Speed Time And Distance - Study Mode

[#261] A train travelling at the speed of x km/h crossed a 300 m long platform in 30 second, and overtook a man walking in the same direction at 6 km/h in 20 seconds. What is the value of x?
Correct Answer

(B) 96

Explanation

Solution: $$eqalign{
& { ext{Train speed}} = x{ ext{ km/h}} cr
& { ext{Length of train}} = L cr
& { ext{Length of platform}} = 300,{ ext{m}} cr
& { ext{Man's speed}} = 6{ ext{ km/h}} cr
& left( {x - 6}
ight) imes frac{5}{{18}} = frac{L}{{20}} cr
& 100x - 600 = 18L{ ext{ }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}left( { ext{i}}
ight) cr
& x imes frac{5}{{18}} = frac{{L + 300}}{{30}} cr
& 150x = 18L + 5400 cr
& 150x - 5400 = 18L{ ext{ }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}left( {{ ext{ii}}}
ight) cr
& { ext{Equation }}left( { ext{i}}
ight),& ,left( {{ ext{ii}}}
ight) cr
& 100x - 600 = 150x - 5400 cr
& 50x = 4800 cr
& x = 96{ ext{ km/ h}} cr} $$

[#262] A man walks from point X to Y at a speed of 20 km/h, but comes back from point Y to X at a speed of 25 km/h. Find his average speed.
Correct Answer

(A) $$22frac{2}{9},{ ext{km/h}}$$
(E) $$22frac{2}{9},{ ext{km/h}}$$

Explanation

Solution: $$eqalign{
& { ext{Average speed}} = frac{{2 imes 20 imes 25}}{{20 + 25}} cr
& = frac{{2 imes 20 imes 25}}{{45}} cr
& = frac{{200}}{9} cr
& = 22frac{2}{9},{ ext{km/h}} cr} $$

[#263] The speed of a bus is 72 kmph. The distance covered by the bus in 5 sec is :
Correct Answer

(C) 100 metres

Explanation

Solution: Speed of bus : = 72 km/hr = $$left( {frac{{72 imes 5}}{{18}}}
ight)$$ xa0m/sec = 20 m/sec Let distance covered by bus in 5 sec be x ∴ Distance = Speed × Time ⇒ x = 20 × 5 ⇒ x = 100 metres

[#264] Deepa rides her bike at an average speed of 30 km/hr and reaches her destination in 6 hours. Hema covers the same distance in 4 hours. If Deepa increases her average speed by 10 km/hr and Hema increases her average speed by 5 km/hr, what would be the difference in their time taken to reach the destination ?
Correct Answer

(C) 54 min

Explanation

Solution: Deepa's original speed = 30 km/hr Deepa's new speed : = (30 + 10) km/hr = 40 km/hr Distance = (30 × 6) km = 180 km Hema's original speed : = $$frac{180}{4}$$ km/hr = 45 km/hr Hema's new speed : = (45 + 5) km/hr = 50 km/hr Difference in time : $$eqalign{
& = left( {frac{{180}}{{40}} - frac{{180}}{{50}}}
ight){ ext{ hrs}} cr
& = frac{9}{{10}}{ ext{ hrs}} cr
& = left( {frac{9}{{10}} imes 60}
ight){ ext{ min}} cr
& = 54{ ext{ min}} cr} $$

[#265] A takes 2 hours more than B to walk d km, but if A doubles his speed, then he can make it in 1 hour less than B. How much times does B required for walking d km ?
Correct Answer

(C) 4 hours

Explanation

Solution: Suppose B takes x hours to walk d km Then, A takes (x + 2) hours to walk d km A's speed = $$left( {frac{d}{{x + 2}}}
ight)$$xa0 km/hr and B's speed = $$left( {frac{d}{{x}}}
ight)$$xa0 km/hr A's new speed = $$left( {frac{2d}{{x + 2}}}
ight)$$xa0 km/hr $$eqalign{
& herefore frac{d}{{left( {frac{d}{x}}
ight)}} - frac{d}{{left( {frac{{2d}}{{x + 2}}}
ight)}} = 1 cr
& Leftrightarrow x - left( {frac{{x + 2}}{2}}
ight) = 1 cr
& Leftrightarrow x - 2 = 2 cr
& Leftrightarrow x = 4 ext{ hours} cr} $$