ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: For the past n days, the average (arithmetic mean) daily production at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n?

Solution:
total production in n days is = 50n

ATQ,
50n + 90 = (n + 1)55
or, 50n + 90 = 55n + 55
or, 55n - 50n = 90 - 55
or, 5n = 35
∴ n = 7

Question: 12 years ago the difference in age between Puja and Sejan was 6 years. What is the difference now?

Solution: 
The difference in age remains the same as they both pass the same period of time together.
so, at present, their age difference is 6 years.

Question: If m is the average of the first 10 positive multiples of 5 and M is the median of the first 10 positive multiples of 5, what is the value of M – m?

Solution: 
The first 10 multiples of 5 are = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
average of these numbers is, m = (5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50)/10
= 27.5

the median of these numbers, M = (25 + 30)/2 = 27.5

∴ M - n = 0

Question: The sum of age of a family consisting of a father, a mother and their only son is 70. Mother is four times the age of the son and father is five times the age of the son. What is the age of the mother?

Solution: 
let, 
the age of the son is x 
so,
mother's age = 4x
father's age = 5x

∴ x + 4x + 5x = 70
or, 10x = 70
∴ x = 7

mother's present age is = (4 × 7) = 28 years

Question: The average of 12 numbers is 16 and the average of the first five is 14. What is the average for the rest?

Solution: 
Average of twelve numbers = 16
Sum of twelve numbers = 16 × 12 = 192
Average of first five numbers = 14
Sum of first five numbers = 14 × 5 = 70

Total of remaining seven numbers = 192 - 70 = 122
Average of rest = 122/7 = 17.43

Question: 9 years ago a brother was 7 years older than his sister. At present their total age is 53. What is the present age of the sister?

Solution: 
Let,
sister was x years old 9 years ago.
∴ brother was (x + 7) years old

ATQ,
x + x + 7 + 9 + 9 = 53
or, 2x = 53 - 25
or, x = 14

present age of the sister is = 14 + 9 = 23 years

Question: For 9 innings, Roman has an average of 70 runs. In the tenth inning, he scores 210 runs, thus increasing his average. His average increased by-

Solution:
Total score for 9 innings = 70 × 9 = 630
Total score after 10th innings = 630 + 210 = 840
So, the new average is 840/10 = 84


So, the increment is 84 - 70 = 14 

Question: The present age of son is half of the present age of his mother. Five years ago, his mother's age was thrice the age of her son. What is the present age of the son?

Solution:
Let the mother's age be 2x years
Then, Son's age = x years

ATQ,
2x - 5 = 3(x - 5)
or, 2x - 5 = 3x - 15
or, x = 10

∴ The present age of the son = 10 years.

Question: The average of five consecutive integers is P. If the next two numbers are added, how shall the average vary?

Solution:
Let the five consecutive numbers be 5, 6, 7, 8, and 9 respectively.

So, the average is = (5 + 6 + 7 + 8 + 9)/5 = 35/5 = 7
Suppose, the average is, P = 7

If the next two numbers are added, then average is = (5 + 6 + 7 + 8 + 9 + 10 + 11)/7 = 56/7 = 8
So, the new average is, P = 7 + 1

So, the average increased by 1

Question: 5 years earlier X was triple the age of Y. At present the sum of their age is 70. What is the present age of X?

Solution: 
Let,
5 years ago Y was a years old
X was = 3a years old.

ATQ,
a + 3a + 5 + 5 = 70
or, 4a = 70 - 10
∴ a = 15

∴ the present age of X is = (3 × 15) + 5
= 50 years

Question: What is the average from 1 to 59?

Solution:
We know,
Average of n natural numbers = (n + 1)/2
Here, n = 59

∴ Average = (59 + 1)/2
= 60/2
= 30

Question: Tanvir’s present age is two-fifths of the age of his father. After 8 years, Tanvir will be one-half of the age of his father. What is the present age of the father?

Solution:
Let,
Present age of father = p year
So, Tanvir's present age = p × (2/5) year
  
After 8 years father will be = (p + 8) year
After 8 years Tanvir will be = {(2p/5) + 8} year

According to the conditions,
(1/2)(p + 8) = {(2p/5) + 8}
or, (p + 8)/2 = (2p + 40)/5
or, 5p + 40 = 4p + 80
or, 5p - 4p = 80 - 40
∴ p = 40

So, Present age of father = 40 years.

Question: Average mark in Math in a class of 40 students is 45. Average mark of all the 30 boys is 48. Then the average mark obtained by the girls is:

Solution: 
Average mark of 40 students is 45
Total mark of 40 students is (45 × 40)
= 1800 

Average mark of all the 30 boys is 48
Total mark of all the 30 boys is (48 × 30)
= 1440

∴ Total marks of all the 10 girls is (1800 - 1440) = 360
The average mark of all the 10 girls is 360/10 = 36

Question: A couple has a son and a daughter. The age of the father is four times the son and the age of the mother is three times the daughter. The mother is 6 years younger to the father and the daughter is 3 years older than the son. What is the age of the mother?

Solution:
Let, father, mother, son and daughter have age of F, M, S and D respectively.
ATQ,
F = 4S ………….(i)
M = 3D………….(ii)
M = F - 6………..(iii)
D = S + 3………..(iv)

From (ii) we get,
M = 3D
Or, M = 3(S + 3)  [from (iv)]
Or, M = 3{(F/4) + 3}  [from (i)]
Or, M = ¾(F + 12)
Or, M = ¾(M + 6 + 12)  [from(iii)]
Or, M = ¾(M + 18)
Or, 4M = 3M + 54
Or, M = 54

∴ the age of the mother is 54 years.

Question: The average of ten numbers is 10. If each number is multiplied by 8, what is the new average? 

Solution:
The sum of ten numbers = 10 × 10 = 100

Now, 
If each number is multiplied by 8, the new average is
= (The sum of ten numbers × 8) / 10
= (100 × 8) / 10
= 80

Question: Of three numbers, the average of the first and third number is greater than the average of the second and third number by 14. What is the difference between the first and the second number?

Solution: 
Let, these numbers are x, y and z respectively.

ATQ,
{(x + z)/2} - {(y + z)/2} = 14
Or, (x + z) - (y + z) = 28
Or, x + z - y - z = 28
Or, x - y = 28

∴ the difference between the first and the second number is = 28.

Question: The average of six numbers is A and the average of three of these is B. If the average of the remaining three is C, then which one is correct?

Solution:
total sum of six numbers = 6A
total sum of three numbers = 3B
total sum of other the numbers = 3C


6A = 3B + 3C
or, A = 3(B + C)/6
or. A = (B + C)/2
∴ 2A = B +C

Question: The average weight of the women in a group is 55 kg and that of the men is 70 kg. If the average weight of the group is 65 kg, what is the ratio of women to men in the group?

Solution: 
Average weight of women = 55 kg
Average weight of men = 70 kg
Average weight of the entire group = 65 kg

Let,
the number of men = M.
and, the number of women = W.
Then, the total number of people in the group is (M + W)

ATQ,
55W + 70M = 65(M + W)
Or, 55W + 70M = 65M + 65W
Or, 70M - 65M = 65W - 55W
Or, 5M = 10W
Or, W : M = 1 : 2

∴ The ratio of women to men in the group is 1 : 2.

Question: The average of 9 consecutive odd numbers is 33. What is the average of the first two and last two numbers?

Solution:
The middle number is = 33
So, the other numbers are = 25, 27, 29, 31, 33, 35, 37, 39, 41

The average of the first two and last two numbers is = (25 + 27 + 39 + 41)/4 = 132/4 = 33

Question: The present age of three persons are in the ratio of 4 : 7 : 9. Eight years ago, their total age was 56 years. In five years what will be the age of the youngest person?

Solution: 
Present age ratio of three persons is 4 : 7 : 9

Let,
Their age is 4X, 7X and 9X respectively.

ATQ,
4X - 8 + 7X - 8 + 9X - 8 = 56
Or, 20X - 24 = 56
Or, 20X = 80
Or, X = 4

∴ The present age of the youngest person is = 4X = 4 x 4 =  16 years.
In five years, his age will be = (16 + 5 ) = 21 years.

Question: The average of five numbers is 16. If one number is excluded, the average becomes 14. The excluded number is -

Solution:
The average of five numbers is 16
The sum of five numbers is (16 × 5) = 80

let, the excluded number is x

so,
(80 - x)/4 = 14
or, 80 - x = 56
or x = 80 - 56
∴ x = 24

Question: X’s age 3 years ago was three times the present age of Y. At present Z’s age is twice the age of Y and also Z is 12 years younger than X. What is the present age of Z?

Solution:
Let,
Their present age is X, Y, Z respectively.

As X’s age 3 years ago was three times the present age of Y.
X - 3 = 3y
X - 3y = 3……….(i)

Z’s age is twice the age of Y
∴ Z = 2Y……….(ii)

Z is 12 years younger than X
∴ X - 12 = Z 
Or, X - 12 = 2Y [ putting Z = 2Y from equation (ii) ]
Or, X - 2Y = 12……..(iii)

(iii) - (i),
X - 2Y - X + 3Y = 12 - 3
Or, Y = 9

So, the present age of Y is 9.
∴ the present age of Z is = 2Y = 2 x 9 = 18 years.

Question: The average age of a group of 20 students is 16 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 20 students = 20 × 16 = 320 years

Total age of 25 students = 25 × 17 = 425 years

Total age of 5 new students = 425 - 320 = 105 years

∴ Average age of 5 new students = 105/5 = 21 years

Question: The average of two numbers x and y is 16. If x, y, z are non-negative integers such that x < z < y, what is the minimum possible average of x, y, z.

Solution: 
According to the questions,
(x + y)/2 = 16
x + y = 32
As x, y, z are non-negative integers and x < z < y.

The minimum possible value of these integers are,
X = 0
Z = 1
Y = 32

∴ Average = (0 + 32 + 1)/3
= 11

Question:  A boy rides his bicycle 10 km at an average speed of 12 km/hr. and again travels 12km at an average speed of 10km/hr. His average speed of for the entire trip is approximately -

Solution: 
Total Distance = 10 + 12 = 22 km
Total time = (10/12) + (12/10)
= (5/6) + (6/5)
= (25 + 36)/30
= 61/30 hours

Average speed = 22/(61/30) = 660/61 = 10.8 km/hour