# Nail the ‘average’ problems for SI exams

Hyderabad: This article is in continuation to the last article on arithmetic that will aid you in your preparation for government recruitment exams. Here are some practice bits along with solutions on the Average topic. 1. If a, b, c, d, e are five consecutive odd numbers, their average is? a) a 4 b) abcde/2 […]

Published Date - 11:10 PM, Mon - 20 June 22

**Hyderabad:** This article is in continuation to the last article on arithmetic that will aid you in your preparation for government recruitment exams. Here are some practice bits along with solutions on the Average topic.

**1. If a, b, c, d, e are five consecutive odd numbers, their average is?**

a) a 4 b) abcde/2 c) 5a d) 5(a 5)

**Ans: a**

**Solution:** a b c d e

Since they are odd consecutive numbers, they could be:

a, a 2, a 4, a 6, a 8

The middle number here is the average = a 4

**2. The average of first 99 even numbers is**

a) 9999 b) 2 c) 100 d) 50

**Ans: c**

**Solution:**

2 × 1 , 2 × 2, ………… 2 × 99

Average = (2 × 1 2 × 99)/ 2 = 100

**3. The average of 1088 real numbers is zero. At most how many of them can be negative?**

a) 1087 b) 595 c) 1000 d) 100

**Ans: a**

**Solution:** The average of 1088 real numbers is zero

Therefore, almost 1086 – 1 = 1087 of them can be negative

**4. The average weight of liquid in 100 bottles is 500gm. The total weight of all the bottles is 20 kg. The average weight of a bottle with liquid is?**

a) 700ml b) 0.7kg c) 0.25kg d) 5 kgs

**Ans: b**

**Solution:** Average weight = 20 50/ 100 = 0.7kg

**5. Average of 10 positive numbers is x. If each number increases by 10%, then x**

a) May decrease b) Remains unchanged

c) Is increased by 10% d) May either increase or decrease

**Ans: c**

**Solution:** Average is increased by 10%

**6. 5 members of a team are weighed consecutively and their average weight calculated after each member is weighed. If the average weight increases by one kg each time, how much heavier is the last player than the first one?**

a) 20 kgs b) 8 kgs c) 10 kgs d) 9 kgs

**Ans: b**

**Solution:** Average1 = x, weight1 = x

Average2 = x 1 weight2 = x 2

Average3 = x 2 weight3 = x 4

Average4 = x 4 weight4 = x 6

Average5 = x 5 weight5 = x 8

Therefore, difference between last and first = 8 kg

**7. Find the average of below series**

1/1×2, 1/2×3, 1/3×4……. 1/(n-1)xn, 1/nx(n-1)

a) 1/(n^3 2) b) 2/(n^2 1) c) n^2/(n 1) d) 1/(n 1)

**Ans: d**

**Solution:** Sum = 1/1 – 1/2 – 1/3 ………… 1/n – 1/n 1

= 1 – 1/n 1 = n/ n 1

Average = (n/ n 1)/n = 1/ n 1

**8. The average of 3 prime numbers lying between 47 and 74 is 191/3. The greatest possible difference between any two out of the 3 prime numbers is?**

a) 17 b) 12 c) 15 d) 18

**Ans: d**

**Solution:** 53, 59, 61, 67, 71, 73

There are only two combinations whose average = 191/3

{ 53, 67, 71 } and { 59, 61, 71 }

Difference between 71 and 53 = 18

**9. The average of the three numbers x, y and z is 45. x is greater than the average of y and z by 9. The average of y and z is greater than y by 2. Then the difference of x and z is?**

a) 7 b) 5 c) 2 d) 11

**Ans: a**

**Solution:** x y z = 45 × 3 = 135

x = (y z)/2 9 =>; x = (135 – x)/2 9

3x = 135 18

x = 153/3 = 51

y z/2 = y 2 =>; x – 9 = y 2

51 – 9 – 2 = y

y = 40

z = 135 – 51 – 40 =>; z = 44

Difference between x and z = 51 – 44 = 7

**To be continued..**