BANK & INSURANCE (AVERAGE) PART 3

Total Questions: 45

31. Sid, Tony, Paul and Ram paid different fees in a coaching. The fee paid by Sid is equal to the average fee paid by remaining three. The fee paid by Tony is 25% of the fee paid by Tony, Paul and Ram together. Ram paid Rs. 10000 less fee than Paul. If the average fee paid by Sid and Ram is Rs. 23000, then find the fee paid by Paul.

Correct Answer: (e) Rs. 32000
Solution:

Let the fee paid by Sid be Rs. ‘x’
Therefore, fee paid by remaining three = Rs. 3x

Fee paid by Tony = 0.25 × 3 = Rs. 0.75x

Fee paid by Paul and Ram = 3x - 0.75x
= Rs. 2.25x

Fee paid by Paul = Rs. (1.125x + 5000)
Fee paid by Ram = Rs. (1.125x - 5000)

According to the question,
1.125x - 5000 + x = 2 × 23000

Or, x = 51000/2.125 = Rs. 24000

Therefore, fee paid by Paul = 1.125x + 5000
= Rs. 32000

32. Four persons ‘A’, ‘B’, ‘C’ and ‘D’ have different amounts with them. ‘D’ has twice amount than that of ‘A’. The amount ‘B’ has is 12.5% more than that of ‘D’. The amount ‘C’ has is equal to the average amount of all four persons have. If the sum of the average amount all four have and the amount ‘A’ has is Rs. 660, then find the amount ‘B’ has.

Correct Answer: (b) Rs. 540 
Solution:

Let ‘A’ has Rs. a
Therefore, amount possessed by ‘D’ = Rs. 2a

Amount possessed by ‘B’ = 1.125 × 2a
= Rs. 2.25a

Let the average amount possessed by all four be Rs. x
Therefore, amount possessed by ‘B’ = Rs. x

According to the question,
2.25a + 2a + a = 4x - x
Or, 5.25a = 3x
Or, x/a = 7/4

Therefore, a = 660 × 4/11 = Rs. 240
Amount possessed by ‘B’ = 2.25a = Rs. 540

33. Bumrah’s bowling average in some matches was 28. After playing one more match (last match) in which he took 6 wickets by giving 102 runs, his bowling average decreased by 1.65. Find the total number of wickets he had taken in all his matches. (Bowling average = total number of runs given ÷ total wickets taken)

Correct Answer: (b) 40  
Solution:

Let total runs given and total wickets taken before the last match be ‘x’ and ‘y’ respectively.

So, x = 28y And, (x + 102) = (28 - 1.65) × (y + 6)

x + 102 = 26.35y + 158.1
28y - 26.35y = 56.1
1.65y = 56.1
y = 34

So, total wickets taken by Bumrah
= 34 + 6 = 40

34. Sunil planned to complete 448 pages assignment in a certain number of days. For the first 9 days, he achieved his planned per day target. However, for the remaining days, he increased his efficiency and completed 6 more pages per day than planned. In this way, he completed 456 pages on one day before the planned finish date. What is the overall average number of pages completed per day by Sunil?

Correct Answer: (a) 30.4 pages  
Solution:

Suppose Sunil has planned to complete a number of pages per day and he planned to finish the assignment in number of days.
∴ 448/a = n

According to the given data.
Pages completed up to 9th day + Pages completed from 10th day to (n - 1)th day = 456

9 × a + (a + 6) × (n - 1 - 9) = 456
⇒ 9a + an + 6n - 10a - 60 = 456
⇒ an - a + 6n = 516

Putting n = 448/a.
⇒ a × (448/a) - a + 6 × (448/a) = 516
⇒ 448 - a + 2688/a = 516
⇒ 448a - a² + 2688 = 516a
⇒ a² + 68a - 2688 = 0
⇒ (a - 28)(a + 96) = 0
⇒ a = 28

∴ Sunil planned to complete the 28 pages per day.
Planned number of days to complete the assignment = 448/28 = 16

But due to increased efficiency, he completes 456 pages in (16 - 1) = 15 days.
∴ Overall average number of pages completed per day by Sunil = 456/15 = 30.4

35. Raj works on day payment method. For the first three days of the week, he got 25% of the amount of the last two days of the week. The fourth and the fifth day collection is 75% of the last two days collection. If the total collection is Rs. 4200 then the average collection of the last two days of week is –

Correct Answer: (b) Rs. 1050  
Solution:

Given,
Total collection of seven days of week = 1st day + 2nd day + 3rd day + 4th day + 5th day + 6th day + 7th day Given,

⇒ 1st day + 2nd day + 3rd day = (25/100) × (6th day + 7th day)
⇒ 4th day + 5th day = (75/100) × (6th day + 7th day)

Then,
⇒ 4200 = (6th day + 7th day)/4 + 3(6th day + 7th day)/4 + (6th day + 7th day)

⇒ 4200 = (6th day + 7th day)/4 + 3(6th day + 7th day)/4 + (6th day + 7th day) Solving,

⇒ 4200 × 4 = (6th day + 7th day) + 3(6th day + 7th day) + 4(6th day + 7th day)

⇒ 4200 × 4 = 8(6th day + 7th day)
⇒ (6th day + 7th day) = 2100

Average collection of the last two days of the week = 2100/2 = Rs. 1050

36. Average of 17 students in a class is X. When their marks are arranged in ascending order it was found to be in Arithmetic Progression. The class teacher found that rank of the students who ranked 15th, 11th, 9th and 7th had copied the exam and hence they are suspended. Now the average of the remaining class is Y. Then

Correct Answer: (c) X < Y
Solution:

17X = 17/2 (2a + 16d)
X = a + 8d

13Y = 17/2 (2a + 16d) - (4a + 26d)
Y = a + 8.46d

37. In an aptitude exam for a company, there is certain average marks. After recovering quantitative mistakes, the average marks of 125 students got decreased from 80 to 68 and the average marks of all the students is decreased by 10 marks. Find the total number of students.

Correct Answer: (b) 150  
Solution:

Let average marks be N. Total marks of 125 students = 80 × 125 = 10000

After recovering quantitative mistakes,
Total marks of 125 students = 68 × 125 = 8500

Average marks are reduced by = 10
Total number of students in an exam
= (10000 - 8500)/10 = 150

∴ Required number of students = 150

38. In a coaching institute in Mukherjee Nagar, there are 252 students, in which the ratio of boys and girls is 2 : 1. Some more girls are enrolled, and the number of girls become equal to that of the boys. The average age of all the students is now 22 years and the average age of boys is 2 years more than the average age of the girls. Find the average age of boys and girls.

Correct Answer: (a) 23, 21  
Solution:

It is given that the ratio of boys and girls is = 2 : 1
Let the number of boys is 2X and number of girls is X

⇒ 3X = 252
⇒ X = 84 Boys = 168, girls = 84

Let K girls enrolled later then ratio = 1 : 1
⇒ 168/(84 + K) = 1/1
⇒ K = 84

Now average, [boys(168) + girls(168)]/336 = 22 ........ (1)

And it is given that, Girls(168)/168 + 2 = boys(168)/168 ........ (2)

By solving these two equations, Girls(168) = 3528
Boys(168) = 3864

Average age of girls = 3528/168 = 21
Average age of boys = 3864/168 = 23

39. The overall change in the average due to the dual change is ______ in which average weight of 5 men is decreased by 3 kg when one of them weighing 150 kg, is replaced by another person. This new person is again replaced by another person whose weight is 30 kg lower than the person he replaced.

Correct Answer: (b) 9 kg  
Solution:

In case of replacement this formula can be applied,
value of new observation = (a + nb) for increment and a( - nb) for decrement

Here, a = 150, n = 5 and b = 3

So, the weight of new person = a - nb = 150 - 5 × 3 = 135 kg

Again, 135 kg is replaced by a new person whose weight is 30 kg.

So, the weight of the next new person = 135 - 30
= 105 kg

Thus, overall change in average = (150 - 105)/5
= 9 kg

40. 10 overs of a cricket match are played by each team A and team B. Average runs of team A is 1.5 times than average runs of B. Combined average of both teams is 25 runs per over. What average of runs is needed to be scored by team A in next 5 overs to score total runs of 385?

Correct Answer: (a) 17 runs  
Solution:

Let the average run of team B be x runs
Then, the average run of team A will be 1.5x runs

(10x + 15x)/20 = 25
⇒ 25x = 25 × 20
⇒ x = 20

Average runs of team A is 20 runs
Average runs of team B is 20 × 1.5 i.e. 30 runs

Total runs of team A = 15 × 20 = 300 runs

Runs required in next 5 overs = 385 - 300
= 85 runs

Average of runs required in next 5 overs
= 85/5 = 17 runs

∴ On an average, 17 runs are required in an over to get 385 runs in next 5 overs