__Practice Problems on Averages – Set 5__**1. The average salary of all the employees in a small organization is Rs 8,000. The average salary of 7 technicians is Rs 12,000 and the average salary of the rest is Rs 6,000. The total number of employees in the organisation is?**

a) 20

b) 21

c) 22

d) 25

**Answer: E)**

**Explanation:**

Let total number of employees be X.

Then, 8000 x X = 7 x 12000 + (X - 7) x 6000

X = 26.

Thus, the total number of employees in the
organization is 26.

**2.**

**Without any stoppage, a person travels a certain distance at an average speed of 42 km/h, and with stoppages he covers the same distance at an average speed of 28 km/h. How many minutes per hour does he stop?**

a)
14 min

b)
15 min

c)
26 min

d)
28 min

e)
None of these

**Answer: E)**

**Explanation:**

Let the total distance to be covered is 48 kms.

Time taken to cover the distance without stoppage =
48/42 hrs = 2 hrs

Time taken to cover the distance with stoppage =
48/28 = 3 hrs.

Thus, he takes 60 minutes to cover the same distance
with stoppage.

Therefore, in 1 hour he stops for 20 minutes.

**3. The average marks of a Suresh in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest?**

a) 55

b) 60

c) 62

d) 65

e) Can’t be determined

**Answer: B)**

**Explanation:**

Total marks of 10 papers = 80 x 10 = 800

Total marks of 8 papers = 81 x 8 = 648

Total marks of two papers = (800 - 648) = 152

If highest total is 92, then the lowest total is

(152 - 92) = 60.

**4. Three maths classes: X, Y and Z take an algebra test. The average score of class X is 83. The average score of class Y is 76. The average score of class Z is 85. The average score of class X and Y is 79 and average score of class Y and Z is 81. What is the average score of classes X, Y, Z?**

a) 81.5

b) 80.5

c) 83

d) 78

e) 75

**Answer: A)**

**Explanation:**

Let the number of students in classes X, Y and Z be
A, B and C respectively.

Then, total score of X= 83A, total score of Y = 76B,
total score of Z = 85C.

Also given that,

(83A + 76B) / (A + B) = 79

=>4A = 3B.

(76B + 85C)/(B + C) = 81

=>4C = 5B,

=>B = 4A/3 and C = 5A/3

Therefore, average score of X, Y, Z = ( 83A + 76B +
85C ) / (A + B + C) = 978/12 = 81.5.

**5. The average of 17 numbers is 10.9. If the average of first nine numbers is 10.5 and that of the last nine numbers is 11.4, the middle number is**

a) 11.8

b) 11.4

c) 10.9

d) 11.7

e) 10.7

**Answer: A)**

**Explanation:**

Sum of first nine numbers (N1 to N9) + Sum of last nine numbers (N9 to N17) = 10.5
x 9 + 11.4 x 9 = 21.9 x 9 = 197.1

Hence, the middle number

= 197.1 - 17 x 10.9

= 197.1 - 185.3 = 11.8

**6. Suraj has a certain average of runs for 12 innings. In the 13th innings he scores 96 runs thereby increasing his average by 5 runs. What is his average after the 13th innings?**

a) 48

b) 64

c) 36

d) 72

e) 68

**Answer: C)**

**Explanation:**

To improve his average by 5 runs per innings he has
to contribute 12 x 5 = 60 runs for the previous 12 innings.

Thus, the average after the 13th innings

= 96 - 60 = 36.

**7. A batsman in his 17th innings makes a score of 85, and thereby increases his average by 3. What is his average after the 17th innings? He had never been ’not out’.**

a) 47

b) 37

c) 39

d) 43

e) 47

**Answer: B)**

**Explanation:**

Average score before 17th innings

= 85 - 3 × 17= 34

Average score after 17th innings

=> 34 + 3 = 37

**8. The sum of three numbers is 98. If the ratio between first and second be 2 : 3 and that between second and third be 5 : 8, then the second number is?**

a) 30

b) 20

c) 58

d) 48

e) 38

**Answer: A)**

**Explanation:**

Let the numbers be X, Y and Z. Then,

X + Y + Z = 98, X/Y = 2/3 and Y/Z = 5/8

Therefore, X = 2Y/3 and Z = 8Y/5. So, 2Y/3 + Y +
8Y/5 = 98.

49Y/15 = 98

Y = 30.

**9. The average weight of 8 sailors in a boat is increased by 1 kg if one of them weighing 56 kg is replaced by a new sailor. The weight of the new sailor is?**

a) 57 kg

b) 60 kg

c) 64 kg

d) 62 kg

e) 68 kg

**Answer: C)**

**Explanation:**

The sailor weighing 56 kg is replaced and the
average is increased by 1 kg.

Hence, the weight of the new sailor is (56 +
increase in total weight)

= 56 +1 x 8

= 56 + 8 = 64 kg.

**10. A number X equals 80% of the average of 5, 7, 14 and a number Y. If the average of X and Y is 26, the value of Y is?**

a) 13

b) 26

c) 39

d) Can’t be determined

e) None of these

**Answer: C)**

**Explanation:**

Average of 5, 7, 14 and Y = ( 5 + 7 + 14 + Y )/ 4

Therefore, X = 80% of ( 5 + 7 + 14 + y )/ 4

= (80/100) x (26 + Y)/4

=> X = (26 + Y)/5

5X - Y = 26----- (i)

Also, (X + Y)/2 = 26 ----- (ii)

From (i)and (ii)

X = 13

Y = 39.