Question

man covered a certain distance at some speed had he moved 3 km/hr faster he would have taken 40 minutes less If he had moved 2 km/hr slower he would have taken 40 minutes more The distance ( in km) 
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Answer 1

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ANSWER

Let the distance x km and usual rate y km/hr =

Then х y X y + 3 40 x(y + 3) — ху 2 60 y(y + 3) 3

3x 2 y(y + 3) 3

+ 2y(y + 3) 9x.. (i)

and X 40 xy - x(y-2) 60 y(y - 2) CON

y - 2 y 2x 2

y(y-2) 3

> y(y 2) — Зх.. = (ii)

On dividing eqn (i) by eqn (ii) we get

2(y +3)

y - 2

= 3 = 2y + 6 = 3y 6 = y = 12

. Putting the value of y in (i) we get 2 x 12 x 15 = 9x

X = 40 km/hr

Answer 2

Let the distance = x km and usual rate = y km/hr

Then yx−y+3x=6040

⇒y(y+3)x(y+3)−xy=32

⇒y(y+3)3x=32

⇒2y(y+3)=9x..

(i)  y−2x−yx=6040

⇒y(y−2)xy−x(y−2)=32 

⇒y(y−2)2x=32

⇒y(y−2)=3x.

(ii)

On dividing eqn

(i) by eqn

(ii) we get

y−22(y+3)=3⇒2y+6=3y−6⇒y=12

∴ Putting the value of y in (i) we get 2 x 12 x 15 = 9x ⇒ x = 40 km/hr