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
• B. 100
• C. 121
• D. 144
• E. none

Answer 1

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

$\Rightarrow 4(5z+4)=5y+1$

$\Rightarrow$ y = 4z+3 and x = 5z+4

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

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.

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

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

$\Rightarrow a=125k$

$\Rightarrow x=125k-1andN=5x+1=625k-4$

$\Rightarrow N<100,thenk=1$

$\Rightarrow N=621$

$\Rightarrow 621=(5\times 124)+3$

$4\times 124=(5\times 99)+1$

$4\times 99=(5\times 79)+1$

$4\times 79=(5\times 63)+1$

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.