ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Let distance travelled by cat before dog catches it be D
We know, time for which Dog and Cat ran is same
∴ T = T
∴ D/5 = (D + 80)/7 [D = S x T]
∴ D = 200 m

Let, distance = x km.
Time taken at 3 kmph : dist/speed = x/3 = 20 min late.
Time taken at 4 kmph : x/4 = 30 min earlier
Difference between time taken : 30 - (-20) = 50 mins = 50/60 hours.
x/3 - x/4 = 50/60
x/12 = 5/6
x = 10 km.

When Rahim meets Shafiq for the third time,
they together would have covered a Distance of 5d, i.e 5 × 30m = 150 m.

The ratio of Speed of Rahim and Shafiq = 2 : 1,
so the total distance traveled by them will also be in the ratio 2 : 1
as the Time is taken is constant.

So the Distance traveled by Rahim will be (2/3) × 150= 100 m.

We are given that two-thirds of the 6 km was covered at 4 km/hr i.e. 4 km distance was covered at 4 km/hr.
Time taken to cover 4 km = 4 km/4 km/hr = 1 hr = 60 minutes.
Time left = 84 – 60 = 24 minutes

Now, the man has to cover the remaining 2 km in 24 minutes or 24/60 = 0.4 hours
Speed required for remaining 2 km = 2 km/0.4 hr = 5 km/hr

Average speed of Joy = 2xy/(x + y)
= (2 × 25 × 4)/(25 + 4)
= 200/29 km/h

Distance traveled = Speed × Time
= 200/29 × 29/5
= 40 Km

Distance between city and town = 40/2 = 20 km.

If Rashed is walking 5/6 of his usual speed that means he is taking 6/5 of using time.

According to the question,
6/5 of usual time - usual time = 10 mins
1/5 of usual time = 10 mins
Usual time = 50 mins.

While A covers 1000 meters, B can cover 900 meters

While B covers 1000 meters, C can cover 900 meters

Let's assume that all three of them are running the same race.
So when B runs 900 meters,
C can run 900 × (9/10)
= 810

So A can beat C by = 1000 - 810 = 190 meters.

Time is taken by A to cover 100 meters = 60 seconds
A gives a start of 4 seconds then time takes by B = 72 seconds
B takes 72 seconds to cover 96 meters
Speed of B = 96/72 = 1.33 m/s

Time is taken in walking both the ways = 7 hours 45 minutes -------- (i)

Time is taken in walking one way and riding back = 6 hours 15 minutes ----------- (ii)
By the equation (ii) × 2 - (i), we have,

Time is taken by the man in riding both ways,
= 12 hours 30 minutes - 7 hours 45 minutes
= 4 hours 45 minutes.

As the speed decreases from 20 kmph to 18 kmph i.e. 10 % increment in usual time.
10% = 10 min
100% = 100 min.

Now,
Distance traveled by him,
= (100/60) × 18
= 30 km.

Let, Speed of the man is x Kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
⇒ 20 × (10/60) = 8/60 × (20 + x)
⇒ 200 = 160 + 8x
⇒ 8x = 40
Hence, x = 5 kmph.

Let the speed of A = x kmph and that of B = y kmph

According to the question,
(x × 6) + (y × 6) = 60
⇒ x + y = 10 --------- (i)
And,
(2x/3) × 5 + (2y × 5) = 60
⇒ 10x + 30y = 180
⇒ x + 3y = 18 ---------- (ii)
From equation (i) × 3 - (ii)
3x + 3y - x - 3y = 30 - 18
⇒ 2x = 12
Hence, x = 6 kmph.

Let the speed of Rizvi be x kmph;
Hence, Amin's speed = (x + 4) kmph;
Distance covered by Amin = 60 + 12 = 72 km;
Distance covered by Rizvi = 60 - 12 = 48 km.

According to question,
⇒ 72/(x + 4) = 48/x
⇒ 3/(x + 4) = 2/x
⇒ 3x = 2x + 8
⇒ x = 8 kmph.

Speed of tiger = 40 m/min
Speed of deer = 20 m/min.
Relative speed = 40 - 20 = 20 m/min.
Initial difference in distance = 50 × 8 = 400 m

Time taken to catch = 400/20 = 20 min.
Distance traveled in 20 min,
= 20 × 40
= 800 m.

Tank filled or work done by P in 1 hour = 1/22
Tank filled or work done by Q in 1 hour (Q takes 11 hrs more than P) = 1/33

Tank filled or work done by both pipes in 1 hour = 1/22 + 1/33 = 5/66
So the entire tank is full in = 66/5 = 13(1/5) hours.

Let Tap A take T minutes to fill the tank alone.
Since Tap A is 5 times faster than Tap B, Tap B takes 5 times more time.
So time taken by Tap B = 5T minutes
Also, 5T-T = 32 ----------- Given
∴ T = 8 minutes = Time taken by A
Time taken by B = 5 x 8 = 40 minutes.

In 1 min, A + B fills = 1/8 + 1/40 = 3/20 parts
So entire tank is filled in = 20/3 hours.

Let the tank get empty in T hours counting from 8 am.
A is on for T hours and work is done by A = Work in 1-hour × T hours = T/1.5 = 2T/3

Similarly, B starts at 9 am i.e. it's on for (T-1) hours & work done is = (T - 1)/2

Similarly, C starts at 10 am i.e. it's on for (T-2) hours & work done is = (T - 2)/(1/2) = 2(T - 2)

Initially, the tank is empty and after T hours too, it is empty. So, the total work done is 0.

According to the question,
2T/3 + (T - 1)/2 - 2(T - 2) = 0
⇒ (4T + 3T - 3 - 12 T + 24)/6 = 0
⇒ -5T + 21 = 0
⇒ 5T = 21
⇒ T = 21/5
= 4.2 hours = 4 hours 12 minutes5
This time is needed for the tank to get empty.
The exact time will be 4 hours 12 min from 8 am = 12.12 pm

Let Tap A fill the cistern completely in A hours.
So in 1 hour, it fills 1/A amount of the cistern

Also in 1 hour in Tap B fills in 1/4 amount of the cistern
Together they fill the cistern in 2.4 hours
So, Also in 1-hour together they fill in (1/2.4)amount of the cistern
So, Also in 1-hour cistern filled by both is given by 1/A + 1/4 = 1/2.4
∴ 1/A = 1/2.4 - 1/4 = 1/6

∴ Pipe A can fill 1/6^th tank in 1 hour
∴ Pipe a fills the tank completely in 6 hours.
It has a rate of 100-litre water per hour,
So, in 6 hours it gives out 6 x 100 = 600 litres
In 6 hours cistern is full, so capacity = 600 litres.

Suppose the slower pipe alone can fill the tank in x minutes.
Then the faster pipe can fill the tank in x/4 minutes.

Part filled by the slower pipe in 1 minute = 1/x
Part filled by the faster pipe in 1 minute = 4/x
Part filled by both the pipes in 1 minute = 1/x + 4/x

Given that both the pipes together can fill the tank in 36 minutes.
Part filled by both the pipes in 1 minute = 1/36

According to the question,
1/x + 4/x = 1/36
5/x = 1/36
x = 180

A tap can fill a tank in 4 hours.
Therefore the tap can fill half the tank in 2 hours.

Remaining = 1/2

After half the tank is filled, three more similar taps are opened.
Hence, the total number of taps becomes 4.

Part filled by one tap in 1 hour = 1/4
Part filled by four taps in 1 hour = 4 × (1/4) = 1
i.e., 4 taps can fill the remaining half in 30 minutes.

Total time taken
= 2 hour + 30 minute = 2 hour 30 minutes.

Let X hours be the time taken to fill a tank by P.
Let Y hours be the time taken to empty the tank by Q.
Then the time taken to fill the tank when P and Q are switched together : XY/(Y - X) hours.
Here, X = 16 minutes And Y = 32 minutes
Therefore,
Required time = (16 × 32)/(32 - 16)
= (32 × 16)/16
= 32 minutes.

Three pipes A, B, and C can fill a tank in 8 hours. A, B, and C’s 1 hour work=1/10
A, B and C's 3 hour work= 3/10 Remaining work= 1 – (3/10) = 7/10

The remaining part will be filled by A and B in 14 hours. Then,
⇒ (7/10) × (A + B) = 14
⇒ (A + B)'s whole work= 14 × (10/7)
= 20 hr (A + B)'s 1-hour work
= 1/20

A, B, and C's 1-hour work = 1/10
C's 1 hour work= (A + B + C) – (A + B)
⇒ (1/10) – (1/20)
⇒ 1/20
∴ C can fill the tank in 20 hours.

Let, B alone filled the pipe by x hours.

Efficiency ratio of B and C =2 : 1
Time ratio of B and C = 1 : 2

Given,
(1/A + 1/B)- (1/A + 1/C) =7/60 - 1/12
⇒ 1/B - 1/C = 2/60 = 1/30
⇒ 1/x - 1/2x=1/30
⇒ 1/2x=1/30
⇒ 1/x=1/15
⇒ x = 15

B alone filled the pipe by 15 hours.

Tank filled by A alone in 1 hr = 1/12
Tank filled by B alone in 1 hr = 1/24
Tank filled by C alone in 1 hr = 1/48
D empty the tank at the rate of 6m/hr

So,
Tank empty by D in 1hr = 6/96 = 1/16
Now, tan is to be filled up to 72m i.e., 72/96 =3/4 of tank
So,
Let the 3/4^th of the tank to be filled in 't' hours time

For 1 hr all are opened then B closed

So, for (t - 1) hr A, C and D opened
(1/12 + 1/24+ 1/48 -  1/16) + (t - 1) (1/12 + 1/48 - 1/16) = 3/4
⇒ (1/12) + (t - 1) (1/24) = 3/4
⇒ (t - 1)/24 = (3/4) - (1/12)
⇒ (t - 1)/24 = 2/3
⇒ 3t - 3 = 48
⇒ 3t = 51
⇒ t = 17 hours.

Given that,
A takes 20 minutes to fill and B takes 12 minutes to empty
Clearly,
tap B is faster than tap A.
And so, the tank will be emptied.
Half of the tank or 1/2 part of the tank is already filled.
Therefore,
we have to find the time taken to empty that 1/2 part.
Part filled by A in 1 minute = 1/20
Part emptied by B in 1 minute = 1/12.
Part emptied by (A + B) in 1 minute
= (1/12) – (1/20)
= (5 - 3)/60
= 2/30.
Therefore,
The time taken by (A + B) to empty the full tank is 15 minutes.
Time taken to empty 1/2 part of the tank is 30/2
= (30/2) × (1/2) minutes.
= 15/2 minutes.

If the tank has 4x liters of total capacity and holds 3x liters of water and if 30 liters of water is taken out, then the tank becomes empty.
It means 3x liters of water is taken out
3x = 30 liters
x = 10 liters
Capacity of tank
= 4x
= 4 × 10
= 40 liters.

Rate of flow of water = x cm/minute
∴ The volume of water that flowed in the in 1 minute
= (5 × 4 × x) = 20x cu.cm.

∴ The volume of water that flowed in the tank in 6 hours 18 minutes.
i.e. (6 × 60) + 18 = 378 minutes
= 2x × 378 cu. cm.

According to question,
20x × 378 = 700 × 400 × 450
⇒ x = (700 × 400 × 450)/(20 × 378) cm/minutes
⇒ x = (700 × 400 × 450 × 60)/(20 × 378 × 100000) km/hours
⇒ x = 10 km/hrs.

Pipe A's work in % = 100/10 = 10%
Pipe B's work in % = 100/20 = 5%
Pipe C's work in % = 100/40 = 2.5%
All of them are opened for 2 hours + after 2 hours,
tap C is closed + After the 4th hour, tap B is also closed = 100

According to the question,
(10 + 5 + 2.5) × 2 + (10 + 5) × 2 + work by tap A alone = 100
⇒ 17.5 × 2 + 15 × 2 + work by tap A alone = 100
⇒ 35 + 30 + work by tap A alone = 100
⇒ work by tap A alone = 100 - 65 = 35%.