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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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26 Jul 2019, 07:59
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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
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26 Jul 2019, 08:12
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
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26 Jul 2019, 08:13
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
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26 Jul 2019, 08:15
A+B=C1 A+D=B2 2C=3D3 Subtracting eq 2 from 1 we get BD=CB ===> 2B=C+D===> 4B=2C+2D====> 4B=5D====> B has to be 5. C is the correct answer.
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Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ
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26 Jul 2019, 08:16
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=CB=BD => 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
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Updated on: 26 Jul 2019, 22:16
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
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26 Jul 2019, 08:18
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
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26 Jul 2019, 08:20
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
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26 Jul 2019, 08:20
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
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26 Jul 2019, 08:20
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.
(C) is our answer.



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Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ
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26 Jul 2019, 08:22
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
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26 Jul 2019, 08:25
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=3B3A since 3D=2C 2A+2B=3B3A 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
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
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26 Jul 2019, 08:30
A+B=3/2 D A+B= 3/2(BA) 2A+2B=3B3A 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
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26 Jul 2019, 08:40
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 > AC=B1 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
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Updated on: 26 Jul 2019, 08:42
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. Answer = C
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
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Updated on: 26 Jul 2019, 11:58
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
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26 Jul 2019, 08:44
Here, 2C=3D So, C=(3D/2)
A+B=C A+D=B So, A+D=CA 2A=CD
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 SO our answer becomes C



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Re: If A + B = C, A + D = B, 2C = 3D and A, B, C, and D are positive integ
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26 Jul 2019, 08:45
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= BA Let’s substitute them into the constraint 2C = 3D 2(A+B) = 3(BA) 2A+2B = 3B3A —> 5A =3B2B 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
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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= BD bD+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
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26 Jul 2019, 08:46
IMO answer is C
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
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26 Jul 2019, 08:46



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