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If p pounds of beef cost d dollars, how many cents would k pounds of
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22 May 2020, 10:11
22 May 2020, 10:11
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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
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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
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Re: If p pounds of beef cost d dollars, how many cents would k pounds of
[#permalink]
22 May 2020, 11:12
22 May 2020, 11:12
1
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Bunuel wrote:
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
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
- Cost of p pounds of beef = $ d = 100*d cents.
- Cost of one pound = \(\frac{100*d}{p}\) cents.
New cost of each pound = \(\frac{100*d}{p} – c = \frac{(100d-cp)}{p}\) cents
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
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Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE:Marketing (Energy and Utilities)
Re: If p pounds of beef cost d dollars, how many cents would k pounds of
[#permalink]
22 May 2020, 18:38
22 May 2020, 18:38
$ rate per pound of beef ; d/p
and in cents ; 100*d/p
after c cents reduction ; new rate ; 100d/p - c
and for k pounds rate shall be \((\frac{100d - cp}{p})k\)
OPTION D
and in cents ; 100*d/p
after c cents reduction ; new rate ; 100d/p - c
and for k pounds rate shall be \((\frac{100d - cp}{p})k\)
OPTION D
Bunuel wrote:
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
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
If p pounds of beef cost d dollars, how many cents would k pounds of
[#permalink]
28 Nov 2020, 13:55
28 Nov 2020, 13:55
1
Kudos
Bunuel wrote:
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
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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Re: If p pounds of beef cost d dollars, how many cents would k pounds of
[#permalink]
28 Aug 2021, 08:40
28 Aug 2021, 08:40
Bunuel wrote:
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
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
If p pounds of beef cost d dollars, how many cents would k pounds of
[#permalink]
28 Aug 2021, 09:06
28 Aug 2021, 09:06
2
Kudos
Expert Reply
Top Contributor
dave13 wrote:
Bunuel wrote:
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
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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Posts: 5191
Location: India
GMAT 1: 690 Q50 V34 
WE:Engineering (Transportation)
Re: If p pounds of beef cost d dollars, how many cents would k pounds of
[#permalink]
28 Aug 2021, 09:11
28 Aug 2021, 09:11
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
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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Affiliations: Target Test Prep
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Location: United States (CA)
Re: If p pounds of beef cost d dollars, how many cents would k pounds of
[#permalink]
13 Apr 2023, 06:41
13 Apr 2023, 06:41
Expert Reply
Bunuel wrote:
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
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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Re: If p pounds of beef cost d dollars, how many cents would k pounds of [#permalink]
13 Apr 2023, 06:41
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