ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

(1 m+ 1w +1 b) 's 1 day work = 1/3
1/6+ 1/x+1/18 = 1/3
⇒ 4/18+ 1/x= 1/3
⇒ 1/x = (1/3- 2/9) = 1/9
⇒ x= 9
∴ 1 woman can finish the job in 9 days.

Work done by (A+B) in 20 days = (1/30×20) = 2/3 ;
remaining work = (1- 2/3 )= 1/3
1/3 work is done by B in 20 days.
Whole work is done by A in (20×3) = 60 days

Let there be x men originally.
X man finish the work in 60 days and (x+8) finish it in 50 days.
X man finish the job in 60 days.
⇒ 1 man can finish it in 50 (x+8) days.
∴ 60 x = 50(X+80)
⇒ 10 x= 400
⇒ x = 40
Hence, there were 40 men originally.

LCM OF 45 and 60 is 180.
let the capacity of tank be 180 liters.
Rate of filling of A is 4 litres per minute and B takes out water at the rate of 3 liters per minute.
So, In first minute A fill 4 units of water
In second minute B empty 3 units of water
After two minutes tanks has 1 units of water.
Note: Decrease the higher value i.e 4 from total capacity 180−4 = 176 units.
1 unit filled in 2 minutes
176 units filled in 352 minutes
Now in next minutes pipe A will fill 4 units. And tank is full so total time taken is 352+1 = 353 minutes or 5 hour 53 minutes.

Time taken by the tap to make the tank half full= 3 hrs.
Remaining part = 1/2
Part filled by 4 taps in 1 hour= (4×1/6) = 2/3
2/3 part is filled in 1 hour.
1/2 part is filled in (3/2×1/2) hr = 3/4 hr = 45 min.
Required time = 3hrs 45 min.

অপশন গুলোতে সঠিক উত্তর নেই। তাই প্রশ্নটি বাতিল করা হলো।
Suppose that one pipe takes x hours to fill the reservoir.
Than the speeder pipes takes (x-10) hours.
∴ 1/x+ 1/(x-10) = 1/12
⇒ 12(x-10+x) = x(x-10)
⇒ x²-34x +120 = 0
⇒ (x-30) (X-4) = 0
⇒ x = 30 or x = 4
∴ The speeder pipes takes (30-10) hours.
So, the speeder/faster pipe takes 20 hrs to fill the reservoir.

In 2 hours Pipe A will fill = 2/3 tank
In 1 hour Pipe B will fill = 1/4 tank
Part of tank filled till 5 PM = (2/3) + (1/4)
= 11/12
Remaining part = 1 - (11/12)
= 1/12
Net part emptied when A, b and C are opened = (1/3) +( 1/4) - 1
= (4+3-12)/12
= -5/12
∴ 5/12 part is emptied in 1 hour.
∴ 11/12 is emptied in (12/5)×(11/12) = 11/5
= 2 hr 12 min
∴ Tank will be emptied at 7:12 PM.

Part of the tank filled by the pump in 1 hour = 1/2
Part of the tank filled by the pump in 1 hour because of the leak = 3/7
∴ Part of the tank emptied by the leak in 1 hour = 1/2 - 3/7 = 1/14
∴ Leak will empty the tank in 14 hours

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.

Fulminate - Cause to explode violently and with loud noise/Come on suddenly and intensely/Criticize severely.
Languish - Become feeble.
Invigorate - Heighten or intensify.
Animate - Heighten or intensify.

Warning: It is a statement or event that serves as a cautionary example.Usually said or given in advance.
Dinigration: Is incongruous in this context as it means criticize unfairly.
Impertinence: Means not showing proper respect.
Reproof: Mild disproval therefore. D option is correct.

Auditory: Connected with hearing.
Audio-visual: using both sound and pictures.
Audition: a short performance given by an actor, singer, etc. So that somebody can decide whether they are sutiable to act in a play, sing in a concert etc.

To reach the winning post A has to cover (500-140)m = 360m
While A covers 3 m, B covers 4 m.
While A covers 360m, B covers (4/3×360)m = 480m
∴ A wins by 20 m.

A: B :C = 100: 90:87
∴ B/C = 90/87 = 90×2/87×2 = 180/174
Thus, while B covers 180m, C covers 174 m.
∴ B beats C by 6m.

A: B :C = 60:45: 40
∴ B:C = 45/40 =9/8 = 9×10/8×10 = 90/80
Thus, if B score 90 points, then C score 80 points.
B can give C 10 points in a game of 90.

Let the speed of the train be x km/hr and (x+10) km/hr.
Then, (600/x) - (600/(x+10))= 3
⇒ 1/x - 1/(x+10) =1/200
⇒ (x+10) - x/x(x+10) =1/200
⇒ x²+10x-2000=0
⇒ (x+50) (x-40) =0
⇒ x=40.
∴ speed of slow train= 40km/hr.

Let the required distance be x km. then,
x/80 - (220-x)/100 = 1/2
⇒ 5x-4 (220-x) = 200
⇒ 9x = 1800
⇒ x = 120
Hence the required distance is 120 km.

Ratio of time taken = 1/2:1/3 = 3:2
Time taken by B = 36 min. let the time taken by A be x min.
∴ x/36 = 3/2
⇒ x = (3×36/2) min. = 54 min

Distance = (speed × time) = (48×50/60) km = 40 km
New time taken = 40/60 hr = 2/3 hr
New speed = (40×3/2) km/hr = 60 km/hr

Due to stoppage it covers 5 km less per hour.
Time taken to cover 5 km= (5/50×60) min. = 6 min.
Hence, the train stops on an average 6 min/hr.

Speed downstream = (5+1) km/hr=6 km/hr
Speed upstream = (5-1) km/hr=4 km/hr.
Let the required distance be x km. Then,
x/6+ x/4=75/60 = 5/4
=> (2x+3x) =15
=> 5x = 15
=> x=3.
Required distance =3km.

Speed in still water = 6 km/hr.
Speed against the current = 6/3 km/hr = 2 km/hr
Let the speed of the current be x km/hr
6-x = 2
=> x = 4 km/hr.

Upstream Speed of motor boat = 56/1 hr 45 min = 32 kmph
Speed of the motor boat in still water = 36 kmph
Speed of the boat in still water = 1/2×(Downstream Speed + Upstream Speed)
36 = 1/2(d + 32)
Downstream Speed (d) = 40 km/hr
t = Distance/speed = 40/36 = 7/5 hr = 1 hr 24 mins

Distance covered upstream in 40min = 4 km.
Distance covered upstream in 60 min = (4/40×60) km= 6 km
Distance covered downstream a in 30 min = 4 km
Distance covered downstream in 60 min = (4/30 × 60) km = 8 km.
Speed upstream =3 km/hr, speed downstream =4 km/hr.
Speed in still water = 1/2 (6 +8) km/hr=7 km /hr

Let the speed in still water be x km/hr. Then,
Speed downstream = (x+ 4) km/hr
speed upstream = (x-4) km/hr.
6/(x+4) + 6 /(x-4) = 2
=> 1/(x+4) +1/(x-4)=2/6 =1/3
=>(x+4)+(x-4)/x²-16=1/3
=>x²-16=6x
=> x² -6x-16=0
=> (x-8) (x+2) = 0
=> x = 8
.∴ Speed of boat in still water = 8 km/hr.

let the length of the train be X km and its speed be y km/hr.
Speed of the train relative to first man = (y-2) km/hr
Speed of the train relative to second man =(y-4)km/hr
∴ x/(y-2) = 9/(60×60) and x/(y-4) =10/(60×60)
⇒ y-2 =400x and y-4 =360x
⇒ 400x+2=360x+4
⇒ 40x=2
⇒ x= 1/20km =(1/20×1000)m= 50m
∴ length of the train =50m

speed of the train = (36×5/18)m/sec. =10m/sec.
Let the length of the second train be x meters. Then, x/10 =10
⇒ x =100m
Time taken by the train to cross the platform = (100+55)/10 sec.= 31/2 sec.

Relative speed
= (36+45)km/hr
= (81×5/18)m/sec.
= 45/2m/sec.
Length of the train = distance covered in 8 sec. at 45/2m sec.
= (45/2×8)= 180 m

Let the distance between meerut and Ghaziabad be x km.
Time taken by y to cover x km =3/2 hours.
Time taken by X to cover x km = 1 hours
Speed of x = x km/hr speed of y = 2x/3 km/hr
Xy+2xy/3 =x
⇒ y (1+2/3) =1
⇒ y = 3/5 hours = (3/5×60) min. =36 min.
Hence, the two train meet at 4.36 p.m