ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: In a certain assembly constituency election, 80% of voters exercised their voting right and the winning candidate got elected with 65% of votes polled. What percent of total votes did he poll? 

Solution: 
Let, the total number of votes be 100
Then voters voted rightly = 80% of 100 = 80
and winning candidate got 65% of 80 = (65/100) × 80 = 52
Then, number of percent of votes polled by winning candidate = 52%

Question: If a man rows at the rate of 7 kmph in still water and his rate against the current is 4 kmph, then the man's rate along the current is-

Solution: 
Speed of current = 7 - 4 = 3 km
so, the speed in downstream = 7 + 3 = 10 kmph 

Question: A company blends two varieties of tea from two different tea gardens, one variety costing Tk. 30 per kg and the other Tk. 20 per kg in the ratio 4 : 3. He sells the blended tea at Tk. 27 per kg. Find the profit or loss percent. 

Solution: 
Let 4 kg of the first variety be mixed with 3 kg of second variety.
Then total cost price of 7 kg of tea = (30 × 4) + (20 × 3) = Tk. 180
Selling price of 7 kg of tea = (27 × 7) = Tk. 189

∴ Profit = Tk. (189 - 180) = Tk. 9 

∴ Percentage of profit = (9 × 100)/180 = 5%

Question: The compound interest on Tk. 3000 for 2 years at 10% is-

Solution:
Givent that,
P = Tk. 3000
n = 2 years
r = 10%

Required amount
= P(1 + r)ⁿ 
= [3000 × {1 + (10/100)² }] 
= [3000 × (11/10) × (11/10) ]
= 3630

Compound Interest = 3630 - 3000 = 630

Question: If a regular square pyramid has a base of side 5 cm and height 45 cm, then what its volume? 
 
Solution: 
Volume of the  square pyramid = (1/3) × a² × h
= (1/3) × 5²  × 45 cm³
= 375 cm³

Question: A committee of 5 members is to be formed by selecting out of 6 man and 7 women. In how many different ways the committee can be formed if it should have 2 men and 3 women?

Solution:
2 men can be selected out of 6 men in  ⁶C₂ ways
3 women can be selected out of 7 women in ⁷C₃ ways

Required number of ways = ⁶C₂ × ⁷C₃ = 15 × 35 = 525

Question: The angle of depression of a point situated at a distance of 50 m from the base of a tower is 30°. What is the height of the tower?

Solution: 

Length of the tower AB = h meter
∠DAC = ∠ACB = 30°
∴ BC = 50 

In △ ABC,
tan30° = AB/BC
⇒ 1/√3 = h/50
∴ h = 50/√3

Question: At the rate of 5% simple interest, a sum of total Tk. 4500 will earn how much interest in 3 years? 

Solution:
Given that,
P = Tk. 4500
r = 5%
n = 3 years

Simple interest 
I = P × n × r
= 4500 × 3 × 5%
= 4500 × 3 × (5/100)
= 675

Question: A train which is moving at an average speed of 30 km/h reaches its destination on time. When its average speed reduces to 25 km/h, then it reaches its destination 10 minutes late. The distance travelled by the train, is - 

Solution:  
Let the distance = x
ATQ,
(x/25) - (x/30) = 10/60
⇒ (6x - 5x)/150 = 1/6
⇒ x/150 = 1/6
⇒ x = 150/6
∴ x = 25

∴ The distance travellede by the train is 25 km

Question: A sum of money is divided among 150 males and some females in the ratio 8 : 11. Individually each male gets Tk. 4 and a female Tk. 3. What is the number of females? 

Solution:
Let the number of females be x
ATQ,
(150 × 4)/3x = 8/11
⇒ x = (150 × 4 × 11)/(3 × 8)
∴ x = 275

Question: The profit earned by selling a chair for Tk. 750 is 1.5 times the loss occurred when the same chair was sold for Tk. 300. What is the cost price of the chair? 

Solution:
Let the cost price of the chair be x
ATQ,
750 - x = 1.5 (x - 300)
⇒ 7500 - 10x = 15x - 4500 
⇒ 15x + 10x = 7500 + 4500
⇒ 25x = 12000
⇒ x = 12000/25
∴ x = 480
 
∴ The price of the chair is Tk. 480

Question: Asif, Sami and Riad started a shop by investing Tk. 2700, Tk. 9000 and Tk. 6300 respectively. At the end of one year, the profit was distributed. If Riad's share was Tk. 2100, what was their total profit? 

Solution: 
Let the total profit is = x

Here, Asif : Sami : Riad = 2700 : 9000 : 6300 
= 3 : 10 : 7 

then, Riad's share = (7/20) × x = 7x/20
ATQ,
7x/20 = 2100
⇒ x = (2100 × 20)/7
∴ x = 6000

∴ The total profit = Tk. 6000 

Question: If 15 men can built a wall 50 meters long in 5 days, what length of a similar wall can be built by 30 men in 2 days? 

Solution:
In 5 days 15 men can build = 50 meters 
∴ In 1 day 1 man can build = 50/(5 × 15) meters
∴ In 2 days 30 men can build = (50 × 30 × 2)/(5 × 15) meters = 40 meters

Question: Amir is 40 years old and Rezwan is 50 years old. How many years ago was the ratio of their ages 3 : 5?

Solution: 
Let x years ago the ratio of their ages were 3 : 5 

ATQ,
(40 - x)/(50 - x) = 3/5
⇒ 200 - 5x = 150 - 3x
⇒ 5x - 3x = 200 - 150 
⇒ 2x = 50 
∴ x = 25

Question: If the arithmetic mean of 70 numbers is calculated, it is 25. If each number is increased by 5, then mean of new number is- 

Solution: 
Sum of 70 numbers = (70 × 25) = 1750

∴ Total increase = (70 × 5) = 350 

∴ Increased some = (1750 + 350) = 2100 

∴ Increased average = 2100/70 = 30

Question: In a class of 60 students, 30 take Bengali, 20 take Hindi, 15 takes no language. How many take both Bengali and Hindi? 

Solution: 
Number of students who took one or both the languages = 60 - 15 = 45

Here,
n(B) = 30
n(H) = 20
n(B∪H) = 45

Now,
n(B∩H) = n(B) + n(H) - n(B∪H) 
= 30 + 20 - 45
= 5

∴ 5 students take both Bengali and Hindi.

Question: 


Solution:
√{1 + (x/144)} = 13/12
⇒ [√{1 + (x/144)}]² = (13/12)² 
⇒ 1 + (x/144) = 169/144
⇒ x/144 = (169/144) - 1
⇒ x/144 = 25/144
∴ x = 25

Question: A tap can fill a tank in 8 hours. After half the tank is filled, two more similar taps are opened. What is the total time taken to fill the tank completely?

Solution:
Time taken by one tap to fill half the tank = 4 hrs
∴ Remaining part after 4 hrs = 1 - (1/2) = 1/2 

1 pipe can fill in 1 hour = 1/8 part 
Part filled by three taps in 1 hour = 3 × (1/8) = 3/8

3/8 is filled by three taps in = 1 hr
1/2 is filled by three taps in = (8/3) × (1/2) hrs 
= (4/3) × 60 min = 80 min = 1 hr 20 min

So, total time taken = 4 hrs + 1 hr 20 min = 5 hrs 20 min

Question: A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work is - 

Solution:
Ratio of rates of working of A and B = 2 : 1 
So, the ratio of time taken = 1 : 2

Since, B takes 12 days, A takes 6 days 

∴ (A + B)'s 1 day's work = (1/6) + (1/12) 
= 3/12
= 1/4 

∴ A + B can finish the work in (4/1) = 4 days

Question: If the area of a triangle is 1125 cm² and its base : corresponding altitude is 2 : 5, what is the base of the triangle?

Solution:
Let, the base = 2x cm
and altitude = 5x cm

ATQ,
(1/2) × 2x × 5x = 1125
⇒ 5x² = 1125
⇒ x² = 225
∴ x = 15

∴ Base of the triangle = (2 × 15) cm = 30 cm

Question: A train crosses a platform 90 m long in 50 seconds at a speed of 36 km/hr. What is the time taken by the train to cross an electric pole?

Solution:
Speed of the train = (36 × 1000)/(60 × 60) = 10 m/sec
So, in 50 sec it goes = (10 × 50) = 500 m

Since the platform is 90 m so, only train = (500 - 90) m = 410 m  [ খুঁটি অতিক্রম করতে হলে শুধু এতটুকু যেতে হবে ] 

∴ Time taken by the train to cross an electric pole = (410/10) sec = 41 sec

Question: The ratio of two numbers is 5 : 7 and their L.C.M is 420. What is the sum of the numbers? 

Solution: 
Let, the numbers be 5x and 7x 
∴ Their L.C.M = 35x  

ATQ,
35x = 420 
⇒ x = 420/35
∴ x = 12

∴ The numbers are (5 × 12) = 60 and (7 × 12) = 84

∴ Sum of the numbers = 60 + 84 = 144

Question: A fraction becomes 1/2 when 1 is added to both its numerator and denominator, and it becomes 1/4 when 1 is subtracted from both the numerator and denominator. Find the fraction. 

Solution: 
Let the required fraction is x/y 
Then,
(x + 1)/(y + 1) = 1/2
⇒ 2x + 2 = y + 1
∴ 2x - y = -1 ........(i)

And
(x - 1)/(y - 1) = 1/4
⇒ 4x - 4 = y - 1
⇒ 4x - y = 4 - 1
∴ 4x - y = 3 ........(ii)

Now, 
Multiplying (i) by 2 then subtracting (ii) from (i) we get, 
-2y + y = - 2 - 3
⇒ -y = - 5
∴ y = 5

Putting the value of y in (i) we get,
2x - 5 = - 1
⇒ 2x = - 1 + 5
⇒ 2x = 4
∴ x = 2 

So, the required fraction = 2/5

Question: In a 300 m race, A beats B by 30 m and C by 40 m. In a race of 540 m, B will beat C by -

Solution:
Given that, 
A : B = 300 : 270 
and A : C = 300 : 260

A/B = 300/270
and A/C = 300/260

∴ B/C = (B/A) × (A/C)
⇒(270/300) × (300/260) = 270/260
⇒ (270 × 2)/(260 × 2) = 540/520
⇒ B/C = 540/520

∴ B : C = 540 : 520

∴ In a 360 m race, B beats C by (540 - 520) m = 20 m

Question: How many 3-digit number are there between 200 and 400, having first and last digit as 3?

Solution:
The numbers between 200 and 400 having first and last digit as 3 are,
303, 313, 323, 333, 343, 353, 363, 373, 383, 393
∴ Total number = 10 

Question: The circumference of a circular plot is 352 meters. What is the area of the circular plot? 

Solution:
Given that,
2πr = 352
⇒ r = (352 × 7)/(2 × 22) 
∴ r = 56 m

Now, 
Area of the circular plot,
= πr²
= {(22/7) × 56 × 56}m²
= 9856 m²