Speed Time And Distance - Study Mode

[#161] When do the two hands of a clock of just after 3 pm make 30°angel between them?
Correct Answer

(B) 3:10:54

Explanation

Solution: $$eqalign{
& frac{{{{left( {90 - 30}
ight)}^ ext{o}}}}{{5.5}} cr
& = frac{{left( {60 imes 2}
ight)}}{{11}} cr
& = 10left( {frac{{10}}{{11}}}
ight) cr
& = 10,min ,54,sec cr
& { ext{Thus}}, cr
& { ext{Required}},{ ext{time}} cr
& = 3:10:54 cr} $$

[#162] Two boats go downstream from point X to Y. The faster boat covers the distance from X to Y, 1.5 times as fast as slower boat. It is known that for every hour slower boat lags behinds the faster boat by 8 km. however, if they go upstream, then the faster boat covers the distance from Y to X in half the time as the slower boat. Find the speed of the faster boat in still water?
Correct Answer

(B) 20 kmph

Explanation

Solution: Given, Speed of the faster boat Downstream = 1.5 × speed of the slower boat downstream ----------(1) [Difference in First hour ] Speed of the Faster Boat Downstream = Speed of the slower boat + 8 ------------- (2) Using Equation (1) and (2), we get Speed of the faster Boat Downstream = 16 kmph Now, $$frac{{{ ext{Time taken by the faster Boat}}}}{{{ ext{Time taken by the Slower boat Upstream}}}}$$ xa0 xa0 xa0 xa0= $$frac{1}{2}$$ Hence, Time taken by the faster Boat Upstream = 2 × Time taken by the slower Boat Upstream . . . . . . . (3) And, Faster boat's speed upstream - 8 = Slower boat's speed upstream . . . . . . . . (4)
Using (4) and (3), we get Speed of the faster Boat upstream = 8 kmph Thus, Speed of the faster Boat in still water = 20 kmph

[#163] A dog after traveling 50 km meets a swami who counsels him to go slower. He then proceeds at $$frac{3}{4}$$ of his former speed and arrives at his destination 35 min late. Had the meeting occurred 24 km further the dog would have reached its destination 25 min late. The speed of dog is:
Correct Answer

(A) 48 kmph

Explanation

Solution: He proceeds at $$frac{3}{4}$$ S where S is his usual speed means $$frac{1}{4}$$ decrease in the speed which will lead to $$frac{1}{3}$$ increase in the time. Now the main difference comes in those 24km and the change in difference of time = 35 - 25 m = 10 m. ⇒ $$frac{1}{3}$$ × T = 10 where T is the time required to cover the distance of (74 - 50) = 24 km. T = 30 min = 0.5 hours. Speed of the dog = $$frac{{24}}{{0.5}}$$xa0 = 48 kmph.

[#164] A girl while walking diametrically across a semicircular playground, takes 3 minutes less than if she had kept walking round the circular path from A to B. If she walks 60 metres a minute, what is diameter of the play ground?
Correct Answer

(D) 315 m

Explanation

Solution: Let the radius be r, then difference in the distance, πr - 2r = r (π - 2) = r [(3.14) -2] = 60 × 3 2r = 315 m.
[πr = semi perimeter, 2r = diameter]

[#165] A man takes 50 minutes to cover a certain distance at a speed of 6 km/hr. If he walks with a speed of 10 km/hr, he covers the same distance in :
Correct Answer

(C) 30 min

Explanation

Solution: $$eqalign{
& { ext{Distance}} = { ext{Speed}} imes { ext{Time}} cr
& { ext{Distance}} = left( {6 imes frac{{50}}{{60}}}
ight){ ext{km}} cr
& { ext{Distance}} = 5{ ext{ km}} cr
& herefore { ext{ Required time}} cr
& = frac{{{ ext{Distance}}}}{{{ ext{Speed}}}} cr
& = frac{5}{{10}}{ ext{ hrs}} cr
& = frac{1}{2}{ ext{ hrs}} cr
& = { ext{30 min}} cr} $$