Here’s and example of a SMART MATH problem for PERCENTAGES.
Problem
The maximum possible marks in an exam is 250. A got 210 and B got 10% less than C and 25% more than D. The marks of C are 90% of that of A. What is the percentage mark of B?
- 92%
- 45%
- 53%
- 68%
- 85%
The Usual Method
[contentblock id=google-adsense-post]
Marks of A = 210 (given)
Marks of C = 90% of A =
Marks of B = 10% less than C = 189 – 18.9 = 170.1
Hence % marks of B = %
(Ans: 4)
Estimated Time to arrive at the answer = 45 seconds.
Using Technique
[contentblock id=google-adsense-post]
% marks of A = %.
From the data, marks of C > B, also marks of A > C, hence marks of A > B. This eliminates options ‘1’ and ‘5’. Also the difference between marks of A and C is 10% and that between B and C is also 10%. Thus % marks B 84 – 10 – 10 = 64%. Thus the answer has to be close to 64%. The closest to it is options ‘4’ = 68%.
(Ans: 4)
Estimated Time to arrive at the answer = 15 seconds.
[starrater tpl=10]
[contentblock id=smartmath-blockquote]