ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Let,
The distance between A and destination be X
The total time taken by the car to cover X = 8 hours
Since X/2 by 80km/hr and remaining X/2 by 100km/hr
Then by Time = distance/speed, we have
(X/2)/80 + (X/2)/100 = 8
⇒ 1/2 {(X/80 + X/100)} = 8
⇒ X/80 + X/100 = 16
⇒ (5X + 4X)/400 = 16
⇒ 5X + 4X = 6400
⇒ 9X = 6400
⇒ X = 6400/9
⇒ X = 711.11
Hence 711.11 km is the required answer.

The man starts with 7 kmph
Distance covered in first 1 hour = 7 km
He increases his speed every hour by 1 km.
Speed in 2nd hour = 8 km/hr
Distance covered in 2nd hour = 8 km
Remaining distance = 19.5 - (7 + 8)
19.5 - 15
= 4.5 km
Speed in the third hour = 9 km/hr
Time taken to cover 4.5 km at 9 km/hr
= 4.5/9
= 1/2 hour.
Therefore,
Total time = 1 + 1 + (1/2) hours = 2 (1/2) hours.
Hence, the answer is 2(1/2) hours.

Here there are 2 cases
Case 1: accident happens at 50 km
Case 2: accident happens at 60 km
Difference between two cases is only for the 10 kms between 50 and 60.
Time difference of 10 minutes is only due to these 10 kms.
In case 1, 10 kms between 50 and 60 is covered at (3/4)^th Speed.
In case 2, 10 kms between 50 and 60 is covered at usual Speed.
So the usual Time 't' taken to cover 10 kms, can be found out as below.
(4/3)t – t = 10 mins [s = vt]
⇒ t = 30 mins, d = 10 kms
so usual
Speed = Distance/Time
= 10/30min
= 10/(1/2) km/hr.
= 20 km/hr.

Let,
The distance between home and office =d
Suppose he reaches the office on Time,
the Time taken = a minutes
Case 1: When he reaches office 20 minutes late,
Time taken = a + 20
Case 2: When he reaches office 10 minutes early,
Time taken = a - 10
As the distance traveled is the same,
The ratio of Speed in case 1 to the Speed in case 2 will be the inverse of the Time taken in both cases.
Ratio of Speed in both cases = 4:6
= 2:3
Ratio of Time in both cases = 3:2
Therefore,
(a + 20)/(a -10)= 3/2
⇒ 2a + 40 = 3a -30
⇒ 3a - 2a = 40 + 30
⇒ a = 70 minutes.
Taking case 1,
4= d/(90/60)
⇒ d= 360/60
= 6 km.

Pipe 1 can fill 1/2 of the cistern in 1 hour
Pipe 2 can fill 1/6 of the cistern in 1 hour
Pipe 3 can empty 1/9 of the cistern in 1 hour
According to the question,
Time taken to full the cistern = 1/2 + 1/6 - 1/9
= 5/9
5/9 of the cistern will be filled in 1 hour.
Full cistern will be filled in = (5/9)/1
= 9/5 x 1
= 1.8 hours

We know that,
Two pipes A and B can fill (or empty) a tank in X and Y minutes respectively, while working alone.
If both the pipes are opened together,
then the time taken to fill (or empty) the cistern is given by XY/(X+Y) minutes.
Here,
X = 30 minutes and Y = 40 minutes
Therefore,
the required time = (30 x 40)/(30 + 40)
= 1200/70
= 120/7
= 17(1/7) minutes.
Hence the answer is 17(1/7) 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.

Let the time taken to fill the cistern by 3 taps A, B and C be X, Y, and Z minutes respectively.
Then the short cut formula for,
Time taken to fill the tank when all the pipes are opened = XYZ/(XY + YZ + ZX) minutes
Here, X = 9 minutes, Y = 6 minutes and Z = 3 minutes.
Now the required time = {(9) × (6) × (3)}/{(9 × 6) + (6 × 3) + (3 × 9)} minutes
= (9 × 6 × 3)/(54 + 18 + 27)
= (9 × 6 × 3)/9{6 + 2 + 3}
= 6 × 3/(6 + 2 + 3)
= 18/11
= 1(7/11) minutes
Hence the answer is 1(7/11) minutes.

Let,
The pipe Z alone takes A minutes to fill the tank.
Given that,
Y is thrice as fast as Z.
Then, Y takes A/3 minutes to fill the tank.
And, X is thrice as fast as Y.
X takes (A/3)/3 = A/9 minutes to fill the tank.
Now,
Part filled by X in 1 minute = 9/A
Part filled by Y in 1 minute = 3/A
Part filled by Z in 1 minute = 1/A
Net part filled by (X+Y+Z) in 1 minute = 9/A + 3/A + 1/A
= 13/A
(X+Y+Z) take 10 minutes to fill the cistern.
Part filled by (X+Y+Z) in 1 minute = 1/10
Thus,
We have,
1/10 = 13/A
⇒ A = 13 × 10
⇒ A = 130
Therefore, Z alone takes 130 minutes
So, X can fill the cistern in 130/9 minutes.
Hence, the correct answer is - ঘ) none of these

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
= 1/30.
Therefore,
The time taken by (A + B) to empty the full tank is 30 minutes.
Time taken to empty 1/2 part of the tank is
= 30/2
= 15 minutes.

Net height ascended in 2 min = (6 - 3) m = 3 m.
Net height ascended in 36 min = (3/2 × 36) = 54m.
In the 37th min,
the monkey ascends 6m and reaches the top.
Hence,
Total time taken = 37 minutes.

Time taken,
7.25 P.M. - 4.10 P.M. = 3 hr. 15 min.
= 3 × (15/60) hrs.
= 3(1/4) hrs.
= (13/4) hrs.
∴ Required distance = {40 × (13/4)} km.
= 130 km.

Let,
The distance be x km.
Then,
Time taken to cover x km by cycle + Time taken to cover x km by scooter = 6 hr 30 min
⇒ (Time taken to cover 2x km by cycle) + (Time taken to cover 2x km by scooter) = 13 hrs
But,
Time taken to cover 2x km by cycle = 8 hr 40 min.
∴ Time Taken to cover 2x km by scooter = 13 hrs - 8 hr 40 min
= 4 hr 20 min.
Hence, required time = 4 hours 20 min.

Let,
A's speed = x km/hr.
Then,
B's speed = (7 - x) km/hr.
So,
24/x + 24/(7 - x) = 14
⇒ {24(7 - x) + 24x} = 14x(7 - x)
⇒ 168 - 24x + 24x = 98x - 14x²
⇒ 14x² - 98x + 168 = 0
⇒ x² - 7x + 12 = 0
⇒ x² - 4x - 3x + 12 = 0
⇒ x(x - 4) - 3(x - 4) = 0
⇒ (x - 3) (x - 4) = 0
⇒ x = 3 or x = 4
Since A is faster than B, so A's speed = 4 km/hr and B's speed = 3 km/hr.

Total distance travelled = (39 + 25)
= 64 km
Total time taken = (45 + 35)
= 80 min.
= (80/60) hr.
= (4/3) hr.
∴ Average speed = {64 × (3/4)} km/hr
= 48 km/hr.
Hence, the average speed of the car is 48 km/hr.

Distance covered by the first plane till 2 P.M.
i.e., in 2 hrs = (500 × 2) miles
= 1000 miles.
Distance covered by the second plane till 2 P.M.
i.e., in 1(1/2) hrs = (400 × 3/2) miles
= 600 miles.
∴ Required distance
= AB =√{(1000)² + (600)²}
=√(1000000 + 360000)
=√1360000
= 200√34 miles
= 200 × 5.83
= 1166 miles.

Average speed = {(2 × 500 × 700)/(500 + 700)} km/hr
= (1750/3) km/hr
= 583(1/3) km/hr.
Hence, The average speed of the aeroplane for the entire journey is 583(1/3) km/hr

Man's speed with the current = 15 km/hr
Speed of the man + Speed of the current = 15 km/hr
Speed of the current is 2.5 km/hr
Hence, Speed of the man = 15 - 2.5 = 12.5 km/hr
Man's speed against the current = (Speed of the man - Speed of the current)
= 12.5 - 2.5
= 10 km/hr.

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) minutes.
= 24 minutes
Now,
The man has to cover remaining 2 km in 24 minutes
or 24/60
= 2/5 hours
Speed required for remaining 2 km
= 2 km/(2/5)hr
= 5 km/hr.

Let,
The speed is x km/hr.
Then,
Distance covered in 1 hr. 40 min
i.e., 1(2/3) hrs. = 5x/3 km.
Remaining Distance = {24 - (5x/3)}
∴ 5x/3 = 5/7{24 - (5x/3)
⇒ 5x/3 = 5/7{(72 - 5x)/3}
⇒ 7x = 72 - 5x
⇒ 12x = 72
⇒ x = 72/12
⇒ x = 6.
Hence, Speed = 6 km/hr
= {6 × (5/18)} m/s
= 5/3 m/s
= 1(2/3) m/s.

Velocity of the river = [100 √{(1/4²) - (1/5²)}] m/min
= [100 √{(1/16) - (1/25)}] m/min
= 100{√(9/400)} m/min
= 100 × (3/20) m/min
= 15 m/min.

Base fare = Tk. 40
Distance Charge = Tk. 3.6 per 0.2 km = Tk. 18 per km
Travelling Time Charge = Tk. 180/1 hour
= 180/60
= Tk. 3 per minute
Total Distance = 6 kms
Total Time = 110 mins
Total Charge = Tk. 40 + Tk. (6 × 18) + Tk. (110 × 3)
= Tk. 478.

ধরি,
একটি জাহাজ A speed - এ যাচ্ছে এবং অপরটি B speed এ যাচ্ছে।
নদীর প্রস্থ w
At time = t₁
A (t₁) = 750
B(t₁) =(w - 750)
⇒ B/A = (w - 750)/750
750B = (w - 750)A
At time = t₂
A(t₂) = w + 250
B(t₂) = w +(w - 250)
= 2w - 250
⇒ B/A = (2w - 250)/(w + 250)
(w + 250)B = (2w - 250)A .... Eq.[1]
750B = (w - 750/A
B = [(w - 750)/750]A
Plug into Eq. [1]
(w + 250)[(w - 750)/750]A = (2w - 250) A
(w + 250)(w - 750)/750 =750(2w -250)
w² - 500w - 187,500 = 1500w - 187500
w2 - 2000w = 0
w2 = 2000w
w = 2000 yards.

ঢাকা থেকে চট্টগ্রামগামী ট্রেনটি 1 ঘন্টায় 62 km. যায়।
চট্টগ্রাম থেকে ঢাকাগামী ট্রেনটি 1 ঘন্টায় 48 km. যায়।
∴ 1 ঘন্টায় দুটি বিপরীতমুখী ট্রেন মোট যায় (62 + 48)
= 110 km. দূরত্ব অতিক্রম করে।
অর্থাৎ মুখোমুখি হওয়ার 1 ঘন্টা আগে তাদের দুরত্ব = 110 km.

ধরি,
Distance of store = x feet
Now, Distance covered by son in one minute = 1 × 20
= 20 ft. and
Distance covered by father = 30 × 1.5
= 45 ft.
∴ Time taken by son = x/20 minutes = Time taken by father = x/45 minutes.
∴ x/20 = x/45 + 10
⇒ (x/20 - x/45) = 10
⇒ (9x - 4x)/180 = 10
⇒ 5x = 1800
⇒ x = 1800/5
⇒ x = 360 ft.

AC = 40 × 4
= 160 km
BD = 35 × 4
= 140 km
BC = 200 - 160
= 40 km
AD = 200 - 140
= 60 km
∴ CD = 200 - BC - AD
= 200 - 40 - 60
= 100 km
Hence, 100 km apart will the two cars be after four hours of continuous travelling.