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# [Smart Math] Ratio Proportion Problem 18

Here’s and example of a SMART MATH problem for RATIO PROPORTION. ### Problem

Of an amount of \$1500 divided between A, B and C, B’s share is \$70 more than C’s and A’s share is \$100 less than C’s. What is A’s share?

1. \$410
2. \$425
3. \$450
4. \$490
5. \$510

### The Usual Method

Let the shares of A, B and C be ‘1’, ‘2’ and ‘3’ respectively.

Hence, a + b + c = 1500

Also,    b = c + 70

a = c – 100

Hence, a + b + c = (c – 100) + (c + 70) + c = 1500 $\therefore$ 3c – 30 = 1500 $\therefore$ c = 510 $\therefore$ a = 510 – 100 = 410

(Ans: 1)

Estimated Time to arrive at the answer = 45 seconds.

### Using Technique

Consider each option. Assume A’s share = 410, C’s will be = 510 as A’s share is \$100 less than C’s. B’s share will be 510 + 70 = 580.

Now check 410 + 580 + 510 = 1500

This satisfies the condition that the total amount = \$1500 and hence, option ‘1’ is the correct option.

(Ans: 1)

Estimated Time to arrive at the answer = 15 seconds.

(Note: Here it is purely coincidental that the first option itself is correct. However, one can mentally check all options by this method and find that it is much easier, if not time saving.)
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