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 : 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$({a+n}/2)$

The correct answers to the above question in:

Answer: (c)

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

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

∴ Required average = ${d+l}/2$

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 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 : 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 first ten prime numbers is

a) 13

b) 12.9

c) 10

d) 10.1

Answer: (b)

Required average

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

= $129/10$ = 12.9

Question : 4

The average of odd numbers upto 100 is

a) 49

b) 49.5

c) 50

d) 50.5

Answer: (c)

Formula,

Average of the first n natural odd numbers = $n/2$

Number of odd numbers upto $100/2$ = 50 = required average

Aliter : Using Rule 7,

Odd numbers are 1, 3, 5, ............., 99

Total odd numbers are= 50

∴ Average = 50

Question : 5

The average of the squares of first ten natural numbers is

a) 38.5

b) 37.5

c) 36

d) 35.5

Answer: (a)

Using Rule 4,

Average of $1^2,2^2,3^2,4^2…x^2$ = ${(n+1)(2n+1)}/6$

${1^2+2^2+3^2+…+n^2}/n$ = ${(n+1)(2n+1)}/6$

∴ ${1^2+2^2+3^2+…+10^2}/10$ = ${(10+1)(2×10+1)}/6$

= ${11×21}/6$ = 38.5

Question : 6

The average of five consecutive positive integers is n. If the next two integers are also included, the average of all these integers will

a) increase by 2

b) remain the same

c) increase by 1

d) increase by 1.5

Answer: (c)

Five consecutive integers are :

x, x + 1, x + 2, x + 3 and x + 4

∴ Their average

= ${x+x+1+x+2+x+3+x+4}/5$

= ${5x+10}/5$ = x+2

New average

= ${(5x+10)+x+5+x+6}/7$

= ${7x+21}/7$= x + 3

Difference = x + 3 – x – 2 = 1

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