Time and Distance - 02

Explanation:-

Ratio of speed of Bus and Train = 15 : 27.
Let speed of the bus is 15X and Speed of the Train is 27X.
Car Covered 720 km in 9 hours.
So, Speed of the Car = ⁷²⁰/₉ = 80 kmph.
Given, Speed of the bus is ³/₄ of Car, So speed of the Bus,
= ^(80*3)/₄ = 60 kmph.
Thus,
15X = 60
X = 4
So, Speed of the train = 27X = 27*4 = 108 kmph.
Hence, Train will cover distance in 7 hours,
= 108 * 7 = 756 km.

Explanation:-

P__150m___T______X m (Let)_______Q.
Let Policeman caught thief at a distance (X + 150)m. And Thief has traveled X m.
Speed of Policeman = 12 kmph = ^(12*5)/₁₈ = ⁶⁰/₁₈ m/sec.
Speed of thief = 10 km = ^(10*5)/₁₈ = ⁵⁰/₁₈ m/sec.
In this case time is constant means Policeman covered (X + 150)m in same time thief covered X m.
Thus,
Speed of the thief /Speed of Policeman = ^X/_(150+X)
⁵⁰/₆₀ = ^X/_(150+X)
7500 + 50X = 60X
10X = 7500
X = 750 m.
So, Thief has traveled 750 m before the caught.

Explanation:-

Let usual speed be X kmph, then new speed will be (X+400)kmph.
Time taken to cover 1600 km with speed X kmph,
= ¹⁶⁰⁰/_X
Time taken to cover 1600 km with Speed (X+400) kmph,
= ¹⁶⁰⁰/_(X+400)
Now,
Time difference = 40 minutes.
(¹⁶⁰⁰/_X)-(¹⁶⁰⁰/_(X+400)) = ⁴⁰/₆₀ hours.
X² +400X - 960000 = 0 On solving,
X = -1200, 800.
Speed cannot be negative, So usual speed will be 800 km per hour.

Explanation:-

let required speed be S km/h.
50*30 = S*20
S = 75 km/h.

Explanation:-

Let speed of the cat be S km/h.
Time taken = Distanced advanced / relative speed
2 = ^{2 * S}/_(30-S)
S = 15 km/h.

Explanation:-

Let length of each train be X m, then $$\eqalign{ & {S_1} = \frac{X}{4}\,{\text{and}} \cr & {S_2} = \frac{X}{5} \cr & {\text{Required time to cross each other,}} \cr & = \left\{ {\frac{{2X}}{{\left[ {\left( {\frac{X}{4}} \right) + \left( {\frac{X}{5}} \right)} \right]}}} \right\} \cr & = \frac{{40}}{9}{\text{seconds}} \cr}$$

Explanation:-

Speed of sound / relative speed of soldier and terrorist = Time taken / Difference in time
¹¹⁸⁸/_S = ³³⁰/₅
S = 18 km/h.

Explanation:-

Downstream (steamer) = 40 min
Downstream (boat) = 60 min
Upstream (steamer) = 60 min
Upstream (boat) = 90 min
Required time = 40+30+45 = 115 min.

Explanation:-

He proceeds at ⁴/₅ S where S is his usual speed means ¹/₅ decrease in speed which will lead to ¹/₄ increase in time. Now the main difference comes in those 12km (30-18) and the change in difference of time = (45-36) min = 9 min

Thus,
¹/₄ * T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km
T = 36 min = ³⁶/₆₀ hours = 0.6 hours.

Speed of the car = ¹²/_0.6 = 20 kmph.

Explanation:-

Let the total distance be 2X.

A_____X km_____M_____X km_____B

Total time taken in the journey = 17.5 hours
Time taken to cover X km at 30 km/h = ^X/₃₀.
Time taken to cover X km at 40 km/h = ^X/₄₀
Now, $$\eqalign{ & \left( {\frac{X}{{30}}} \right) + \left( {\frac{X}{{40}}} \right) = 17.5 \cr & \left[ {\frac{{\left( {40X + 30X} \right)}}{{1200}}} \right] = 17.5 \cr & 70X = 17.5 \times 1200 \cr & X = 300\,km \cr & {\text{Total}}\,{\text{distance}},\,2x \cr & = 600\,km \cr}$$