Question

A bus covers a distance of 120km in 4 hours with an uniform speed what distance will it cover 2 hours with the same speed?​

Answer 1

Answer:

According to the given problem,

A bus covers a certain distance at a speed of 120 km/hr in 5 hrs.

Let S denotes the speed at which the bus must travel in order to cover the same distance in [5 (1/2) =] 11/2 hrs.

Hence we get the following relation,

S*(11/2) = 120*5

or S = 1200/11 = 109 (1/11) = 109.09 (km/hr) [Ans]

Answer 2

HERE IS UR ANSWER MATE!

ANSWER

Uniform speed =x km/h

Uniform speed =x km/hdistance =240 km

Uniform speed =x km/hdistance =240 kmtime =

Uniform speed =x km/hdistance =240 kmtime = Speed

Uniform speed =x km/hdistance =240 kmtime = SpeedDistance

Uniform speed =x km/hdistance =240 kmtime = SpeedDistanceDue to heavy rain speed =(x−10) km/h

Uniform speed =x km/hdistance =240 kmtime = SpeedDistanceDue to heavy rain speed =(x−10) km/hNow time =

Uniform speed =x km/hdistance =240 kmtime = SpeedDistanceDue to heavy rain speed =(x−10) km/hNow time = x−10

Uniform speed =x km/hdistance =240 kmtime = SpeedDistanceDue to heavy rain speed =(x−10) km/hNow time = x−10240 hours

Uniform speed =x km/hdistance =240 kmtime = SpeedDistanceDue to heavy rain speed =(x−10) km/hNow time = x−10240 hoursAccording to question

Uniform speed =x km/hdistance =240 kmtime = SpeedDistanceDue to heavy rain speed =(x−10) km/hNow time = x−10240 hoursAccording to questionx−10

Uniform speed =x km/hdistance =240 kmtime = SpeedDistanceDue to heavy rain speed =(x−10) km/hNow time = x−10240 hoursAccording to questionx−10240x

x240

x240 =2

x240 =2x(x−10)

x240 =2x(x−10)240x−240(x−10) =2

x240 =2x(x−10)240x−240(x−10) =2x

x240 =2x(x−10)240x−240(x−10) =2x 2

x240 =2x(x−10)240x−240(x−10) =2x 2 −10x

x240 =2x(x−10)240x−240(x−10) =2x 2 −10x240x−240x+2400 =2

⇒2x

⇒2x 2

⇒2x 2 −20x−2400=0

⇒2x 2 −20x−2400=0⇒x

⇒2x 2 −20x−2400=0⇒x 2

⇒2x 2 −20x−2400=0⇒x 2 −10x−1200=0

⇒2x 2 −20x−2400=0⇒x 2 −10x−1200=0x

⇒2x 2 −20x−2400=0⇒x 2 −10x−1200=0x 2

⇒2x 2 −20x−2400=0⇒x 2 −10x−1200=0x 2 −40x+30x−1200=0

x(x−40)+30(x−40)=0

x(x−40)+30(x−40)=0⇒(x−40)(x+30)=0

x(x−40)+30(x−40)=0⇒(x−40)(x+30)=0x=40 and x=−30 (speed can not be negative)

x(x−40)+30(x−40)=0⇒(x−40)(x+30)=0x=40 and x=−30 (speed can not be negative)∴ uniform speed =40 km/h.