Speed Time And Distance - Study Mode

[#156] A minibus takes 6 hour less to cover 1680 km distance, if its speed is increased by 14 kmph ? What is the usual time of the minibus ?
Correct Answer

(D) 30 hours

Explanation

Solution: Let
Usual Speed = x kmph. Now, Speed of bus after increasing the speed = (x + 14)kmph. . . . . . . (1) A ____________1680km ____________B In first case,
Time taken to covered the distance 1680 km = $$frac{{1680}}{x}$$ . . . . . . . (2) In Second Case, Time Taken to covered the distance 1680 km = $$frac{{1680}}{{x + 14}}$$ Time difference = 6 Hours. So, $$eqalign{
& Rightarrow {frac{{1680}}{x} - frac{{1680}}{{ {x + 14} }}} = 6 cr
& Rightarrow 1680 {frac{{14}}{{ {{x^2} + 14x} }}} = 6 cr
& Rightarrow 280 imes 14 = {x^2} + 14x cr
& Rightarrow {x^2} + 70x - 56x - 3920 = 0 cr
& Rightarrow xleft( {x + 70}
ight) - 56left( {x + 70}
ight) = 0 cr
& Rightarrow x = - 70,56. cr
& { ext{hence,}},{ ext{speed}},{ ext{of}},{ ext{minibus}},{ ext{is}},56,km/h cr
& { ext{put}},x = 56,{ ext{in}},{ ext{equation}},(2) cr
& T = frac{{1680}}{{56}} cr
& ,,,,,, = 30,{ ext{hours}} cr} $$

[#157] A man reduces his speed from 20 kmph to 18 kmph. So, he takes 10 minutes more than the normal time. what is the distance traveled by him.
Correct Answer

(A) 30 km

Explanation

Solution: Usual time = 9 × 10 = 90 min = $$frac{3}{2}$$ h Thus, Distance traveled = Speed × time = $$20 imes frac{3}{2}$$xa0 = 30 km. Alternatively, As the speed decreases from 20 kmph to 18 kmph i.e. 10 % increment in usual time. 10% = 10 min 100% = 100 min. Now, Distance traveled by him, = $$frac{{100}}{{60}} imes 18$$ = 30 km.

[#158] A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speed is :
Correct Answer

(B) 4 : 1

Explanation

Solution: Let the original speed be S 1 and time t 1 and distance be D. Now, $$eqalign{
& frac{{ {frac{D}{2}} }}{{2{t_1}}} = {S_2} cr
& {S_2} = frac{D}{{4{t_1}}},{ ext{and}},{S_1} = frac{D}{{{t_1}}} cr
& { ext{Thus}}, cr
& frac{{{S_1}}}{{{S_2}}} = frac{4}{1} = 4:1 cr} $$

[#159] A is twice fast as B and B is thrice fast as C. The journey covered by C in 78 minutes will be covered by A in :
Correct Answer

(A) 13 min

Explanation

Solution: The ratio of speeds of A, B, C = 6 : 3 : 1
The ratio of time taken by A, B, C = $$frac{1}{6}$$ : $$frac{1}{3}$$ : 1 To simplify it, we will multiply it by LCM of ratio of speeds given. Hence, the ratio of time taken by A, B, C = 1 : 2 : 6
[Speed is inversely proportional to time, means if speed increase time decreases. So, ratio of time would be reciprocal of the ratio of speed given. ] Time taken by C to covered given distance = 78 = 6 × 13 The ratio of time of A and C = 1 : 6 Thus, time taken by A = 13 min.

[#160] Vinay and Versha run a race with their speed in the ratio of 5 : 3. They prefer to run on a circular track of circumference 1.5 km. What is the distance covered by Vinay when he passes Versha for the seventh time?
Correct Answer

(B) 26.25 km

Explanation

Solution: Since, the speeds of Vinay and Versha are in the ratio 5 : 3 i.e. when Vinay covers 5 rounds, then Versa covers 3 rounds, but first time Vinay and Versha meet when Vinay completes $$left{ {2frac{1}{2} = 2.5}
ight}$$ xa0 round and Versha completes $$frac{1}{2}$$ round.
For Vinay to pass Versha seventh time, Vinay would have completed,
= 7 × 2.5 rounds
Since, each round is 1.5 km, the distance covered by Vinay is,
= 7 × 2.5 × 1.5 = 26.25 km.