# Difficult Ratios & Proportion Solved QuestionAptitude Discussion

Common Information

There are $N$ numbers of gold biscuits in the house, in which four people are lived. If the first men 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.

 Q. Common Information Question: 3/3 In the previous question, what is the sum of the number of biscuits hidden by the last 2 men?
 ✖ A. 142 ✔ B. 144 ✖ C. 180 ✖ D. 178

Solution:
Option(B) is correct

Suppose $N=5x+1$

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 coins 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

$⇒ a=125k$
$⇒ x=125k−1$ and $N=5x+1=625k−4$
$⇒ N<100$, 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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