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Total space =140

Speed of P =140/9 which can be approximated as (140/10) = 14
Speed of Q = 140/5= 28

Relative speed (sum since they are going in opposite direction) =14+28=42
Space to cover =140

Relative Time = Space/Relative speed = 140/42 =approximately 3 (time P and Q encounter each other)

For t=3 calculate Space travelled by P

Thus space= speed * relative time = 14 *3 =42 (approximate) (distance travelled by P when it encounters Q)

Since we approximated the speed of Q by increasing the denominator to 10 the overall result is underestimated, it should be a little more that 14
In addition we estimated T=3 but it should be a little more than 3

Thus the product 14*3 should be higher and the closed answer is A
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Method 1: Time required to meet, t = 140/(140/9+140/5)
t=45/14 hours

Distance travelled by P= 140/9 * 45/14
= 50 km

Method 2: Time at which they meet is constant. Hence,
Dp/Sp= Dq/Sq
Sp/Sq= Dp/Dq.

Let x be the distance travelled by P.
(140/9)/(140/5)= x/(140-x)
5/9=x/(140-x)
Solve for x or check from options.
x=50 km
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