ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?

Solution:
Average = 32
So, total = 32 × 10 = 320.

Now increase in average = 4
So, new average = 32 + 4 = 36

Total runs after next innings =  36 × 11 = 396
∴ Runs made in the 11th inning = 396 - 320 = 76

Question: In 10 years, Alif will be twice as old as Bashir was 10 years ago. If Alif is now 9 years older than Bashir, the present age of Bashir is-

Solution:
Let Bashir's present age = x years.
Then, A's present age = (x + 9) years.

ATQ,
(x + 9) + 10 = 2(x - 10)
⇒ x + 19 = 2x - 20
⇒ x = 39.

Question: A grocer has a sale of Tk. 6430, Tk. 6932, Tk. 6850, Tk. 7235 and Tk. 6560 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Tk. 6500?

Solution:
Total sale for 5 months = Tk. (6430 + 6932 + 6850 + 7235 + 6560) = Tk. 34007.

Required sale = Tk. [(6500 × 6) - 34009]
= Tk. (39000 - 34007)
= Tk. 4993

Question: The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A?

Solution:
Here,
(A + B) - (B + C) = 12
∴ A - C = 12.

∴ C is younger than A by 12 years.

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

Question: The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.

Solution:
As the ratio of their present ages is 3 : 4 ,
let their present ages be 3X and 4X.
So, 5 years ago, as the ratio of their ages was 5 : 7,
we can write,
(3x - 5) : (4x - 5) = 5 : 7
⇒ (3x - 5)/(4x - 5) = 5/7
⇒ 21x - 35 = 20x - 25
⇒ x = 10

Hence, their present ages are 3X = 30 and 4X = 40

Question: The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

Solution:
Average of 20 numbers = 0
Sum of 20 numbers = (0 × 20) = 0
It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

Therefore, at the most 19 numbers can be greater than zero.

Question: The age of a man is 4 times of his son. Five years ago, the man was nine times old as his son was at that time. The present age of man is?

Solution:
Let the son's age be x years and the father's age be 4x years

ATQ,
4x - 5 = 9(x - 5)
⇒ 4x - 5 = 9x - 45
⇒ 5x = 40
∴ x = 8

∴ Present age of the father = 4 × 8 = 32 years

Question: A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is-

Solution:
Let there be x pupils in the class.
Total increase in marks = x × (1/2) = x/2

Here,
x/2 = (83 - 63)
⇒ x/2 = 20 
⇒ x = 40

Question: The sum of the present ages of a father and his son is 60 years. five years ago, father's age was four times the age of the son. So now the son's age will be-

Solution:
Let the present ages of son and father be x and (60 - x) years respectively.

Then,
(60 - x) - 5 = 4(x - 5)
⇒ 55 - x = 4x - 20
⇒ 5x = 75
∴ x = 15

Question: Find the average of the square of first 5 consecutive odd numbers starting from 1 to 9, where the last odd number is 9.

Solution:
The average of square of first n consecutive odd numbers starting from 1 to X, where the last odd number is X, is given by = X(X + 2)/3
Here,
X=9
So, average = 9(9 + 2)/3
= (9 × 11)/3
= 99/3
= 33

Question: The average age of a group of 10 students is 15 years. When 5 more students join the group, the average age increase by 1 year. The average age of the new students is?

Solution:
Total age of 10 students = 10 × 15 = 150 years

Total age of 15 students = 15 × 16 = 240 years

Total age of 5 new students = 240 - 150 = 90 years

∴ Average age of 5 new students = 90/5 = 18 years

Question: The average weight of P, Q and R is 45 kg. If the average weight of P and Q is 40 kg and that of Q and R is 43 kg, what is the weight of Q?

Solution:
Let
P, Q, R represent their respective weights. Then, we have:
P + Q + R = (45 × 3) = 135  ........ (i)
P + Q = (40 × 2) = 80 ....... (ii)
Q + R = (43 × 2) = 86 ....... (iii)

Adding (ii) and (iii), we get: P + 2Q + R = 166 ...... (iv)
Subtracting (i) from (iv), we get: Q = 31

Question: If the ratio of present age of Ashik and Vishal is 9 : 4 and their sum of present age is 52 years, find the present age of Vishal.

Solution:
Let, age of Ashik be 9x and age of Vishal be 4x
Then, sum of both ages = 9x + 4x = 13x
⇒ 13x = 52
⇒ x = 4 years

∴ Age of Vishal = 4x = 4 × 4 = 16 years

Question: In a class average age of 15 boys is 11. If 5 boys each of age 9 years are added, what would be the new average?

Solution:
Sum of ages of 15 boys = 15 × 11= 165
Sum of ages of 5 boys = 5 × 9 = 45
Total age of 20 boys = 165 + 45 = 210
∴ Average of ages of 20 boys = 210/20 = 10.5 years

Question: Product of present age of Rahim and Latif is 2223 years and their present age ratio is 19 : 13 find the difference age of Rahim and Latif.

Solution:
Let, age of Rahim be 19x and age of Latif be 13x
Then, product of their ages = 19x × 13x = 247x²
⇒ 247x² = 2223
⇒ x² = 9
∴ x = 3

Hence, required difference = 19x - 13x = 6x = 6 × 3 = 18 years

Question: In Rakib's opinion, his weight is greater than 65 kg but less than 72 kg. His father does not agree with Rakib, and he thinks that Rakib's weight is greater than 60 kg but less than 70 kg. His sister's view is that his weight cannot be greater than 68 kg. If all are correct in their estimation, what is the average of the different possible weights of Rakib?

Solution:
Let Rakib's weight by X kg.
According to Rakib: 65 < X < 72
According to Rakib's father: 60 < X < 70.
According to Rakib's sister: X <= 68

The different possible weights of Rakib or the values that satisfy all the above conditions are 66, 67 and 68.
So, the Average of different possible weights of Rakib = (66 + 67 + 68)/3
= 201/3
= 67 kg.

Question: The ratio of present age A and B is 13 : 10 after 2.5 years their ratio will be 32 : 25 then find the present age of A.

Solution:
Let, present age of A = 13x and present age of B = 10x

According to question:
(13x + 2.5)/(10x + 2.5) = 32/25
⇒ (13x + 2.5) × 25 = (10x + 2.5) × 32
⇒ 325x + 62.5 = 320x + 80
⇒ 5x = 17.5
⇒ x = 3.5

∴ Present oge of A = 13 × 3.5 = 45.5

Question: Find the average of the numbers 32, 18, 95, 52 and 86.

Solution:
Average = Sum of all observations/ Number of observations
∴ Average = (32 + 18 + 95 + 52 + 86)/5
= 283/5
= 56.6
∴ The correct answer is 56.6.

Question: If 5 years ago, the ratio of age of Mridul and Lovely was 1 : 2 and after 15 years from present their ratio would be 5 : 6. Find the age of Lovely after 20 years.

Solution:
Let, present age of Mridul be x and present age of Lovely be y.

Then, according to question
(x - 5)/(y - 5) = 1/2
⇒ 2x - 10 = y - 5
⇒ x = (y + 5)/2 .............(1)

Also,
(x + 15)/(y + 15) = 5/6
⇒ 6x + 90 = 5y + 75
⇒ 6x + 15 = 5y

Putting value of x from equation 1, we get
3y + 15 + 15 = 5y
⇒ 2y = 30
⇒ y = 15

∴ Age of Lovely after 20 years = 15 + 20 = 35 years.

Question: The average age a family consists of six members is 20 years. If one person is removed, the average becomes 15 years. Find the age of the person removed?

Solution:
The average age of a family consists of 6 members is 20 years.
The total age of all family members 20 × 6 = 120

one person is removed then the average becomes 15 years.
The sum of the age of 5 members 15 × 5 = 75 years.

The age of the person removed 120 - 75 = 45 years.
∴ The average age of the person removed is 45 years.

Question: Ages of two persons differ by 16 years. If 6 year ago, the elder one be 3 times as old the younger one, find their present age?

Solution:
Let the age of an younger person be A
age of elder person = (A + 16)

According to the question, we have
3(A - 6) = (A + 16 - 6)
⇒ 3A - 18 = A + 10
⇒ 2A = 28
⇒ A = 14

∴ The younger and elder person age is 14 years old and 30 years old.

Question: The average of six consecutive numbers A, B, C, D, E and F is 62. What is the sum of B and F?

Solution:
Let the Numbers A, B, C, D, E, F be x, x + 1, x + 2, x + 3, x + 4, x + 5.
According to question, 
x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 62 × 6
⇒ 6x + 15 = 372
⇒ 6x = 372 - 15 = 357
⇒ 6x = 357
⇒ x = 59.5

∴ B = x + 1 = 60.5 
∴ F = x + 5 = 64.5

∴ B + F = 60.5 + 64.5 = 125

Question: Twice the age of X is thrice the age of Y. 8 years back, the difference between the ages of X and Y was 18 years. What is the present age of X?

Solution:
2 times the age of X = 3 times the age of Y
Difference between X and Y 8 years back = 18 years

2X = 3Y
⇒ X : Y = 3 : 2

Let the present age of X and Y be 3R and 2R respectively
⇒ 3R - 2R = 18
⇒ R = 18

∴ 3R = 3 × 18 = 54
∴ The present age of X is 54 years

Question: The average of nine numbers is 60, that of the first five numbers is 55 and the next three is 65. The ninth number is 10 less than the tenth number. Then, tenth number is-

Solution:
Average of nine numbers = 60
Average of first five numbers = 55 and
average of next three numbers = 65
Tenth number = Ninth number + 10 

The sum of nine numbers = 60 × 9 = 540
The sum of the first five numbers = 55 × 5 = 275
The sum of the next three numbers = 65 × 3 = 195
Ninth number = (540 - 275 - 195) = (540 - 470) = 70

∴ Tenth number = 70 + 10 = 80