Quantitative Aptitude & Data Interpretation

Mixture concentration calculation:

• In the initial $150\text{ L}$ mixture:
  Volume of Water $= 20\% \times 150 = 30\text{ L}$.
  Volume of Wine $= 150 - 30 = 120\text{ L}$.
• Let $x$ liters of water be added. The volume of wine remains constant at $120\text{ L}$.
• In the new mixture, water is $25\%$, so wine is $75\%$ of the total volume:
  $75\% \times (150 + x) = 120$
  $\frac{3}{4} \times (150 + x) = 120$
  $150 + x = 120 \times \frac{4}{3} = 160$
  $x = 160 - 150 = 10\text{ litres}$.

Average of an arithmetic progression:

• Two-digit numbers divisible by 5 range from 10 to 95 ($10, 15, 20, \dots, 95$).
• Since the numbers form an arithmetic progression with a constant difference of 5, the mean/average equals the average of the first and last terms:
  $\text{Average} = \frac{\text{First term} + \text{Last term}}{2} = \frac{10 + 95}{2} = \frac{105}{2} = 52.5$.

Partnership share calculation proportional to worker-months:

• Effective labor contribution of A $= 10 \times 7 = 70\text{ worker-months}$.
• Effective labor contribution of B $= 12 \times 5 = 60\text{ worker-months}$.
• Effective labor contribution of C $= 15 \times 3 = 45\text{ worker-months}$.
• Ratio of their shares:
  $\text{Ratio } A : B : C = 70 : 60 : 45 = 14 : 12 : 9$.
  Sum of ratio parts $= 14 + 12 + 9 = 35$.
• Share of C:
  $\text{C's Share} = \frac{9}{35} \times 17500 = 9 \times 500 = 4500$.

Age ratio problem solution:

• Let the present ages of P and Q be $3x$ and $4x$ years, respectively.
• Five years ago, their ages were $(3x - 5)$ and $(4x - 5)$.
• Equating to the past ratio:
  $\frac{3x - 5}{4x - 5} = \frac{5}{7}$.
• Cross-multiplying:
  $7(3x - 5) = 5(4x - 5)$
  $21x - 35 = 20x - 25$
  $21x - 20x = 35 - 25$
  $x = 10$.
• Present ages:
  Age of P $= 3x = 3 \times 10 = 30\text{ years}$.
  Age of Q $= 4x = 4 \times 10 = 40\text{ years}$.