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Re: There 3 kinds of books in the library fiction, non-fiction and biology [#permalink]
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Given: F:N=3:2 and N:B=4:3.

Equate N's by finding the LCM of 2 and 4, so that to be able to write down one ratio for all three kinds of books: F:N=3:2=6:4 and N:B=4:3 --> F:N:B=6:4:3. Now, 6+4+3=13, which means that we have total of 13 parts, thus the total number of books must be divisible by 13, only answer divisible by 13 is 1001.

Answer: A.

you can equate N's in any other way (LCM is just easiest one). For example: F:N=3:2=30:20 and N:B=4:3=20:15 --> F:N:B=30:20:15=6:4:3, the same reduced ratio. Or the way you were doing: F:N=3:2=12:8 and N:B=4:3=8:6 --> F:N:B=12:8:6=6:4:3, the same reduced ratio.
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Re: There 3 kinds of books in the library fiction, non-fiction and biology [#permalink]
Carcass wrote:
Given: F:N=3:2 and N:B=4:3.

Equate N's by finding the LCM of 2 and 4, so that to be able to write down one ratio for all three kinds of books: F:N=3:2=6:4 and N:B=4:3 --> F:N:B=6:4:3. Now, 6+4+3=13, which means that we have total of 13 parts, thus the total number of books must be divisible by 13, only answer divisible by 13 is 1001.

Answer: A.

you can equate N's in any other way (LCM is just easiest one). For example: F:N=3:2=30:20 and N:B=4:3=20:15 --> F:N:B=30:20:15=6:4:3, the same reduced ratio. Or the way you were doing: F:N=3:2=12:8 and N:B=4:3=8:6 --> F:N:B=12:8:6=6:4:3, the same reduced ratio.


I think I knew my mistake. I was trying to find the value of N not the sum of all. Thanks.


I have another question regarding the ratios,
Now we are given that F : N is 3:2
if we want to write it as an equation, we will write
\(\frac{F}{N}= \frac{3}{2}\)
or the opposite \(\frac{2}{3}\) ??
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Re: There 3 kinds of books in the library fiction, non-fiction and biology [#permalink]
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first one
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