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Help with a question...

by seal4913 » Mon Apr 02, 2012 2:40 pm
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.
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by pemdas » Mon Apr 02, 2012 3:03 pm
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
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.
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by seal4913 » Mon Apr 02, 2012 4:34 pm
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
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by pemdas » Mon Apr 02, 2012 7:22 pm
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.
We are given only: "3/4 of the of the paperback books are biographies".

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
we have the following ratios P/H=3/1 and P+H=b=4 with b=4H
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.
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by seal4913 » Mon Apr 02, 2012 7:39 pm
I'm still not getting it. I think it is time to call it a night. I'm not understanding anything anymore. I will review again when rest and hope it makes sense. Thanks a lot
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by Bill@VeritasPrep » Mon Apr 02, 2012 7:52 pm
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
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by Scott@TargetTestPrep » Tue Dec 12, 2017 7:00 am
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 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.

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

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