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

**Problem**

**Problem**

Of a certain sum, 1/3^{rd} is invested at 3%, 1/6^{th} at 6% and the rest at 8%. If the annual simple interest from this entire amount is $600, what is the original sum?

- $3333
- $6000
- $6666
- $7500
- $10000

**The Usual Method**

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Let the original sum be $*x*.

Thus, invested at 3% gives an interest of

Similarly, invested at 6% gives an interest of and the remaining, = is invested at 8% gives an interest of

Hence, total earnings =

Since,

*x* = $10,000

**(Ans: 5)**

*Estimated Time to arrive at the answer = 60 seconds.*

**Using Technique**

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Instead of taking the initial figure as ‘*x*’, to avoid working with fractions, assume the figure as $300.

Thus 1/3^{rd} of $300 = $100 @ 3% gives interest of $3

And 1/6^{th} 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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