BTGModeratorVI wrote: ↑

Thu Jul 23, 2020 6:34 am

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

Answer: E
Source: Veritas Prep

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

BTGModeratorVI wrote: ↑

Thu Jul 23, 2020 6:34 am

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

Answer: E
Source: Veritas Prep

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