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Bradley owns b video game cartridges. If Bradley’s total is one-third
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21 May 2021, 09:38
21 May 2021, 09:38
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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 cartridges do the three of them own altogether, in terms of b?
A. (16/3)b
B. (17/4)b
C. (13/4)b
D. (19/12)b
E. (7/12)b
A. (16/3)b
B. (17/4)b
C. (13/4)b
D. (19/12)b
E. (7/12)b
Senior Manager
Joined: 23 Jan 2021
Posts: 294
Given Kudos: 81
Concentration: , International Business
Re: Bradley owns b video game cartridges. If Bradley’s total is one-third
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21 May 2021, 18:44
21 May 2021, 18:44
Bradley owns b video game cartridges. then,
If Bradley’s total is one-third the total owned by Andrew,
b= 1/3a
and,
four times the total owned by Charlie
b=4c,
altogether they have 3b+b+b/4.
on calculation
17/4*b
, in terms of b..
If Bradley’s total is one-third the total owned by Andrew,
b= 1/3a
and,
four times the total owned by Charlie
b=4c,
altogether they have 3b+b+b/4.
on calculation
17/4*b
, in terms of b..
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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Bradley owns b video game cartridges. If Bradley’s total is one-third
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22 May 2021, 05:09
22 May 2021, 05:09
1
Carcass 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 cartridges do the three of them own altogether, in terms of b?
A. (16/3)b
B. (17/4)b
C. (13/4)b
D. (19/12)b
E. (7/12)b
A. (16/3)b
B. (17/4)b
C. (13/4)b
D. (19/12)b
E. (7/12)b
# of cartridges owned by Bradley = b
# of cartridges owned by Andrew = 3b
# of cartridges owned by Charlie = b/4
TOTAL # of cartridges = b + 3b + b/4
Rewrite with a common denominator of 4 to get: 4b/4 + 12b/4 + b/4
Combine terms to get: 17b/4
Answer: B
Cheers,
Brent
