Here’s and example of a **SMART MATH** problem for **ARITHMETIC.**

**Problem**

**Problem**

C’s age is 1.5 times the averages age of A, B and C. A’s age is half the average age of the three. What was their average age two years ago if the present age of B is 10 years?

- 10 years
- 8 years
- 14 years
- 7 years
- 5 years

**The Usual Method**

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Let the present average age of the three be ‘*x*’ years.

C’s present age = 1.5*x* years

A’s present age = 0.5*x* years

B’s present age = 10 years

Average age =

*x* = 10 years

C’s present age = 1.5 x 10 = 15* *years

A’s present age = 0.5 x 10 = 5 years

C’s age two years ago = 15 – 2 = 13* *years

A’s age two years ago = 5 – 2 = 3 years

B’s age two years ago = 10 – 2 = 8 years

Average age two years ago = years

** **

**(Ans: 2)**

*Estimated Time to arrive at the answer = 75 seconds.*

**Using Technique**

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Since, C’s present age is 50% more than the average age and A’s present age is 50% less than the average age, the average age of the three will be same as the average age of C and A.

Since,

Also, since B’s age is equal to 10 years, the present average age of all the three is 10 years. So, the average age of all the three two years ago is 10 – 2 = 8 years.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 10 seconds.*

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