NIMCET 2008 — Reasoning PYQ

Q: How do I solve NIMCET 2008 Reasoning questions?

Practice NIMCET 2008 Reasoning previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Reasoning questions?

NIMCET Reasoning questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2008 — Reasoning PYQ

NIMCET | Reasoning | 2008

A woman took a certain number of eggs to the market and sold some of them. The next day through her poultry industry, the number left over had been doubled, and she sold the same number as the previous day. On the third day, the new remainder was tripled, and she sold the same number as before. On the fourth day, the remainder was quardrupled, and her sale were the same as before. On the fifth day, what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions and so disposed of her entire stock. What is the smallest number of eggs she could have taken to the market the first day, and how many did she sell daily?

Choose the correct answer:

Explanation

1. Define the Variables:

Let $x$ be the initial number of eggs taken to the market.

Let $y$ be the constant number of eggs sold every day.

2. Formulate the Daily Remainders:

We will track the remaining stock at the end of each day:

Day 1: She starts with $x$ and sells $y$ .

$\text{Remainder}_1 = x - y$
Day 2: The remainder doubles, then she sells $y$ .

$\text{Remainder}_2 = 2(x - y) - y = 2x - 3y$
Day 3: The remainder triples, then she sells $y$ .

$\text{Remainder}_3 = 3(2x - 3y) - y = 6x - 10y$
Day 4: The remainder quadruples, then she sells $y$ .

$\text{Remainder}_4 = 4(6x - 10y) - y = 24x - 41y$
Day 5: The remainder quintuples, then she sells $y$ . This exhausts the stock ( $0$ ).

$\text{Remainder}_5 = 5(24x - 41y) - y = 0$

3. Solve the Final Equation:

Simplify the equation from Day 5:

5(24x - 41y) - y = 0

120x - 205y - y = 0

120x = 206y

\frac{x}{y} = \frac{206}{120}

Reduce the fraction to its simplest form to find the smallest number of eggs:

\frac{x}{y} = \frac{103}{60}

4. Determine the Values:

Since we need the smallest integer values for eggs:

$x = 103$ (Initial eggs)
$y = 60$ (Daily sales)

Final Answer:

The woman took 103 eggs to the market and sold 60 eggs daily.

Correct Option: (d)