Question

There are N numbers of gold biscuits in the house, in which four people are lived. 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.If N<1000, how many biscuits were left after the fourth man took his share?

• A. 152
• B. 225
• C. 252
• D. 282
• E. none

Answer 1

The correct option is C

252

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 $\Rightarrow$ 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 [(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 of5:4 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-1 and N=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$63)+1 After the fourth man took his share(5$\times$63+1), the biscuits lefts is4$\times$63=252