BANK & INSURANCE (AVERAGE) PART 3

Total Questions: 45

21. The average weight of the boys in the class is 30 kg and the number of girls in the class is 18. If the average weight of the class is 36 kg and the one new student joined the class, then the average weight of the class is increased by 4 kg. If the weight of the new student is 160 kg, then find the average weight of the girls in the class?

Correct Answer: (c) 40 kg
Solution:

Number of girls = 18
Number of students in the class = x
Number of boys in the class = x - 18

Average weight of the class = 36 kg
Total weight of the class = x × 36 = 36x

36x + 160 = (x + 1) × 40
36x + 160 = 40x + 40
4x = 120
x = 30

Number of students = 30
Number of girls = 18
Number of boys = 30 - 18 = 12

Total weight of the class = 36 × 30 = 1080 kg
Total weight of the boys = 30 × 12 = 360
Total weight of the girls = 1080 - 360 = 720

Average weight of the girls = 720/18 = 40 kg

22. If the ratio of the number of boys to girls in the class is 4:3 and the average age of the boys is 2 years more than the average age of girls. If the difference between the number of boys and girls in the class is 4 and the total ages of the girls in the class is 120 years less than the total ages of the boys in the class, then what is the approximate average age of the class?

Correct Answer: (b) 23  
Solution:

Number of boys = 4 × 4 = 16
Number of girls = 3 × 4 = 12

Average age of boys = x
Average age of girls = y

x - y = 2 ----(1)

Total age of girls = y × 12
Total age of boys = x × 16

x × 16 - y × 12 = 120
4x - 3y = 30 ----(2)

(1) × 4 - (2)
-y = -22
y = 22
x = 24

Required average = (24 × 16 + 22 × 12)/28
= 23

23. There are 7 consecutive odd and 5 consecutive even numbers. The average of seven consecutive odd numbers is 43. The lowest even number is twice the highest odd number. Find the difference between the sums of seven consecutive odd numbers to that of five consecutive even numbers?

Correct Answer: (e) None of these
Solution:

The average of seven consecutive odd numbers = 43

If the average of consecutive number/odd/even number is always the middle number, So,
The 7 consecutive odd numbers are,
37, 39, 41, 43, 45, 47, 49

The lowest even number = 49 × 2 = 98

The 5 consecutive even numbers are,
98, 100, 102, 104, 106

Required difference = (98 + 100 + 102 + 104 + 106) - (43 × 7)
⇒ 510 - 301 = 209

24. The average weight of the class is 50 kg and the average weight of the boys in the class is 42 kg more than that of the average weight of the girls in the class. If one student absent the class, then average weight of the class is decreased by 2 kg and the total number of students in the class is 20, then find the weight of the girls in the class?

Correct Answer: (e) Can’t be determined
Solution:

Weight of the boys = x
Weight of the girls = y

Total weight of the class = 20 × 50 = 1000 kg
Total weight of 19 students = 19 × 48 = 912 kg

x - y = 42

Weight of the absentees = 1000 - 912 = 88
We can’t find the answer.

25. Ravindra Jadeja played ‘n’ number of T20 internationals in 2020. His batting average increased by 10 runs if first 5 innings not included, and his batting average is decreased by 10 runs if last five innings not included. If his average score of first 5 T20 innings is 80 and last T20 innings is 120, then find the value of (20% of a + n² + 25).

Correct Answer: (d) 270 
Solution:

Let average score of n innings of Ravindra Jadeja = a
Total score of n innings = n × a

If first 5 innings not included then average score of (n - 5) innings = (a + 10):
So, n × a = 80 × 5 + (n - 5) × (a + 10):
So, n × a = 400 + n × a + 10n - 5a - 50
5a - 10n = 350
So, a - 2n = 70 .......... (1)

If last 5 innings not included then average score of (n - 5) innings = (a - 10)
So, n × a = 120 × 5 + (n - 5) × (a - 10)
So, n × a = 600 + n × a - 10n - 5a + 50
10n + 5a = 650
So, a + 2n = 130 .......... (2)

On adding both equations, we get
Value of a = 100
Value of n = 15

Required value = (20% of a + n² + 25) = (20% of 100 + 15² + 25) = 270

26. In a company, 60% of the workers are apprentice. All the remaining workers are company employees. The monthly income of each apprentice is Rs.4000. The monthly income of each company employee is Rs.8000. What is the average monthly income of all the workers in the company together?

Correct Answer: (d) Rs.5600 
Solution:

Let e be the number of workers.
60% of the workers are apprentice, 60/100×e = 0.6e
Number of company employees = (The number of workers) - (The number of apprentice)
= e - 0.6e = 0.4e

The monthly income of each apprentice is Rs.4000
The monthly income of each employee is Rs.8000

Hence, the total income of the workers = (0.6e × 4000) + (0.4e × 8000)
= 2400e + 3200e
= 5600e

Required average = Total income/Number of workers = 5600e/e = Rs.5600

27. The average weight of ‘x’ persons is 51 kg. The average weight of (x – 18) women is 50 kg, the average weight of (x – 20) children is 45 kg, and the average weight of (x – 22) men is 60 kg. Find the value of ‘x’, if total person is the sum of men, women and children.

Correct Answer: (a) 30  
Solution:

Total weight of x persons = 51 × x = 51x kg
Total weight of (x - 18) women = 50 × (x - 18) = 50x - 900 kg
Total weight of (x - 20) children = 45 × (x - 20) = 45x - 900 kg
Total weight of (x - 22) men = 60 × (x - 22) = 60x - 1320 kg

According to the question,
51x = 50x - 900 + 45x - 900 + 60x - 1320
155x - 3120 = 51x
104x = 3120, x = 30

28. The average of four numbers is twice the second number. The fourth number is 25% more than that of 1st number. The third number is 52 more than that of the average of all the four numbers. If the 3rd number is 56 more than the 4th number, and 2nd number is 16 less than the 1st number then find the average of all the four numbers.

Correct Answer: (c) 64
Solution:

Let the average of all the four numbers be 2x.
Therefore, 3rd number = 2x + 52
2nd number = x
1st number = (x + 16)
4th number = 1.25(x + 16)

According to the question,
2x + 52 + x + (x + 16) + 1.25(x + 16) + 20 = 4 × 2x
Or, x = 88/2.75 = 32

Therefore, average = 2x = 64

29. The average marks obtained by all the students of a class in Maths is 75. If the marks of 2 students had been less by 18 and 26 respectively, then the average marks had been 2 less. If the marks of all the students arranged in A.P., then the difference between the consecutive terms came out to be same. If the highest marks obtained is 50% more than the lowest marks obtained, then which of the following can be determined using the given data? I. Number of students in the class II. The highest marks obtained by a student III. The marks obtained by the 15th student

Correct Answer: (d) All I, II and III  
Solution:

Let there are n students in the class

For I: According to the question, (x₁ + x₂ + ……… + xₙ) = 75n ........ (1)

Also, (x₁ + x₂ + ……… + xₙ) = 73n + (18 + 26) ........ (2)

On solving equation (1) and (2), we get
Number of students = n = 22

For II: Also, let the lowest marks be ‘y’.
Then, highest marks = 1.5y So, (y + 1.5y)/2 = 75, y = 60
So, highest marks = 90

For III: According to the question, 75 × 22
= (22/2)[2 × 60 + (22 - 1)d]
Or, d = 30/21 = 10/7

Marks obtained by the 15th student
= 60 + (15 - 1) × (10/7) = 80

30. Four different types (A, B, C and D) of boxes were sold. The number of box ‘D’ sold is 15 more than the average number of all the boxes sold. The number of box ‘B’ sold is 120 more than that of ‘D’. The number of box ‘C’ sold is 60 more than that of ‘A’. If the average number of box ‘A’, ‘B’, ‘C’ sold is 220, then find the number of box ‘A’ sold.

Correct Answer: (a) 120  
Solution:

Let the average number of boxes of all the four types sold be ‘x’
Therefore, number of box ‘D’ sold = (x + 15)

Total number of box A, B and C sold
= 220 × 3 = 660

Therefore, (660 + x + 15)/4 = x
Or, 3x = 675 Or, x = 675/3 = 225

Therefore, number of box ‘D’ sold = 225 + 15 = 240

Number of box ‘B’ sold = 240 + 120 = 360

Let the number of box ‘A’ sold be ‘y’
Therefore, number of box ‘C’ sold = (y + 60)

(y + 60) + y = 660 - 360
2y = 240 Or, y = 120