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Que: With the increase of 20% in the price of milk, a housewife can buy 5 liters less quantity for $60...........

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by Max@Math Revolution » Thu Jan 28, 2021 11:56 pm

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Que: With the increase of 20% in the price of milk, a housewife can buy 5 liters less quantity for $60 than she was buying before the increase. What was the initial price per liter of milk?

(A) $2.00
(B) $2.50
(C) $2.75
(D) $3.00
(E) $3.50
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Re: Que: With the increase of 20% in the price of milk, a housewife can buy 5 liters less quantity for $60...........

by Max@Math Revolution » Wed Feb 03, 2021 10:29 pm
Solution: The price of milk is increased by 20%.

Let the original price of milk per liter be $10.

Increased price: \(\frac{120}{100}\) * $10 = $12[/m]

Number of liters of milk: \(\frac{Total\ price\ }{price\ per\ litre}\)

Number of liters of milk before increase: \(\frac{60}{10}\)=6

Number of liters of milk after increase: \(\frac{60}{12}\)=5

The difference in the quantity of milk obtained: 6 - 5 = 1

Thus, for the $10 initial price, the difference is 1 liter.

For 5 liters difference, \(\frac{10}{5}\) = $2

Therefore, A is the correct answer.

Answer A
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Re: Que: With the increase of 20% in the price of milk, a housewife can buy 5 liters less quantity for $60...........

by Scott@TargetTestPrep » Fri Apr 16, 2021 8:22 am
Max@Math Revolution wrote: ↑
Thu Jan 28, 2021 11:56 pm
 Que: With the increase of 20% in the price of milk, a housewife can buy 5 liters less quantity for $60 than she was buying before the increase. What was the initial price per liter of milk?

(A) $2.00
(B) $2.50
(C) $2.75
(D) $3.00
(E) $3.50
Solution:

We can let p be the initial price per liter of the milk and create the equation:

60/(1.2p) = 60/p - 5

50/p = 60/p - 5

50 = 60 - 5p

5p = 10

p = 2

Answer: A

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