Ration Q - A fruit seller buys fresh grapes containing 80 percent wate

Question

A fruit seller buys fresh grapes containing 80 percent water. It is known that the price per kg of grapes vary inversely to the square of the percent water content. The water content decreases with time. What can be the final percentage of water content if the fruit seller makes a loss of 100/9 percent?

A. 25
B. 35
C. 40
D. 45
E. 60

A. 25
B. 35
C. 40
D. 45
E. 60

Kinshook · Answer

@KavicogsciGiven: A fruit seller buys fresh grapes containing 80 percent water. It is known that the price per kg of grapes vary inversely to the square of the percent water content. The water content decreases with time.
Asked: What can be the final percentage of water content if the fruit seller makes a loss of 100/9 percent?
Price per kg of grapes = k * 1/(Water content %)^2
Purchase price per kg of grapes = k * 1/(.8)^2 = 25k/16

Selling price per kg of grapes = k / p^2

Loss % = (Selling price  - Purchase price )/Purchase price * 100% = -100%/9
(Selling price - Purchase price)/Purchase price = -1/9
Selling price/purhcase price = 8/9

(k/p^2 ) / (25k/16) = 8/9
p^2 = 16*9/25*8 = 18/25 
 p = 3\sqrt{2}/5 = 2*1.414/5 = 2.828/5 = 56.56% = 60% approx

IMO E

AbhiS101 · Answer

You made mistake in last step of calculation.
It should be (3*1.41)/5, you have taken 2 inspite of 3

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maink · Answer

I did not understand the solution, if the official explanation can be posted or an expert can post the solution, that would be wonderful.

Kavicogsci · Answer

mikemcgarry   KarishmaB   CrackverbalGMAT  pls help

KarishmaB · Answer

Price and square of water content are inversely related. When water content reduces, we expect price to increase but he makes a loss. Why? Because when water content reduces, the amount of grapes he has to sell reduces too. The algebra here will perhaps require multiple variables. I would just try out the options. 40 and 60 seem to be the easiest to try. Say price of 1 kg grapes with 80% water is $1.

Try 40: 
 1*(80)^2 = p_2 * (40)^2 
 p_2 = $4  per kg

When water content is 80%, pulp is 20%. (Say 800 gm and 200 gm in 1 kg of grapes)
If water content becomes 40%, pulp is 60%. Pulp is still 200 gms. Since the pulp percentage has become 3 times of before, it means the weight of the grapes has become 1/3 kg. Recall C1*V1 = C2*V2

So price obtained on selling 1/3 kg of grapes =  $4*\frac{1}{3} = $\frac{4}{3}  which is a profit.
Not the answer.

Try 60: 
 1*(80)^2 = p_2 * (60)^2 
 p_2 = $\frac{16}{9}  per kg

When water content is 80%, pulp is 20%. (Say 800 gm and 200 gm in 1 kg of grapes)
If water content becomes 60%, pulp is 40%. It is still 200 gms. Since the pulp percentage has become 2 times of before, it means the weight of the grapes has become 1/2 kg.

So price obtained on selling 1/2 kg of grapes =  \frac{$16}{9}*\frac{1}{2} = $\frac{8}{9}  which is a loss of 1/9 or 100/9%.

Answer (E)

I seriously doubt that the question is suitable for GMAT. Though no harm in doing it to understand the concept.

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Ration Q - A fruit seller buys fresh grapes containing 80 percent wate

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Ration Q - A fruit seller buys fresh grapes containing 80 percent wate

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