Categories
Ratio Proportion Smart Math

[Smart Math] Ratio Proportion Problem 44

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

Ratio Proportion

Problem

Three parts of alcohol A is mixed with two parts of alcohol B and sold at $110 per liter at a 10% profit. If alcohol A cost $20 more per liter than alcohol B, what is the price of alcohol A per liter?

  1. $104
  2. $115
  3. $106
  4. $108
  5. $110

The Usual Method

[contentblock id=google-adsense-post]

Let price of alcohol A be $a / liter

And that of alcohol B be $b / liter

\therefore a-b=20

\therefore b=a-20

Also cost of mixture = \frac{110}{1.1} = $100 (1.1 is due to 10% profit)

\therefore By Alligation rule:

\frac{100-b}{a-100}=\frac{3}{2}

\therefore \frac{120-a}{a-100}=\frac{3}{2}

\therefore 240-2a=3a-300

\therefore 5a=540

\therefore a=\frac{540}{5} = $108

(Ans: 4)

Estimated Time to arrive at the answer = 60 seconds.

Using Technique

[contentblock id=google-adsense-post]

Note that 3 : 2 can also be written as 6 : 4 or 60 liter and 40 liters. Hence total of 100 liters. Assuming that 100 liters costs $100 ($110 less 10% profit). Of this $100, $60 belongs to alcohol A and $40 to alcohol B. (Since difference of cost of the two alcohols is $20)

This means that the price of alcohol A is a multiple of 6. The only value from the options which is a multiple of 6 is $108 of option ‘4’.

(Ans: 4)

Estimated Time to arrive at the answer = 15 seconds.
[starrater tpl=10]

[contentblock id=smartmath-blockquote]

Leave a Reply

Your email address will not be published. Required fields are marked *