Question

A man covered a certain distance at some speed. He had moved 3km\h faster, he would have taken 40 minutes less. If he had moved 2km\h slower, he would taken 40 minutes more. The distance is _____. A) 35km. B) 3623km. C) 3712km. D) 40km.​

Answer 1

Solution :-

Let the speed of the man be 's' Km/h

Let time taken be 't' hours

We know that

Speed = Distance/Time

So Distance covered by man = s * t = 'st' Km

Case 1

Speed of the man = 3 Km/h faster than the original = (s + 3) Km/h

Time taken = 40 minutes less = (t - 40 min) = (t - 40/60) = (t - 2/3) hours

Distance covered i.e st = (s + 3)(t - 2/3)

st = s(t - 2/3) + 3(t - 2/3)

st = st - 2s/3 + 3t - 2

st - st = - 2s/3 + 3t - 2

0 = - 2s/3 + 3t - 2

2s/3 - 3t + 2 = 0

(2s - 9t + 6)/3 = 0

2s - 9t + 6 = 0(3)

2s - 9t + 6 = 0 ---eq(1)

Case 2

Speed of the man = 2 Km/h slower = (s - 2) Km/h

Time taken = 40 minutes more = (t + 40 minutes) = (t + 40/60) = (t + 2/3) hours

Distance covered i.e st = (s - 2)(t + 2/3)

st = s(t + 2/3) - 2(t + 2/3)

st = st + 2s/3 - 2t - 4/3

st - st = 2s/3 - 2t - 4/3

0 = 2s/3 - 2t - 4/3

2s/3 - 2t - 4/3 = 0

(2s - 6t - 4)/3 = 0

2s - 6t - 4 = 0(3)

2s - 6t - 4 = 0 ---eq(2)

Substracting (2) from eq(1)

2s - 9t + 6 - (2s - 6t - 4) = 0

2s - 9t + 6 - 2s + 6t + 4 = 0

- 3t + 10 = 0

10 = 3t

10/3 = t

t = 10/3

Substitute t = 10/3 in (1)

2s - 9t + 6 = 0

2s - 9(10/3) + 6 = 0

2s - 3(10) + 6 = 0

2s - 30 + 6 = 0

2s - 24 = 0

2s = 24

s = 24/2

s = 12

Speed of the man = 12 Km/h

Time taken = 10/3 hours

So Distance covered = st = 12(10/3) = 4(10) = 40 Km

So the answer is 40 Km i.e Option D

Note :-

Here distance covered by man is same in all cases.

Answer 2

Explanation:-

Given :-

If A man moves 3 km/hr he would takes 40 minutes less.

If he moves 2 km/hr slower he would taken 40 minutes more.

To find :-

Distance covered.

Solution:-

Let the speed be x km/hr

and the time be y hr

• Case :- 1

• Distance = XY

→$xy = ( x + 3) \left( y - \dfrac{40}{60}\right)$

→$xy = ( x +3) \left( y - \dfrac{2}{3}\right)$

→$xy = ( x+3) \left( \dfrac{3y -2}{3}\right)$

→$xy = \dfrac{3xy -2x +9y -6}{3}$

→$3 xy = 3xy -2x +9y -6$

→$3xy - 3xy +2x -9y =-6$

→$2x -9y = -6 ------1)$

• Case :- 2

→$xy = ( x -2) \left( y + \dfrac{40}{60}\right)$

→$xy = (x-2) \left( y + \dfrac{2}{3}\right)$

→$xy = (x-2) \left( \dfrac{3y+2}{3}\right)$

→$xy = \dfrac{3xy +2x -6y -4}{3}$

→$3xy = 3xy +2x -6y -4$

→$3xy -3xy -2x +6y = -4$

→$-2x +6y = -4----2)$

• Adding equation.1 and equation 2.

→$(2x -9y ) + (-2x +6y) = -6 -4$

→$2x -2x -9y +6y = -10$

→$-3y = -10$

→$y = \dfrac{10}{3}hr$

• Put the value of y in eq.1

→$2x - 9 \times \dfrac{10}{3}= -6$

→$2x -30 = -6$

→$2x = 24$

→$x = 12 km/hr$

Distance travelled is given by = xy

→$S = 12 \times \dfrac{10}{3}$

→$S = 4 \times 10$

→$S = 40 km$

hence,

The distance travelled will be 40 km.