Quantitative Aptitude & Data Interpretation

The average of the first $n$ natural numbers ($1, 2, 3, \dots, n$) is given by:
$\text{Average} = \frac{n + 1}{2}$
For $n = 100$:
$\text{Average} = \frac{100 + 1}{2} = \frac{101}{2} = 50.5$.

Option analysis:

• 50: Ignores the $+1$ term in median calculation.
• 50.5: Correct average of numbers 1 to 100.
• 51: Calculation error.
• 49.5: Average of numbers 0 to 99.

Initial sum of 15 observations $= 15 \times 35 = 525$.
Incorrect values read $= 23 + 14 = 37$.
Correct values to add $= 33 + 24 = 57$.
Corrected sum $= 525 - 37 + 57 = 545$.

Total mixture = 42 litres in ratio $4:3$.
Total parts $= 4 + 3 = 7$.
Chemical A $= \frac{4}{7} \times 42 = 24\text{ litres}$.
Chemical B $= \frac{3}{7} \times 42 = 18\text{ litres}$.
When 12 litres of Chemical B is added:
New quantity of Chemical B $= 18 + 12 = 30\text{ litres}$.
New ratio $A : B = 24 : 30 = 4 : 5$.

Let total profit be $P$.
Management commission to Wajid $= 0.25P$.
Remaining profit to be shared $= 0.75P$.
Ratio of profit sharing (Investment $\times$ Time period):
Sajid: $70000 \times 12 = 840,000$
Wajid: $52500 \times 8 = 420,000$

Sajid's share of profit $= \frac{2}{3} \times 0.75P = 0.50P$.
Wajid's share of profit $= \frac{1}{3} \times 0.75P = 0.25P$.
Total received by Wajid $= \text{Commission} + \text{Profit share} = 0.25P + 0.25P = 0.50P$.
Given $0.50P = 60,000 \implies P = 120,000$.
Amount received by Sajid $= 0.50P = \text{Rs. } 60,000$.