Aptitude 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. |
If $N > 1000$, what could be the least value of $N$? |

✖ A. |
1249 |

✖ B. |
1023 |

✖ C. |
1202 |

✔ D. |
1246 |

**Solution:**

Option(**D**) 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 $N=5x+1=625k−4$

As, $N>1000$, the least value of $N$ is when $k=2$

$⇒ N=$ **1246 **

**Saikiran**

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**Gaurav Karnani**

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how can you take the ratio A:B as 5:4 when A is taking additional 1 biscuit.

for example consider 21 gold coins. so if A took 5 coins , left is 16 with B and the ratio becomes 21:16.

**Vardhan**

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Help me out here....

how did $x$ become $5z+4$?

The solution is way easier.

The first person split the total into 5 parts & there was one extra biscuit..

Look @ the options, only 1246-1 is divisible by 5.. so obviously rest all options cant be split into 5 parts equally - 1..