Speed Time And Distance - Study Mode

[#131] Rahul and Mithun travel a distance of 30 km. The sum of their speeds is 70 km/h and the total time taken by both to travel the distance is 2 hours 6 minutes. The difference between their speeds is:
Correct Answer

(A) 30 km/h

Explanation

Solution: $$eqalign{
& x + y = 70 cr
& frac{{30}}{x} + frac{{30}}{y} = frac{{21}}{{10}} cr
& 30left[ {frac{{x + y}}{{xy}}}
ight] = frac{{21}}{{10}} cr
& frac{{30 imes 70}}{{xy}} = frac{{21}}{{10}} cr
& xy = 1000 cr
& {left( {x - y}
ight)^2} = {left( {x + y}
ight)^2} - 4xy cr
& {left( {x - y}
ight)^2} = 4900 - 4000 cr
& {left( {x - y}
ight)^2} = 900 cr
& left( {x - y}
ight) = 30,{ ext{km/hr}} cr} $$

[#132] The distance between two places A and B is 140 km. Two cars x and y start simultaneously from A and B respectively. If they move in the same direction, they meet after 7 hours. If they move towards each other, they meet after one hour. What is the speed (in km/h) or car y if its speed is more than that or car x?
Correct Answer

(B) 80

Explanation

Solution: Distance = 140 km y > x Same direction: y - x = $$frac{{140}}{7}$$ = 20 . . . . . .(i) Opposite direction: y + x = $$frac{{140}}{1}$$ = 140 . . . . . . (ii) Add equation (i) and (ii) 2y = 160 y = 80 km/hr

[#133] A car covered 150 km in 5 hours. If it travels at one-third its usual speed, then how much more time will it take to cover the same distance?
Correct Answer

(B) 10 hours

Explanation

Solution: $$eqalign{
& underleftrightarrow {,,,,,,,,,,150{ ext{ km}},,,,,,,,,,} cr
& { ext{Initial speed}} = frac{{150}}{5} = 30{ ext{ km/h}} cr
& { ext{New speed}} = frac{1}{3} imes 30 = 10{ ext{ km/h}} cr
& { ext{Time taken to cover 150 km}} = frac{{150}}{{10}} = 15{ ext{ h}} cr
& { ext{Time difference}} = 15 - 5 = 10{ ext{ hour}} cr} $$

[#134] Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minutes late. His usual time (in hours) to reach the destination is:
Correct Answer

(B) $$2frac{1}{2}$$

Explanation

Solution: $$60\% = frac{3}{5}$$ Let the speed of the man be 5x 60% of the speed = 5x × $$frac{3}{5}$$ = 3x Ratio of speed of man before and after = 5x : 3x As we know, speed is inversely proportional to time. Time ratio of man before and after = 3x : 5x According to the question 5x - 3x = 1 hr 40 min 5x - 3x = $$left( {1 + frac{{40}}{{60}}}
ight){ ext{hr}}$$ 2x = $$frac{5}{3}$$ x = $$frac{5}{{3 imes 2}}$$ x = $$frac{5}{6}$$ hr Required time = 3x = 3 × $$frac{5}{6}$$ = $$2frac{1}{2}$$ hr

[#135] A train without stoppage travels with an average speed of 65 km/h and with stoppage, it travels with an average speed of 52 km/h. For how many minutes does the train stop on an average per hour?
Correct Answer

(C) 12

Explanation

Solution: The distance of 13 km covered by the speed of 65 km/hr in time $$eqalign{
& o frac{{13}}{{65}} = frac{1}{5} imes 60 = 12,{ ext{min/hr}} cr
& cr
& {x08f{Alternate}},{x08f{solution:}} cr
& left( {frac{{{ ext{Faster speed}} - { ext{Slower speed}}}}{{{ ext{Faster speed}}}} imes 60}
ight){ ext{min/hr}} cr
& = frac{{65 - 52}}{{65}} imes 60 cr
& = 12,{ ext{min/hr}} cr} $$