ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: Three numbers are in the ratio 1 : 2 : 3, and the sum of their cubes is 7776. The smallest number will be -

Solution:
Given,
The numbers be in the ratio 1 : 2 : 3
So, let:
Smallest number = x
Middle number = 2x
Largest number = 3x

ATQ,
x³ + 8x³ + 27x³ = 7776
⇒ 36x³ = 7776
⇒ x³ = 7776/36 = 216
∴ x = 6

So the smallest number is 6

Question: The least number which when divided by 4, 6, 8, 12 and 16 leaves remainder of 2 in each case is = ?

Solution: 
LCM of (4, 6, 8, 12, 16)
⇒ 16 × 3 = 48

∴ The number when divided by (4, 6, 8, 12, 16) leaves remainder 2 is
= 48 + 2
= 50

Question: The average age of a family of 5 members is 20 years. If the age of the youngest member be 10 years then what was the average age of the family at the time of the birth of the youngest member?

Solution: 
At present the total age of the family = 5 × 20 =100
The total age of the family at the time of the birth of the youngest member,
= 100 - 10 - (10 × 4)
= 50

Therefore, average age of the family at the time of birth of the youngest member,
= 50/5
= 10

Question: 6 years ago, the ratio of the ages of Kamrul and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present ?

Solution:
Let the ages of Kamrul and Sagar 6 years ago be 6x and 5x years
Then,
{(6x + 6) + 4}/{(5x + 6) + 4} = 11/10
⇒ (6x + 10)/(5x + 10) = 11/10
⇒ 10(6x + 10) = 11(5x + 10)
⇒ 60x + 100 = 55x + 110
⇒ 5x = 10
∴ x = 2

∴ Sagar's present age,
= (5x + 6) years
= (5 × 2 + 6) years
= 16 years

∴ Sagar's present age is 16 years. 

Question: A train running at the speed of 60 kmph crosses a 200 m long platform in 27 seconds. What is the length of the train?

Solution: 
Given that,
Time = 27 sec
∴ Speed = (60 × 5/18)m/sec = 50/3 m/sec

Let the length of the train be x metres.
Then,
(x + 200)/27 = 50/3
⇒ x + 200 = (50/3) × 27 
⇒ x + 200 = 450
⇒ x = 450 - 200
∴ x = 250 metres

So the length of the train is 250 meters.

Question: 3 pumps, working 8 hours a day, can empty a tank in 2 days, How many hours a day must 4 pumps work to empty the tank in 1 day?

Solution:
3 pumps, working 8 hours a day, can empty a tank in 2 days
Formula used: M₁ × T₁ = M₂ × T₂
Where M₁ and M₂ is men and T₁ and T₂ is time

Calculation:
Let H hours be the number of hours required Applying the above formula
⇒ 3 × 8 × 2 = 4 × 1 × H
⇒ H = 48/4
⇒ H = 12 hours

∴ 4 pump need to work 12 hours to complete the work in 1 day.

Question: A, B and C start together from the same place to walk round a circular path of length 12 km. A walks at the rate of 4 km/h, B 3 km/h and C 3/2 km/h. They will meet together at the starting place at the end of-

Solution:
Given that,
Circumference of circular path = 12 km
Speeds, A = 4 km/h
B = 3 km/h
C = 3/2 = 1.5 km/h

We know, 
Time = Distance/Speed

Than, 
Time for A = 12/4 = 3 hours
Time for B = 12/3 = 4 hours
Time for C = 12/1.5 = 8 hours

∴ Required time,
= LCM of 3, 4, 8.
= 24 hours.

∴ They will meet together at the starting place at the end of 24 hours.

Question: Three taps A, B, C can fill an overhead tank in 4, 6 and 12 hours respectively. How long would the three taps take to fill the tank if all of them are opened together?

Solution:
Work with rates (fraction of tank per hour).
A fills 1/4 per hour.
B fills 1/6 per hour.
C fills 1/12 per hour.
∴ Combined rate = 1/4 + 1/6 + 1/12
= (3 + 2 + 1)/12 
= 6/12
= 1/2 (tank per hour).

∴ Time to fill one tank = 1 ÷ (1/2) = 2 hours.

∴ The three taps would take 2 hours to fill the tank if all of them are opened together.

Question: With an average speed of 50 km/hr, a train reaches its destination in time. If it goes with an average speed of 40 km/hr, it is late by 30 min. The total journey is-

Solution:
Difference between timings = 30 min = 30/60 hr = 1/2 hr.
Let the length of the journey be x km.

Then,
(x/40) - (x/50) = 1/2
(5x - 4x)/200 = 1/2
x/200 = 1/2
x = 200/2
∴ x = 100 km.

∴ The total journey is 100 km.

Question: In one hour, a boat goes 11 km/hr along the stream and 7 km/hr against the stream. The speed of the boat in still water (in km/hr) is-

Solution:
Speed in still water = (11 + 7)/2 kmph
= 18/2 kmph
= 9 kmph.

Question: The number 2272 and 875 are divided by a 3 digit number N, giving the same remainders. The sum of the digit is-

Solution: 
Let the remainder in each case be x
Then, (2272 - x) and (875 - x) are exactly divisible by three digit number
Difference :
= (2272 - x) - (875 - x)
= 1397
Factor of 1397 = 11 × 127
Since, both 11 and 127 are prime number
Three digit number is 127

∴ Sum of digits = 1 + 2 + 7 = 10

Question: The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is = ?

Solution: 
Given that, 
LCM = 495, HCF = 5
And,
The sum of the numbers is, a + b = 100
The product of two numbers is equal to the product of their LCM and HCF. 
Product(a × b) = LCM × HCF = 495 × 5 = 2475

We know, 
(a - b)² = (a + b)² - 4ab
⇒ (a - b)² = 100² - 4 × 2475 ; [where, a + b = 100 and  ab = 2475 ]
⇒ (a - b)² = 10000 - 9900
⇒ (a - b)² = 100
⇒ a - b = √100 = 10
∴ a - b = 10 

So the difference between the numbers is  10.

Question: In the first 10 overs of a cricket game, the run rate was only 3.2 runs per over. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?

Solution:
Given that, 
Target = 282 runs
Runs scored in first 10 overs = 10 × 3.2 = 32 runs
∴ Runs remaining = 282 - 32 = 250 runs

∴ Remaining overs = 50 - 10 = 40 overs

∴ Required run rate in the remaining 40 overs
= 250/40
= 6.25 runs per over

Question: If 12 men can do a piece of work in 14 days, the time taken by 14 men to do the same piece of work will be - 

Solution: 
from 12 men can do a piece of work in 14 days, we get - 
M₁ = 12
D₁ = 14

The time taken by 14 men to do the same piece of work will be
M₂ = 14
D₂ = ?

we know that,
M₁ × D₁ = M₂ × D₂
D₂ = (12 × 14)/14
∴ D₂ = 12 days

Question: P's age 3 years ago was three times the present age of Q. At present R's age is twice the age of Q. Also R is 12 years younger than P. What is the present age of R?

Solution:
Let the present age of Q be a years
Three years ago P's age = 3a years
Then, present age of P is (3a + 3)
R's present age = 2a

According to question
(3a + 3) - 2a = 12
a = 9 year

∴ Present age of R = 2a = 2 × 9 = 18 years

Question: A train 110 metres long is running at a speed of 60 km/h. In what time will it pass a man who is running at 6 km/h in the direction opposite to that in which the train is going?

Solution:
Given that,
Length of train = 110 meters
Speed of train = 60 km/h
Speed of man = 6 km/h (in opposite direction)

Now, Since the train and the man are moving in opposite directions, their relative speed is the sum of their individual speeds.
 S_relative = S_train + S_man = (60 + 6)km/h = 66 × (5/18) = 55/3 m/s

Than, 
The train passes the man when it covers its own length (110 m) at relative speed.

∴ Time = Distance/Relative speed
= 110/(55/3)
= 110 × (3/55)
= 2 × 3
= 6 seconds

∴ The train will pass the man in 6 seconds.

Question: If 5 workers can collect 75 kg wheat in 3 days, how many kilograms of wheat will 8 workers collect in 5 days ?

Solution: 
5 workers 3 days collection = 75 kg
1 worker 1 day collection = (75/15) kg
8 workers 5 days collection = (75 × 40)/15 kg
= 200 kg

∴ 8 workers will collect 200 kg in 5 days.

Question: In a race of 1km, A can beat B by 100m. In a 400m, B beats C by 40m. In a race of 500m. A will beat C by-

Solution:
We know, 
1km = 1000m

∴ While A covers 1000 B covers 900
∴ while A covers 500 B covers 450m

∴ While B covers 400, C covers 360m
∴ While B covers 450, C covers (360 × 450)/400 = 405m

∴ in a 500m race A will beat C  by = (500 - 405) = 95m

That means when A runs 500 meter then B can run 450m then C runs 405m.

Question: Pipe A can fill the tank in 8 hours and pipe B can fill it in 12 hours. If pipe A is opened at 8:00 AM and pipe B is opened at 10:00 AM, then at what time will the tank be full ?

Solution: 
A opened 2 hours early to B
In 2 hours A can do 3 × 2 = 6 unit work
Remaining work = 24 - 6 = 18
A + B can do it in
= 18/5 hours
= 3 hours 36 minutes
∴ Tank will be full in 10 A.M. + 3 hours 36 minutes = 1 : 36 P.M.

Question: The speeds of three cars are the ratio 2 : 3 : 4. The ratio of the taken by these cars to travel the same distance is-

Solution:
Given that,
Speed ratio of three cars,
v₁ : v₂ : v₃ = 2 : 3 : 4 
Let, 
v₁ = 2k, v₂ = 3k, v₃​ = 4k (for some constant k)

We know,
Time = distance​/Speed
∴ t₁​ = d/2k​, t₂ ​= d/3k​, t₃ ​= d/4k

∴ Ratio of time = t₁ : t₂ : t₃ = d/2k : d/3k : d/4k 
= 1/2 : 1/3 : 1/4    ; [Cancel d and k (since d,k ≠ 0)]​
= 12/2 : 12/3 : 12/4.  ; [LCM of 2, 3, 4 = 12]
= 6 : 4 : 3

So the ratio of the time taken is 6 : 4 : 3

Question: The speed of a boat in still water is 15 km/h, and the speed of the current is 5 km/h. In how much time (in hours) will the boat travel a distance of 60 km upstream and the same distance downstream?

Solution: 
Given that, 
Speed of boat in still water = 15 km/h
Speed of current = 5 km/h
Distance (each way) = 60 km

∴ Downstream speed = 15+ 5 = 20 km/h
∴ Upstream speed = 15 - 5 = 10 km/h

∴ Time to travel 60 km downstream = 60/20 = 3 hours
∴ Time to travel 60 km upstream = 60/10 = 6 hours

∴ Total time = 3 + 6 = 9 hours

∴ The boat will take 9 hours in total.

Question: The mean of the five observations x, x + 2, x + 4, x + 6, x + 8 is 11. Then the mean of the first three observations is?

Solution:
Given that,
The mean of the five observations = 11
Number of observations = 5
The observations = x, x + 2, x + 4, x + 6, x + 8

We know, 
Mean or Average = Sum of observations ÷ Number of observations
Sum of observations = x + x + 2 + x + 4 + x + 6 + x + 8
⇒ 5x + 20 = 5(x + 4)

∴ Mean = [5(x + 4)] ÷ 5
⇒ 11 = x + 4
⇒ x = 7

∴ First three observations,
x = 7, x + 2 = 9, x + 4 = 11

∴ Mean of first three = (7 + 9 + 11)/ 3
= 27/3
= 9

Question: Two positive numbers are in the ratio 3 : 2. The product of their HCF and LCM is 3456. Find the sum of both the numbers.

Solution:
Let two numbers are 3a and 2a.

We know,
HCF × LCM = 1st no. × 2nd no.
⇒ 3456 = 3a × 2a
⇒ 3456 = 6a²
⇒ a² = 576
⇒ a = 24 

∴ Sum of both the numbers = 3a + 2a = 5a = 5 × 24 = 120

Question: Find the mother age after 8 years, if the ratio of age of mother and son is 5 ∶ 2 and the product of their ages in years is 1000.

Solution:
Given that,
Mother : Son = 5 ∶ 2
Product of ages = 1000

Let ages of mother and son be 5x, 2x

According to problem,
5x × 2x = 1000
⇒ 10x² = 1000
⇒ x² = 100 = 10²
∴ x = 10

So, age of mother after 8 years = (5x + 8) 
= (5 × 10 + 8)
= 50 + 8
= 58 years

∴ The mother age is 58 years.