NIMCET 2007 — Reasoning PYQ

Q: How do I solve NIMCET 2007 Reasoning questions?

Practice NIMCET 2007 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 2007 — Reasoning PYQ

NIMCET | Reasoning | 2007

A girl starts to paint a fence on one day. On the second day two more girls join her and on the third day three more girls join the group and so on and so forth. If the fence is completely painted in this manner in 20 days, then the number of days, in which 10 boys painting together can paint fence completely given a boy can paint twice as fast as a girl can, is

Choose the correct answer:

Explanation

Step 1: Establish Efficiencies

Let the daily work done (efficiency) by $1$ girl be $g$ .

According to the problem, a boy paints twice as fast as a girl. Therefore, the daily work done by $1$ boy ( $b$ ) is:

$b = 2 g$

Step 2: Determine the Pattern of Girls Joining Daily

Let's analyze the number of girls working on the project day by day:

Day 1: $1$ girl works. $\to$ Work done = $1 g$
Day 2: $2$ more girls join, so $1 + 2 = 3$ girls work. $\to$ Work done = $3 g$
Day 3: $3$ more girls join, so $3 + 3 = 6$ girls work. $\to$ Work done = $6 g$
Day 4: $4$ more girls join, so $6 + 4 = 10$ girls work. $\to$ Work done = $10 g$

The number of girls working on any day $n$ follows the formula for the $n^{th}$ triangular number:

$Girls on day n = \frac{n ( n + 1 )}{2}$

Step 3: Calculate the Total Work Built Up over 20 Days

The fence is completely painted at the end of $20$ days. Therefore, the total work ( $W$ ) is the sum of the work done across all $20$ days:

$W = n = 1 \sum 20 [\frac{n ( n + 1 )}{2}] \cdot g$

$W = \frac{g}{2} n = 1 \sum 20 (n^{2} + n)$

$W = \frac{g}{2} [n = 1 \sum 20 n^{2} + n = 1 \sum 20 n]$

Now, we use standard summation formulas:

Sum of first $n$ natural numbers: $\sum n = \frac{n ( n + 1 )}{2}$
Sum of squares of first $n$ natural numbers: $\sum n^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

For $n = 20$ :

$n = 1 \sum 20 n = \frac{20 \times 21}{2} = 210$

$n = 1 \sum 20 n^{2} = \frac{20 \times 21 \times 41}{6} = 10 \times 7 \times 41 = 2870$

Substitute these values back into our equation for Total Work ( $W$ ):

$W = \frac{g}{2} [2870 + 210]$

$W = \frac{g}{2} [3080] = 1540 g$

Step 4: Calculate the Number of Days Needed by 10 Boys

Now, we need to find out how long it takes for $10$ boys working together to complete this same amount of work ( $W = 1540 g$ ).

First, let's find the total efficiency of $10$ boys:

$Total efficiency of 10 boys = 10 \times b$

$Total efficiency of 10 boys = 10 \times (2 g) = 20 g$

Now, calculate the required number of days ( $D$ ):

$D = \frac{Total Work}{Total Efficiency of 10 boys}$

$D = \frac{1540 g}{20 g}$

$D = \frac{1540}{20} = 77 days$

Final Answer

The correct option is (b) 77.