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# [Smart Math] Percentages Problem 6

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?

1. 92%
2. 45%
3. 53%
4. 68%
5. 85%

### The Usual Method

Marks of A = 210 (given)

Marks of C = 90% of A = $\frac{210\times 90}{100}=189$

Marks of B = 10% less than C = 189 – 18.9 = 170.1

Hence % marks of B = $\frac{170.1\times 100}{250}\approx 68$%

(Ans: 4)

Estimated Time to arrive at the answer = 45 seconds.

### Using Technique

% marks of A = $\frac{210\times 100}{250}=84$%.
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 $\approx$ 84 – 10 – 10 = 64%. Thus the answer has to be close to 64%. The closest to it is options ‘4’ = 68%.