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# [Smart Math] Percentages Problem 13

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

### Problem

Divide $1000 in two parts, so that if the two parts are invested @ 4% and 5% simple interest, the total yearly income may be$46.50.

1. $350 @ 4%;$650 @ 5%
2. $650 @ 4%;$350 @ 5%
3. $400 @ 4%;$600 @ 5%
4. $600 @ 4%;$400 @ 5%
5. $500 @ 4%;$500 @ 5%

### The Usual Method

Let the investment @ 4% be ‘x’.

Hence, investment @ 5% = (1000 – x)

Interest earned = $\frac{x\times 4\times 1}{100}+\frac{(1000-x)\times 5\times 1}{100}=46.50$

$\therefore \frac{4x}{100}+\frac{5000}{100}-\frac{5x}{100}=46.50$

$\therefore \frac{x}{100}=50-46.50=3.50$

$\therefore x=350$

Hence, amount invested @ 4% = $350 And amount invested @ 5% =$650

(Ans: 1)

Estimated Time to arrive at the answer = 60 seconds.

### Using Technique

Should the two parts invested be equal in size i.e. $500 each, the earning will be$45 (average of 40 and 50). But since the actual earning is more than $45 (46.50 > 45.00), the amount invested at 5% is more than the amount invested at 4%. (Using the concept of weighted average, where average tends to shift towards the higher weight). Hence, amount invested @ 4% is less than$500 and that invested @ 5% is more than \$500. The only options satisfying this are ‘1’ and ‘3’.

Option ‘3’ is eliminated by working backwards:

$\frac{400\times 4\times 1}{100}+\frac{600\times 5\times 1}{100}=16+30=46\ne 46.50$

Hence, option ‘1’ is the right option.

(Ans: 1)

Estimated Time to arrive at the answer = 15 seconds.
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