ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.
∴ 1/x + 2/x + 4/x = 1/5
⇒ 7/x = 1/5
⇒ x = 35 hours

৩ জন শ্রমিক একটি কাজ ১২ দিনে করতে পারে।
এই ৩ জনের মধ্যে ২ জন ৩য় জনের তুলনায় দ্বিগুণ কাজ করতে পারে।
প্রথম দুজনের ১ জন শ্রমিকের কাজ = ২ × ৩য় শ্রমিকের কাজ।
সুতরাং ৩ জন শ্রমিকের কাজ = ৫ × ৩য় শ্রমিকের কাজ।
৫ × ৩য় শ্রমিক কাজটি করতে পারে = ১২ দিনে
৩য় শ্রমিক কাজটি করতে পারে = ৫ × ১২ দিনে = ৬০ দিনে।
২ × ৩য় শ্রমিক কাজটি করতে পারে = ৬০/২ দিনে = ৩০ দিনে
নির্ণেয় সময় = ৩০ দিন

A takes 20 minutes and it is 4 times faster than B,
it means B will take 80 minutes to fill the tank.
(1/20 + 1/80) × t = 1.
So, We get t = 16.

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.

Since the clock takes 24 hours or 1 day to slow down by 1 minute.
So, it will take 60 days to slow down 60 minutes or 1 hour.

Let man's rate upstream be x kmph
Then, his rate downstream = 2x kmph
∴ (speed in still water) : (Speed of stream)
=(2x+x)/2 : (2x−x)/2
= 3x/2 : x/2 = 3 : 1

Let,
The time taken by Y alone to fill the tank be A minutes.
Given that, the size of the pipe X is thrice as Y.
Then, X fills the tank in A/3 minutes.
Part filled by X in 1 minute = 1/(A/3)
= 3/A
Part filled by Y in 1 minute = 1/A
Since, X and Y together take 72 minutes.
Part filled by (X + Y) in 1 minute = 1/72
i.e.,
(1/A + 3/A) = 1/72
⇒ 4/A = 1/72
A = 288 minutes
= (288/60) hours
= 4(48/60)
= 4(4/5) hours
= 4 hours and (4/5 x 60) minutes
= 4 hours and 48 minutes.
Hence, the pipe Y alone takes 4 hours and 48 minutes to fill the tank.

Speed = 45 km/hr = 45 × 5/18
= 25/2 m/s
Distance travelled = Length of the train + Length of the platform
= 360 + 140
= 500 metre.
Time taken to cross the platform = 500/(25/2)
= 40 seconds

In 1 min both pipes can fill = 1/20 + 1/30 = 1/12
In 10 min second pipe can fill = (1/30)×10 = 1/3 part
Part of cistern filled by both the pipes = 1 - 1/3 = 2/3
1/12 part is filled in 1 min
∴ 2/3 part will be filled in 12×2/3 = 8 min
Hence, first first pipe should be turned off after 8 min.

Ratio of speeds = 3:4
Distance remaining constant, the ratio of time taken = 4:3
A takes 0.5 hours more than B
Hence time taken by A = 4 times 0.5 = 2 hours

Speed of track = 550 per minute.
Speed of bus
= 33km/45m
= 33000/45

∴ Ratio of their speeds
= 550 : 33000/45
= 24750 : 33000
= 8250 × 3 : 8250 × 4
= 3 : 4

Part of the tank filled in 1 min by A, B and C.
1/20 + 1/15 + 1/12
= (3 + 4 + 5)/60
= 12/60
= 1/5

Therefore, Time taken by A, B and C to fill the tank = 5 min.
Therefore, Time taken by A, B and C to fill 40% of the tank
= 40% of 5 = 40/100 × 5 = 2 minutes

Train’s speed = 240/24 = 10 m/s
The train has to cover = (240 + 650) = 890 m.
∴ Required time = 890/10 = 89 seconds

Time taken by man if he did not stop = 5 km/10 kmph = 1/2 hr = 30 min
∵Man takes rest for 5 minutes on each km Total rest time= 5×4= 20 min
Total travelling time: = 30 min+20 min = 50 min

Speed = 9 km/hr = 9 x (5/18) m/sec = 5/2 m/sec
Distance = (35 x 4) m = 140 m
Time taken = 140 x (2/5) sec = 56 sec

Speed = 132 km/hr = 132 × 5/18 m/s = 110/3 m/s
time = distance/speed = (110 + 165)/(110/3) = 7.5 seconds

If t₁ and t₂ are the upstream and down stream times.
Then time taken in still water is given by
(2 × t₁ × t₂) / (t₁ + t₂)
= (2 × 12 × 24) / 36
= 16h

Suppose B takes x days. Then,
A takes 150x/100 days, i.e. 3x/2 days
∴1/x+ 2/3x= 1/18 ⇒5/3x= 1/18 ⇒3x = 90 ⇒x= 30
Hence B takes 30 days.

Let length of the train be x m
Speed of train,
(x+162)/18 = (x+120)/15
∴ x = 90 m

Pipe B will fill the tank in 24/7 minutes as it is 7 times as fast as Pipe A.
Together, the two pipes will fill 1/24 + 7/24
= 8/24
= 1/3rd of the tank in a minute.

So, it will take 3 minutes for the tank to overflow.

Required average speed
= (3 × 15 × 20 × 30)/(15 × 20 + 20 × 30 + 30 × 15)
= 20 km/hour
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Alternative way:
Distance 30 km, Speed 30 km/h, Time 1 Hour
Distance 30 km, Speed 15 km/h, Time 2 Hours
Distance 30 km, Speed 20 km/h, Time 1 hour & 30 minutes.
Total Distance 90 km, Total Time 4.5 Hours,
Average Speed 90/4.5 km/h = 20 km/h
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Alternative way:
Let, equal distances travelled by the man be ’s’.
Time taken to travel first distance, s at 15 kmph = s/15
Time taken to travel second distance, s at 20 kmph = s/20
Tine taken to travel last distance, s at 30 kmph = s/30
Therefore, total time taken by the man = s(1/15 +1/20 +1/30)
Total distance travelled by the man = 3s
Hence, average speed of the man
= 3s / [ s(1/15 +1/20 +1/30) ] =3/(1/15 +1/20 +1/30)
= 20 kmph
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Alternative way:
Avg speed formula= 3 ÷ (1/x +1/y +1/z)
Now we have 1/30 +1/20 +1/15
L.C.M. of 20, 30 and15 is 60.
Then (3 + 4 + 8)/60=9/60 = 3/20
From above formula, average speed = 3 ÷ 3/20=20 km/h

Let the speed of car be x km/hr
Distance= Speed × Time
Distance = 8x km
According to the question,
⇒ (x+4)×7.5= 8x
⇒ 7.5x+30= 8x
⇒ 8x−7.5x= 30
⇒ 0.5x= 30
⇒x= (30/0.5)= 60 km/hr
Required distance: = 8 × 60 = 480 km

Let the time in which he traveled on foot = x hour
Time for travelling on bicycle = (9-x) hr
Distance = Speed×Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4
So distance traveled on foot = 4x4 = 16 km

Distance = Speed × time
Here time = 2hr 45 min = 11/4 hr
Distance = 4× (11/4)= 11 km
New Speed =16.5 kmph
Therefore time = D/S = 11/16.5 = 40 min