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Joined: 06 Jun 2010
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M23-29 [#permalink] New post 01 Mar 2013, 06:55
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0% (00:00) correct 100% (00:00) wrong based on 1 sessions
If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?
(1) When the books were stacked in rows of 7, 1 book was left

(2) When the books were stacked in rows of 8, 4 books were left

I solved this using Remainder approach.
From question stmt: N/5

1.N/7 Remainder is 1
so find a number such that when N/5 and N/7 gives remainder of 1. Found N=36.Sufficient

2.N/8 Remainder is 4
so N/5 gives remainder 1 and N/8 gives remainder 4.Found N =36.Sufficient.

Can someone xplain whats wrong with my approach?
[Reveal] Spoiler: OA
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Re: M23-29 [#permalink] New post 01 Mar 2013, 09:28
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shreerajp99 wrote:
If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?
(1) When the books were stacked in rows of 7, 1 book was left

(2) When the books were stacked in rows of 8, 4 books were left

I solved this using Remainder approach.
From question stmt: N/5

1.N/7 Remainder is 1
so find a number such that when N/5 and N/7 gives remainder of 1. Found N=36.Sufficient

2.N/8 Remainder is 4
so N/5 gives remainder 1 and N/8 gives remainder 4.Found N =36.Sufficient.

Can someone xplain whats wrong with my approach?


B = 5k + 1

STAT1
B = 7t + 1
5k + 1 =7t + 1
k = 7t/5
So, Those values of t which will give you integer value are the common solution.
So, t=5,10,15,20
k = 7,14,21,28,35,42,49,56,63,...
So, B = 36, 71,...

So, Not Sufficient as there are many possible values!

STAT2
B = 8s + 4
5k + 1 = 8s + 4
k = (8s + 3)/5
Those values of s which will give you integer value of k are solution.
s = 4, 9,14,19,24,...
k = 7, 15, 23,31,39, 47, 55, 63
So, B = 36, 76,...

So, Not Sufficient as there are many possible values!

Trying for C
k = 7 and 63 are the two solutions which are possible (actually there are more...)
B = 36, 316

So, Answer is E
Hope it helps!
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Re: M23-29   [#permalink] 01 Mar 2013, 09:28
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