Two cars (A and B) start at the starting line at the same time on a 3-mile long circular track, going in opposite directions. As they drive around the course, they pass each other many times. Exactly one hour after starting, they pass each other for the 33rd time. At this point car A has completed exactly 20 laps. What is the average speed of Car B?

Ram, Shyam and Ghanshyam start running towards a circular track from same place and at same time in same directions but with different speeds. Ram and Shyam meet after every 2 minutes while Shyam and Ghanshyam meet every 3 minutes. If Ram and Ghanshyam meet after every ‘t’ minutes, then which of the following cannot be value of ‘t’?

1.2

4.8

?

Hi Utkarsh,

I can suggest you this

We don't know whose speed is greatest so we have to assume cases

Let speed be R ,G,S

Case 1 - R>G>S

R-S= D/2 { let circle be of distance D and we find relative speed of RAM w.r.t to shyam by assuming shyam to be still }

G-S =D/3

subtracting both

R-G= D/2-D/3= D/6

so the time t is 6 in this case which is in option

Case 2- G>S>R

S-R =D/2

G-S= D/3

adding both

G-R = 5D/6

t= 6/5= 1.2

So this is also possible hence option 4.8 is the answer

Note-If you will take other cases either you will get time negative or you will get same values as you got in these cases

**Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B and C?**

5:4:2

4:3:2

5:4:3

3:2:1

Hi rachit

Let total length of circular track is d

And B is x away from end of the circular track so C is 2x away from end of the track

Let's compare ratio of speed of A and C

In first case A covers d distance and C covers d-2x and in second case A covers 3 lap means 3d distance and C covers 1 whole lap and comes to the position of B in same time so distance traveled is d+ d-x

d/3d= (d-2x)/(d+d-x)

Solving this we get x=d/5

Let d be 5 so x= 1

In first lap distanced traveled in same time by A ,B ,C are d , d-x d-2x putting values we get ratio as 5:4:3

smjhe to ni na.

Three sprinters X, Y and Z are initially standing at the vertex A of an equilateral triangle ABC of side length ‘a’ units. They start running simultaneously from the point A. X runs from A to B and continue towards C after reaching B. Y runs along the median AD and then continues towards C after reaching D. Z runs from A to C and then towards B after reaching C. If is also known that X reached D, Y reached D and Z reached C all at the same time. What is the distance of Y from D, the moment when X meets Z.

(√3-1)a/2

(√3a)/10

(√3 -1)a/4

3a/8

since x,y,z reach at D, D, C so the ratio of distance travelled by them will be the ratio of speeds

x:y:z= a+a/2 : √3a/2 : a

= 3:√3:2

now we have to find out the time in which X and Z meet for that we have to take relative speed of X=3+2=5 and assume Z to be still then X has to cover 3a distance to catch Z

time= distance/speed=3a/5

y will cover distance in 3a/5 time= speed *time= √3*3a/5= 3√3a/5

Distance of Y from D= 3√3a/5- √3/2 = √3a/10 (answer)

I am getting this and I am pretty sure I did it right

Hello sir , kindly share the solution to ques 25 -

Is it 420 km ?

Don't have pen - paper right now.

yes, the answer is 420

A |_____d_______|B

let the time taken by B be t min ,

Time taken by A =. t - 8 min

d/ [d/t + d/(t-8)] = 3

t(t-8)/(t-8+t) = 3

t^2 - 8t = 6t - 24

t = 12 or 2 , 2 is inadmissible

Hence , time taken by A to cover the distance 'd' = 12 - 8 = 4min

P ---------------------------Q

Time taken by Deepak to travel 20km : 2hours

Time taken by Sandeep to travel 20km : 1 hour

Time taken by Pawan to travel 20km: 2/3 hour

LCM : [ 2,1,2/3 ] = 2

They meet at P after every 2 hours Hence , 12hours /2 = 6 times .

They never meet at Q.

Hence , 6 times

Hello sir , kindly share the solution to this problem-

A started to chase B when B was 27 steps ahead of A. B takes 8 steps for every 5 steps that A takes. 2 steps of A is equal to 5 steps of B. After how many steps did A catch up with B?

A started to chase B when B was 27 steps ahead of A

here whose 27 steps is given ? A's or B's ?

A left pune at noon sharp. Two hours later, H startesd from pune in same direction. B overtook A at 8pm. Find the average speed of two trains over this journey if the sum of their average speeds is 70km/hr.

from 12 to 8

A and H covered same distance in 8 and 6 hours respectively

so the ratio of their speed will be 6:8 (s1/s2=t2/t1)

=>

S_{A}/S_{H}=3/4

let's say their speeds are 3x and 4x respectively

sum of their speed is 7x = 70 (given)

x=10

hence speeds are 30 and 40

average of speed is 35 kmph

Two trains, A and B , start at the same time from stations P And Q respectively towards each other. After passing each other , they take 12 hours and 3 hours too reach Q and P respectively. If A is moving at speed of 48km/hr, then what is speed of B?

Let's say they meet at a point x, after t hours

speed of A is a and speed of B is b.

so distance PX = at

distance QX=bt

now after meeting at x train A still needs to cover 'bt' distance and

and train B needs to cover 'at' distance

ratio of time taken by both trains to cover respective distance = (bt/a) to (at/b) = 12/3

so

(b/a)^{2}= (12/3)

a=48 (given)

so b= 96 kmph

Walking at 3/4 of hus normal speed , a man takes 2.5 hours more than the normal time. Find the normal time.

st = 3/4 s(t+2.5)

4t = 3t+7.5

t =7.5

hence normal time is 7.5 hours

Let earlier speed is 4 now it becomes 3

So time ratio becomes 3:4, difference 1x=>2.5

Actual i.e,3x=7.5

Speed is reducing by 1/4th, hence the time will be increased by 1/3rd. This increase is given to be 2.5 hours. => 1/3 of the actual time = 2.5 hours => Actual time =7.5 hours.

There are two clocks, which are set to correct time on sunday at 12:00 noon. The first clock gains 5/2 minutes every hour while the second clock loses 3/2 minutes every hour. When will they be 2 hours apart?

1. Monday 9:00 pm

2. Tuesday 12:00 mid-night

3. Monday 6:00 pm

4. Tuesday 6:00 am

first clock goes 2.5 mins ahead and second loses 1.5 mins in an hour.

Effectively both clocks are at difference of 4 mins in an hour.

so for 120 mins difference time taken will be 120/4 = 30 hours.

Hence on Monday 6.00 pm

The speed of the boat in still water is 6kmph. What is the speed of the stream (in kmph) if the boat can cover 32km downstream or 16 km upstream in the same time?

I got the answer, thank you 🙂