Delhi University MCA 2020 Previous Year Questions (PYQs) – Page 1 of 4

Let rates be $a,b,c$. Given: $a+b=\tfrac{1}{15}$, $b+c=\tfrac{1}{20}$, $a+b+c=\tfrac{1}{10}$.

Using $(a+b)+(a+c)+(b+c)=2(a+b+c)$:

$\tfrac{1}{15}+(a+c)+\tfrac{1}{20}=\tfrac{1}{5}\ \Rightarrow\ a+c=\tfrac{1}{5}-\left(\tfrac{1}{15}+\tfrac{1}{20}\right)=\tfrac{1}{12}$

Time by $A+C = \dfrac{1}{1/12}=12$ days.

Answer: $\boxed{12\text{ days}}$ ✅

Speed of faster train $= 50$ km/h $= 13.89$ m/s

Speed of slower train $= 40$ km/h $= 11.11$ m/s

Relative speed (opposite directions) $= 13.89 + 11.11 = 25$ m/s

Distance to be covered $= 500$ m

Time $= \dfrac{500}{25} = 20$ seconds

Answer: $\boxed{20\text{ seconds}}$ ✅

Speed in still water $= 12$ km/hr

Rate of current $= 3$ km/hr

Downstream speed $= 12 + 3 = 15$ km/hr

Time $= 30$ min $= \dfrac{1}{2}$ hr

Distance $= 15 \times \dfrac{1}{2} = 7.5$ km

Answer: $\boxed{7.5\text{ km}}$ ✅