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

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=C-B=B-D => 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

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

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

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

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.

Answer is (C)

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.

ANSWER: C-5

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

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

Algebra PS

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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

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.

Answer: C

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?

Soln:

We are Given
i) A+B=C therefore A=C-B ( We want to find B so we do this )
ii) A+D =B
iii) 2C =3D

Replacing A in (ii) A+D = B we get C-B+D=B , simplifying it further we get C+D=2B

2C=3D , we replace in above equation C+D=2B , we get, 5D=4B

B= 5/4 D

We need least possible value of B. Also note it is positive integer. D has to be 4 to get integer value of B.

Therefore least possible value of B=5

And choice 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?

A. 1
B. 4
C. 5
D. 6
E. 10

2C = 3D
thus minimum value of D = 3/2C
hence D = 3 and C =2
but C can not be 2 as a+b>2
let us take another smallest value of C = 6, D= 4
now a+b =C
a+d = b
1+4 = 5
and
A+B=C
1+5 = 6
hence B = 5
thus C is the answer

given abcd positive integers
we have
c/d=3/2=> c,d can be (3,2) (6,4) etc
b=c-a
b=a+d
=>b=(c+d)/2


3,2 not possible so option least number is option d

A+B=C
A+D=B ==> D=B-A ( so we only get As and Bs in our equation)

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

Therefore the least value of B is 5 for A=1

Answer C)

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