NIMCET 2007 — Reasoning PYQ

The easiest and fastest way to solve this type of problem is by working backward from the final remaining amount.

Let's look at the general rule for what happens when a thief takes half of the bread plus half a bread:

If the remaining bread at any stage is $R$, and the amount before that stage was $B$, the thief takes $\frac{B}{2}+\frac{1}{2}$.

The remaining amount after the theft is:

$$R=B-\left(\frac{B}{2}+\frac{1}{2}\right)=\frac{B}{2}-\frac{1}{2}$$

Rearranging this formula to find the previous amount ($B$) from the remaining amount ($R$):

$$R+\frac{1}{2}=\frac{B}{2}$$

$$B=2\cdot \left(R+\frac{1}{2}\right)=2R+1$$

So, to find the initial amount before any thief stole, we simply double the remaining amount and add 1 ($2R+1$) for each step backward.

Step 1: Before the $3^{\text{rd}}$ Thief

• Breads left at the very end ($R_3$) = 3

• Breads before the $3^{\text{rd}}$ thief arrived ($B_3$):

  $$B_3=2\cdot 3+1=6+1=7$$

Step 2: Before the $2^{\text{nd}}$ Thief

• Breads left after the $2^{\text{nd}}$ thief ($R_2$) = 7

• Breads before the $2^{\text{nd}}$ thief arrived ($B_2$):

  $$B_2=2\cdot 7+1=14+1=15$$

Step 3: Before the $1^{\text{st}}$ Thief (Initial Amount)

• Breads left after the $1^{\text{st}}$ thief ($R_1$) = 15

• Breads before the $1^{\text{st}}$ thief arrived ($B_1$):

  $$B_1=2\cdot 15+1=30+1=31$$

Correct Answer

The correct option is (b) 31.