MAH MBA CET Speed Time Distance Work Questions

Tank filled in an hour when all of them are working simultaneously = 1/3 + 1/6 - 1/9 = 7/18
Hence time taken to fill up the tank = 18/7 h

Let the volume of the tank = V
Let the efficiency of the tap A = V/A = V/6
And efficiency of the tap B = V/B = V/8
Efficiency of tap C to empty the tank = V/C = V/12
Let T be the total time taken by the system to fill the tank.
T*(V/6 + V/8 - V/ 12) = V
==> T = 24/5 = 4.8 hours.
So the correct option to choose is B - 4.8 hours.

Let the total work be 200 units.
A’s rate of work = 200 units/10 days = 20 units per day
B’s rate of work = 200 units/20 days = 10 units per day
So, work done by A and B together in 4 days = (20+10)*4 = 120 units
Work left = 200 - 120 = 80 units
Time taken by B to complete the remaining work = 80 units/10 units per day = 8 days

Let the distance between A and B be D.

Time taken = D/58 + D/52 = 110*D/(58*52)

So, average speed for the whole journey = 2D/Time = 58*52*2D/110D = 54.8363636 kmph = 55kmph approximately

Average speed = (Total Distance travelled)/(Total time taken)
= (100+200)/(100/25 + 200/40) = 300/(4+5) = 100/3 = 33.33km/hr

Relative speed of train A with respect to B = 5 km/h
Distance to be covered = 200 km
Time taken = $$\frac{200}{5}$$ = 40 h

Let the speed of the first train be v m/s
Then
400/10 = v
=> v = 40 m/s
Now the speed of the second train = 72 km/hr = 20 m/s
Since both the trains are running in the same direction, the relative speed between the two trains is 40 - 20 = 20 m/s
Distance to be covered = 400 + 300 = 700m
Hence the required time = 700/20 = 35 seconds.

Sunil checks 1000 items in 5 hours, so Sunil's rate = 1000/5 = 200 items/hour

Nilesh completes 75% of the same job (1000 items) in 3 hours. 75% of 1000 is 750 items. = 750/3 = 250 items/hour

In the first 2 hours, both Sunil and Nilesh are working together.  Combined Rate: 200 + 250 = 450 items / hour

Items checked in 2 hours: 450*2 = 900 

The total target is 1300 items. Remaining Items = 1300 - 900 = 400 items. 

Now, Nilesh stops checking after the first 2 hours, so Sunil must finish the remaining 400 items alone.

Time required by Sunil = 400 items / 200 items/hour = 2 hours

Total Time = 2 hours + 2 hours = 4 hours.

Let the speed of stream be 's' km/hr.
$$\ \frac{\ 20}{5-s}=\ 3\times\frac{20}{5+s}$$
5+s=15-3s
s=2.5km/hr
Option B is the answer.

The ship admits 2.5 metric tons of water in 6 minutes and 30 seconds.
Water Influx rate = $$\frac{2.5}{\ \frac{\ 6.5}{60}}$$ metric tons/hr = 23.07 metric tons/hr
Water outflux rate = 15 metric tons/hr
Net influx rate = 23.07-15 = 8.07 metric tons/hr
Time after which ship starts sinking = $$\ \frac{\ 68}{8.07}$$ = 8.42 hours
Average rate of sailing=$$\frac{156}{8.42}$$=18.53kmph
Closest option is option C.

Let the total portion of tank to be filled be '3x' units.
Let the amount of units filled per minute by A,B and C be a,b and c respectively.
We are given that, at+10b+(t-12)c = 3x
Also, each pipe fills equal share of tank i.e. x units
at=x, 10b=x, (t-12)c=x. 
$$\therefore\ $$ b=$$\ \frac{\ x}{10}$$
Also, at+bt=3x
bt=2x, substitute b=$$\ \frac{\ x}{10}$$, t=20 min.
(t-12)c=x
c=$$\ \frac{\ x}{8}$$
Let k be time taken by c to fill the tank alone,
ck=3x
k=$$\ \ \frac{\ 3x}{c}$$=24 minutes.
Option B is the answer.

Given: 

Rate = x revolutions per minute and Time = k seconds

Convert time into minutes: $$\dfrac{k}{60}$$ minutes

Number of revolutions: $$x\times\dfrac{k}{60}$$ = $$\dfrac{xk}{60}$$

Assume that the capacity of the tank is $$x$$ litres, the efficiency of an inlet pipe is $$y$$ litres per hour, and that of an outlet pipe is $$z$$ litres per hour.

So, given that an inlet pipe can fill an empty tank in 12 hours or, 

$$\dfrac{x}{y}=12$$

$$y=\dfrac{x}{12}$$ . . . (1)

Also, given that an outlet pipe can drain a filled tank in 10 hours or,

$$\dfrac{x}{z}=10$$

$$z=\dfrac{x}{10}$$ . . . (2)

Now, all the pipes are opened together to an empty tank,

so the time taken to fill the tank is $$\dfrac{x}{3y-2z}$$

Putting the values of $$y$$ and $$z$$ from equations (1) and (2) we get

Time taken = $$\dfrac{x}{3y-2z}=\dfrac{x}{\dfrac{3x}{12}-\dfrac{2x}{10}}=20$$ hours

Hence, the answer is 20 hours.

It is said that the task was finished in 15 days, so if we assume that one man does one unit of work each day, the amount of work done by the men after n days can be found by the formula, $$\frac{n\left(n+1\right)}{2}$$

It is given that it takes them 15 days, substituting 15 for n in that formula, we get 120 units of work. 

Now these 120 units are work done by men, it is given that women are 40% less efficient, that means $$0.4\left(M\right)=W$$
The same 120 units of men's work are equivalent to $$\frac{120}{0.4}=300\ units$$ of Women's work. 

From the options we see, 24 days, which will equal $$\frac{24\left(25\right)}{2}=300$$
Hence, the answer is 24. 

Let the total work to be done is 300 units.

So, P does = 300/10 = 30 units per day

Q does = 300/15 = 20 units per day,

R does = 300/20 = 15 units per day.

P and Q start working on the take and worked for 3 days. So the work done in 3 days = (30 + 20)*3 = 150 units.

Now, R joins and P left. So, the work done by R and Q in the next 3 days = (20 + 15)*3 = 105 units

Remaining work = 300 - 150 - 105 = 45 units

Fraction of the work remaining = 45/300 = 3/20

Let us assume the speeds of Raju and Ravi as kx, ky.

The ratio of their speeds are x:y.

x+y = 11

x-y = 5

x = 8, y=3

We are given that Raju travels 1200m in 15 sec. His speed will be 80 m/s.

kx = 80 => k = 10.

The speed of Ravi is 30m/s. 

Time Ravi takes to travel 1200m will be 40s.

Anuj took 4 hours after meeting Bharat. Hence, Anuj met Bhrat after 6 hours. 

The time before their meeting can be calculated as $$\sqrt{\ t_1t_2}$$, where $$t_1$$ and $$t_2$$ are time taken by the two people to cover the remaining distance after crossing each other. 

We get the equation $$6=\sqrt{\ 4t_B}$$
$$36=4t_B$$
$$t_B=9$$

Hence, it took Bharat 9 more hours after meeting Anuj to complete his journey and reach P. 

This, in addition to the 6 hours before meeting Anuj, gives the total time taken by Bharat to be 15 hours. 

Therefore, Option D is the correct answer. 

Let the speed of the train be ‘v’ m/s and it’s length be ‘l’

Then
l/v = 20
20v = l
Also,
(l + 1000)/v = 100
=> 100v = 1000 + l
=> Solving both the equations, we get
v = 12.5 m/s and l = 250 m
Hence the length of the train is 250 m and speed = 12.5 m/s

In the first half an hour, the slower train covers some distance, which is 

$$80\times\dfrac{30}{60}=40\ $$ km

Hence, now the distance between the two trains is $$940-40\ =900$$ km.

The time after which both trains meet is $$\dfrac{900}{80+100}=\dfrac{900}{180}=5$$ hrs

Hence, the distance travelled in this time by the slower train will be $$5\times80=400$$ km

But it also covered 40 km before this, so the total distance travelled is 40+400 = 440 km.

Hence, the answer is 440 km.

Average speed = $$\ \frac{\ Total\ Distance}{Total\ time}$$
Total distance = 50+50+30 = 130km
Total time = $$\ \frac{\ 40}{60}$$+$$\ \ \ \frac{\ 50km}{120kmph}$$+$$\ \ \ \frac{\ 30km}{60kmph}$$ = $$\ \frac{\ 19}{12}hr$$
Avg speed = $$\ \frac{\ 130}{\frac{19}{12}}=82.1\ kmph$$
So, option E is the answer.