Fraction and Simplification, Average and Mean

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,
The 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: Over 27 innings, a cricket player averages 47 runs per innings. His maximum score is 157 runs more than his minimum score. Excluding these two innings, the average of the other 25 innings drops to 42 runs. Find the player’s highest score.

Solution:
Given that,
The batting average for 27 innings of a cricket player is 47 runs.
His highest score exceeds his lowest score by 157 runs.
If these two innings are excluded, the average of the remaining 25 innings is 42 runs.

Now,
Sum of runs for 27 innings of a cricket player = 47 × 27 = 1269
Sum of runs for 25 innings of a cricket player = 42 × 25 = 1050
∴ Sum of remaining 2 innings = 1269 - 1050 = 219

Let,
The minimum score be x and the maximum score be x + 157

According to the question,
x + x + 157 = 219
⇒ 2x = 219 - 157
⇒ 2x = 62
∴ x = 31

So, highest score = 157 + 31 = 188

60 ও 80 এর মধ্যে বৃহত্তম মৌলিক সংখ্যা 79
60 ও 80 এর মধ্যে ক্ষুদ্রতম মৌলিক সংখ্যা 61
সংখ্যা দুটির সমষ্টি = 140
∴ সংখ্যা দুটির গড় = 140/2 = 70

Let P, Q and R represent their respective monthly incomes. Then, we have:

P + Q = (5050 x 2) = 10100 .... (i)
Q + R = (6250 x 2) = 12500 .... (ii)
P + R = (5200 x 2) = 10400 .... (iii)

Adding (i), (ii) and (iii), we get:
2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv)

Subtracting (ii) from (iv), we get,
P = 4000.
∴ P's monthly income = TK. 4000.

Question: Which of the following is the smallest?

Solution:
ক) 1/2 = 0.5
খ) 3/8 = 0.375
গ) 11/25 = 0.44
ঘ) 5/12 = 0.4167

∴ The smallest value is 3/8.

Required average = (50.25 × 16 + 45.15 × 8)/(16 + 8)
= (804 + 361.20)/24
= 1165.20/24
= 48.55

Question: If √x + √3 = √48, than what is the value of x.

Solution: 
Given that, 
√x + √3 = √48
⇒ √x + √3 = √(16 × 3)
⇒ √x + √3 = 4√3
⇒ √x = 4√3 - √3
⇒ √x = 3√3
⇒ (√x)² = (3√3)²   ;[Square both sides]
∴ x = 27

Question: The average age of the children in a tour group is 8 years, and that of the adults is 30 years. If the average age of the entire tour group is 15 years, find the ratio of children to adults in the group.

Solution:
Here,
Average age of children = 8 years
Average age of adults = 30 years
Average age of the entire group = 15 years

Let the number of children = m
and the number of adults = n

Then, the total number of people in the group is (m + n)

ATQ,
8m + 30n = 15(m + n)
⇒ 8m + 30n = 15m + 15n
⇒ 15m - 8m = 30n - 15n
⇒ 7m = 15n
⇒ m/n = 15/7
⇒ m : n = 15 : 7

∴ The ratio of children to adults in the group is 15 : 7

Question: A sum of Tk. 1800 has been divided among A, B and C such that A gets 1/2 of what B gets and B gets 1/3 of what C gets. C's share is-

Solution:
Let, C's share = Tk. x
Then, B's share = Tk. x/3,
A's share = Tk. (1/2)(x/3) = Tk. x/6

∴ x/6 + x/3 + x = 1800
⇒ (x + 2x + 6x)/6 = 1800
⇒ 9x/6 = 1800
⇒ x = (1800 × 6)/9
⇒ x = Tk. 1200

∴ C's share = Tk. 1200

Question: (0.1 × 0.01 × 0.001 × 10⁷) is equal to: 

Solution:
Given expression,
(0.1 × 0.01 × 0.001 × 10⁷)
= (1/10) × (1/100) × (1/1000) × 10⁷
= 10⁷/10⁶
= 10 ^{(7 - 6)}
= 10

x + y = 2 × 18 = 36
x + y + z = 36 + 12 = 48
Average: 48/3 = 16

Question: Which of the following fractions is smaller than 7/8 and greater than 3/7?

Solution:
7/8 = 0.875
3/7 = 0.4286

ক) 11/12 = 0.917  (> 0.875)
খ) 5/9 = 0.5556   (satisfies the condition)
গ) 4/11 = 0.3636 (< 0.4286)
ঘ) 1/3 = 0.3333  (< 0.4286))

∴ অপশন (খ) সঠিক উত্তর।

Total attendance for first 3 days = 325×3 = 975
Total attendance for first 4 days = 320×4 = 1280
∴ Present on the 4th day = 1280 - 975 = 305

Required average = total sum of multiple of 9/5
= (9 + 18 + 27 + 36 + 45)/5
= 27

Note that, the average of 9 and 45 is also 27.
An average of 18 and 36 is also 27.

Question: The least number by which 180 must be multiplied to make it a perfect square is:

Solution: 
Prime factorization of 180 = 2 × 2 × 3 × 3 × 5
= 2² × 3² × 5¹
For a number to be a perfect square, every exponent in the prime factorization must be even.
2² ; even power 
3² ; even power 
5¹ ; odd power 
The only odd exponent is 5. To make it even, multiply by 5.
= 180 × 5
= 900
= 30²

So the least number by which 180 must be multiplied to make it a perfect square is 5.

Average of 15 numbers = 15, Average of 5 numbers = 14, Average of 9 numbers = 16

Average = Total numbers/Number of Numbers
15 = Total numbers/15
Therefore, total numbers = 15 x 15 = 225.

Middle number = (Total numbers) – [(Average of 5 num x no of num) + ( Average of 9 num x no of num)]
= (225) – [(14 x 5) + (16 x 9)]
= (225) – (214)
= 11.

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: {(2.39)² - (1.61)²}/(2.39 - 1.61) = ?

Solution: 
Let, 2.39 = a 
And 1.61 = b

Now, 
(a² - b²)/(a - b)
= (a + b)(a - b)/(a - b)
= a + b
= 2.39 + 1.61
= 4

Question: In a group of 135 persons, 50% people contributed tk. 40 each, 30% contributed Tk. 60 each and the remaining persons contributed Tk. 80. Find the average contribution for the group?

Solution:
total people=135

Since
50% of them contributed Tk. 40 
∴ amount contributed by these 50% = (135/2) × 40 = 2700

Since 30% of them contributed TK. 60
∴ amount contributed by these 30% = (30/100) × (135) × 60 = 2430 

and
remaining 20% of them contributed Tk. 80
∴ amount contributed by these 20% = (20/100) × 135 × 80 = 2160

∴ Total amount contributed = 2700 + 2430 + 2160 = 7290 

∴ Average contribution = 7290/135 = 54

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 average income of A, B and C is Tk. 12000 per month and average income of B, C and D is Tk. 15000 per month. If the average salary of D be twice that of A, then the average salary of B and C is in Tk. 

Solution:
Given that,
Average income of A, B, C = 12,000
∴ Total income of A + B + C = 3 × 12000 = 36000 … (1)
Average income of B, C, D = 15,000
 ∴ Total income of B + C + D = 3 × 15000 = 45000 … (2)
And, D = 2A ....(3)

Now, Subtract equation (1) from (2) than we get,
B + C + D - (A + B + C) = 45000 - 36000
⇒ D - A = 9000
⇒ 2A - A = 9000   ; [from (3)]
∴ A = 9000

Now put a = 9,000 in (i)
9000 + B + C = 36000
⇒ B + C = 36000 - 9000 = 27000
∴ B + C = 27000

∴ Average salary of B and C = (B + C)/2 = 27000/2 = 13500

So the average salary of B and C is Tk. 13500.

15 দিনের কাজের মোট বেতন (15 × 90) = 1350 টাকা
১ম 7 দিন ও শেষ 7 দিনের কাজের মোট বেতন (7×87 + 7×92) = 609 + 644 = 1253 টাকা।
∴ অষ্টম দিনের কাজের বেতন = (1350 - 1253) = 97 টাকা

Question: 


Solution:

Question: The sum of the digits of two-digit number is 8. If the digits are reversed the number is decreased by 54. What is the number?

Solution: 
Let the two-digit number be 10x + y,
where, x = tens digit and y = ones digit.

Given,
1st conditions,
x + y = 8
x = 8 - y .......(1)

And 2nd conditions,
10x + y - (10y + x) = 54 
⇒ 9x - 9y = 54
⇒ 9(8 - y) - 9y = 54
⇒ 72 - 9y - 9y = 54
⇒ - 18y = 54 - 72
⇒ y = - 18/- 18
∴ y = 1

From equation (1) we get,
x = 8 - y = 8 - 1 = 7
∴ x = 7

So the number is = 10 x + y = 10(7) + 1 = 71

Students in batch1 = 55
Average marks of batch 1 = 50
Total marks of batch 1 = 55 × 50 = 2750

Students in batch 2 = 60
Average marks of batch 2 = 55
Total marks of batch 2 = 60 × 55 = 3300

Students in batch 3 = 45
Average marks of batch3 = 60
Total marks of batch 3 = 45 × 60 = 2700

Total marks = 2750 + 3300 + 2700 = 8750
Total students = 55 + 60 + 45 = 160

Average marks of all students = 8750/160 = 54.68

Question: Which one of the following numbers can be removed from the set S = {2, 4, 5, 9} without changing the average of set S?

Solution:
- S = {2, 4, 5, 9}
- Number of elements 4,
- Total = (2+4+5+9)= 20.
∴ Average = 20/ 4= 5

Try removing each number and check if the new average is still 5,
After removing 2, we get S = {4,5,9}
- Summation = 18
- Number of elements = 3
∴ Average = 18/3 =6. Which is not equal to 5.

Again, removing 4 from set S, we get S = {2,5,9}
- Summation = 16
- Number of elements = 3
- Average = 16/3 = 5.33

Again, removing 5 from set S, we get S = {2,4,9}
- Summation = 15
- Number of elements = 3
- Average = 15/3 = 5.

- Final Answer: 5 can be removed without changing the average.

The average of 11 results = 60
∴ The total of 11 results = 60 × 11 = 660
Average of first six results = 58
∴ Total of first six results = 58 × 6 = 348
Average of last six results = 63
∴ Total of last six results = 63 × 6 = 378
∴ sixth results = total of first and last six results - total of 11 results 
                     = (348 + 378) - 660
                     = 726 - 660
                     = 66
Answer: 66

Question: 


Solution:

Sum of 6 numbers = (6 × 25) = 150.
Sum of 3 additional numbers = (3 × 22) = 66.
Sum of (6 + 3) =9 numbers = (150 + 66) = 216
∴ average of the combined 9 numbers = 216/9 = 24

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.

Sum of the present ages of husband, wife and child
= (27 x 3 + 3 x 3) years
= 90 years
Sum of the present ages of wife and child
= (20 x 2 + 5 x 2) years
= 50 years
∴ Husband's present age
= (90 - 50) years
= 40 years

Question: A sum of money is distributed equally among 15 persons, but if 5 more persons were included, each person would get Tk. 50 less. What was the total sum?

Solution:
Let the total sum be Tk. x.

When the sum is distributed among 15 persons, each person gets x/15.
If 5 more persons are included, making it 20 persons, each person would get x/20.

According to the question,
(x/15) - (x/20) = 50

⇒ (4x - 3x)/60 = 50
⇒ x/60 = 50
⇒ x = 50 × 60
⇒ x = 3000

∴ The total sum of money is Tk. 3000.

Question: If the average of m numbers is n² and the average of n numbers is m², find the average of all (m + n) numbers.
​
​Solution:
​Given that,
​The average of m numbers is n²
​Sum of m numbers = m⋅n²
​
​And
The average of n numbers is m²
​Sum of n numbers = n⋅m²

​∴ ​Total sum = mn² + nm² = mn(m + n)
​
​∴ ​​Average = mn(m + n)/(m + n) = mn​