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# [Smart Math] Ratio Proportion Problem 44

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