Jack has a total of b hardback and paperback books in his library. If the number of hardback books is 1/3 the number of paperback books, and 3/4 of the of the paperback books are biographies, how many biographies, in terms of b, are in Jack's library?
A)(1/9)b
B)(3/20)b
C)(3/16)b
D)(1/3)b
E)(9/16)b
I honestly don't think any of the answers are correct. I will give the OA and explanation after some discussion.
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Help with a question...
This topic has expert replies
-
pemdas
- Legendary Member
- Posts: 1085
- Joined: Fri Apr 15, 2011 2:33 pm
- Thanked: 158 times
- Followed by:21 members
this is simple ratio problem, b is total and H (hardback)=P/3 (paperback), 3P/4=Biographies. P=3H and Biographies=3*3H/4=9H/4. Since b is total and consists of H and P, we have the following ratios P/H=3/1 and P+H=b=4 with b=4H --> Biographies=9H/4, H=b/4 and accordingly Biographies=9(b/4):4=9b/16
e
e
seal4913 wrote:Jack has a total of b hardback and paperback books in his library. If the number of hardback books is 1/3 the number of paperback books, and 3/4 of the of the paperback books are biographies, how many biographies, in terms of b, are in Jack's library?
A)(1/9)b
B)(3/20)b
C)(3/16)b
D)(1/3)b
E)(9/16)b
I honestly don't think any of the answers are correct. I will give the OA and explanation after some discussion.
Success doesn't come overnight!
That is the OA and how they explain it. Here is my question, if you use the plug in method I don't see where I am making my mistake.
So say if you have 4 books that means 3 are soft and 1 is hard. Then the number of soft that are bios is 3 x (3/4) which is 9/4. The number of hard that are bios is 9/4s = bio so it's (9/4) x (1/3) which is 1/4. So is not my bio equal to 1/4 plus 9/4 which is 10/4?
So (9/16) * 4 = 9/4 but 9/4 is only the bios he has in soft... I fail to see it... please help
So say if you have 4 books that means 3 are soft and 1 is hard. Then the number of soft that are bios is 3 x (3/4) which is 9/4. The number of hard that are bios is 9/4s = bio so it's (9/4) x (1/3) which is 1/4. So is not my bio equal to 1/4 plus 9/4 which is 10/4?
So (9/16) * 4 = 9/4 but 9/4 is only the bios he has in soft... I fail to see it... please help
-
pemdas
- Legendary Member
- Posts: 1085
- Joined: Fri Apr 15, 2011 2:33 pm
- Thanked: 158 times
- Followed by:21 members
We are given only: "3/4 of the of the paperback books are biographies".seal4913 wrote:The number of hard that are bios is 9/4s = bio ? so it's (9/4) x (1/3) which is 1/4.
Distinctively, we must have integer value for bios, i.e. this could not be 9/4 and b isn't 4. If you noticed I supplied
not culminating by 'b=4' and leaving with 'b=4H'. The distinction made is b=4H not b=4. Hence b could be 16, as bios=(9/16)*b=(3^2/2^4)*b and bios might be 9. If b=32, b would be 18, etc. We were required to find bios expressed through 'b' only.we have the following ratios P/H=3/1 and P+H=b=4 with b=4H
Success doesn't come overnight!
-
Bill@VeritasPrep
- GMAT Instructor
- Posts: 1248
- Joined: Thu Mar 29, 2012 2:57 pm
- Location: Everywhere
- Thanked: 503 times
- Followed by:192 members
- GMAT Score:780
The total of hardbacks and paperbacks is B; H + P = B
Hardbacks are 1/3 of the number of paperbacks: H = P/3
3/4 of paperbacks are biographies: 3P/4 = X
To solve for X, we need P in terms of B, which we can find by substituting our second equation into the first equation:
H + P = B
P/3 + P = B
4P/3 = B
4P = 3B
P = 3B/4
We can then substitute into the third equation:
3P/4 = X
3(3B/4) / 4 = X
(9B/4)/4 = X
9B/16 = X
Hardbacks are 1/3 of the number of paperbacks: H = P/3
3/4 of paperbacks are biographies: 3P/4 = X
To solve for X, we need P in terms of B, which we can find by substituting our second equation into the first equation:
H + P = B
P/3 + P = B
4P/3 = B
4P = 3B
P = 3B/4
We can then substitute into the third equation:
3P/4 = X
3(3B/4) / 4 = X
(9B/4)/4 = X
9B/16 = X
Join Veritas Prep's 2010 Instructor of the Year, Matt Douglas for GMATT Mondays
Visit the Veritas Prep Blog
Try the FREE Veritas Prep Practice Test
Visit the Veritas Prep Blog
Try the FREE Veritas Prep Practice Test
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7294
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We are given that Jack as a total of b hardback and paperback books in his library. We can let h = the number of hardback books and p = the number of paperback books, and thus h + p = b.seal4913 wrote:Jack has a total of b hardback and paperback books in his library. If the number of hardback books is 1/3 the number of paperback books, and 3/4 of the of the paperback books are biographies, how many biographies, in terms of b, are in Jack's library?
A)(1/9)b
B)(3/20)b
C)(3/16)b
D)(1/3)b
E)(9/16)b
We know that the number of hardback books is 1/3 the number of paperback books, and thus:
h = (1/3)p
We can now substitute (1/3)p for h in the equation h + p = b:
(1/3)p + p = b
Multiplying the entire equation by 3, we have:
p + 3p = 3b
4p = 3b
p = 3b/4
Finally, we are given that 3/4 of the paperback books are biographies. Since p = 3b/4, we have:
(3/4)(3b/4) = biographies
9b/16 = biographies
Answer: E
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
