The only items in a container A are 150 pencils and 725 pens. The ratio of the number of pencils to the number of pens in container B is 2 to 3. If all the pencils and pens in container B are placed in container A, then the ratio of the number of pencils to the number of pens in container A would be 3 to 5. What is the total number of pencils and pens in both container A and container B ?
a.5,600
b.6,725
c.7,125
d.7,275
e.8,000
I've set up the following equations from the question stem:
1. c/n=2/3 (where c=# of pencils in cont. B and n=# of pens in cont. B)
2. (c+150)/(n + 725)=3/5
then, solving for c = 1425, and n=4275. Adding them together, I got 7125, which is choice c. But, the answer is e. 8000. What am i doing wrong?
Thanks.
proportions q.
This topic has expert replies

 Senior  Next Rank: 100 Posts
 Posts: 52
 Joined: 23 Jul 2009
 Location: Irvine, CA
 Thanked: 1 times
This is how I did the problem.
Case A: 150/725 (real number)
Case B: 2/3 (ratio)
Combined: (150 + 2x) / (725 + 3x) = 3/5
The reason for the x is because case B is a ratio and we don't know what the real number is.
Now we cross multiply:
3(720 + 3x) = 5(150 + 2x)
2175 + 9x = 750 + 10x
1425 = x
2x + 3x + 150 + 725
2(1425) + 3(1425) + 150 + 725 = 8000
Case A: 150/725 (real number)
Case B: 2/3 (ratio)
Combined: (150 + 2x) / (725 + 3x) = 3/5
The reason for the x is because case B is a ratio and we don't know what the real number is.
Now we cross multiply:
3(720 + 3x) = 5(150 + 2x)
2175 + 9x = 750 + 10x
1425 = x
2x + 3x + 150 + 725
2(1425) + 3(1425) + 150 + 725 = 8000

 Master  Next Rank: 500 Posts
 Posts: 148
 Joined: 03 Jun 2009
 Thanked: 18 times
 Followed by:1 members
We know that the total number of Pens to pencil ratio for the combined container
The ratio = 3:5 hence the total number must be a multiple of 8
so b , c and d out
Left with a and e
a= 5600 and e = 8000 , but we know the ratio of pens to pencils in container B so a is too low hence the answer is E
The ratio = 3:5 hence the total number must be a multiple of 8
so b , c and d out
Left with a and e
a= 5600 and e = 8000 , but we know the ratio of pens to pencils in container B so a is too low hence the answer is E

 Junior  Next Rank: 30 Posts
 Posts: 22
 Joined: 30 Jan 2008
 Thanked: 5 times
Dear Godemol,
You forgot to add the 150 & 725 of container A to your 7125 Ans.
7125 + 875= 8000
Regards
You forgot to add the 150 & 725 of container A to your 7125 Ans.
7125 + 875= 8000
Regards
GMAT/MBA Expert
 [email protected]
 GMAT Instructor
 Posts: 5881
 Joined: 25 Apr 2015
 Location: Los Angeles, CA
 Thanked: 43 times
 Followed by:24 members
We are given that there are 150 pencils and 725 pens in container A.godemol wrote:The only items in a container A are 150 pencils and 725 pens. The ratio of the number of pencils to the number of pens in container B is 2 to 3. If all the pencils and pens in container B are placed in container A, then the ratio of the number of pencils to the number of pens in container A would be 3 to 5. What is the total number of pencils and pens in both container A and container B ?
a.5,600
b.6,725
c.7,125
d.7,275
e.8,000
We are also given that the ratio of pens to pencils in container B is 2 to 3. Thus:
pencils : pens = 2x : 3x
When all of the pens and pencils from container B are placed in container A, the number of pencils in container A becomes 150 + 2x, and the number of pens in container B becomes 725 + 3x. Since the new ratio of pencils to pens is 3 to 5, we have:
(150 + 2x)/(725 + 3x) = 3/5
3(725 + 3x) = 5(150 + 2x)
2175 + 9x = 750 + 10x
1,425 = x
There are 2(1425) = 2850 pencils and 3(1425) = 4275 pens in container B.
Thus, we have a total of 150 + 725 + 2850 + 4275 = 8,000 pens and pencils.
Answer: E
Scott WoodburyStewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews