model 2 average of consecutive numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 9 EXERCISES

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The following question based on average topic of quantitative aptitude

Questions : The average of first ten prime numbers is

(a) 13

(b) 12.9

(c) 10

(d) 10.1

The correct answers to the above question in:

Answer: (b)

Required average

= ${2+3+5+7+11+13+17+19+23+29}/10$

= $129/10$ = 12.9

Practice average (model 2 average of consecutive numbers) Online Quiz

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Read more consecutive numbers Based Quantitative Aptitude Questions and Answers

Question : 1

The average of 6 consecutive natural numbers is K. If the next two natural numbers are also included, how much more than K will the average of these 8 numbers be?

a) 1.8

b) 2

c) 1

d) 1.3

Answer: (c)

x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 6K

⇒ 6x + 15 = 6K

⇒ x + $15/6$ = K

⇒ x + $5/2$ = K ..... (i)

Again, ${x+(x+1)+…+(x+6)+(x+7)}/8$

= ${8x}/2$ + $28/8$ = x +$7/2$ ........ (ii)

Now, x + $7/2$ – x – $5/2$ = 1

Question : 2

The average of seven consecutive positive integers is 26. The smallest of these integers is :

a) 26

b) 25

c) 23

d) 21

Answer: (c)

Using Rule 1,

Average of numbers = ${x_1+x_2+…+x_n}/n$

x+x+1+x+2+x+3+x+4+x+5+x+ 6 = 26 × 7

⇒ 7x + 21 = 182

⇒ 7x = 182 – 21 = 161

⇒ x = $161/7$ = 23

Shortcut Approach

Fourth number = 26

∴ First number = 23

Question : 3

The average of 5 consecutive natural numbers is m. If the next three natural numbers are also included, how much more than m will the average of these 8 numbers be?

a) 1.5

b) 1.4

c) 1

d) 2

Answer: (a)

Using Rule 1,

Average of numbers = ${x_1+x_2+…+x_n}/n$

The average will increase by 1.5.

As, ${1+2+3+4+5}/5$ = 3,

${1+2+3+4+5+6+7+8}/8$ = 4.5

⇒ 4.5 - 3 = 1.5

Question : 4

a, b, c, d, e, f, g are consecutive even numbers. j, k, l, m, n are consecutive odd numbers. The average of all the numbers is

a) $({j+c+n+g}/4)$

b) $({a+b+m+n}/4)$

c) $({l+d}/2)$

d) 3$({a+n}/2)$

Answer: (c)

Average of a, b, c, d, e, f, g = d

Average of j, k, l, m, n, = l

∴ Required average = ${d+l}/2$

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