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. If N>1000, what could be the least value of N?
Correct answer: B
Explanation
Let, each equal part is x and N be the total number of biscuits.
Thus, 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.
As, N>1000, the least value of N is when k=2.
=>N= 1246.
B
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AQuA-RAT items are crowdsourced algebra problems, not ACT questions. They carry five answer choices in the pre-2025 ACT Mathematics style and no ACT reporting category. Useful for drilling Preparing for Higher Math content; not a substitute for a real form.