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A man spends $48 to buy 6 hamburgers and 8 colas for his office worker
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03 Mar 2021, 02:26
03 Mar 2021, 02:26
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Question Stats:
81% (01:32) correct
18% (02:49) wrong based on 11 sessions
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A man spends $48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4 colas and spends $32. Assuming the prices of hamburgers and colas remain constant, what is the price of one hamburger and one cola?
A. $6
B. $7
C. $8
D. $9
E. $10
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A. $6
B. $7
C. $8
D. $9
E. $10
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: A man spends $48 to buy 6 hamburgers and 8 colas for his office worker
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03 Mar 2021, 03:44
03 Mar 2021, 03:44
Carcass wrote:
A man spends $48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4 colas and spends $32. Assuming the prices of hamburgers and colas remain constant, what is the price of one hamburger and one cola?
A. $6
B. $7
C. $8
D. $9
E. $10
A. $6
B. $7
C. $8
D. $9
E. $10
6H + 8C = 48 ....... (1)
5H + 4C = 32 ....... (2)
Multiplying (2) with 2 and subtracting (1) from it;
10H + 8C = 64
6H + 8C = 48
4H = 16
H = 4
Now, 6(4) + 8C = 48
8C = 24
C = 3
So, H + C = 4 + 3 = 7
Hence, option B
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Re: A man spends $48 to buy 6 hamburgers and 8 colas for his office worker
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03 Mar 2021, 07:30
03 Mar 2021, 07:30
1
Carcass wrote:
A man spends $48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4 colas and spends $32. Assuming the prices of hamburgers and colas remain constant, what is the price of one hamburger and one cola?
A. $6
B. $7
C. $8
D. $9
E. $10
A. $6
B. $7
C. $8
D. $9
E. $10
Let H = price of one hamburger
Let C = price of one cola
A man spends $48 to buy 6 hamburgers and 8 colas for his office workers.
6H + 8C = 48
The next day, he buys 5 hamburgers and 4 colas and spends $32.
5H + 4C = 32
Assuming the prices of hamburgers and colas remain constant, what is the price of one hamburger and one cola?
We have:
6H + 8C = 48
5H + 4C = 32
Take TOP equation and divide both sides by 2 to get:
3H + 4C = 24
5H + 4C = 32
Subtract the bottom equation from the top equation to get: -2H = -8
Solve: H = 4
Now plug H = 4 into 3H + 4C = 24 to get: 3(4) + 4C = 24
Simplify: 12 + 4C = 24
So: 4C = 12
Solve: C = 3
So the price of one hamburger and one cola = H + C = 4 + 3 = $7
Answer: B
Cheers,
Brent
