Here’s and example of a SMART MATH problem for RATIO PROPORTION.
Problem
In an alloy of three metals A, B and C, the quantity of metal A is 4/5th of B and that of B is 3/4th 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]