# SSC CPO Averages Quiz 1

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# SSC CPO Averages Quiz 1

### Introduction

Average is a straight- forward concept and it can be solved easily by equal distribution method. Average is the sum of all the elements in a given data set divided by the total number of elements in the data set. The most commonly denoted term for average is Arithmetic Mean, simply termed as mean. The article SSC CPO Averages Quiz 1 very useful for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO, IBPS RRB Exams and etc. SSC CPO Averages Quiz 1 is very useful to crack the Quantitative Aptitude sections in several exams.

### Quiz

1. In an exam of 100 marks, the average marks of a class of 40 students are 76. If the top 3 scorers of the class leave, the average score falls “down by 1. If the other two toppers except “the highest topper scored not more than 85. “then what is the minimum score the topper can score?

A. 86
B. 98
C. 95
D. 92
E. None of these

Explanation:
Total score of 40 students = (40 × 76) = 3040

Total score of top 3 scorers = 3040 – (37 × 75) = 265

To minimize the score of the top scorer, we assume the other two top scorers score the maximum they can = 85 marks each.

So, the top scorer scored = 265 – 170 = 95 marks.

2. Average weight of three friends Amar, Visera and Daman is 70 kg. Another person Vishal joins the group and now the average is 66 kg. If another person Tahir whose weight is 6 kg more than Vishal, joins the group replacing Amar, then average weight of Visera, Daman, Vishal and Tahir becomes 75 kg. What is the weight of Amar (in kg)?

A. 18
B. 20
C. 22
D. 24
E. None of these

Explanation: Total weight of Amar, Visera and Daman = 70 × 3 = 210 kg

Again, Amar + Visera + Daman + Vishal = 66 × 4 = 264 kg ……… (i)

∴ Weight of Vishal = 264 – 210 = 54 kg

∴ Weight of Tahir = 54 + 6 = 60 kg

Now, as per the question

Visera + Daman + Vishal + Tahir = 75 × 4 = 300 kg. ………………. (ii)

Subtracting (i) from (ii), we get

Tahir – Amar = 60 – Amar = 300 – 264 = 36

Therefore, weight of Amar = 60 – 36 = 24 kg

3. The average salary of a company increases by 100 when the salary of the manager, which is Rs. 9500, is included. If the number of employees excluding the manager is the smallest cube divisible by 16, what is the final average of the company?

A. Rs. 4000
B. Rs. 3400
C. Rs. 3700
D. Rs. 3100
E. None of these

Explanation:
The smallest cube divisible by 16 is 64.

Let’s assume the average salary before the manager’s salary is included is x

After addition of the Manager’s salary the average increases by 100

We can write down the above information in form of an equation as:

64x + 9500 = 65 × (x + 100)

Solving for x, we get x = 3000

The final average is 3000 + 100 = Rs. 3100

4. An exam was conducted in a state over 222 centres. The average number of applicants per centre was found to be 1560. However, it was later realized that in one centre, the number of applicants was counted as 1857 instead of 1747. What was the correct average number of applicants per centre (upto two decimals)?

A. 1557.87
B. 1558.20
C. 1558.92
D. 1559.51
E. None of these

Explanation:
Number of applicants that have been counted extra = 1857 − 1747 = 110

Hence, decrease in average = $$\frac{110}{222}$$ = 0.495

∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51

5. In Champions league, Rohit scored an average of 120 runs per match in the first 3 match and an average of 140 runs per match in the last four match. What is Rohit’s average runs for the first match and the last two match if his average runs per match for all the five match is 122 and total number of matches are 5?

A. 100
B. 200
C. 150
D. 50
E. None of these

Explanation:
Rohit’s average score in the first 3 exams = 120

Let the scores in the 5 exams be denoted by M1, M2, M3, M4, and M5

M1 + M2 + M3 = 120 × 3 = 360 …….(i)

Average of last 4 matches = 140
$$\frac{M2 + M3 + M4 + M5}{4}$$ = 140
⇒ M2 + M3 + M4 + M5 = 560 ……(ii)
Average of all the exams
$$\frac{M1 + M2 + M3 + M4 + M5}{5}$$ = 122
M1 + M2 + M3 + M4 + M5 = 122 × 5 = 610 …….(iii)

From solving above equation, we get M1 + M4 + M5 = 300
$$\frac{300}{3}$$ = 100

1. The average salary of 65 workers is Rs. 5680 out of which average salary of 31 workers is Rs. 2356 and that of 23 workers is Rs. 4589. What is the average salary of remaining workers?(value in approximate)

A. Rs. 19832
B. Rs. 19732
C. Rs. 17329
D. Rs. 18329

Explanation:
Total salary of 65 workers = 65 × 5680 = Rs. 369200

Total salary of 31 workers = 31 × 2356 = Rs. 73036

Total salary of 23 workers = 23 × 4589 = Rs. 105547

No. of remaining workers = 65 – 31 – 23 = 65 – 54 = 11

Total Salary of 11 workers = 369200 – 73036 – 105547 = 369200 – 178583 = Rs. 190617
Required average = $$\frac{190617 }{11}$$ = 17329(approx)

2. There are 7 friends. If salaries of these friends are in order-
10K = S1 ≥ S2 ≥ S3 ≥ S4 ≥….. ≥ S7 = 5K

Which of the following can’t be the average of these 7 friends.

A. 5
B. 6.71
C. 9.42
D. 6

Explanation:
Maximum average could be
$$\frac{(10 + 10 + 10 + 10 + 10 + 10 + 5) }{7}$$ = 9.28
Which is being satisfied by option C i.e. 9.42.

3. The average age of A and B is 20 years. If A is to be replaced by C, the average would be 19 years. The average age of C and A is 21 years. The ages of A, B and C in order (in years) are

A. 18, 22, 20
B. 18, 20, 22
C. 22, 18, 20
D. 22, 20, 18

Explanation:
A + B = 2 × 20 = 40 yr

B + C = 2 × 19 = 38 yr

C + A = 2 × 21 = 42 yr

2 (A + B + C) = 40 + 38 + 42 = 120

⇒ A + B + C = 60

∴ A = (A + B + C) – (B + C) = 60 – 38 = 22 yr

Similarly,

B = (A + B) – A = 40 – 22 = 18 yr

C = (C + A) – A = 42 – 22 = 20 yr

4. The average monthly income of a family of four earning members was Rs. 15130. One of the daughters in the family got married and left home, so the average monthly income of the family came down to Rs. 14660. What is the monthly income of the married daughter ?

A. Rs. 15350
B. Rs. 12000
C. Rs. 16540
D. Can’t be determined

Explanation:
Total monthly income of 4 members = 15130 × 4 = Rs. 60520

Except daughter, total monthly income of 3 members = 14660 × 3 = Rs. 43980

∴ Monthly income of married daughter = 60520 – 43980 = Rs. 16540

5. A man purchases milk for three consecutive years. In the first year, he purchases milk at the rate of Rs. 7.50 per litre, in the second year, at the rate of Rs. 8.00 per litre and in the third year, at Rs. 8.50 per litre. If he purchases milk worth Rs. 4,080 each year, the average price of milk per litre for the three years is

A. Rs. 7.68
B. Rs. 7.98
C. Rs. 7.54
D. Rs. 7.83
E. None of these

Explanation:
Price of milk in first year = Rs. 7.50 per litre.
Quantity of milk in the first year = $$\frac{4080}{7.50}$$ = 544 litres
Price of milk in second year = Rs. 8.00 per litre.
∴ Quantity of milk in second year = $$\frac{4080}{8.00}$$ = 510 litres

Price of milk in the third year = Rs. 8.50 per liter.

∴ Quantity of milk in second year = $$\frac{4080}{8.50}$$ = 480 litres
Required average = $$\frac{3 × 4080 }{544 + 510 + 480}$$ = $$\frac{12250 }{1534}$$ = = Rs. 7.98

1. The average weight of 21 boys was recorded as 64 kg. If the weight of the teacher was added, the average increased by one kg. What was the teacher’s weight ?

A. 86 Kg
B. 64 Kg
C. 72 Kg
D. 98 Kg

Explanation:
Total weight of 21 boys = 64 × 21 = 1344 kg

Given that if the weight of the teacher was added, the average increased by one kg

∴ Total weight along with the teacher = 65 × 22 = 1430 kg

Now, teacher’s weight = 1430 – 1344 = 86

2. How many marks did Tarun secure in English?

I. The average mark obtained by Tarun in four subjects including English is 60.

II. The total marks obtained by him in English and Mathematics together are 170.

III. The total marks obtained by him in Mathematics and Science together are 180.

A. I and II only
B. II and III only

C. I and III only

D. None of these

Explanation:
I gives, total marks in 4 subjects = (60 x 4) = 240.

II gives, E + M = 170

III gives, M + S = 180.

Thus, none of (A), (B), (C), (D) is true.

3. In a cricket team, the average age of eleven players in 28 years. What is the age of the captain?

I. The captain is eleven years older than the youngest player.

II. The average age of 10 players, other than the captain is 27.3 years.

III. Leaving aside the captain and the youngest player, the average ages of three groups of three players each are 25 years, 28 years and 30 years respectively.

A. Any two of the three
B. All I, II and III
C. II only or I and III only

D. II and III only

Explanation:
Total age of 11 players = (28 x 11) years = 308 years.

I. C = Y + 11 C – Y = 11 …. (i)

II. Total age of 10 players (excluding captain) = (27.3 x 10) years = 273 years.

Age of captain = (308 – 273) years = 35 years.

Thus, C = 35. …. (ii)

From (i) and (ii), we get Y = 24

III. Total age of 9 players = [ (25 x 3) + (28 x 3) + (30 x 3)] years = 249 years.

C + Y = (308 – 249) = 59 …. (iii)

From (i) and (iii), we get C = 35.

Thus, II alone gives the answer.

Also, I and III together give the answer.

4. The average age of P, Q, R, and S is 30 years. How old is R?

I. The sum of ages of P and R is 60 years.

II. S is 10 years younger than R.

A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

Explanation:
P + Q + R + S = (30 x 4) P + Q + R + S = 120 …. (i)

I. P + R = 60 …. (ii)

II. S = (R – 10) …. (iii)

From (i), (ii) and (iii), we cannot find R.

5. How many candidates were interviewed every day by panel A out of the three panels A, B, and C?

I. The three panels on average interview 15 candidates every day.

II. Out of a total of 45 candidates interviewed every day by the three panels, the number of candidates interviewed by panel A is more by 2 than the candidates interviewed by panel c and is more by 1 than the candidates interviewed by panel B.

A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

Explanation:
I. Total candidates interviewed by 3 panels = (15 x 3) = 45.

II. Let x candidates be interviewed by C.

Number of candidates interviewed by A = (x + 2).

Number of candidates interviewed by B = (x + 1).

x + (x + 2) + (x + 1) = 45

3x = 42

x = 14