ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: A student obtained 78, 82, 69, 91 marks in four subjects. What should be the 5th subject's mark to get an average of 80? 

Solution:
Desired average = 80
Number of subjects = 5

Total marks needed = 80 × 5 = 400
Sum of the first four subjects obtained = 78 + 82 + 69 + 91 = 320

∴ Required marks in the fifth subject = 400 - 320 = 80

Therefore, the student must obtain 80 in the fifth subject's to achieve an average of 80.

Question: The average mark of 40 students is 62. One mark wrongly entered as 72 instead of 47. Corrected average? 

Solution:
Original total marks
= 62 × 40
= 2480
One student’s mark was recorded as 72, but the correct mark was 47.
So the error = 72 - 47 = 25

Since 25 extra marks were added, subtract them from the total.
= 2480 - 25
= 2455

Corrected average
= 2455/40
= 61.3

Question: A father is 30 years older than his son. In 10 years, the father will be twice the son’s age. Find the present age of father.

Solution:
Let the present age of the son be x years.
Then the present age of the father = x + 30 years.

In 10 years,
Son’s age = x + 10
Father’s age = x + 30 + 10 = x + 40

According to the problem, father’s age will be twice the son’s age,
x + 40 = 2(x + 10)
⇒ x + 40 = 2x + 20
⇒ 2x - x = 40 - 20
⇒ x = 20

∴ Present ages of father = x + 30 = 20 + 30 = 50

Question: The average of a and b is 30, and the average of b and c is 40. If b = 36 than find the value of (a + c).

Solution: 
Given b = 36

Average of a and b = 30
a + b = 30 × 2
⇒ a + b = 60
⇒ a + 36 = 60
⇒ a = 60 - 36
∴ a = 24

Average of b and c = 40
b + c = 40 × 2
⇒ b + c = 80
⇒ c = 80 - 36
∴ c = 44

Now,
a + c = 24 + 44 = 68

Question: Ten years ago, the ratio of ages of S and T was 4 : 3. Now the ratio is 5 : 4. Find the present age of T.

Solution: 
Let the ages of S and T ten years ago be 4x and 3x.
Present ages of
S = 4x + 10 and T = 3x + 10

Now ratio = 5 : 4 
⇒ (4x + 10)/(3x + 10) = 5/4
⇒ 4(4x + 10) = 5(3x + 10)
⇒ 16x + 40 = 15x + 50
⇒ 16x - 15x = 50 - 40
⇒ x = 10

∴ Present age of T = 3x + 10 
= (10 × 3) + 10
= 30 + 10 
= 40

Question: The average age of a group is 35. If 5 new members with an average age of 25 join, the overall average becomes 33. How many members were originally in the group?

Solution:
Let
The original number of members be x.
Total age of original members = 35 × x = 35x

Total age of 5 new members = 25 × 5 = 125
Total number of members after adding 5 = x + 5

New average = 33,
So total age after adding 5 members = 33 × (x + 5) = 33x + 165

Set up the equation for total age:
Original total age + New members’ total age = New total age
⇒ 35x + 125 = 33x + 165
⇒ 35x - 33x = 165 - 125
⇒ 2x = 40
⇒ x = 20

Question: The mean of 8 numbers is 22. Five of them are 20, 18, 24, 16, and 26. Find the mean of the remaining three numbers.

Solution:
Mean of 8 numbers = 22 
Total sum of 8 numbers,
= 8 × 22 
= 176

Sum of the given 5 numbers,
= 20 + 18 + 24 + 16 + 26
= 104

Sum of the remaining 3 numbers:
= 176 - 104
= 72

∴ Mean of the remaining 3 numbers,
= 72/3
= 24

Question: The difference of ages between two brothers is 6 years. If the ratio of their ages is 7 : 5, find their ages

Solution:
Let
The ages of the two brothers be 7x and 5x. (elder : younger).
Difference of ages = 6 

Accordingly, 
7x - 5x = 6
⇒ 2x = 6  
⇒ x = 3.

∴ Elder = 7x = 7 × 3 = 21 
∴ Younger = 5x = 5 × 3 = 15

So Elder brother 21 years and younger brother 15 years.

Question: The average age of a group is 28. If a 42-year-old person leaves, the average becomes 26. How many people are in the group?

Solution: 
Let
The original number of people be x.

Total age of the group:
= 28 × x
= 28x

After the 42-year-old leaves, total age:
= 28x - 42

Number of people remaining = x - 1
New average = 26

Accordingly, 
(28x - 42) / (x - 1) = 26
⇒ 28x - 42 = 26(x - 1)
⇒ 28x - 42 = 26x - 26
⇒ 28x - 26x = - 26 + 42
⇒ 2x = 16
⇒ x = 8

Question: A cricketer’s average after 20 innings is 45 runs. If he scores 108 runs in the next innings, what is his new average?

Solution:
Average after 20 innings = 45 
Total runs after 20 innings:
= 20 × 45
= 900 runs.

Runs scored in the 21st innings = 108.
Total runs after 21 innings:
= 900 + 108
= 1008 runs.

New average = Total runs / Number of innings
= 1008 / 21
= 48 runs.

Question: The mean of x, x + 4, x + 8, x + 12 is 20. Find x.

Solution:
Given,
Total numbers = 4
Mean = 20

Sum of numbers: 
x+ (x + 4) + (x + 8) + (x + 12) 
= 4x + 24

ATQ,
(4x + 24)/4 = 20
⇒ 4x + 24 = 80
⇒ 4x = 56
∴ x = 14

Question: The average of five numbers is 32. The sum of the first three numbers is 75. If the fourth number is 28, what is the fifth number?

Answer:
Average of five numbers = 32
Total sum = 5 × 32
= 160

Sum of first three numbers = 75.
Fourth number = 28.

Sum of first four numbers = 75 + 28 = 103.

∴ Fifth number = total sum - sum of first four 
= 160 - 103
= 57

So the fifth number is 57.

Question: The average weight of 12 students is 52 kg. If a new student joins, the average becomes 53 kg. What is the weight of the new student?

Solution:
Average weight of 12 students = 52 
Total weight = 12 × 52
= 624 kg.

Let the weight of the new student = x kg.
New total number of students = 13

New average = 53 
Total weight = 13 × 53
= 689 kg.

Weight of new student = new total - original total
= 689 - 624
= 65 kg

∴ The weight of the new student is 65 kg.

Question: The present age of a father is 5 times that of his son. After 8 years, the ratio will be 3 : 1. Find their age gap.

Solution:
Let the present age of the son = x years.
Present age of the father = 5x years.

After 8 years:
Son = x + 8 
Father = 5x + 8
Given ratio after 8 years = 3 : 1

Accordingly,
(5x + 8​) / (x + 8) = 3
⇒ 5x + 8 = 3(x + 8)
⇒ 5x + 8 = 3x + 24
⇒ 5x - 3x = 24 - 8
⇒ 2x = 16
∴ x = 8

Present ages of
Son = x = 8 years
And Father = 5x = 40 years

Age gap = 40 - 8 = 32

So the age gap is 32 years.

Question: The average of 15 numbers is 48. If the average of first 7 numbers is 44 and the average of last 7 numbers is 50, what is the middle number?

Solution:
Let
The middle number be x

Total sum of 15 numbers,
= 15 × 48
= 720

Sum of first 7 numbers,
= 7 × 44
= 308

Sum of last 7 numbers,
= 7 × 50
= 350

∴ Sum of first 7 numbers + x + Sum of last 7 numbers = 720
⇒ 308 + x + 350 = 720 
⇒ 658 + x = 720
⇒ x = 720 - 658
∴ x = 62

So the middle number be 62.