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?
- 30 years
- 50 years
- 60 years
- 70 years
- 80 years
The Usual Method
[contentblock id=google-adsense-post]
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
[contentblock id=google-adsense-post]
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.
[starrater tpl=10]
[contentblock id=smartmath-blockquote]