A Speedy One For The Road

• X = distance between A & B (miles)
• Time for Chip to meet Ash is $\frac { X }{ 120+80 }$ (hour)
• Time for Chip to meet Bob is $\frac { X }{ 120+60 }$ (hour)
• Time from Chip's passing by Bob to Chip's passing by Ash is $\frac {2}{60}$ = $\frac {1}{30}$ (hour)

Solving equation of $\frac { X }{ 120+60 } - \frac { X }{ 120+80 } = \frac {1}{30}$ gives $X = 60$ (miles)

Let the initial time be t=0

Bob meets Chip after t hr. so the distence between two town is (120+80) * t = 200 * t

when Bob meets Chip, at that time distance between Ash and Bob is (80-60) * t = 20 * t

This distance is covered by Ash and Chip in 2 mins = (1/30) hr

(60+120) * (1/30) = 20 * t

=> t= 3/10

so the distance is 200 * t = ( 200 * 3/10) =60 miles.

a , b, c travel at 1, 4/3, 2 mi pm; let c meet b at t=t min; c meets a at t=t+2 min; 4/3 t - 1(t+2) = distance driven by c between meeting b and a = 2*2 mi; t = 18 min; now the distance between a and b = distance driven by a in time t + distance driven by c in time t = (2 + 4/3)18 = 60 mi

Another way to solve--

Let p2 be point at which Chip passes Bob and p1 be point at which Chip passes Ash.

Now we know distance between p2 and p1 is 4 miles since Chip at 120mph is covering it in 2 minutes.

Let p0 be the point at which Ash was when Chip and Bob cross. We know that p0 to p1 distance is 2miles since Ash is at 60mph. so p0to p2 is 6 miles.

If distance from town A to p0 is d then Bob will have covered 4/3 (d) at 80 mph in same time as Ash covers d at 60 mph.

we can now equate d+6= 4/3 (d)

4d= 3d+18

d=18.

Chip is twice the speed of Ash so wiill cover 2d that 36 miles.

now just add 18+36 +6= 60 miles

I'm not completely sure of my answer...

Take into consideration relative velocity(speed or whatever>>>>). driver 1 rel. vel. = 60+120=180, driver 2 rel vel. = 80+120=200, driver 3 is assumed to be at rest. (This is the whole point of assuming relative velocity- by assuming one body to be at rest.)

Now, Distance traveled by Driver 1= Driver 2 , 180 * (t+2)= 200 * t . :

( because driver 1 takes 2 mins more to travel same distance.) Which gives t= 18 mins. Therefore, distance = 180 * (18+2)/60 = 60 miles .

this was a fine question...can be given in jee mains or advaned

Let the distance between A and B be 240( LCM of 60, 80, 120) . If 240 is the distance between A and B then bob and chip will meet after 240/200 hr (200 is the relative velocity of bob and chip), at this point the distance between ash and bob will be (240*20/200)=24. If the distance between them is 24 then it will take 24/180=2/15 (180 is the relative velocity of ash and chip), but the answer should be 1/30 hr (since it will take 2 min for them to meet), so the answer should be 240/4 which is 60.

distance between Ash & Bob calculated by using speed and time interval. d = 2x180/60 miles (min to hrs is 1/60) D = 6 miles. So, At What time will 20mph speed lead will make 6miles lead. 0.3 hrs.. So, Dis = vxt = 80 0.3 = 24 miles... Dis traveled by chip at same time = 120 0.3 = 36 miles... Then, we can find the total distance by adding these two 24+36 = 60 miles. :)

2 mins , relative speed 180 between chip and ash, so 6 mile between bob and ash........................... 6 mile , relative speed between bob and ash is 20 mph so after 18 mins bob and chip met............. relative speed 200 mph between chip and bob and met in 18 mins so distance is 60 m....

Let Bob and Chip meet after t1 hrs, so distance travelled by them (i.e distance between A and B point) in t1 hrs is 80t1+120t1 = 200t1 ...Eq (1)... Let Ash and Chi meet after t2 hrs, so distance travelled by them is 60t2+120t2 = 180t2 which is also distance between A and B ...Eq (2)... Equating 1 and 2 we get 10t1 = 9t2 ...Eq(3).... Also we have t1+ 2/60 = t2 ...Eq(4)... Solve Eq (3) and Eq (4) and we get t1= 0.3 hrs, substitute this in Eq1 and we get distance = 200t1 = 200 * 0.3 = 60 miles

Let the distance b/w the two towns A and B is l

Time taken for Chip to meet Bob = l/(their relative velocity)=l/(120+80)

Time taken for Chip to meet Ash = l/(their relative velocity) =l/(120+60)

According to question; difference b/w the time of their meetings=l/(120+60)-l/(120+80)=2/60(hr) solve above equation and get: l=60Km

Good one

Relative speed L = 200 t1 L=180 t2 L/200 = t1 L /180 = t2 L/180 - L/200 = t1 - t2 = ( 2 / 60 ) solving L = 60

In two minutes Chip can travel 120/60(since one hour = 60 min) * 2 = 4 miles in two minutes Ash can travel 60/60 * 2 = 2 miles

The difference between them was 20 miles when Bob and Ash traveled 1 hour(60 min) the difference between them was 1 mile when they travel 60/20 or 3 min the difference between them was 6 miles ( the distance between Ash and Bob when Chip passed Bob) when they traveled 3* 6 or 18 minutes

in 18 minutes Bob traveled 80/60 * 18 = 24 miles and in 18 minutes Chip traveled 120/60 * 18 = 36 miles

The total distance = 24 + 36 = 60 miles

first convert all the speeds into miles/minute. let distance between the towns be x and let bob and chip meet at time t. So distance traveled by Bob = 4t/3 distance traveled by chip = 2t so 4t/3 + 2t = x which implies that [ x = total distance] x = 10t + 20 = 10x/3 _ _ (i) multiplying by 10 and dividing by 3
chip meets ash at t+2 (time) so distance traveled by chip + distance traveled by ash = x or 2(t+2) + (t+2) = x or 3(t+2) = x (ii) solve equation (i) and (ii) by eliminating t to get 10x/3 -3x = 10t + 20 -10t which gives x = 60 miles.

Using ratios Speed A: C 1:2 i.e Distance covered by A:C when they meet is :D/3 and 2D/3 Similarly B:C Speed 2:3 Distance covered 40%:60%. and C travels in 2 minutes 4 miles ie distance between meeting B and A while traveling in the other direction. This leads to distance work out of 60 miles.

• Distance between A and C = relative velocity of A w.r.t B x time(2 sec) = 180 X (2/60)= 6 miles
• Time taken by B to overtake A by 6 miles = 6/(80-60)=3/10 hours
• Distance travelled by B in 3/10 hours = 80 x 3/10 = 24 miles
• Distance travelled by c during same time = 120 x 3/10 = 36 miles
• Distance between A and B = 24 + 36 = 60 miles

D=distance between A and B, t = time in hours. D/200 = t ( distance travelled( AB) when Chip and Bob meet at time t)....eq.1 D/180=t+ 2/60 ( distance travelled(AB when Chip and Ash meet at time t+2/60 hour) ...eq2

Solve this simultaneous Eq. and you can solve for the distance.