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nisha

· started a discussion

· 1 Months ago

q66 mock3 (Question:108110.0)

in the given solution ,in 6th line
if time is same
then we must have
y/20=x-y/30
as time=distance/speed
pleasse clear my doubt

Question:

P and Q starting simultaneously from two different places proceed towards each other at a speed of 20 km/hour and 30 km/hour respectively. By the time they meet each other, Q has covered 36 km more than P. The distance (in k.m.) between the two places is:

Options:
A) 108
B) 144
C) 162
D) 180
Solution:
Ans: (d)

Let the distance = x km

Let they meet when P travels y km.

then, Q travels in the same time  x – y km

from question,

(x – y) – y = 36 

x – 2y = 36                 .......(i)

Also, 20 × y = 30 (x–y)       [\(\because\)Time is same]

\(\therefore \cfrac{y}{x-y}=\cfrac{3}{2}\)

2y = 3x – 3y

5y = 3x                     ........(ii)

Putting the value of y from (ii) to (i),

\(x-\cfrac{6x}{5}=36 \Rightarrow\cfrac{-x}{5}=36\Rightarrow x =180\) km

Aliter


Knowledge Expert

· commented

· 1 Months ago

Dear Student
If the solution is not clear you can use the approach of the relative speed. From that the relative speed of P=20-(-30)=50km/hr. Lets take the distance to be travelled as "d". So time of meeting will be d/50hr. Given Q has travelled 36km extra than P. So 30*d/50-20*d/50=336. From here d=180km.
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