ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The average of the largest and smallest 3 digits numbers formed by 0, 2, and 4 would be:

Solution:
ATQ,
Largest number = 420
Smallest number = 204

∴ Average = (420 + 204)/2
= 624/2
= 312

Question: Tomal is younger than Jamal by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Jamal?

Solution:
Let Jamal's age be x years.
Then, Tomal's age = (x - 7) years.

 ATQ,
(x - 7)/x = 7/9
⇒ 9((x - 7) = 7x
⇒ 9x - 63 = 7x
⇒ 9x - 7x = 63
⇒ 2x = 63
∴ x = 31.5

Jamal's age is 31.5 years.

Question: The average of 2, 7, 6, and m is 5 and the average of 18, 1, 6, m and n is 10. What is the value of n? 

Solution: 
Given that
average of 2, 7, 6, m is 5

Therefore,
5 = (2 + 7 + 6 + m​)/4
⇒ 20 = 15 + m
⇒ m = 20 - 15
∴ m = 5

Therefore,
10 = (18 + 1 + 6 + m + n​)/5
⇒ 50 = 25 + 5 + n
⇒ n = 50 - 30 
∴ n = 20

Question: The ratio of the present ages of Rahim and Karim is 9 : 10 respectively. Twenty years ago, the ratio of their ages was 4 : 5 respectively. What is the present age of Rahim?

Solution:
The ratio of the present ages of Rahim and Karim = 9 : 10
Let's consider the present ages of Rahim and Karim as 9x and 10x

According to the question, 
(9x - 20) : (10x - 20) = 4 : 5
⇒ (9x - 20)/(10x - 20) = 4/5
⇒ 5(9x - 20) = 4(10x - 20)
⇒ 45x - 100 = 40x - 80
⇒ 45x - 40x = 100 - 80
⇒ 5x = 20
∴ x = 4 

The present age of Rahim = (9 × 4) years.
= 36 years.

Question: The mean of the first 12 even natural number numbers is-

Solution:
First 10 even natural numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

Mean = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24​)/12
= 156/12
= 13

Question: Mitu is 60 years old and Ritu is 80 years old. How many years ago their age ratio was 4 : 6?

Solution:
Let us assume x years ago.
At present, Mitu is 60 years old
and Ritu is 80 years old.
X years ago, Mitu’s age = (60 - x) and Ritu’s age (80 - x)

ATQ, 
(60 - x)/(80-x) = 4/6
⇒ 6(60 - x) = 4(80 - x)
⇒ 360 - 6x = 320 - 4x
⇒ 6x - 4x = 360 - 320
⇒ 2x = 40
∴ x = 20

Therefore, 20 years ago their age ratio was 4 : 6.

Question: The mean weight of a class of 20 students is 48 kg. Two more students weighing 60 kg and 58 kg respectively join the class. What is the mean weight of the class now?

Solution:
The mean Weight of 20 students = 48 
∴ Total weight = 20 × 48 = 960

Add two students weighing 60, 58 to 960 = 960 + 60 + 58 = 1078

Therefore, Mean = 1078/22
= 49 kg

So the new mean weight of the class is 49 kg.

Question: Four numbers are in the ratio 1 : 2 : 3 : 4 and their HCF 12. The average of four numbers is -

Solution:
Let the four numbers are x, 2x, 3x, 4x respectively 
Their HCF = x

ATQ,
x = 12

So, the four numbers are 12, 24, 36 and 48 respectively 
The average of the four number = (12 + 24 + 36 + 48)/4 = 30

Question: Mofiz got married 10 years ago. His present age is 6/5 times his age at the time of his marriage. What is the present age of Mofiz?

Solution:
Let, Mofiz's age 10 years ago be x years
His present age is = x + 10 years

ATQ,
x + 10 = 6x/5
⇒ 5x + 50 = 6x
⇒ x = 50

His present age is = 50 + 10 = 60 years

Question: The average of 25 results is 18. The average of the first 12 of those is 14 and the average of the last 12 is 17. What is the 13th result?

Solution:
Sum of 1st 12 results = 12 × 14 = 168
Sum of last 12 results = 12 × 17 = 204
Let,
13th result = x 

ATQ,
168 + 204 + x = (25 × 18)
⇒ 372 + x = 450
⇒ x = 450 - 372
∴ x = 78

Question: The difference between the ages of the two men is 10 years. 15 years ago, the elder one was twice as old as the younger one. The present age of the elder man is-

Solution:
Let, their ages be x years and (x + 10) years.

ATQ,
(x + 10 - 15) = 2(x - 15)
⇒ x - 5 = 2x - 30
⇒ 2x - 30 - x + 5 = 0
⇒ x - 25 = 0
∴ x = 25

∴ Present age of the elder man = x + 10 = 25 + 10 = 35 years.

Question: Find the average of all the numbers between 11 and 54 which are divisible by 5.

Solution:
Numbers between 11 and 54 divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50.
Required average =(15 + 20 + 25 + 30 + 35 + 40 + 45 + 50​)/8
= 260/8
= 32.5

Question: The sum of the ages of X, Y, and Z is 120 years. Ten years ago, the ratio of their ages was 2 : 3 : 4. What is the present age of Y?

Solution:
Let, X, Y, and Z's ages 10 years ago be 2a, 3a, and 4a years respectively

ATQ,
(2a + 10) + (3a + 10) + (4a + 10) = 120
⇒ 9a  + 30 = 120
⇒ 9a = 90
∴ a = 10

So, the age of Y = 3 × 10 + 10 = 40 years.

Question: The average of six numbers is p and the average of three of these is q. If the average of the remaining three is r, then which one is correct?

Solution:
ATQ,
6p = 3q + 3r
⇒ p = 3(q + r)/6
⇒ p = (q + r)/2
∴ 2p = q + r

Question: Five years ago, the average age of A and B was 18 years. At present the average age of A, B, and C is 24 years. What would be the age of C after 5 years?

Solution:
The sum of the ages of A and B, 5 years ago = 18 × 2 = 36 years.
The sum of the present age of A and B = 5 + 5 + 36 = 46 years.
Sum of the present ages of A, B, and C = 24 × 3 = 72 years.

∴ Present age of C = 72 - 46 = 26 years.

∴ C's age after 5 years = 26 + 5 = 31 years.

Question: The average age of 9 students and their teacher is 16 years. The average age of the first four students is 19 years and that of the last five is 10 years. The teacher's age is -

Solution:
ATQ,
The average age of nine students and teachers = 16 years
Then, the total average age of students and teachers = 16 × 10 = 160
And, the average age of the first 4 students = 19 × 4 = 76
Average age of last 5 students = 10 × 5 = 50

∴ Teacher's age = 160 - 76 - 50 = 34 years

Question: The sum of the ages of a daughter and her mother is 56 years. After 4 years, the age of the mother will be three times that of the daughter. At present their ages are?

Solution:
Let, the daughter's age be x years.
Then, mother's age = (56 - x) years.

ATQ,
(56 - x) + 4 = 3(x + 4)
⇒ 60 - x = 3x + 12
⇒ 4x = 48
⇒ x = 12

∴ Daughter's age = 12 years
And the mother's age = (56 - 12) = 44 years.

Question: The average mark obtained by 22 candidates in an examination is 43. The average marks of the first ten are 45 and the last eleven are 40. The number of marks obtained by the 11^th candidate is-

Solution:
Total marks scored by 22 candidates = 22 × 43
= 946
Total marks scored by first 10 candidates =10 × 45
= 450
Total marks scored by last 11 candidates = 11 × 40
= 440

∴ Marks scored by 11th candidate = 946 - (450 + 440)
= 56

Question: Asif got married 8 years ago. Today his age is 9/7 times his age at the time of his marriage. At present his daughter's age, is one-sixth of his age. What was his daughter's age 3 years ago?

Solution:
Let, Asif's age 8 years ago be x years.
Then, his present age (x + 8) years.

ATQ,
x + 8 = 9x/7
⇒ 9x = 7x + 56
⇒ 9x - 7x = 56
⇒ 2x = 56
∴ x = 28

Asif's age now = x + 8 = 28 + 8 = 36 years.
His daughter's age now = 36/6 = 6 years.
His daughter's age 3 years ago = (6 - 3) = 3 years.

Question: 13 chairs and 5 tables were bought for 8280. If the average cost of a table be Tk. 1227, what is the average cost of a chair?

Solution:
The total cost of 5 tables = (1227 × 5) = Tk. 6135
The total cost of 13 chairs = 8280 - 6135 = Tk. 2145

∴ Average cost of a chair = 2145/13
= Tk. 165

Question: Tamim has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is:

Solution:
Let Tamim's average be x for 9 innings.
So, Tamim scored 9x runs in 9 innings.

In the 10th inning, he scored 100 runs then the average became (x + 8).
And he scored (x + 8) × 10 runs in 10 innings.

ATQ,
9x + 100 = 10 × (x + 8)
or, 9x + 100 = 10x + 80
or, x = 100 - 80
or, x = 20

New average = (x + 8) = 28 runs.

Question: At present, the ratio between the ages of Hassan and Hossain is 4 : 3. After 6 years, Hassan's age will be 26 years. What is the age of Hossain at present?

Solution:
Let, the present ages of Hassan and Hossain be 4x years and 3x years respectively. 

ATQ,
4x + 6 = 26
⇒ 4x = 20
∴ x = 5

∴ Hossain's age = 3x = 15 years.

Question: The average of 10 numbers is 40.2. Later it was found that two numbers had been wrongly added. The first one is 18 greater than the actual number and the second number added is 13 instead of 33. Find the correct average- 

Solution:
The correct sum = (40.2 × 10) - 18 + (33 - 13)
= 402 - 18 + (33 - 13)
= 404

∴ Correct average = 404/10 = 40.4

Question: A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?

Solution:
Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years. 

ATQ,
(2x + 2) + 2x + x = 27
⇒ 5x = 25
∴ x = 5.
Hence, B's age = 2x = 10 years.

Question: If the mean of 5 observations x, x + 2, x + 4, x + 6, and x + 8 is 11, then the mean of the last three observations is -

Solution:
ATQ,
{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}/5 = 11
⇒ 5x + 20 = 55
⇒ 5x = 55 - 20
⇒ x = 35/5
∴ x = 7

So, the numbers are 7, 9, 11, 13, 15

Required mean = (11 + 13 + 15)/3
= 39/3
= 13

Question: Shakib's bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was -

Solution: 
Let, Shakib take x wickets before the last match. 

Total run = 12.4x + 26 
New average = (12.4x + 26)/(x + 5)

ATQ,
(12.4x + 26)/(x + 5) = 12.4 - 0.4 
⇒ (12.4x + 26) = 12(x + 5)
⇒ 12.4x - 12x = 60 - 26
⇒ 0.4x = 34
⇒ x = 34/0.4
∴ x = 85 

The number of wickets taken by him till the last match was = 85 wickets 

Question: A father is twice as old as his daughter. If 20 years ago, the age of the father was 10 times the age of the daughter, what is the present age of the father?

Solution:
Let, the present age of the father be 2x
So, the present age of the daughter = x

ATQ,
2x - 20 = 10(x - 20)
⇒ 2x - 20 = 10x - 200
⇒ 10x - 2x = 200 - 20
⇒ 8x = 180
⇒ x = 180/8
∴ x = 22.5
 
∴ The present age of father = 22.5 × 2 = 45 years

Question: In a hostel, 52 students are living. If 18 students are joined in this hostel then the average expenditure will be 3 tk. less whenever the total expenditure is 510 tk. will be increased. What is the total expenditure initially?

Solution:
Let, the average expenditure of initial students be x,

ATQ,
52x + 510 = (52 + 18) × (x - 3)
⇒ 52x + 510 = 70x - 210
⇒ 70x - 52x = 510 + 210
⇒ 18x = 720
∴ x = 40

Hence, the total expenditure of 52 students = 52 × 40 = 2080 tk.

Question: A person’s age is 3 times the age of his friend. The total of their ages is 64 years. How old is the person?

Solution:
A person’s age is 3 times the age of his friend.
The total of their ages is 64 years.

Let, the age of the friend be x
Let the age of the person be 4x
Total of their ages = x + 3x

ATQ,
x + 3x = 64
⇒ 4x = 64
∴ x = 16

∴ The age of the friend = 16,
The age of the person = 16 × 3 = 48 years.

∴ The age of the person and friend is 48 years.