Speed Time And Distance - Study Mode

[#241] A student walks from his house at a speed of $$2frac{1}{2}$$ km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hours and reaches 6 minutes before school time. How far is the school from his house ?
Correct Answer

(B) $$frac{7}{4}$$ km

Explanation

Solution: Difference between his reaching time : = (14 - 10) hrs = 4 hrs = 4 hrs → 6m + 6m (late + before) = 4 hrs → 12 minutes = 1 unit = $$frac{12}{4 × 60}$$ xa0 km ($$x08ecause $$ 1 m = 60 seconds) 1 unit = $$frac{1}{20}$$ km Then, 35 units : = 35 × $$frac{1}{20}$$ km = $$frac{7}{4}$$ km Then the distance between his house and school is = $$frac{7}{4}$$ km

[#242] A train 150 metres long takes 20 seconds to cross a platform 450 metres long. The speed of the train in, km per hour is :
Correct Answer

(C) 108 km/hr

Explanation

Solution: $$eqalign{
& { ext{Speed of train :}} cr
& = frac{{450 + 150}}{{20}} cr
& = 30{ ext{ m/s}} cr
& { ext{Speed (in km/hr):}} cr
& = 30 imes frac{{18}}{5} cr
& = 108{ ext{ km/hr}} cr} $$

[#243] Mr A started half an hour later than usual for the marketplace. But by increasing his speed to $$frac{3}{2}$$ times his usual speed he reached 10 minutes earlier than usual. What is his usual time for this journey?
Correct Answer

(B) 2 hours

Explanation

Solution: Let the usual speed be S.
So, increased speed = $$frac{{3{ ext{S}}}}{2}$$ Let usual time taken be T Thus, S × T = D, and also $$frac{{3{ ext{S}}}}{2}$$ × (T - 40) = D. (T - 40, Because he started 30 minutes later and reached 10 minutes earlier, means he saved 40 minutes) $$frac{{3{ ext{S}}}}{2}$$ × (T - 40) = ST $$frac{{3 imes { ext{ST}}}}{2}$$xa0 - 60S = ST $$frac{1}{2}$$ × ST = 60S T = 120 minutes = 2 hours Alternatively, As, ST = D [S ∝ $$frac{1}{{ ext{T}}}$$ , speed is inversely proportional to time] Speed increase to $$frac{{3}}{{2}}$$ times so time decreases to $$frac{{2}}{{3}}$$ times . Net decrease = $$frac{{1}}{{3}}$$ = 40 minutes. Hence actual time taken = 3 × 40 = 120 mnts. = 2 hours

[#244] Page and Plant are running on a track AB of length 10 metres. They start running simultaneously from the ends A and B respectively. The moment they reach either of the ends, they turn around and continue running. Page and Plant run with constant speeds of 2m/s and 5m/s respectively. How far from A (in metres) are they, when they meet for the 23 rd time?
Correct Answer

(A) 0 meters

Explanation

Solution: The ratio of speeds is 2 : 5. So when slower one completes 2 one way journeys (and reaches its starting point), faster one travels 5 one way journeys (and reaches the other end). So that means after the faster one has traveled 5 one way journeys, both of them have reached same end and in next 5 one-way journey of faster runner, both reach their starting position simultaneously. Now most important to observe is that FASTER one will always meet the SLOWER one EXACTLY ONCE in each of its one-way journey, except when both of them have started with the same starting point.
Once you reduce that for the 5 th time, they'll meet at A, and the next 5 rounds will have 4 meetings. Then, just add 5 + 4 + 5 + 4 + 5 = 23 or just go by the position of Plant after every 5 rounds : A, B, A, B, A. The cycle repeats. So, total distance will be 0 meters.

[#245] Two cars start simultaneously from cities A and B, towards B and A respectively, on the same route. Once the two cars reach their destinations they turned around and move towards the other city without any loss of time. The two cars continue shuttling in this manner for exactly 20 hours. If the speed of the car starting from A is 60km/hr and the speed of the car starting from B is 40km/hr and the distance between the two cities is 120 km, find the number of times the two cars cross each other?
Correct Answer

(A) 8

Explanation

Solution: Suppose both moves with the same speed 60 km/h then they will meet max 10 times. So answer will be less than 10. Thus correct answer will be 8.