UPSC CDS 2 2020 Math Paper

Let X takes t hr. to travel a distance with speed of 4 kmph.
Then distance travelled by X, d=4tkm.
Now Y takes (t−2) hr. to complete the same distance with speed of 5 kmph.
Then distance d=4t=5(t−2)
4t=5t−10
t=10hr
And d=40km
Let Z starts t0 time after Y starts then
6(10−2−t0)=40
60−12−6t0=40
6t0=8
t0=

4

3

hr.