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

**Problem**

**Problem**

8 mangoes and 6 apples can be purchased for $9.60 and 6 mangoes and 8 apples for $10. How much does each apple cost?

- 50p
- 60p
- 70p
- 80p
- 90p

**The Usual Method**

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Let the cost of 1 mango be ‘*m*’ cents and one apple be ‘*n*’ cents.

Hence 8*m* + 6*n* = 960 ..…. Eq. 1.

And 6*m* + 8*n* = 1000 ..…. Eq. 2.

Solving both these equations simultaneously as follows:

Multiplying Eq. 1 with 6 and Eq. 2 with 8, we get:

48*m* + 36*n* = 5760 ..…. Eq. 3.

And 48*m* + 64*n* = 8000 ..…. Eq. 4.

Subtraction Eq. 3 from Eq. 4, we get:

28*n* = 2240

*n* = 80p

**(Ans: 4)**

*Estimated Time to arrive at the answer = 45 seconds.*

**Using Technique**

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Let the cost of 1 mango be ‘*m*’ cents and one apple be ‘*n*’ cents.

Hence 8*m* + 6*n* = 960 ..…. Eq. 1.

And 6*m* + 8*n* = 1000 ..…. Eq. 2.

Subtraction Eq. 1 from Eq. 2, we get:

2*m* + 2*n* = 40

– *m* + *n* = 20

This means that the price of the apple is 20 p more than that of a mango.

Assuming that m = n (i.e. the price of mango = price of apple)

6*m* + 8*n* = 1000 becomes 6*n* + 8*n* = 1000

14*n* = 1000

p

Now since *n* > *m* by 20p, *n* > average of *n* and *m* by 10p.

p

**(Ans: 4)**

*Estimated Time to arrive at the answer = 15 seconds.*

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