Speed Time And Distance - Study Mode

[#206] A bus moving at a speed of 45 km/hr catches a truck 150 metres ahead going in the same direction in 30 seconds. The speed of the truck is :
Correct Answer

(A) 27 km/hr

Explanation

Solution: Let the speed of truck is = x km/hr Their relative speed in same direction = (45 - x) km/hr [Here (45 - x) has been written because bus crosses the truck which is running 150 metres ahead of it. i.e., Truck speed will be lower than that of bus] According to the question, $$eqalign{
& { ext{Time}} = frac{{{ ext{Distance}}}}{{{ ext{Speed}}}} cr
& Rightarrow frac{{150}}{{left( {45 - x}
ight) imes frac{5}{{18}}}} = 30 cr
& Rightarrow frac{{150 imes 18}}{{left( {45 - x}
ight) imes 5}} = 30 cr
& Rightarrow x = 27{ ext{ km/hr}} cr} $$

[#207] A bus covers three successive 3 km stretches at speed of 10 km/hr, 20 km/hr and 60 km/hr respectively. Its average speed over this distance is :
Correct Answer

(C) 18 km/hr

Explanation

Solution: $$eqalign{
& { ext{Average speed }} = frac{{{ ext{Total distance }}}}{{{ ext{Total time}}}} cr
& { ext{Average speed}} = frac{{60 imes 3}}{{frac{{60}}{{10}} + frac{{60}}{{20}} + frac{{60}}{{60}}}} cr
& { ext{Average speed}} = frac{{180{ ext{ km}}}}{{left( {6 + 3 + 1}
ight){ ext{ hr}}}} cr
& { ext{Average speed}} = frac{{180{ ext{ km}}}}{{{ ext{10 hr}}}} cr
& { ext{Average speed}} = 18{ ext{ km/hr}} cr} $$ Short trick formula : $$left( {{ ext{Average speed}} = frac{{3xyz}}{{xy + yz + zx}}}
ight)$$

[#208] A man walks a certain distance in certain time, if he had gone 3 km per hour faster, he would have taken 1 hour less than the scheduled time. If he had gone 2 km per hour slower, he would have taken one hour longer on the road. The distance (in km) is :
Correct Answer

(A) 60

Explanation

Solution: Let the speed = x km/hr And, time = y hr According to the question, x × y = (x + 3) (y - 1) xy = xy + 3y - x - 3 x - 3y = - 3 ..... (i) Also, x × y = (x - 2)(y + 1) xy = xy - 2y + x - 2 x - 2y = 2 ..... (ii) Solve equation (i) and (ii) x = 12 y = 5 Distance = Speed × Time Distance = 12 × 5 = 60 km

[#209] A man travels the distance of his journey $$frac{3}{4}$$ by bus, $$frac{1}{6}$$ by rickshaw and remaining 2 km on foot. The total distance travelled by the man is :
Correct Answer

(D) 24 km

Explanation

Solution: Let the man travels 1 unit distance So, remaining distance $$eqalign{
& = 1 - left( {frac{1}{6} + frac{3}{4}}
ight) cr
& = 1 - frac{{22}}{{24}} cr
& = frac{1}{{12}} cr
& x08ecause frac{1}{{12}}{ ext{ unit}} = { ext{2 km}} cr
& { ext{So, 1 unit}} = 24{ ext{ km}} cr} $$

[#210] A train passes two persons walking in the same direction at a speed of 3 km/hr and 5 km/hr respectively in 10 seconds and 11 seconds respectively. The speed of the train is :
Correct Answer

(C) 25 km/hr

Explanation

Solution: Let the train's speed = x km/hr $$x08ecause ,$$ When train will cross a man then it covers only its length. $$eqalign{
& herefore left( {x - 3}
ight) imes frac{{10}}{{60}} = left( {x - 5}
ight) imes frac{{11}}{{60}} cr
& x = 25{ ext{ km/hr}} cr} $$ Note : Products of time × speed is always subtracted if both the men are running in the same direction and the products of time × speed is added only if the men are running opposite direction. ⇒ Here train's direction is not considered. But attention please ⇒ Always divided by the difference of time.