We can use the formula for when two people moving towards each other and meeting in a straight line.&nbsp;$t^2=t_1\times\ t_2$Where $t$ is time taken for them to meet each other.&nbsp;$t_1$ is the time taken by person 1 to reach the destination after meeting and&nbsp;$t_2$&nbsp;is the time taken by person 2 to reach the destination after meetingWe are told they meet each other in 1 and a half hours, that is 90 minutes.&nbsp;And if Sunil takes X minutes to reach A, Anil will take X+75 minutesSince it is given that,&nbsp;Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A$90^2=x\left(x+75\right)$$8100=x^2+75x$$x^2+75x-8100=0$$\dfrac{\left(-75\pm\sqrt{5625+32400}\right)}{2}$$\dfrac{\left(-75+195\right)}{2}$$x=60$Total time of travel of Anil is,&nbsp;$90\ \min\ +60\ \min\ +75\ \min$Total of 225 minutes.&nbsp;Time in hours will be 3.75 hours.&nbsp;Speed is&nbsp;$\dfrac{45}{3.75}=12\ \dfrac{km}{hr}$

Let's take the scheduled time taken by the bus to be tFrom the first statement (bus travelling at 60 kmph), we can get the total distance travelled by bus to 60(t+3.5)The second scenario gives us that the bus covered two-thirds of the distance in one-third of the time, meaning that the remaining one-third distance was covered in two-thirds of the time, giving us the relation&nbsp;$\frac{1}{3}st$ covered in&nbsp;$\frac{2}{3}t$ giving the speed to be&nbsp;$\frac{s}{2}$ which is given as 40 km/h, thereby giving the usual speed of the bus to be 80 km/hrNow the first relation we get 60(t+3.5)=80tGiving us t=10.5 hoursThus, the bus usually takes 10.5 hours on its journey.&nbsp;Staring at 9:00, it will complete the journey at 7:30 pmTherefore, Option A is the correct answer.&nbsp;

Let us assume that the distance is D, speed of the train is S and time taken by the train is t.&nbsp;t is nothing but&nbsp;$\frac{D}{S}$Statement 1:&nbsp;Had the speed been 6 km per hour more, it would have needed 4 hours less$\frac{D}{S+6}=t-4$ $\frac{D}{S+6}=\frac{D}{S}-4$ $4=\frac{D}{S}-\frac{D}{S+6}$ $\frac{S+6-S}{S\left(S+6\right)}=\frac{4}{D}$ $\frac{6}{S\left(S+6\right)}=\frac{4}{D}$ $D=\frac{2S\left(S+6\right)}{3}$Statement 2:&nbsp;Had the speed been 6 km per hour less, it would have needed 6 hours more$\frac{D}{S-6}=t+6$ $D\left[\frac{1}{S-6}-\frac{1}{S}\right]=6$ $\frac{S-S+6}{S\left(S-6\right)}=6$ $D=S\left(S-6\right)$Equating the two equations for distance,&nbsp;$S\left(S-6\right)=\frac{2S\left(S+6\right)}{3}$ $3S-18=2S+12$ $S=30$Hence the speed is 30 kmphWe know that the distance D = S(S - 6) = 30*24 = 720km

CAT Time, Speed and Distance

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CAT Time, Speed and Distance

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<div><p>A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is</p></div>

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