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Manager  S
Joined: 07 Dec 2018
Posts: 113
Location: India
Concentration: Technology, Finance
GMAT 1: 670 Q49 V32 Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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)
Manager  G
Joined: 28 Feb 2014
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Location: India
Concentration: General Management, International Business
GPA: 3.97
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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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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.
Senior Manager  P
Joined: 16 Jan 2019
Posts: 490
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WE: Sales (Other)
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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 $$5$$

Director  G
Joined: 22 Nov 2018
Posts: 557
Location: India
GMAT 1: 640 Q45 V35 GMAT 2: 660 Q48 V33 Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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
_________________
Give +1 kudos if this answer helps..!!
Manager  S
Joined: 26 Mar 2019
Posts: 100
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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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 $$B = \frac{5}{4}D = 5$$.

Senior Manager  P
Joined: 30 Sep 2017
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GMAT 1: 720 Q49 V40 GPA: 3.8
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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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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.

Manager  B
Joined: 24 Jun 2019
Posts: 111
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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.

Manager  S
Joined: 12 Mar 2019
Posts: 160
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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
Manager  S
Joined: 11 Feb 2018
Posts: 80
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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.
Manager  G
Joined: 08 Jan 2018
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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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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)
Manager  S
Joined: 18 Sep 2018
Posts: 100
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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
Manager  S
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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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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.

Please hit kudos if you like the solution.
Director  P
Joined: 24 Nov 2016
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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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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

Manager  G
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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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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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Senior Manager  G
Joined: 13 Feb 2018
Posts: 450
GMAT 1: 640 Q48 V28 Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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
Manager  S
Joined: 06 Aug 2018
Posts: 98
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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 !!

Posted from my mobile device
Intern  B
Joined: 11 Jun 2014
Posts: 23
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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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
Manager  G
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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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If A + B = C, A + D = B, 2C = 3D

Solving Using Proportions

A + B = C----- from question stem
A + D = B----- from question stem
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

Senior Manager  P
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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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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. _________________
My SC approach flowchart

(no one is ideal, please correct if you see any mistakes or gaps in my explanation, it will be helpful for both of us, thank you)

___________________
"Nothing in this life is to be feared, it is only to be understood"
~ Marie Curie
Intern  B
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Posts: 12
Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ  [#permalink]

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

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