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# [Smart Math] Percentages Problem 10

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

Of a certain sum, 1/3rd is invested at 3%, 1/6th at 6% and the rest at 8%. If the annual simple interest from this entire amount is $600, what is the original sum? 1.$3333
2. $6000 3.$6666
4. $7500 5.$10000

### The Usual Method

Let the original sum be $x. Thus, $\frac{1}{3}x$ invested at 3% gives an interest of $\frac{1}{3}x\times \frac{3}{100}\times 1=\frac{1}{100}x$ Similarly, $\frac{1}{6}x$ invested at 6% gives an interest of $\frac{1}{6}x\times \frac{6}{100}\times 1=\frac{1}{100}x$ and the remaining, = $\left[ 1-\left( \frac{1}{3}+\frac{1}{6} \right) \right]x=\frac{1}{2}x$ is invested at 8% gives an interest of $\frac{1}{2}x\times \frac{8}{100}\times 1=\frac{4}{100}x$ Hence, total earnings = $\frac{1}{100}x+\frac{1}{100}x+\frac{4}{100}x=\frac{6}{100}x$ Since, $\frac{6}{100}x=600$ x =$10,000

(Ans: 5)

Estimated Time to arrive at the answer = 60 seconds.

### Using Technique

Instead of taking the initial figure as ‘x’, to avoid working with fractions, assume the figure as $300. Thus 1/3rd of$300 = $100 @ 3% gives interest of$3

And 1/6th of $300 =$50 @ 6% gives interest of $3 And the balance Rs300 –$100 – $50 =$150 @ 8% gives interest of $12 Thus total interest earned = 3 + 3 + 12 =$18.

For a total amount of $300, the interest earned =$18,

For what total amount would the interest = $600 Doing cross multiplication, we get$10,000

(Ans: 5)

Estimated Time to arrive at the answer =30 seconds.
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