BANK & INSURANCE (AVERAGE) PART 3

Total Questions: 45

1. In a class test, average score of 42 students is 76. If the highest and lowest score is excluded, then the average score of the remaining students drops by 1. If the difference between highest and lowest score is 108, then find the lowest score of a student.

Correct Answer: (e) 42
Solution:

Sum of marks scored by 42 students
= 42 × 76 = 3192 marks
Sum of marks when highest and lowest scores are excluded = 75 × 40 = 3000
Sum of highest and lowest marks = 3192 - 3000 = 192

Let the highest score be ‘x’
So, lowest score = (x - 108)

ATQ:
(x - 108) + x = 192
2x = 300
So, x = 150
So, lowest score of a student = 150 - 108 = 42

2. The average of the income of ‘A’ in 2019 and 2020 is Rs. 40000. He saves 25% of his income in 2019 such that his savings in both the year is same. If his expenditure in 2020 was equal to his average income of 2019 and 2020, then find the expenditure of ‘A’ in 2019.

Correct Answer: (a) Rs. 24000  
Solution:

Sum of incomes of ‘A’ in 2019 and 2020
= 2 × 40000 = Rs. 80000

Let the income of ‘A’ in 2019 be Rs. x
Therefore, his income in 2020 = Rs. (80000 - x)

Savings of ‘A’ in given two years = Rs. 0.25x each

According to the question,
0.25x = 80000 - x - 40000
Or, 0.25x = 40000 - x
Or, 1.25x = 40000
Or, x = 32000

Therefore, expenditure of ‘A’ in 2019
= 0.75x = Rs. 24000

3. The average number of students in each class of a school of classes up to 1 to 12 was 84, in the previous year. In the present year, average number of new students who took admission up to class 5 was 18 and average number of students who left school up to class 5 was also 18. If in this year the average number of students who left the school from class 6 to class 12 was 12 and the present total number of students in school is 1116 then, find the total number of new students who took admission this year.

Correct Answer: (b) 282  
Solution:

Number of students earlier = 12 × 84 = 1008
Number of students left the school
= 7 × 12 + 5 × 18 = 84 + 90 = 174

Number of students taken admission in class 1 to 5
= 18 × 5 = 90

So, number of students taken admission in class 6 to 12
= 1116 - (1008 + 90 - 174) = 192

Therefore, total new admission
= 90 + 192 = 282

4. The average number of cycles sold in a month by a shopkeeper from January to June was 96 and that from March to July was 102. The total number of cycles sold in July was 126. The number cycles sold in February was thrice the number of cycles sold in January. What is the average number of cycles sold in a month from February to July?

Correct Answer: (a) 109  
Solution:

Total number of cycles sold from January to June
= 96 × 6 = 576

Total number of cycles sold from March to July
= 102 × 5 = 510

Total number of cycles sold in July = 126

Total number of cycles sold from March to June
= 510 - 126 = 384

Total number of cycles sold in January and February together = 576 - 384 = 192

Let, total number of cycles sold in January = x
Then, total number of cycles sold in February = 3x

So, x + 3x = 192
⇒ 4x = 192
⇒ x = 192/4
⇒ x = 48

So, total number of cycles sold in February
= 3x = 144

Total number of cycles sold from February to July
= 144 + 384 + 126 = 654

Required average = 654/6 = 109

5. There are two groups of persons, A and B, and each group has 5 persons. The average age of persons of group A is 22 years while the average age of persons of group B is 19 years. A person Mohan from group A joins group B and then average age of group A becomes 24 years. Then, Rahim, a person from group A joins group B and the average age of group B becomes 20 years. Now, a person Ram of group B joins group A and the average age of group A becomes 22 years then find the ratio of age of Ram to the age of Mohan?

Correct Answer: (a) 23:14  
Solution:

Sum of total age of 5 persons of group A
= 22 × 5 = 110 years

Sum of total age of 5 persons of group B
= 19 × 5 = 95 years

Case 1:
Mohan joins group B
Resultant total age of persons of group A
= 24 × 4 = 96 years
Age of Mohan = 110 - 96 = 14 years

Case 2:
Rahim joins group B from group A
Sum of age of persons of group B = 20 × 7 = 140
Age of Rahim = 140 - (95 + 14) = 31 years
Sum of age of person of group A = 96 - 31 = 65 years

Case 3:
Ram joins group A
Sum of total age of person of group A
= 22 × 4 = 88 years
Age of Ram = 88 - 65 = 23 years

Required ratio = 23 : 14

6. Mr.Suresh’s average monthly expenditure for the first four months of the year was Rs.260 For the next five months,the average monthly expenditure was Rs.40 more than what it was during the first four months. If he spent Rs.560 in all during the remaining three months of the year, Find what percentage of his annual income of Rs.5000 did he save in the year?

Correct Answer: (c) 38%
Solution:

Suresh’s average monthly expenditure for the first four months of the year = Rs. 260.
260 × 4 = Rs. 1040

For the next five months, the average monthly expenditure was Rs.40 more than what it was during the first four months. He spent 260 + 40 for one month

In 5 months he spent 300 × 5 = 1500
He spent Rs.560 in all during the remaining three months of the year.
Total expenditure = 1040 + 1500 + 560 = 3100
Savings = 5000 - 3100 = 1900
% savings = 1900/5000 × 100 = 38%

7. A sequence P, contains six consecutive even numbers whose average of lowest and highest term is 83. There is another sequence Q, containing 5 consecutive odd numbers whose lowest term is 40 less than the 50% of sum of second lowest and third lowest number of sequence P. Find (2T1 + 3T3 – T5), where Tn is the nth term of sequence Q?

Correct Answer: (d) 168  
Solution:

Let terms in P = 2n - 4, 2n - 2, 2n, 2n + 2, 2n + 4, 2n + 6
Now, [(2n - 4) + (2n + 6)]/2 = 83
2n + 1 = 83
2n = 82,
So terms in sequence P = 78, 80, 82, 84, 86, 88
Lowest term in sequence Q = 50% of (80 + 82) - 40 = 41
So required value = (2 × 41 + 3 × 45 - 49) = 168

8. The ratio of the average of two series of five numbers each is 28: 44. All five terms of the former series are consecutive multiples of 7, while all five terms of the latter series are consecutive multiples of 11. If the sum of their first term is 36, then find the difference between the greatest term of the series divisible by 11 and the second greatest term of the series divisible by 7?

Correct Answer: (c) 31
Solution:

Let, the first series be x, (x + 7), (x + 14), (x + 21) and (x + 28):
Also second series be y, (y + 11), (y + 22), (y + 33) and (y + 44)
Average of series divisible by 7X = (x + (x + 7) + (x + 14) + (x + 21) + (x + 28))/5 = (5x + 70)/5 = x + 14
Average of series divisible by 11
= (y + (y + 11) + (y + 22) + (y + 33) + (y + 44))/5
= (5y + 110)/5 = y + 22

Also, given,
(x + 14) = 28
(y + 22) = 44

44x + 616 = 28y + 616
28y = 44x ----(1): Given x + y = 36) × 44
44y + 44x = 1584 ----(2)

From 1 and 2
y = 22 and x = 14
Series divisible by 7 is 14, 21, 28, 35, 42
Series divisible by 11 is 22, 33, 44, 55, 66
Required difference = 66 - 35 = 31

9. Total number of students in class A and B are same and average number of boys in both the classes are 80 and average number of girls in both the classes are 40. Boys in class A is __ while difference between boys and girls in class B is __ and ratio of boys to girls in class A is 7: 5. Which of the following value will fill the blanks given in the question in the same order?

A: 80 and 70  B: 60 and 50  C: 70 and 60

Correct Answer: (e) Only C
Solution:

Let total number of students in class A and B is ‘12x’ each.
Boys in class A = 12x × (7/12) = 7x
Girls in class A = 12x × (5/12) = 5x
Total boys in both the classes together = 80 × 2 = 160
Total girls in both the classes together = 40 × 2 = 80

Boys in class B = 160 - 7x
Girls in class B = 80 - 5x

Now,
(160 - 7x) + (80 - 5x) = 12x
240 = 24x
x = 10

Boys in class A = 7x = 70
Girls in class A = 5x = 50
Boys in class B = 160 - 7x = 90
Girls in class B = 80 - 5x = 30

Difference between boys and girls in class B = 90 - 30 = 60

10. Average of five integers is 52 and the sum of the first three integers is 160. The average of the first and the last integer is 187.5% of the third integer. If the fourth integer is 60, which is 28 more than the third integer, find the second integer.

Correct Answer: (c) 48
Solution:

Let the five integers be P, Q, R, S and T respectively.
P + Q + R + S + T = 52 × 5 = 260 ---- (1)
Value of S = 60

Value of R = 60 - 28 = 32
Also, P + T = 2 × 187.5% × 32 = 120 ---- (2)
So value of Q = 260 - (P + R + S + T) = 260 - 212 = 48