The sum of three integers A, B and C is 120. A is one third of the su

Question

The sum of three integers A, B and C is 120. A is one third of the sum of B and C and B is one fifth of the sum of A and C. What is C?

A. 20
B. 30
C. 45
D. 50
E. 70

BrentGMATPrepNow · Accepted Answer

The sum of three integers A, B and C is 120  
We can write: A + B + C = 120

A is one third of the sum of B and C   
We can write: A = (1/3)(B + C)
Multiply both sides by 3 to get:  3A = B + C

B is one fifth of the sum of A and C.  
We can write: B = (1/5)(A + C)
Multiply both sides by 5 to get:  5B = A + C

There are MANY different ways to solve this system of 3 equations. 
Here's ONE solution.

Take A + B + C = 120 and replace  B + C  with  3A  to get: A +  3A  = 120
Solve to get A = 30

Take A + B + C = 120 and replace  A + C  with  5B  to get: B +  5B  = 120
Solve to get B = 20

Now that we now A = 30, and B = 20, AND we know that A + B + C = 120, we can conclude that C = 70

Answer: E

Cheers,
Brent

chetan2u · Answer

Hi,
FROM:-
1) A is one third of the sum of B and C..
or A=1/3(B+C)...
But A+B+C=120..
so 1/3(B+C)+B+C=120..
B+C=120*3/4=90...
A=120-90=30

2)B is one fifth of the sum of A and C
or B=1/5(A+C)...
But A+B+C=120..
so 1/5(A+C)+A+C=120..
A+C=120*5/6=100...

but A=30, so C=100-30=70..

LakerFan24 · Answer

This is what I did:

Given:  (I) A+B+C=120 ;  (II) A=1/3(B+C) ;  (III) B=1/5(A+C)

Recognize:
- You can rewrite  (I) as: A=120-B-C . Therefore, set both equations equal to each other: 1/3(B+C)=120-B-C --> (1/3)B+(1/3)C=120-B-C. Isolate 120 --> (4/3)B+(4/3)C=120. --> (4/3)B+C=120 --> 4(B+C)=360. -->  B+C=90

- Same method above applied to B. --> B=120-A-C. (1/5)A+C=120-A-C --> (6/5)A+C=120.  A+C=100 .

NOW:
A+B+C=120.
B+C=90
A+C=100.

Plug in A/C:
E)  C=70 . Means that: B=20, A=30. 
- Check:  A+B+C=120 . Correct.

Nunuboy1994 · Answer

Thank you for the solution- this is very clear

Nunuboy1994 · Answer

I don't understand the step where you isolate 40 and divide by 40?

Nevermind- just needed to drill into the question deeper.

Mislead · Answer

The quicker way:

A=1/3(B+C)
B+C=3A 
Substituting the value of B+C in Equation A+B+C=120
A+3A=120===> A=30

B=1/5(A+C)
A+C=5B
Substituting the value of A+C in Equation A+B+C=120
5B+B=120===> B=20

To find C, substitue the value of A and B in the equation A+B+C=120
30+20+C=120===> C=70

Ans: E

ScottTargetTestPrep · Answer

Solution:

We can create the equations:

A + B + C = 120

A = ⅓(B + C)

and

B = ⅕(A + C)

Multiplying the second equation by 3 and the third equation by 5, we have:

3A = B + C

and

5B = A + C

Substituting B + C = 3A in A + B + C = 120, we get:

A + 3A = 120

4A = 120

A = 30

Substituting A + C = 5B in A + B + C = 120, we get:

B + A + C = 120

B + 5B = 120

6B = 120

B = 20

Thus, A + B = 30 + 20 = 50 and C = 120 - (A + B) = 120 - 50 = 70.

Answer: E

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The sum of three integers A, B and C is 120. A is one third of the su

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