We know that, if two equal distances are covered at two different speeds A and B, then
Average speed = 2AB/(A + B)

Correct Option: A

We know that, if two equal distances are covered at two different speeds A and B, then
Average speed = 2AB/(A + B) = (2 x 58 x 52)/(58 + 52)
= 54.8 km/h
= 55 km/h (Approx)

Walking time + Cycling time = 5 h 45 min = 345 min ....(i)
If he had cycled both way he would have gained 2 h (120 min).
∴ 2 x Cycling time = 345 - 120 = 225 min ...(ii)
Walking time = 2 x 345 - 225 = 690 - 225 = 465 min

Correct Option: B

Walking time + Cycling time = 5 h 45 min = 345 min ....(i)
If he had cycled both way he would have gained 2 h (120 min).
∴ 2 x Cycling time = 345 - 120 = 225 min ...(ii)
Walking time = 2 x 345 - 225 = 690 - 225 = 465 min
Time taken by him to walk both ways = 7 h 45 min

Let w be the time taken in one way by walking and c be the time taken in one way by car.
Now, according to the question,
In first case, w + c = 6 h 45 min
or
2w + 2c = 13 h 30 min ..(i)
In second case, 2c = 4 h 45 min ....(ii)
From Eqs. (i) and (ii), we get
2w + 2c = 13 h 30 min

Correct Option: D

Let w be the time taken in one way by walking and c be the time taken in one way by car.
Now, according to the question,
In first case, w + c = 6 h 45 min
or
2w + 2c = 13 h 30 min ..(i)
In second case, 2c = 4 h 45 min ....(ii)
From Eqs. (i) and (ii), we get
2w + 2c = 13 h 30 min
⇒ 2w + 4h 45 min = 13 h 30 min
⇒ 2w = 13 h 30 min - 4 h 45 min
⇒ 2w = 8 h 45 min
∴ If he walks both ways, then time taken = 8 h 45 min

Since, the speed of X and Y are 60 km/h each. The distance between A and B is 700 km. Distance traveled by X upto 11 am is 60 km. Since, X stops at 10 am and distance traveled by up to 11 am is 120 km.
Now, the distance between them = 700 - 180 = 520 km
Now, let they will meet at a distance of x km from X's position.
∴ Distance traveled by X = x km
and distance traveled by Y = (520 - x) km
∴ x/60 = (520 - x)/60

Correct Option: B

Since, the speed of X and Y are 60 km/h each. The distance between A and B is 700 km. Distance traveled by X upto 11 am is 60 km. Since, X stops at 10 am and distance traveled by up to 11 am is 120 km.
Now, the distance between them = 700 - 180 = 520 km
Now, let they will meet at a distance of x km from X's position.
∴ Distance traveled by X = x km
and distance traveled by Y = (520 - x) km
∴ x/60 = (520 - x)/60
⇒ 2x = 520
⇒ x = 260
∴ T = 260/60 = 4¹/₃ h
Thus, they will cross each other after, 4¹/₃ h after 11 am i.e., at 3 : 20 pm.

Let T h, he travels on foot.
∴ 4T + 10(8 - T) = 50

Correct Option: B

Let T h, he travels on foot.
∴ 4T + 10(8 - T) = 50
⇒ 80 - 6T = 50
⇒ 6T = 30,
∴ T = 5 h
∴ The distance travelled on foot = 4 x 5 = 20 km