Bradley owns b video game cartridges. If Bradley's total is one-third the total
owned by Andrew and four times the total owned by Charlie, how many video
game cartirdges do the three of them own altogether, in terms of b?
(16/3)b
(17/4)b
(13/4)b
(19/12)b
(7/12)b
OA after sometime!
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Anurag@Gurome
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Number of video game cartridges owned by Bradley = bGmatKiss wrote:Bradley owns b video game cartridges. If Bradley's total is one-third the total
owned by Andrew and four times the total owned by Charlie, how many video game cartirdges do the three of them own altogether, in terms of b?
Number of video game cartridges owned by Andrew = 3b
Number of video game cartridges owned by Charlie = b/4
Hence, total = (b + 3b + b/4) = 17b/4
The correct answer is B.
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Let a = 12.GmatKiss wrote:Bradley owns b video game cartridges. If Bradley's total is one-third the total
owned by Andrew and four times the total owned by Charlie, how many video
game cartirdges do the three of them own altogether, in terms of b?
(16/3)b
(17/4)b
(13/4)b
(19/12)b
(7/12)b
OA after sometime!
Then b = (1/3)*12 = 4.
Since Bradley owns 4 times the number owned by Charlie, c = 1.
Total = 12+4+1 = 17. This is our target.
Now we plug b=4 into the answers to see which yields our target of 17.
Only answer choice B works:
(17/4)*4 = 17.
The correct answer is B.
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Bradley's total is one-third the total owned by AndrewAkiB wrote:Can some one please explain this problem in Greater detail. I am a little confused
Shouldnt it be
Bradly- b
1/3b - Andrew
4b- Charlie
So, Andrew owns MORE cartridges than Bradley owns.
In fact, Andrew owns 3 TIMES as many cartridges as Bradley owns.
So, Andrew owns 3b cartridges.
If Bradley's total is ... four times the total owned by Charlie
So, Bradley owns MORE cartridges than Charlie owns.
If Bradley owns 4 TIMES as many cartridges as Charlie owns, we can also say that Charlie owns 1/4 as many cartridges as Bradley owns
So, Charlie owns (1/4)b cartridges.
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