Fraction and Simplification, Average and Mean

Question: A batsman makes a score of 80 runs in the 16th innings and increases average by 3. What is his average after 16th innings?

Solution: 
Assume his initial average after 15 innings = x
His total runs after 15 innings = 15x

After scoring 80 runs his average got increased by 3 to x + 3
So his total runs after 16 innings= 16 × (x +3)

But it was given that the difference in the total scores after 15 innings and 16 innings =80
16×(x + 3) - 15X=80
16x + 48 - 15x = 80 
x + 48 = 80
x = 32 

His average after 16th innings is = 32 + 3 = 3 = 35 

The number of boys = B
The number of Girls = G
According to question, 5G + 5.7B = 5.5(B + G)
Or, 0.2B = 0.5G
Or, B:G = 5:2
Now, sum of the ratio= 7.
Sum of one option is divisible by 7 that is = 50 + 20
= 70

এখানে,
দূরত্বকে একটি ধারা মনে করে পাই,
ধারাঃ     100m        108m       114m             118m
পার্থক্যঃ           8m            6m              4m
সুতরাং সঠিক উত্তর 118m

Question: If the average of p numbers is 2q² and the average of q numbers is 2p², what is the average of the combined (p + q) numbers?

Solution:
দেওয়া আছে,
p সংখ্যার গড় = 2q²
∴ p সংখ্যার সমষ্টি = p × 2q²

q সংখ্যার গড় = 2p²
∴ q সংখ্যার সমষ্টি = q × 2p²

∴ মোট সমষ্টি = (p × 2q²) + (q × 2p²)
= 2pq(q + p)

∴ তাদের গড় = মোট সমষ্টি / (p + q)
= 2pq(p + q)/(p + q)
= 2pq

Average = Sum of Quantities/Number of Quantities

1) First calculate the total age of 40 students
Total age of 29 students = ( Average age x No. of students)
= (20 x 29) = 580 years

2) Average age of 29 students + 1 teacher = 20 years + 3 months = 81/4 years

3) Finally the total age of 29 students + 1 teacher = 81/4 × 30 = 607.5 years

Therefore, age of teacher = (Total age of 30 members - Total age of 29 students)
= (607.5 – 580)
= 27.5 years.

Question: If the average of 'a' numbers is b² and the average of 'b' numbers is a², what is the average of the combined (a + b) numbers?

Solution:
দেওয়া আছে:
'a' সংখ্যার গড় = b² 
∴ a সংখ্যার সমষ্টি = a × b²

'b' সংখ্যার গড় = a² 
∴ 'b' সংখ্যার সমষ্টি = b × a²

∴ মোট সমষ্টি = (a × b²) + (b × a²)
= ab(a + b)

∴ তাদের গড় = মোট সমষ্টি/(a + b)
= ab(a + b)/(a + b)
= ab

Question: 


Solution: 

Let the x, x + 2, x + 4 and x + 6 be the 4 consecutive odd integers.
We have to find the lowest integer x.
It is given that the average of these numbers is 24.
i,e., (x + x + 2 + x + 4 + x + 6)/4 = 24
⇒ 4x + 12 = 24x4
⇒ 4x = 84
⇒ x = 84/4
= 21
Hence, the answer is 21.

Question: A sum of Tk. 3300 is divided among A, B and C such that A gets 2/5 of what B gets and B gets 1/3 of what C gets. B’s share is: 

Solution:
Let,
C’s share = Tk. x
Then,
B’s share = Tk. x/3
A’s share = Tk. (2/5) × (x/3) = Tk. 2x/15

∴ 2x/15 + x/3 + x = 3300
⇒ (2x + 5x + 15x)/15 = 3300
⇒ 22x/15 = 3300
⇒ 22x = 3300 × 15
⇒ 22x = 49500
⇒ x = 49500/22
⇒ x = 2250

∴ B’s share = 2250/3 = 750 Tk.

Question: The average age of a group of 20 employees is 30 years. If 10 more employees join the group, the average age increases by 3 years. Find the average age of the new employees?

Solution:
Here,
Total age of the 20 employees,
= 20 × 30=600 years

After joining 10 new employees, average age increases by 3 years.
So, New average age = 30 + 3 = 33 years

Total age of the 30 employees,
= 30 × 33 = 990 years

∴ Total age of the 10 new employees = (990 - 600) years
= 390 years

∴ Average age of the 10 new employees = 390/10 = 39 years

Question: The average monthly income of P and Q is Tk. 5000. The average monthly income of Q and R is Tk. 6050 and the average monthly income of P and R is Tk. 5400. Calculate the monthly income of Q.

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

P + Q = (5000 x 2) = 10000 .... (i)
Q + R = (6050 x 2) = 12100 .... (ii)
P + R = (5400 x 2) = 10800 .... (iii)

Adding (i), (ii) and (iii), we get:
 2(P + Q + R) = 32900  
 ∴ P + Q + R = 16450 .... (iv)

Subtracting (iii) from (iv),
We get Q = 5650

∴ Q's monthly income = TK. 5650

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

Solution:
Let, 
the second number be x.
Then first number = 2x, third number = 4x.

∴ 2x + x + 4x = 56 × 3
⇒ 7x = 168
⇒ x = 168/7
⇒ x = 24

Required difference:
= 4x - 2x
= 2x
= 2 × 24
= 48.

Question: A man earns N dollars a month and spends S dollars a month on rent. If he then spends 3/8 of the remainder on food, how much, in dollars, is left over for other expenses, in terms of N and S?

Solution:
Given that,
Monthly income = N dollars
Rent = S dollars

∴ Remaining after rent = N - S

And, He spends 3/8​ of the remainder on food.
∴ Food expense = (3/8)(N - S)

∴ Left for other expenses = Remaining after rent - Food expense
= (N - S) - {(3/8)(N - S)}
= (8/8)(N - S) - {(3/8)(N - S)}
= {(8 - 3)/8}(N - S)
= (5/8)( N - S)

Total salary of 65 workers = 65 × 5680 = Tk. 369200
Total salary of 31 workers = 31 × 2356 = Tk. 73036
Total salary of 23 workers = 23 × 4589 = Tk. 105547
No. of remaining workers = 65 - 31 – 23 = 65 – 54 = 11
Total Salary of 11 workers = 369200 – 73036 – 105547
= 369200 – 178583
= Tk. 190617
Required average = 190617/11
= Tk. 17328.81
Hence, the required average is Tk. 17328.81

Question: If , then x = ?

Solution:

According to the question,
(x + y)/2 = 6.5,
or, x + y = 13 and
xy = 36,
Now substitute the value of x and y from the option and find the answer.

Question: The average age of all the students in a driving class is 22 years. The average age of the boys is 24 years and that of the girls is 18 years. If there are 15 girls in the class, find the number of boys.

Solution:
Let the number of boys in the class be x.

According to the question,
22 × (x + 15) = 24x + (18 × 15)
⇒ 22x + 330 = 24x + 270
⇒ 24x + 270 = 22x + 330
⇒ 24x - 22x = 330 - 270
⇒ 2x = 60
⇒ x = 30

∴ The number of boys in the class is 30.

Let average of 17 innings = x
Total runs scored in 17 innings = 17x
Average of 16 innings = (x - 3)
Total runs scored in 16 innings = 16 (x -3)
Total runs scored in 16 innings + 87 = Total runs scored in 17 innings
⇒ 16(x - 3) + 87 = 17x
⇒ 16x - 48 + 87 = 17x
⇒ x = 39.

Question: The average salary of all the workers in a workshop is Tk. 8,000. The average salary of 7 technicians is Tk. 12,000 and the average salary of the rest is Tk. 6,000. The total number of workers in the workshop is:

Solution:
Let the number of rest workers = x

Now, According to the question,
(7 + x) × 8000 = 12000 × 7 + 6000x
⇒ 56000 + 8000x = 84000 + 6000x
⇒ 2000x = 28000
⇒ x = 14

So the total number of worker
= 14 + 7
= 21

Question: Among 80 students, the average marks in Mathematics is 65. If the 50 girls scored an average of 68, determine the average score of the remaining boys.

Solution:
Let,
the average marks of the boys = k
Total marks of 80 students = 80 × 65 = 5200 
Total marks of 50 girls = 50 × 68 = 3400

According to the question,
3400 + (80 − 50) × k = 5200
⇒ 3400 + 30k = 5200
⇒ 30k = 5200 − 3400
⇒ 30k = 1800
⇒ k = 1800 / 30
⇒ k = 60

∴ Average marks of the remaining 30 boys = 60

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

Average = 12 × (1 + 2 + 3 + 4 + 5) × (1/2)
= 12 × 15 × (1/2)
= 12 × 3
= 36.

∴ The first five multiples of 12 is 36.

Question: The sum of the digits of a 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 condition: x + y = 8
⇒ x = 8 - y   .......(1)

2nd condition:
(10x + y) - (10y + x) = 54
⇒ 9x - 9y = 54
⇒ 9(8 - y) - 9y = 54
⇒ 72 - 9y - 9y = 54
⇒ 72 - 18y = 54
⇒ - 18y = 54 - 72
⇒ - 18y = - 18
⇒ y = 1

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

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

Question: If X + Y = 174, and X is half of Y, then find the value of X.

Solution:
Given that,
X + Y = 174 …… (i)
Y = 2X …… (ii)
On solving (i) and (ii), we get,
⇒ X + 2X = 174
⇒ 3X = 174
⇒ X = 174/3
∴ X = 58

Let N be the no. of persons in the group.
Required number of person is given by;
Member in group × aged increased = difference of replacement
N × 5 = 38 - 18
Or, 5N = 20
Or, N = 4

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.

Total sum of remaining two
= (8 × 14 – 6 × 16) = 16
∴ Average of these two numbers is = 16 / 2 = 8

Question: Among 90 students, the average marks in science is 72. If the 60 girls scored an average of 75, determine the average score of the remaining 30 boys.

Solution:
Let, the average marks of the boys = k
Total marks of 90 students = 90 × 72 = 6480
Total marks of 60 girls = 60 × 75 = 4500

According to the question,
4500 + 30k = 6480
⇒ 30k = 6480 - 4500
⇒ 30k = 1980
⇒ k = 1980/30
⇒ k = 66

∴ Average marks of the remaining 30 boys = 66.