Question

.P started from town X at 10 a.m. and went towards
town Y. Q started from Y at 12 p.m. and went towards
X. They met at 2 p.m. Preached his destination
20 minutes after Q did. At what time did Q reach his
destination?
(A) 4:20 p.m.
(B) 4:40 p.m.
(C) 5 p.m.
(D) 5:20 p.m.​

Answer 1

Step-by-step explanation:

Let Z be the point where P and Q met. Let x be the elapsed time for Q to travel from Z to X, so the elapsed time for P to travel from Z to Y is x+1/3.I am confused here. 4.Sx/Sx=2 . Sy/Sy and I also did after this 4/x=2/1+1/3x but I am getting no where to the result.

Answer 2

Answer:

B

Step-by-step explanation:

Let's take the distance between X and Y as D km.

and "a km" be the distance travelled by P in 4 hours (i.e., P travelled from 10 am to 2 pm) ... speed of P = a/4 kmph.

and by 2 PM, Q must have traveled D-a kms in 2 hours. Speed of Q = (D-a)/2 kmph

Now distances left for P and Q to travel are D-a and a km. and P takes t+1/3 hours and Q takes t hours to reach their destionations.

i.e., for P  =>   $\frac{D-a}{t+1/3} = \frac{a}{4}$ (distance/time = speed)

for Q => $\frac{a}{t} = \frac{D-a}{2}$

$\frac{D-a}{a} = \frac{t+1/3}{4}$ ----(i)

$\frac{D-a}{a} = \frac{2}{t}$ -----(ii)

by solving (i) and (ii), we get,

$3t^{2}+t-24 = 0$

3t(t+3)-8(t+3) = 0;

t = 8/3 as t should be greater than zero.

now total time taken by Q is 2 hours + 8/3 hours = 14/3 = 4hr 40 minutes.