Speed Time And Distance - Study Mode

[#146] A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities ?
Correct Answer

(A) 420 km

Explanation

Solution: Let distance = x km and usual rate y kmph Then, $$eqalign{
& Rightarrow frac{x}{y} - frac{x}{{y + 10}} = 1 cr
& Rightarrow yleft( {y + 10}
ight) = 10x.....(i) cr} $$ And, $$eqalign{
& Rightarrow frac{x}{y} - frac{x}{{y + 20}} = frac{7}{4} cr
& Rightarrow yleft( {y + 20}
ight) = frac{{80x}}{7}.....(ii) cr} $$ On dividing (i) by (ii) we get : y = 60 Substituting y = 60 in (i), we get: x = 420 km

[#147] Amit travelled black to home in a car, after visiting his friend in a distant village. When he stated at his friend's house the car had exactly 18 litres of petrol in it. He travelled along at a steady 40 kilometres per hour and managed a 10 kilometres per litre of petrol. As the car was old, the fuel tank lost fuel at the rate of half a litre per hour. Amit was lucky as his car stopped just in front of his home because it had rum out of fuel and he only just made it. How far was it from his friend's home to Amit's home ?
Correct Answer

(D) None of these

Explanation

Solution: Quantity of petrol consumed in 1 hour: $$eqalign{
& = left( {frac{{40}}{{10}} + frac{1}{2}}
ight){ ext{ litres}} cr
& = 4frac{1}{2}{ ext{ litres}} cr} $$ Time for which the fuel lasted : $$eqalign{
& = left[ {frac{{18}}{{left( {4frac{1}{2}}
ight)}}}
ight]{ ext{ hrs}} cr
& = left( {18 imes frac{2}{9}}
ight){ ext{ hrs}} cr
& = 4{ ext{ hrs}} cr} $$ ∴ Required distance = (40 × 4) km = 160 km

[#148] A distance of 425 km separates two trains moving towards each other at a speed of 200 km/hr each. What will be the distance between them after 1 hr 30 min, if they reduce their speed by half, every half an hour ?
Correct Answer

(A) 75 km

Explanation

Solution: Relative speed = (200 + 200) km/hr = 400 km/hr Distance covered in 1 hr 30 min $$eqalign{
& = left( {400 imes frac{1}{2} + 200 imes frac{1}{2} + 100 imes frac{1}{2}}
ight){ ext{km}} cr
& = left( {200 + 100 + 50}
ight){ ext{km}} cr
& = 350{ ext{ km}} cr} $$ [$$x08ecause $$ Speed reduces by half every half an hour] Hence, distance between the trains : = (425 - 350) km = 75 km

[#149] Ashok left from place A for place B at 8 am and Rahul left place B for place A at 10.00 am the distance between place A and B is 637 km. If Ashok and Rahul are travelling at a uniform speed of 39 kmph and 47 kmph respectively, at what time will they meet ?
Correct Answer

(B) 4.30 pm

Explanation

Solution: Speed of Ashok = 39 km/ph Speed of Rahul = 47 km/ph Distance between place A and place B = 637 km Distance covered by Ashok (from 8 am to 10 am) in 2 hours = 2 × 39 = 78 km ∴ Remaining distance = 637 - 78 = 559 Relative speed = 39 + 47 = 86 km/ph ∴ Time taken to travel 559 km = $$frac{559}{86}$$ = 6.5 hours So, they meet at = (10 am + 6.5 hrs) = 4.30 pm

[#150] A boy is running at a speed of p kmph to cover a distance of 1 km. But, due to the slippery ground, his speed is reduced by q kmph (p > q). If he takes r hours to cover the distance, then:
Correct Answer

(A) $$frac{1}{r} = p - q$$

Explanation

Solution: $${ ext{Speed }} = frac{{{ ext{Distance }}}}{{{ ext{Time}}}}$$ xa0xa0 or $$frac{{{ ext{Distance }}}}{{{ ext{Time}}}} = { ext{Speed}}$$ ⇒ $$frac{1}{r} = p - q$$