ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The average age of Mona, Neha, Oishi, and Roza is 22 years. The average age of Mona, Neha, and Oishi is 20 years and the average age of Neha, Oishi, and Roza is 24 years. Find the average age of Neha and Oishi.

Solution: 
Given,
Mona + Neha + Oishi + Roza = (22 × 4) = 88 years.......(1)
Mona + Neha + Oishi = 20 × 3 = 60 years.............(2)
Neha + Oishi + Roza = 24 × 3 = 72 years...............(3)

now from (1) - (2)
Roza’s age = (88 - 60) = 28 years

from (3) ⇒ Neha + Oishi = (72 - 28) = 44 years
∴ the average age of Neha and Oishi = 44/2 = 22 years

Question: The average of 3, 8, 7, and x is 6 and the average of 19, 2, 7, x and y is 9. What is the value of y? 

Solution: 
Given that
average of 3, 8, 7, x is 6

Therefore,
6 = (3 + 8 + 7 + x​)/4
⇒ 24 = 18 + x
⇒ x = 24 - 18
∴ x = 6

Therefore,
9 = (19 + 2 + 7 + x + y​)/5
⇒ 45 = 28 + 6 + y
⇒ y = 45 - 34 
∴ y = 11

Question: The mean of the first 15 even natural numbers is-

Solution:
First 15 even natural numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.

Mean = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30)/15
= 240/15
= 16

Question: Ali got married 7 years ago. His present age is 6/5 times his age at the time of his marriage. What is the present age of Ali?

Solution:
Let,
Ali's age 7 years ago be x years
His present age is = (x + 7) years

ATQ,
x + 7 = 6x/5
⇒ 5x + 35 = 6x
⇒ x = 35

His present age is = (35 + 7) = 42 years

Question: Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, then the average of the first and the third number is -

Solution:
Let,
the second number be = x
So, the first number is = 2x
and the third number is = 2x × 2 = 4x (since the first is half of the third)

ATQ,
(2x + x + 4x)/3 = 56
⇒ 2x + x + 4x = 56 × 3
⇒ 7x = 168
∴ x = 24

∴ the second number = 24
So, the first number is = 2 × 24 = 48
and the third number is = 4 × 24 = 96

Then, the average of the first and the third number is = (48 + 96)/2 = 72

Question: Find the average of all the numbers between 10 and 50 which are divisible by 4.

Solution:
Numbers between 10 and 50 divisible by 4 are = 12, 16, 20, 24, 28, 32, 36, 40, 44, 48.

Required average = (12 + 16 + 20 + 24 + 28 + 32 + 36 + 40 + 44 + 48​)/10
= 300/10
= 30

Question: At present, the ratio between the ages of Nehal and Rahat is 5 : 4. After 6 years, Nehal's age will be 26 years. What is the age of Rahat at present?

Solution:
Let,
the present ages of Nehal and Rahat be 5x years and 4x years respectively. 

ATQ,
5x + 6 = 26
⇒ 5x = 20
∴ x = 4

∴ Rahat's age = 4x = (4 × 4) = 16 years

Question: The sum of 10 numbers is 490. If the average of their first 4 numbers is 53 and that of the last five is 42, then what is the 5^th number?

Solution:
Given,
the average of their first 4 numbers = 53
∴ The total of their first 4 numbers = 4 × 53 = 212
and,
the average of their last five = 42
The total of the last 5 numbers = 5 × 42 = 210

∴ The sum of the (4 + 5) = 9 numbers = (212 + 210) = 422

∴ The 5^th number = 490 - 422 = 68

Question: The average of the first five multiples of 5 is:

Solution:
We know,
The first five multiples of 5 = 5, 10, 15, 20, 25.

∴ Average = (5 + 10 + 15 + 20 + 25)/5
= 75/5
= 15

Question: If the average of four consecutive odd numbers is 42, find the largest numbers.

Solution:
Let
the first number is x,
then the next three odd numbers would be (x + 2), (x + 4), (x + 6)

ATQ,
{x + (x + 2) + (x + 4) + (x + 6)}/4 = 42
⇒ (4x + 12)/4 = 42
⇒ 4x + 12 = 168
⇒ 4x = 156
∴ x = 39

Largest number would be = 39 + 6 = 45

Question: The average age of family of 6 members is 30 years. A new member is added then the average age becomes 26 years. Find the age of the new member.

Solution:
Total age of 6 family members = (6 × 30) = 180 years

Total age of 7 members (after adding the new one) = (7 × 26) = 182 years

∴ Age of the new member = (182 - 180) = 2 years

The new member is 2 years old.

Question: A factory employs 130 workers on Fridays and 250 workers on other days. If a month has 30 days and starts on a Friday, what is the average number of workers per day over the month?

Solution:
Since,
the month begins with a Friday, so there will be 5 Fridays and 25 other days in this month.

Total workers on Fridays = 5 × 130 = 650
Total workers on other days = 25 × 250 = 6250

∴ Total workers in the whole month = (650 + 6250) = 6900

∴ Average number of workers per day of the month = 6900/30 = 230

Question: A student's marks were wrongly entered as 75 instead of 60. Due to this, the average marks of the class increased by 0.3. Find the number of students in the class.

Solution:
Let
the number of students be x
Total increase in marks = x × 0.3
= x × (3/10)
= 3x/10

ATQ,
3x/10 = (75 - 60)
⇒ 3x/10 = 15
⇒ 3x = 150
⇒ x = 150/3
∴ x = 50

So the number of students in the class = 50

Question: The average age of 12 girls in a class is 14 years. If 4 new girls, each aged 10 years, join the class, what will be the new average?

Solution:
Given,
The average age of 12 girls in a class = 14 years
Sum of ages of 12 girls = (12 × 14) years
= 168 years

Sum of ages of 4 girls = (4 × 10) years
= 40 years

∴Total age of 16 girls = (168 + 40) years
= 208 years

∴ Average of ages of 16 girls = 208/16 years
= 13 years

Question: For 8 innings, Miraz has an average of 60 runs. In the 9th inning, he scored 6 runs, thus decrease his average. How much does his average decrease?

Solution:
Total score for 8 innings = 60 × 8 = 480
Total score after 9th innings = 480 + 6 = 486

∴ the new average is = 486/9 = 54

So, his average decrease = 60 - 54 = 6

Question: Of three numbers, the average of the first and second numbers is 12 more than the average of the second and third numbers. What is the difference between the first and third numbers?

Solution: 
Let,
these numbers are x, y and z respectively.

ATQ,
{(x + y)/2} - {(y + z)/2} = 12
⇒ {(x + y) - (y + z)}/2 = 12
⇒ (x + y - y - z)/2 = 12
∴ x - z = 24

∴ the difference between the first and the third number is = 24.

Question: The average of 10 numbers is 23. If each number is increased by 4, what will the new average be?
 
Solution:
Given,
Average of 10 numbers = 23
⇒ Sum/Total numbers = 23
⇒ Sum/10 = 23
∴ Sum of the 10 numbers = 230


If each number is increased by 4, the total increase = 4 × 10 = 40
New sum = 230 + 40 = 270


Therefore, the new average = 270/10 = 27

Question: The mean of 7 numbers is 24. If one of the numbers is removed, the mean increases by 3. Find the value of the removed number.

Solution:
Mean of 7 numbers = 24
Sum of these 7 numbers = (24 × 7) = 168

Mean of the remaining 6 numbers = (24 + 3) = 27
Sum of these remaining 6 numbers = (27 × 6) = 162

Removed number = (sum of the given 7 numbers) - (sum of the remaining 6 numbers)
= (168 - 162)
= 6

Hence, the removed number is 6

Question: The product of the present ages of Rina and Mina is 1500 years. The ratio of their present ages is 5 : 3. What is the difference of their present ages?

Solution:
Let,
the present age of Rina be 5x and the present age of Mina be 3x

ATQ,
5x × 3x = 1500
⇒ 15x² = 1500
⇒ x² = 100
∴ x = 10

∴ the present ages of Rina = (5 × 10) = 50 years
∴ the present ages of Mina = (3 × 10) = 30 years

So, the difference of their present ages = (50 - 30) = 20 years

Question: A is 50 years old and B is 35 years old. How many years ago was the ratio of their ages 3 : 2?

Solution: 
Let,
'x' years ago the ratio of their ages was 3 : 2

ATQ,
(50 - x) : (35 - x) = 3 : 2
⇒ (50 - x)/(35 - x) = 3/2
⇒ 105 - 3x = 100 - 2x
⇒ 3x - 2x = 105 - 100
∴ x = 5

Question: The average weight of 25 students in a class was found to be 48 kg. Later it was discovered that one student's weight was recorded as 35 kg instead of 60 kg. What is the correct average weight of the 25 students?

Solution:
Given,
The average weight of 25 students in a class was found to be 48 kg
Incorrect sum of the weight of 25 students = (48 × 25) kg
= 1200 kg

Correct sum of the weight of 25 students = (incorrect sum) - (wrongly copied item) + (actual item)
= (1200 - 35 + 60) kg
= 1225 kg

Correct mean = correct sum/number of students
= (1225/25) kg
= 49 kg

∴ Hence, the correct mean weight is 49 kg

Question: Average mark in a class test of 40 students is 40. Average mark of all the 25 boys is 46. Then the average mark obtained by the girls is

Solution: 
Given,
Average mark of 40 students = 40
Total mark of 40 students = (40 × 40)
= 1600 

Average mark of all the 25 boys = 46
Total mark of all the 25 boys = (46 × 25)
= 1150

∴ Total marks of all the 15 girls is (1600 - 1150) = 450

So the average mark of 15 girls = 450/15 = 30

Question: The sum of the present ages of a son and his father is 70 years. After 5 years, the age of the father will be three times that of the son. At present their ages are?

Solution:
Let,
the son's age be x years.
Then, father's age = (70 - x) years.

ATQ,
(70 - x) + 5 = 3(x + 5)
⇒ 75 - x = 3x + 15
⇒ 4x = 60
∴ x = 15

∴ Son's age = 15 years
And the father's age = (70 - 15) = 55 years.

Question: The average temperature on Monday, Tuesday, and Wednesday was 26°C. The average temperature on Tuesday, Wednesday, and Thursday was 25°C. If the temperature on Monday was 28°C, what was the temperature on Thursday?

Solution:
The total temperature on Monday, Tuesday, and Wednesday = 26° × 3 = 78°
The total temperature on Tuesday, Wednesday, and Thursday = 25° × 3 = 75°
 
ATQ,
(Mon + Tue + Wed) - (Tue + Wed + Thu) = 78° - 75°
⇒ Mon - Thu = 3°
⇒ Thu = Mon - 3°
⇒ Thu = 28° - 3°
∴ Thu = 25°

∴ the temperature on Thursday = 25°