Test: Time, Distance & Work - CAT MCQ

Let the total journey time taken by Amal be 3t hours.
Hence, total distance = 10t + 20t + 30t = 60t kms Bimal took each mode of transport 1/3 of the total distance.
Therefore, total time taken by him =hours
The percentage by which Bimal’s travel time exceeds Amal’s travel time

Let the distance of the common destination = d
and time taken by first car to reach the destination = t
then time taken by second car to reach the destination = t – 1.
∴ Speed of the first  car, S₁ = d / t
and speed of the second car, S₂ =
Percentage increase in the speed of the second car

It is given that t ≥ 6 ⇒ t – 1 ≥ 5
Now percentage increase in the speed of second car will be maximum when (t – 1) will be minimum i.e. 5.
∴ maximum percentage increase in the speed of the second car.
= 100 / 5 = 20

Let the first, second and third horses be A, B and C respectively. Also, let the length of the race course be x.

Distance = Speed × Time
Now, DB = SB × t₁
From first given condition, x – 11 = SB × t₁ ...(i) and from second given condition, x = SB × t₂ ...(ii)
Similarly, x – 90 = SC × t1 ...(iii)
and x –  80 = SC × t₂ ...(iv)
Now dividing equation (i) by (iii), we get
...(v)
And dividing equation (ii) by (iv), we get

From equation (v) and (vi), we get

⇒ x² – 91x + 880 = x² – 90x ⇒ x = 880

Distance travelled by bicycle A in one revolution = 2πra = 2π × 30 = 60π cm
Distance travelled by bicycle B in one revolution = 2πrb = 2π × 40 = 80π cm
Let B makes ‘n’ revolutions to cover the distance. Then, A would make (n + 5000) revolution to cover the same distance.
∴ n × 80π = (n + 5000) × 60π ⇒ n = 15000
Distance travelled by B = n × 80π cm =
= 12π km
Time taken by B = 45 min =
Hence the speed of B =

LCM of 9 and 12 = 36
Let the amount of work = 36 units
Let the amount of work done in one day by A and B with their normal efficiencies be x and y units respectively.  
∴ (x + y) × 12 = 36 ⇒ x + y = 3 ⇒ y = 3 – x ...(i)
and (x/2 + 3y) × 9 = 36 ⇒ x/2 + 3y = 4 ⇒ x + 6y = 8 ...(ii)
Putting the value of y from eq. (i) into eq. (ii), we get
x + 6(3 – x) = 8 ⇒ –5x = 8 – 18 ⇒ x = 2
Therefore, A alone would take 36/x = 36/2 = 18 days to complete the work with her normal efficiency.

Let one machine completes 1 unit of work per day.
Since, two machines can finish the job in 13 days
∴ Amount of work = 2 × 1 × 13 = 26 units.
Also, let a man completes M units of work per day.
From the given conditions 3M + 8 × 1 = 2(8M + 3 × 1) ⇒ M = 2 / 13 units
Let it require ‘x’ number of men to complete the work in 13 days.
xM × 13 = 26 units ⇒
hence, required number of men = 13.

Speed of John = 6 kmph
Speed of Mary = 7.5 kmph
Let the track length of A and B be x and y respectively
Given, x + y = 325  ...(i)
Time taken by John to cover one round of track

Therefore, time taken to cover 9 rounds of track A

Time taken by Mary to cover one round of track
Therefore, time taken to cover 5 rounds of track B 

According to the question,
..(ii)
Put the value of y in (i)

⇒ 13x = 1300 ⇒ x = 100
∴ Time taken by Mary to cover one round of track A

LCM of 20 and 40 = 40
Let the work be of 40 units
Amount of work done by Anil in one day = 40/20 = 2 units
Amount of work done by Sunil in one day = 4
40 / 40 = 1 units
Bimal does 10% work i.e. 4 units work
Rest 40 – 4 = 36 units is done by Anil and Sunil.
Let Anil took x days. Therefore, Sunil took (x – 3) days.  
Therefore 2 × x + 1 × (x – 3) = 36 ⇒ x = 13 days.
Hence, the job was done in 13 days.

Time taken by cyclist to cover the distance AB = 60 min Given, starting from 10:01 am, every minute a motor cycle leaves A and moves towards B.
Forty-five such motor cycles reach B by 11 am.
Also, the speed of all the motor cycles is same This means that the 45th motor cycle, which started at 10:45 am, reached B exactly at 11 am, Rest all reached B some time before B.
Therefore, each motor cycle takes 15 min to cover the distance AB.
Now, if the cyclist doubles his speed, then he will reach B in 30 min i.e. at 10:30 am.
So, the 15th motor cycle (started at 10:15 am from A) would be the last motor cycle that reach to point B at 10:30 am.
Hence, when the cyclist had doubled his speed, then there will be 15 motor cycles would have reached B by the time the cyclist reached B.

Let the track length be 10x
When they meet at 10 am, ant A travelled 6x of the distance and ant B travelled 4x of the distance.


The distance covered by ant A from meeting point to point P was 4x. Similarly, the distance covered by ant B from meeting point to point P was 6x.
Given, ant A took 12 min to reach P.

Time taken by ant B to travelled the 6x distance to reach at P

Therefore ant B reaches P at 10:27 am.

Let the number of hours for regular and overtime work be x and y respectively.
∴ x + y = 172 ...(i)

⇒ 40y = 3x
⇒ 3x – 40y = 0 ...(ii)
On solving equations (i) and (ii), we get x = 160 and y = 12
∴ overtime work = 12 hours

Let the rates of work of each human and each robot be H  units per day and R units per day respectively.
∴ 15H + 5R = 1 / 30 ...(i)
and 5H + 15R = 1 / 60 ..(ii)
⇒ 3 (i) – (ii) ⇒ 40H = 1 / 12
H = 1 / 480
In a day, 15 humans can complete 15H i.e. 1/32th of the job.
∴ 15 humans can complete the job in 32 days.

Let the rates at which each filling pipe and each draining pipe works be F units/hr and D units/hr.
∴ 6F – 5D = 1 / 6 ..(i)
and 5F – 6D = 1 / 60 ..(ii)
on solving (i) and (ii), we get


Hence one draining and two filling pipes can fill the tank in 10 hours.

Let the time taken by A to finish the job be “a” days.
Time taken by B to finish the job = 5a/4 days
Part of the job completed when A and B worked together for 4 days =

Time taken by B alone to complete the entire job = 5a/4 = 20 days.

Let the time taken for car 1 to reach P from A be x hours.
Speed of car 1=AP/x BP = 3AP.
Car 2 starts from B to A and reaches P one hour after car 1 reaches P.
Speed of car 2 =
Now speed of car 2 =

∴  Time taken for car 1 to reach P from A is 12 min.

Let the time taken by S to reach Z be t hours.
Let the speed of T be St. Distance between X and Z is 3/5 of the distance between X and Y.
XZ : ZY = 3 : 2

t = 8
S takes 8 hours to cover YZ.
T would take 8 × (3/4) i.e. 6 hours to cover ZY.
T would take t + 1 i.e. 9 hours to cover XZ.
T would take 15 hours to reach Y.

Let the speeds of Partha and Narayan be S_p and S_n respectively.
...(i)
...(ii)
Subtraction, equation (ii) from (i), we get

Let the speed of the boat in still water and the speed of the river be u and v respectively
u = x
v = y
According to question,

Let BC = 5 k
Given, by the time Q reaches C, P was halfway to C, i.e., AC/2 = 3k and AC = 6k.
As Q met P, 165 km away from A, the distance to the meeting point from A is 3k


∴ Distance between A and C is 240 km and that between B and C is 200 km.
From the data, as distance between A and B is twice that between B and C, it is 400 km.

∴ Time taken by Q to reach C from 

Let's take there are two ants.
Considering worst possible case we can see easily the required time is same as the time taken by an ant to reach one extreme point to another extreme point.
Which will be same when there are 1000 ants.
i.e.maximum time = 

The distance covered by R in 2 hours (3 PM to 5 PM) is same as the distance covered by P in 4 hours (1 PM to 5 PM).
Ratio of speeds of P and R = 1 : 2.
Now, R takes over Q in another hour by doubling his speed, so R has covered a total distance in 3 hours (3 PM to 6 PM) which he would have covered in 4 hours without changing his speed.
To cover the same distance, a takes 5 hours (1 PM to 6 PM) Ratio of speed of Q and R = 4: 5
⇒ Ratio of speed of P and Q = 5: 8.

The rabbit overstretched his sleeping time by  min.

Let the speed (in km/min) of the rabbit and tortoise be 6x and x respectively.
Had the rabbit not overstretched his hap, he would have beaten the tortoise by 13 min.

Hence the time taken by the tortoise to complete the race = 

Let the efficiencies of Anirudh and Anushka be 'a' and 'b' units/day respectively.
according to question,

So, the required ratio is 1 : 1.

Let the time at which Mr. Clockilal entered the museum be 'm' minutes past 12 noon.

As he spent more than 3 hours and less than 4 hours in the museum, and the angle between the minute and hour hands at the time of leaving was 220°, he must have left the museum between  p.m. and 4 p.m.
Let the time at which he left the museum be 'n' minutes past 3 pm.

Hence, the time spent by Mr. Clcckilal in the museum was either 3 hours  minutes or 3 hours   minutes.

Area of cross section for P₁ pipe

Volume of water flowing through P1 in one second = 154 × 10 = 1540 m³
Volume of tank = V = 1540 × 2 × 3600 = 11088 × 10³ m³
Area of cross section for P₂ pipe

Let the rate of water flowing through P₂ = s m/s
Volume of water flowing through P₂ in one second = 616s m³
Volume of water flow in the tank when P₁ is used as inlet and P₂ is used as outlet pipe: (1540 – 616s)
Time taken to fill = 4 hours = 4 × 3600 seconds
So, (1540 – 616s) × 4 × 3600 = V
⇒ (1540 – 616s) × 4 × 3600 = 11088 × 10³
or 1540 – 616s = 770
or s = 1.25 m/s
∴  Rate of water flow through P₂ = 1.25 m/s

Let the length of train = x metres

Distance covered in crossing the platform = (x + 100) m

or   2x + 200 = 1500 or x = 650
Now, time taken to cross the pole

Let the slowest runner covers a distance of x ms. from the starting point.
Then the fastest runner will cover a distance of 1000 + x from the starting point.
Let the speed of the slowest runner be y.
Then, speed of the fastest runner will be 2y.

But time taken for both is same .
Hence, 
∴ Fastest runner completes a distance of 2000m in 5 mins.
∴ Fastest runner can complete a distance of 4000m in 10 min.

Rate of admission of water
= 2/6 tonnes / min. = 1/3 tonnes/ min
Rate of pumping out of water

Rate of accumulation =   tonnes / min.
Time to accumulate 80 tonnes of water

= 10 hours
∴ Average sailing rate so as avoid sinking

Total time taken to build the server = 60 man hours.
6 of them starts at 11 : 00 am and works till 5 pm They will complete 6 × 6 = 36 man hours of work.
At 5 pm they will have 24 more man hours of work to complete.
Between 5 pm and 6 pm they will complete 7 man hours.
Between 6 pm and 7 pm they will complete 8 man hours.
Between 7 pm and 8 pm they will complete 9 man hours.
So, totally they will complete 36 + 7 + 8 + 9 = 60 man hours by 8 pm.

The watch gains (5 + 10) = 15 min in 30 hours (12 Noon to 6 PM next day).
This means that it will show the correct time when it gains 5 min in 10 hours or at 10 PM on Monday.