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

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

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

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

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

Answer: $\boxed{{\displaystyle 12\text{ days}}}$ ✅

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 $={\displaystyle \frac{500}{25}}=20$ seconds

Answer: $\boxed{{\displaystyle 20\text{ seconds}}}$ ✅

Speed in still water $=12$ km/hr

Rate of current $=3$ km/hr

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

Time $=30$ min $={\displaystyle \frac{1}{2}}$ hr

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

Answer: $\boxed{{\displaystyle 7.5\text{ km}}}$ ✅