In a library, all the books were arranged on shelves of 15 books each

Question

In a library, all the books were arranged on shelves of 15 books each, with no books left over. After 90 new books arrived, all the books in the library were arranged on shelves of 18 books each, with no books left over. Suppose, the total number of books in the library after the 90 new books arrived was at most 250, and the initial number of books in the library was non-zero. How many books were there in the library before the 90 additional books arrived?

(a) 60
(b) 90
(c) 150
(d) 180
(e) None

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(a) 60
(b) 90
(c) 150
(d) 180
(e) None

chetan2u · Answer

Since, the books earlier could be arranged in book shelves of 15 each, the number of books will be a multiple of 15, say 15x.

After 90 books are added, the books could be arranged in book shelves of 18 each, the number of books will be a multiple of 18, say 18y. 
=> 15x+90=18y
15x=18(y-5)
5x=6(y-5)
Thus, x is a multiple of 6 = 6a
15x=15*6a=90a
But 90a+90<250………90a<160
Only possibility is when a=1, so initial books = 90*1 or 90.

B

PyjamaScientist · Answer

Is this a possible way to do it-

We need to find the value of 15x. The original number of books on the shelf.

Given (following your nomenclature)-
15x+90=18y ---------(1)
and 15x + 90 <= 250 ----------- (2)
Simplifying (2), 15x <= 160 --------- (3)

From (1),
15x = 18y - 90
15x = 18 (y - 90), this means that the value of 15x is a multiple of "18".

Among the given choices, only make the cut 90 and 180.
180 is rejected by (3) so we have the only possible option 90 left. And that is the correct answer choice too.

rmahe11 · Answer

why is the answer not 150 , initially books can be 150 and later 90 since 18*5 . atmost will be 250 hence should be around 240 right ?

parag08 · Answer

If you take intial as 150 and final as 240 then if we try to calculate number of shelves in the end it becomes 240/18 = 13.33.. which is not possible as shelves has to be integer

hemanthPA · Answer

I could try to understand the solution but I really told myself this is a waste of time, rather go through options.
Let x – initial number of books – multiple of 15
Let y – final number of books – multiple of 18

To find? Initial number of books.

Solution:
60: multiple of 15 but 60+90 = 150 which isn’t divisible by 18, hence no
90: multiple of 15, 90+90 = 180, divisible by 18
150: multiple of 15, but 150+90 = 240 which isn’t divisible by 18, hence no
180 wont work, bcs max no of books is 250.”

I felt this approach very easy and it did work for me for similar sums.

hemanthPA · Answer

Solution:
r - estimated reading speed (pages per hour).
t - estimated time (hours).
So, r×t = 450.
Actual time taken t+3 hours.
Actual speed r−5 pages per hour.
Given: (r−5)×(t+3)=450.

You’re looking for a value of t where the decrease in speed and increase in time results in reading 450 pages.

Check if the option fits the equation
r×t =450 where

Plug the option into the adjusted equation to check consistency.
Let’s try with the options:
Option: 15 hours (t=15)
Estimated speed r = 450/15 = 30 pages/hour.
Actual speed 30−5=25 pages/hour.
Actual time 15+3=18 hours.
Actual pages read: 25×18= 450 (this works!)
Other Options: t=18, t=20, t=23 – plug these in similarly, and you’ll find they won’t satisfy both conditions.

Answer: 15 hours.

This method is quick and allows me to test each option with simple calculations, eliminating unnecessary complexity.

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In a library, all the books were arranged on shelves of 15 books each

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