<B and even, then A=2. So, C=4+2=6. Then, D=4. However, we know D<B so this is not true. Eliminate answer choice b.

Try B=5. Then, A must be odd and smaller, so A=3. Then, C equals 8 and D is not an integer. so eliminate answer choice C.

Try B=6. Then, A must be even and either is 2 or 4. If A=2, then C equals 8 and the above problem happens. So try A=4, then C=10 and D is not an integer. so eliminate answer choice d.

Try B=10. Then, try A=2. so, then C=12 and D=8. Since 8<10 then D<B and all the above conditions are met i.e. A and D < B < C and all are integers so the right answer choice is e.
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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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Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ   [#permalink] 26 Jul 2019, 22:27

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