Fraction and Simplification, Average and Mean

Median is the middle element or the mean of two middle elements in a numerical data set with the elements ordered by their value.
The mode is the value in a set of data that has the most occurrences.
The mode is not unique and a data set may have more than one mode of no mode as well
Given set of numbers are 10, 70, 20, 40, 70, 90
Arrange in order: 10, 20, 40, 70, 70, 90
So, Median M = ( 40 + 70)/2 = 55
and, Mode m = 70

∴ average of M and m = (55 + 70)/2 = 62.5

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 of 7 consecutive numbers is 21. The largest of these numbers is-

Solution:
Let the numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6,
Then,
(x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6)/7 = 21
⇒ 7x + 21 = 147
⇒ 7x = 126 
∴ x = 18

∴ Largest number = x + 6 = 18 + 6 = 24

Let there be x pupils in the class.
The total increase in marks = (x × 1/2) = x/2
∴ x/2 = (83 - 63)
⇒ x/2 = 20
⇒ x = 40
Answer: 40

Question: A cake is divided into 24 pieces. Rahim takes 1/4 of the cake. Karim takes 1/3 of the rest. How many pieces are left?

Solution:
Given that, 
A cake is divided into 24 pieces.

Now, 
Rahim takes 1/4 of the cake = 24 × (1/4) = 6 pieces
So Rahim takes 6 pieces.
∴ Remaining after Rahim = 24 - 6 = 18 pieces

And
Karim takes 1/3 of the rest = 18 × (1/3) = 6 pieces

∴ Pieces left after Karim = 18 - 6 = 12 pieces

So, 12 pieces are still left.

Question: A water tank contains 24 liters of water. Rahim uses 1/4 of the total water. Then Karim uses 1/2 of the remaining water. How many liters of water are left in the tank?

Solution:
Total water = 24 liters
Water used by Rahim = (1/4) × 24 = 6 liters

∴ Remaining water = 24 - 6 = 18 liters

Water used by Karim = (1/2) × 18 = 9 liters

∴ Water left in the tank = 18 - 9 = 9 liters

Question: Solve: (81.84 + 118.16) ÷ 5³ = 1.2 × 2 + ?

Solution:
Given that,
(81.84 + 118.16) ÷ 5³ = 1.2 × 2 + ?
⇒ 200 ÷ 5³ = 1.2 × 2 + ?
⇒ 200 ÷ 125 = 1.2 × 2 +?
⇒ 1.6 = 2.4 + ?
∴ ? = - 0.8

Question: Find the value of x, if 3^{(2x + 1)} = 243.

Solution:
3^{(2x + 1)} = 243
⇒ 3^{(2x + 1)} = 3⁵ (since 243 = 3⁵)
⇒ 2x + 1 = 5
⇒ 2x = 5 - 1
⇒ 2x = 4
⇒ x = 4/2
∴ x = 2

Let the Father be x years when he died.
Average Age 10 years ago be A.
Total Age 10 years ago = 6 × A
Total Age after 10 years(Just before father's Death) = 6A + 6 × 10 = 6A + 60
Father Died and Baby was born => the Total number of people in the family is Same (6)
Baby born today so age of baby = 0
(6A + 60 - x)/6 = 6A/6
=> A + 10 -(x/6) = A
=> x/6 = 10
=> x = 60
Therefore we can conclude that the father was 60 years old when he died.

Question: What will come in the place of the question mark in the following question- 40% of 50% of 3/4 of 1800 = ?

Solution:
40% of 50% of 3/4 of 1800 = ?
⇒ 40% × 50% × (3/4) × 1800 = ?
⇒ (40/100) × (50/100) × 3/4 × 1800 = ?
⇒ (2/5) × (1/2) × (3/4) × 1800 = ?
⇒ (6/40) × 1800 = ?
⇒ ? = 270

∴ The question mark is replaced by 270

Question: A cricket team consists of 11 players. The captain is 24 years old, and the wicketkeeper is 1 years older than the captain. When these two players are excluded, the average age of the remaining players becomes one year less than the average age of the whole team. What is the average age of the team?

Solution:
Let,
the average age of the whole team is x years.
⇒ 11x - (24 + 25) = 9(x - 1)
⇒ 11x - 49 = 9x - 9 
⇒ 11x - 9x = 40
⇒ 2x = 40
⇒ x = 20.

So, the average age of the team is 20 years.

Question: 


Solution: 

Speed of aeroplane is 200, 400, 600 and 800 km/h respectively
Let the side of side be LCM of (200, 400, 600 and 800) = 2400
Time taken by aeroplane to travel the side at the speed of 200 km/hr 
⇒ 2400/200 = 12 hours
Time taken by aeroplane to travel the side at the speed of 400 km/hr 
⇒ 2400/400 = 6 hours
Time taken by aeroplane to travel the side at the speed of 600 km/hr 
⇒ 2400/600 = 4 hours
Time taken by aeroplane to travel the side at the speed of 800 km/hr 
⇒ 2400/800 = 3 hours

Average speed = (Total Distance travelled)/(Total time taken)
∴ Average speed = (4×2400)/25 = 384 km/hr

Question: In a football team of 12 players, the captain is 30 years old and the goalkeeper is 4 years older than the captain. If these two players are excluded, the average age of the remaining players becomes 2 years less than the team’s overall average age. Find the overall average age of the team?

Solution:
Let the average age of the whole team be x years.

Captain’s age = 30 years
Goalkeeper’s age = 30 + 4 = 34 years

Total age of all 12 players = 12x

After excluding the captain and goalkeeper,
Total age of remaining players = 12x - (30 + 34)

Given that the new average age = (x - 2) years,
Total age of remaining players = 10(x - 2)

So,
12x - 64 = 10(x - 2)
⇒ 12x − 64 = 10x - 20
⇒ 12x - 10x = 64 - 20
⇒ 2x = 44
⇒ x = 22

∴ The average age of the team is 22 years.

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 sum of Tk. 1360 has been divided among A, B and C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. B's share is-

Solution:
Let,
C′s share = Tk. x
Then,
B′s share = Tk. x/4,
A′s share = Tk.(2/3) × (x/4) = Tk. x/6

∴ x/6 + x/4 + x = 1360
⇒ 17x/12 = 1360
⇒ x = (1360 × 12)/17
∴ x = Tk. 960

∴ B′s share = Tk.(960/4) = Tk. 240

The average of 6 players = X
Average increases by 10, when the seventh player makes a score of 112 runs.
Therefore, average of 7 players = X + 10
Average =Sum of Scores/Number of Players -------(1)
Here, average = X, number of players = 6
Hence,
Sum of scores = 6X
Score of 7 players = (Score of 6 players + score of 7 player) = (6X + 112)------- (2)
Total average = (X + 10) --------- (3)
Substitute (2) and (3), in (1)
(X + 10) = (6X + 112)/7
⇒ 7X + 70 = 6X + 112
⇒ 7X - 6X = 112 - 70
⇒ X = 42.

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: What is the difference between the biggest and the smallest fraction among 2/3, 3/4, 4/5 and 5/6 ?

Solution:
Converting each of the given fractions into decimal form, we get,
2/3 = 0.66
3/4 = 0.75
4/5 = 0.8
5/6 = 0.833

Since 0.833 > 0.8 > 0.75 > 0.66
So, 5/6 > 4/5 > 3/4 > 2/3

∴ Required difference = (5/6) - (2/3) = 1/6

If a month beings with Sunday then there are 5 Sundays in that month.
Total number of visitors come on Sunday = 510×5 = 2550
Total number of visitors come on other days = 240×25 = 6000
∴ Average number of visitors per day = (2550+6000)/30 = 8550/30 = 285

Question: In a zoo, each pigeon has 2 legs, and each rabbit has 4 legs. The head count of the two species together is 12, and the leg count is 32. How many pigeons and how many rabbits are there in the zoo?

Solution:
Let,
Number of pigeons = P
Number of rabbits = R

From the problem we get two equations,
Heads,  P + R = 12 ......(1)

And,
Legs, 2P + 4R = 32 ; [pigeons 2 Legs and rabbits 4 Legs]
⇒ 2(P + 2R) = 32
⇒ P + 2R = 32/2
∴ P + 2R = 16 .......(2)

Now, (2) - (1) Then we get,
⇒ P + 2R - P - R = 16 - 12
∴ R = 4

From (1) we get, 
⇒ P + 4 = 12
⇒ P = 12 - 4
∴ P = 8

There are 8 pigeons and 4 rabbits.

After 5 test player's total score will be = 5×80 = 400
Total score after 4 test was = 4×78 = 312
So, required score is = 400 - 312 = 88

Total annual income
= TK (10000 X 4 + 12000 X 8 + 26000)
= TK. 162000
Average monthly income = TK 16200012 = TK. 13500