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Algebra Smart Math

[Smart Math] Algebra Problem 12

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

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

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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

\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

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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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