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# [Smart Math] Algebra Problem 12

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

1. 30 years
2. 50 years
3. 60 years
4. 70 years
5. 80 years

### The Usual Method

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 $\therefore \left( r-10 \right)=3\left( s-10 \right)$                                 …… Eq. 1.

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

And S’s age 10 years hence = (s + 10) $\therefore \left( r+10 \right)=2\left( s+10 \right)$                               …… Eq. 2.

From equations 1 and 2 we get: $r-10=3s-30$     and $r+10=2s+20$ $r-3s=-20$                     and $r-2s=10$

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

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

i.e. $\left( r-10 \right)=3\left( s-10 \right)$ 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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