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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
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
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26 Jul 2019, 08:47
A+B=C A+D=B, > 2A + D = C
C=3/2 x D > A = (CD)/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
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26 Jul 2019, 08:48
\(A+B=C\) , \(A+D=B\), \(2C=3D\)
Rearranging
\(B=CA\) \(B=D+A\)
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
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26 Jul 2019, 08:56
Given: (1)A+B=C (2)A+D=B (3)2C=3D and A, B, C, and D are positive integersEasy 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
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26 Jul 2019, 08:58
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
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26 Jul 2019, 09:02
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): \(AB=D\)
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: \(D=4;\) \(C=\frac{3}{2}D = 6;\) \(B=5;\) \(AB=D\) > \(A5=4\) > \(A=1\) 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
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26 Jul 2019, 09:04
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 eachotherA+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<CA 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 = 5Any other value of A will result in a higher value of B. ANSWER: C5



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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, 09:12
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
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26 Jul 2019, 09:13
Given:
A+B=C >(Eq1) A+D=B >(Eq2) 2C=3D >(Eq3)
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 Eq3 only has two variables C and D and Eq1 tells us that the sum of A & B is C, so it looks like a good place to start: 2C=3D > C=\(\frac{3D}{2}\) >(Eq4).
 Before we substitute Eq4 into Eq1, we can remove the D by considering Eq2: A+D=B > D=BA >(Eq5)
 We can now substitute Eqs 4 & 5, to give us another equation with only A & B as the variables: A+B=C > A+B=\(\frac{3}{2}\)*(BA) > 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
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26 Jul 2019, 09:15
IMOC
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
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26 Jul 2019, 09:18
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) => BD=CB => 2B=C+D => B=5D/4 [By applying (3)] From (1) + (2) => 2A = CD => 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
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26 Jul 2019, 09:21
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.



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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, 09:23
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=CB A=BD CB=BD,…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
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26 Jul 2019, 09:31
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
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26 Jul 2019, 09:39
A+B=C > B=CA 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
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26 Jul 2019, 09:50
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
CA=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
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26 Jul 2019, 10:25
ANSWER :C GIVEN, A+B=C,A+D=B,2C=3D CB+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
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26 Jul 2019, 10:43
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=CD
C = 1.5D from question stem 2A=1.5DD=0.5D
2A=(1/2)D 4A=D A/D=1/41
A + D = B 5A=B
A/B=1/52
C/D = 3/2=6/43
A:B:C:D = 1:5:6:44
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
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26 Jul 2019, 10:45
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. 1A+1=C A+D=1 cannot be true because A,B,C,D are integers. B. 4A+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. 5A+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
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26 Jul 2019, 10:46
Given : B+A=C BA=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
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