Delhi University MCA 2019 Previous Year Questions (PYQs) – Page 5 of 5

Y alone can finish the work in 9 days.

⇒ Y’s 1-day work $= \dfrac{1}{9}$

X works twice as fast as Y.

⇒ X’s 1-day work $= 2 \times \dfrac{1}{9} = \dfrac{2}{9}$

Together, their 1-day work $= \dfrac{2}{9} + \dfrac{1}{9} = \dfrac{3}{9} = \dfrac{1}{3}$

Time taken to complete the whole work $= \dfrac{1}{(1/3)} = 3 \text{ days}$

Answer: $\boxed{3\text{ days}}$ ✅

Average of 10 numbers $= 25$

⇒ Sum of 10 numbers $= 10 \times 25 = 250$

New average $= 25 + 5 = 30$

⇒ New sum $= 10 \times 30 = 300$

Increase in sum $= 300 - 250 = 50$

Let the new number be $x$.

Then $x - 15 = 50$

⇒ $x = 65$

Answer: $\boxed{65}$ ✅

Total solution $= 10 \text{ liters}$

Percentage of acid $= 60\%$

⇒ Amount of acid $= \dfrac{60}{100} \times 10 = 6 \text{ liters}$

Answer: $\boxed{6 \text{ liters}}$ ✅

Given series: $2,\,7,\,14,\,23,\,34,\,\_\_$

Find the differences:

$7-2=5$,   $14-7=7$,   $23-14=9$,   $34-23=11$

Difference pattern: $5,\,7,\,9,\,11$ (increases by 2 each time)

Next difference $= 11 + 2 = 13$

Next term $= 34 + 13 = 47$

Answer: $\boxed{47}$ ✅

Logic: Each letter is shifted forward by 5 positions (D→I, O→T, G→L).

Apply to ITL: I→N, T→Y, L→Q

Answer: NYQ ✅