[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?

$104
$115
$106
$108
$110

The Usual Method

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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

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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.
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Problem

The Usual Method

Using Technique

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