Time and Distance - 03

Explanation:-

Downstream speed,
= ⁹/₂ = 4.5 kmph
Upstream speed,
= ⁹/₆ = 1.5 kmph
Thus, speed of boat in still water,
= ^{(4.5 +1.5)}/₂ = 3 kmph.
Speed of current = ^(4.5-1.5)/₂ = 1.5 km/h.

Explanation:-

Let downstream speed be D and Upstream speed be U. Then
(⁴⁸/_D + ⁷²/_U) = 12 -------- (1)
and
(⁴⁸/_D + ⁷²/_U) = 13 ------------ (2) On solving the eqn (1) and (2), we get
D = 12 kmph and
U = 8 kmph
Thus, speed of the current = ^(D-U)/₂ = ⁴/₂ = 2 kmph.

Explanation:-

Let the speed of the current be x km/h. Now, according to the question,
^(15+x)/_(15-x) = ²/₁.
15 + x = 30 - 2x
3x = 30 - 15
3x = 15
x = 5 km/h.

Explanation:-

Let the radius be r, then difference in the distance,
∏r - 2r = r (∏ - 2)
= r [(3.14) -2] = 60 *3
2r = 315 m. [∏r = semi perimeter, 2r = diameter]

Explanation:-

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,

Time taken by the faster Boat / Time taken by the Slower boat Upstream = 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

Explanation:-

$$\eqalign{ & \frac{{{{\left( {90 - 30} \right)}^0}}}{{5.5}} \cr & = \frac{{\left( {60 \times 2} \right)}}{{11}} \cr & = 10\left( {\frac{{10}}{{11}}} \right) \cr & = 10\,\min \,54\,\sec \cr & {\text{Thus}}, \cr & {\text{Required}}\,{\text{time}} \cr & = 3:10:54 \cr}$$

Explanation:-

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 {2(¹/₂)= 2.5} round and Versha completes 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.

Explanation:-

The ratio of speeds of A, B, C = 6 : 3 : 1.
The ratio of time taken by A, B, C = (¹/₆) : (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.

Explanation:-

Usual time = 9 * 10 = 90 min = ³/₂ h
Thus,
Distance traveled = Speed * time = 20 * (³/₂) = 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,
= (¹⁰⁰/₆₀ h) *18
= 30 km.

Explanation:-

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 = 1680/X ----------- (2)
In Second Case,
Time Taken to covered the distance 1680 km = 1680/(X + 14).
Time difference = 6 Hours.
So,
$$\eqalign{ & \Rightarrow \left[ {\frac{{1680}}{x} - \frac{{1680}}{{\left( {x + 14} \right)}}} \right] = 6 \cr & \Rightarrow 1680\left[ {\frac{{14}}{{\left( {{x^2} + 14x} \right)}}} \right] = 6 \cr & \Rightarrow 280 \times 14 = {x^2} + 14x \cr & \Rightarrow {x^2} + 70x - 56x - 3920 = 0 \cr & \Rightarrow x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & \Rightarrow x = - 70,56. \cr & {\text{hence,}}\,{\text{speed}}\,{\text{of}}\,{\text{minibus}}\,{\text{is}}\,56\,km/h \cr & {\text{put}}\,x = 56\,{\text{in}}\,{\text{equation}}\,(2) \cr & T = \frac{{1680}}{{56}} \cr & \,\,\,\,\,\, = 30\,{\text{hours}} \cr}$$