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Bunuel
If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price had been reduced by c cents per pound?


A. \((\frac{d - cp}{p})k\)

B. \((\frac{d - cp}{100p})k\)

C. \((\frac{d - 100cp}{100p})k\)

D. \((\frac{100d - cp}{p})k\)

E. \((\frac{100d - 100cp}{p})k\)



PS20402

\(p\) \(pounds = d \) \(dollars\)

\(p pounds = 100d cents\)

Reduction = c * p (c cents multiplied by the number of pounds)

How many cents would k pounds of beef cost after the price had been reduced by c cent per pounds?

\(( \frac{100d - cp}{p} ) * k\)

Answer is D.
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Bunuel
If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price had been reduced by c cents per pound?


A. \((\frac{d - cp}{p})k\)

B. \((\frac{d - cp}{100p})k\)

C. \((\frac{d - 100cp}{100p})k\)

D. \((\frac{100d - cp}{p})k\)

E. \((\frac{100d - 100cp}{p})k\)


PS20402



BrentGMATPrepNow any idea why these numbers dont work
D = 2
P=2
K=3
C=0.5
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Bunuel
If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price had been reduced by c cents per pound?


A. \((\frac{d - cp}{p})k\)

B. \((\frac{d - cp}{100p})k\)

C. \((\frac{d - 100cp}{100p})k\)

D. \((\frac{100d - cp}{p})k\)

E. \((\frac{100d - 100cp}{p})k\)


PS20402



BrentGMATPrepNow any idea why these numbers dont work
D = 2
P=2
K=3
C=0.5


Those numbers work (although c = 0.5 means the price per pound has been reduced by 0.5 cents, which is kind of strange but irrelevant :-)
When we input those values, the answer is 298.5.
In other words, k pounds of beef will cost 298.5 cents

When we plug the 4 values into each answer choice, only D evaluates to be 298.5
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Asked: If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price had been reduced by c cents per pound?

Price per pound of beef = d/p
Price after reduction of c cents per pound = (d/p - c/100)
Price in cents of k pounds of beef after price reduction = k100(d/p - c/100) = k/p (100d - cp)

IMO D
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Bunuel
If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price had been reduced by c cents per pound?


A. \((\frac{d - cp}{p})k\)

B. \((\frac{d - cp}{100p})k\)

C. \((\frac{d - 100cp}{100p})k\)

D. \((\frac{100d - cp}{p})k\)

E. \((\frac{100d - 100cp}{p})k\)



PS20402

The price of beef in dollars, per pound, is d/p. In cents, the price per pound is 100d/p. When we reduce the price by c cents per pound, we have:

100d/p - cp/p = (100d - cp)/p

Thus, k pounds costs:

k(100d - cp)/p

Answer: D
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Not sure why the numbers approach is not working here.

I took p = 5 and d = $50, so the per-pound cost is $10. Next, I took c = 50 (so half a dollar) and k = 3. If the price per pound is reduced by half, 3 pounds should cost, 3*10/2 = $15. The answer should be 1500 since it is asking in cents but plugging the numbers gives a different figure.
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sarthak1701
If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price had been reduced by c cents per pound?

A. \((\frac{d - cp}{p})k\)

B. \((\frac{d - cp}{100p})k\)

C. \((\frac{d - 100cp}{100p})k\)

D. \((\frac{100d - cp}{p})k\)

E. \((\frac{100d - 100cp}{p})k\)


Not sure why the numbers approach is not working here.

I took p = 5 and d = $50, so the per-pound cost is $10. Next, I took c = 50 (so half a dollar) and k = 3. If the price per pound is reduced by half, 3 pounds should cost, 3*10/2 = $15. The answer should be 1500 since it is asking in cents but plugging the numbers gives a different figure.


After the reduction of 50 cents per pound, or $0.50 per pound, the price would become $9.50, not $5.
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Bunuel
sarthak1701
If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price had been reduced by c cents per pound?

A. \((\frac{d - cp}{p})k\)

B. \((\frac{d - cp}{100p})k\)

C. \((\frac{d - 100cp}{100p})k\)

D. \((\frac{100d - cp}{p})k\)

E. \((\frac{100d - 100cp}{p})k\)


Not sure why the numbers approach is not working here.

I took p = 5 and d = $50, so the per-pound cost is $10. Next, I took c = 50 (so half a dollar) and k = 3. If the price per pound is reduced by half, 3 pounds should cost, 3*10/2 = $15. The answer should be 1500 since it is asking in cents but plugging the numbers gives a different figure.


After the reduction of 50 cents per pound, or $0.50 per pound, the price would become $9.50, not $5.
Oh shoot, you are right! Thankssss!!
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