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

**Problem**

**Problem**

A is 2 years younger than B. If seven years back, A was 5/6^{th} of B’s present age, what is half of B’s present age?

- 25 years
- 27 years
- 28 years
- 29 years
- 32 years

**The Usual Method**

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Let the present age of B be 2*x* years.

Hence half of present age = *x* years.

Hence A’s age =

Age of B 7 years back =

Age of A 7 years back =

Also A’s age 7 years back = of B’s present age.

Hence = (2*x*)

27 years

**(Ans: 2)**

*Estimated Time to arrive at the answer = 60 seconds.*

**Using Technique**

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A’s age 7 years back = of B’s present age. All options being natural numbers, B’s present age is also a natural number and so is A’s present age. Hence A’s age 7 years back is also a natural number.

Now if (A’s age 7 years back) a natural number is equated to (B’s present age) another natural number, than B’s present age should be evenly / completely divisible by 6 to have A’s age 7 years back as a natural number.

Since the options are half of B’s present age, just double them up and check which of these numbers is divisible by 6. Better still, just check which from amongst the options are divisible by 3 (half of 6).

Only option ‘2’ is divisible by 3.

**(Ans: 2)**

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

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