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Question

There are N number of gold biscuits in the house, in which four people live. If the first man woke up and divided the biscuits into 5 equal piles and found one extra biscuit. He took one of those piles along with the extra biscuit and hid them. He then gathered the 4 remaining piles into a big pile, woke up the second person and went to sleep. Each of the other 3 persons did the same one by one i.e. divided the big pile into 5 equal piles and found one extra biscuit. Each hid one of the piles along with the extra biscuit and gathered the remaining 4 piles into a big pile.
What is the sum of the number of biscuits hidden by the last 2 men?


A

72

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B

100

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C

121

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D

144

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E

none

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Solution

The correct option is D

144


Initially, N=5x+1

A took x+1 biscuit.

Now, 4x is of the form 5y+1 then x must be in the form 5z+4

4(5z+4)=5y+1

y = 4z+3 and x = 5z+4

The ratio of number of biscuits that A and B took is [(5z+4)+1]:[(4z+3)+1]=5:4

So, we can say that any two successive persons A, B, C and D take gold biscuits in the ratio of 5:4.

Let the number of biscuits that A, B, C and D took be a, b, c and d respectively.

a:b=b:c=c:d=5:4

a:b:c:d=125:100:80:64

a=125k

x=125k1 and N=5x+1=625k4

N<100,then k=1

N=621

621=(5×124)+3

4×124=(5×99)+1

4×99=(5×79)+1

4×79=(5×63)+1

The number of biscuits hidden by 3rd and the 4th men is 79+1=80 and 63+1 = 64 i.e. a total of 144.


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