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

**Problem**

**Problem**

10 years ago R was thrice as old as S was, but 10 years hence, he will be twice as old. What is R’s present age?

- 30 years
- 50 years
- 60 years
- 70 years
- 80 years

**The Usual Method**

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Let the present age of R be ‘*r*’ years.

Hence, R’s age 10 years ago = (*r* – 10) years

Let the present age of S be ‘*s*’ years.

Hence, S’s age 10 years ago = (*s* – 10) years

…… Eq. 1.

Similarly R’s age 10 years hence = (*r* + 10)

And S’s age 10 years hence = (*s* + 10)

…… Eq. 2.

From equations 1 and 2 we get:

and

and

Solving simultaneously, we get:

*s* = 30 and *r* = 70

Hence, R’s present age = 70 years.

**(Ans: 4)**

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

**Using Technique**

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We know that R’s age 10 years ago is three times S’s age.

i.e. where ‘*r*’ and ‘*s*’ are present ages of R and S respectively.

Hence, (r – 10) = multiple of 3.

From the options, subtract 10 to get the age of R 10 years ago and see which amongst these is a multiple of 3.

30 => 30 – 10 = 20 (Not a multiple of 3)

50 => 50 – 10 = 40 (Not a multiple of 3)

60 => 60 – 10 = 50 (Not a multiple of 3)

70 => 70 – 10 = 60 (is a multiple of 3)

80 => 80 – 10 = 70 (Not a multiple of 3)

Since, only option ‘4’ satisfies the conditions, 70 years is the present age of R.

**(Ans: 4)**

*Estimated Time to arrive at the answer = 15 seconds.*

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