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

**Problem**

**Problem**

In an alloy of three metals A, B and C, the quantity of metal A is 4/5^{th} of B and that of B is 3/4^{th} of C. What is the quantity in lb of each metal in 47 lbs of the alloy? (In the order of A, B and C.)

- 13, 22, 12
- 12, 13, 22
- 12, 15, 20
- 20, 15, 12
- 10, 20, 18

**The Usual Method**

[contentblock id=google-adsense-post]

A = 4/5 B B = 3/4 C

5A = 4B 4B = 3C

A : B = 4 : 5 B : C = 3 : 4

A : B : C

4 : 5

aaaaaaaaaa 3 : 4

12 : 15 : 20

Also, 12 + 15 + 20 = 47 lbs

Hence, A = 12 lbs, B = 15 lbs and C = 20 lbs.

**(Ans: 3)**

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

**Using Technique**

[contentblock id=google-adsense-post]

The first step is to eliminate those options whose sum 47. Thus option ‘5’ is eliminated as sum of 10 + 20 + 18 = 48.

The next step is to identify the metal with maximum or minimum quantity. Because A = 4/5 B and B = 3/4 C; the metal C is of maximum quantity and A is of minimum quantity. Thus options ‘1’ and ‘4’ are eliminated. This leaves us with options ‘2’ and ‘3’ only. Since, A = 4/5 B; 5A = 4B.

For option ‘2’, 5 x 12 4 x 13, hence ‘2’ is also eliminated. Thus answer is option ‘3’.

**(Ans: 3)**

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

[starrater tpl=10]

[contentblock id=smartmath-blockquote]