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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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Posts: 93
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, 22:55
1
IMO C.

Given:
A+B=C.
A+D=B.
2C=3D

C = $$\frac{3}{2}* D$$

A = B-D.

B-D+B = $$\frac{3}{2}$$ * D.

2B = ($$\frac{3}{2}$$ + 1) * D

2B = $$\frac{5}{2}$$* D

OR B= $$\frac{5}{4}$$ * D.

The least value that that D can take is 4, making B=5. Hence, option C is the right 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, 23:16
1
A + B= C —>A = C—B
A + D= B —>A=B—D
—> C—B=B—D —> C=2B—D

2C=3D —> 2(2B—D)=3D
—> 4B—2D=3D
4B=5D—> B=5D/4

A,B and C are all positive integers.

To get the least value of B as an integer, D must be 4.
B=5*4/4= 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, 23:44
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

This problem may be solved by 2 ways.
First, we can simply check all the options. Soon we will find out that the least possible value of B is 5.

The second way is to combine the equations.

We will get:
A+B-C = 0
A+D-B = 0

The sum will be:
2A+D-С = 0

We know that 2C=3D, so C =1.5D
2A+D-1.5D = 0
2A-0.5D = 0
4A = D

A and D are both positive integers, so the minimum value of A is 1.
Thus D = 4, B = 5, C = 6.

The minimum 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, 23:53
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

To find the least possible integer value of ‘B’ first we have to calculate what values is ‘B’ function of.

From A+B=C, A+D=B we have
2A + D = C OR 4A + 2D = 2C → Eqn. ①

Using 2C=3D in Eqn. ① we have
4A + 2D = 3D
 D = 4A
 3D = 12A
 2C = 12 A
 C = 6A → Eqn. ②

Now using A + B = C in Eqn. ② we have
A + B = 6A
 B = 5A

Since ‘A’ is a positive integer i.e.
A ≥ 1

Hence 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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27 Jul 2019, 00:13
1
A+B=C...1
A+D=B...2
2C=3D...3

Rewriting 2
A+2C/3 =B

Substituting the value of C for ..1
A+2(A+B)/3=3

Simplifying this we get
B=5A

Hence, B can either be 5 or 10 .

Let's test for 5

B=5 then A=1 (as B=5A..we just calculated that)
C=A+B=6
D=B-A=4

So all positive Integer values for A, B, C, D (So B=5 is the least value of B)

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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, 00:19
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?

given A+B = C , A +D = B , 2C = 3D
A = C-B =B-D ==> 2B = C + D
==> 4B = 2C + 2D
==> 4B = 3D +2D ==> D = 4B/5

A= B-D = B-4B/5 = B/5 ===> A = B/5
C = 3/2D = 3/2 *4B/5 = 6B/5 ===> C = 6B/5

so A , B , C ,D can be written as --B/5 , B , 6B/5 , \$B/5

since all are positive integers ,B should be a multiple of 5 , least possible value for 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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27 Jul 2019, 00:25
1
Given: A+B=C, A+D=B, 2C=3D where A,B,C & D are positive integers
To Find: Min. or least possible value of B

Solution: Substract A+B=C from A+D=B
we get, B-D=C-B=> 2B=C+D ------eq.(1)
Now, 2C=3D=>C=3D/2

Substituting C in eq.(1), we get B=5D/4
Since B is a positive integer, so D should be divisible by 4 to make (5D/4) an intger as 5 is not divisible by 4.
As Min. value of D can be 4, so least possible 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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27 Jul 2019, 00:40
1
A+B=C -- (1)
A+D=B --> A = B-D -- (2)

substitute (2) in (1)
B-D+B = C --> 2B-D = C -- (3)

2C = 3D --> C = 3D/2 -- (4)

substitute (4) in (3)
2B-D = 3D/2 --> 2B = 5D/2
B = 5D/4, since B & D are integers, we need to find for which values of D, 5D/4 is an integer..
Possible values for D are 4, 8, 12, .. etc. So, the least value of B is 5 when D is 4

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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, 01:19
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?

Given:
1. A+B=C
2. A+D=B
3. 2C=3D -----> C=$$\frac{3D}{2}$$

Rearrange the first one: A=C-B
Substitute A in second one:
C-B+D=B ---> C+D=2B

Substitute C with $$\frac{3D}{2}$$

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

3D+2D=2B
5D=4B, since all variables are positive integers, B at least has to be 5 and D=4 to equal to each other.
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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, 03:05
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?

Given:
A,B, C>=1 & integers
A+B=C --(i)
A+D=B--(ii)
2C=3D---(iii)
replace C from (iii) in (i)
A+B= 3D/2
=> 2A+2B=3D ---(iv)
replace D from (ii) in (iv)
2A+2B=3(B-A)
=> B=5A
A >=1
so B >=5
so minimum value of B is 5

A. 1
B. 4
C. 5--> correct
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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27 Jul 2019, 03:43
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+B = C ---> Eq. 1
A+D = B ---> Eq. 2
2C= 3D ---> Eq. 3

Eq. 1 - Eq. 2 gives B-D=C-B --> 2B = C+D
i.e. B = C/2+D/2 = C/2+2C/3/2 (Substituting D = 2C/3 from Eq. 3).
i.e. B = 5C/6

Now, B and C are both positive integers, therefore C needs to be multiple of 6.
Least value of 6 will give least value of B.
Therefore if C = 6, then B = 5.
Hence least value of B = 5.

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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27 Jul 2019, 04:17
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

here, 2C=3D or, 2(A+B)=3(B-A) or, 2A + 2B =3B -3A or, B = 5A
hence, C = 6A, D=4A
A:B:C:D = A: 5A: 6A: 4A
so, B =5

so, the correct answer choice 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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27 Jul 2019, 04:18
1
Given,

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

Let us check the options,

A. 1

This cannot be the answer as it is given that A,B,C,D are positive integers. Hence, we cannot have A and D integer values which will add up to 1.

Incorrect.

B. 4

Consider 3 cases here,

A B C D
1 4 5 3
3 4 7 1
2 4 6 2

None of the 3 cases satisfy the given equations.

Incorrect.

C. 5

Consider 4 cases here,

A B C D
1 5 6 4 --> This satisfies all the 3 equations.
4 5 9 1
2 5 7 3
3 5 8 2

Since this is the smallest of the remaining options, this is the answer.

Correct.

D. 6

Incorrect.

E. 10

Incorrect, although note that the following values of A,B,C,D also satisfy the given equations but B's value here is not the smallest.

A B C D
2 10 12 8

Incorrect.

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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, 04:19
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, B, C, and D are positive integers
A+B=C
A+D=B
2C=3D

A = C - B
C -B +D = B
C+D=2B
2C+2D=4B
3D+2D=4B
5D=4B
B=5D/4

First possible minimum is when D = 4 then B = 5
Let's check values of A and C if D = 4 and B = 5
A + 4 = 5
A = 1
C = 1+5
C = 6
All condition hold true

ANWER C

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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27 Jul 2019, 04:42
1
If
A+B=C
A+B=C
,
A+D=B
A+D=B
,
2C=3D
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

Solution:

I will solve this by hit and trial method.
Since in the question its given that A,B,C,D are positive numbers, so there cannot be 0 for any value.
As the question asks the least value so we will start by taking B=1
Put Value in Equation:
A+1=C
A+D=1
Substract both
1-D=C-1
or, 2=C+D
Multiply above by 2 ,we get
4=2C+2D
4=3D+2D
D=4/5 , But this is not possible as D has to be integer value.

Take B=4
Again put the value in the equation given:
A+4=C
A+D=4
Substract both
4-D=C-4
or, 8=C+D
Multiply by 2
16=2C+2D
D=16/5 Again not posible as D has to be an integer value.

Take B=5
Put the value in the given equation:
A+5=C
A+D=5
Substract both , we get
5-D=C-5
10=C+D
Multiply both sides by 2
20=2C+2D
We get,
20=5D
or D=4
Now this is perfect as this also satisfy 2C =3D (C=6,2C=12,D=4 and 3D=12)
Hence all the conditions are getting satisfied.

Hence B=5 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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27 Jul 2019, 05:06
1
Lets consider each option one by one
A) can't be the answer as
A+D =B
if B =1 then one of them has to be 0 which is not possible as all are positive integers. Hence not the answer

B) Since 2C = 3D ( we know C is multiple of 3 and D is multiple of 2 (as all num are integers))
If B =4
then to satisfy A+D = B ( A and D both would 2 (as D has to be a multiple of 2))
if A=2
D=2
C would be 3
which does not satisfy 1st eqn i.e A+B =C (2+4 is not equal to 3)
Hence 4 cannot be the value for B

C) if B=5

A=1
D=4 (D has to be multiple of 2)
C =6

A+B =C ( value of A, B and C satisfies the eqn 1+5 =6)

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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, 05:26
1
A+B=C...1
A+D=B....2
2C=3D....3

and also subtract 1 and 2

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

C=3/2D
CASE 1 C=3,D=2
Bbut we know C+D and C-D is even so not possible.

now C=6 and D=4 satistfies..
therefore 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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27 Jul 2019, 05:59
1
-B=A-C
B=A+D
Solving above 2 equations we get, 2B=C+D; Also we have 2C=3D; if C has to be a positive integer least value D must take is 2. When D=2, C= 3; With these values B= 2.5, which is not an integer.
Next possible value is D=4, which means C=6. With these values B= 5.

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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27 Jul 2019, 06:15
1
Solution:

The question stem is asking for the least possible value of B.

It seems that it'd be easier to solve this question by reserve solving, that is through answer choices. We are concerned with the least possible value of B, Hence lets take B= 1 since it's the least value in the answer choice.

Therefore A+ 1 = C..(i)

A+ D = 1...(ii)

Subtracting ii from i

We get 1 - D = C - 1,

C + D = 2
since we know that 2C = 3D,
we must multiply this equation by 2,
we get 2C + 2D = 4
Therefore 5 D = 4 and D = 4/5.

We know from the stem that these numbers are integers hence the value of B cannot be 1.

Lets try the mid option, C

Taking least value as 5,

we get A+ 5 = C..(i)
A + D = 4 (ii)

from one and two we get,5 - D = C - 5
10 + C + D
Multiplying both sides by 2,
20= 2C + 2D
since 2C = 3D,
5D = 20 & D will be 4.

Taking the value B = 5 satisfies all the values of the equations above.
we only have one value to test that is B = 4, which when solved dosen't lead into a integer value, and 6 & 10 are out because even if 6 & 10 satisfy the above conditions, they wouldn't be the least.

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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, 06:15
1
A+B=C

A+D=B

2C=3D

Using above three-
C+D=2B
5D=4B or 5C=6B
Minimum value of B can be 5. (i.e.D=4,c=6)
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ   [#permalink] 27 Jul 2019, 06:15

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