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 7 consecutive numbers is 20. The largest of these numbers is :

(a) 20

(b) 22

(c) 23

(d) 24

The correct answers to the above question in:

Answer: (c)

Average of 7 consecutive numbers is 20.

Since the numbers are consecutive, they form an arithmetic series with common difference 1.

Since, 7 is odd, 20 must be the middle number.

We can write the series as below,

17, 18, 19, 20, 21, 22, 23

∴ The largest of these numbers is 23.

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Read more consecutive numbers Based Quantitative Aptitude Questions and Answers

Question : 1

The average of three consecutive odd numbers is 12 more than one third of the first of these numbers. What is the last of the three numbers ?

a) Data inadequate

b) 19

c) 17

d) 15

Answer: (b)

If the smallest number be x, then

$x/3$+12 = x+2

⇒ x + 36 = 3x + 6

⇒ 3x – x = 36 – 6

⇒ 2x = 30 ⇒ x = 15

∴ Third number = 15 + 4 = 19

Question : 2

The average of 7 consecutive numbers is 20. The largest of these numbers is

a) 26

b) 24

c) 23

d) 20

Answer: (c)

Average of 7 consecutive numbers = 20

∴ Fourth number = 20

∴ Largest number = 20 + 3 = 23

Question : 3

The sum of three consecutive even numbers is 28 more than the average of these three numbers. Then the smallest of these three numbers is

a) 16

b) 14

c) 12

d) 6

Answer: (c)

Let three consecutive even numbers be x, x + 2 and x + 4. According to the question,

(x + x + 2 + x + 4) – ${x+x+2+x+4}/3$= 28

⇒ (3x + 6) – ${3x+6}/3$ = 28

⇒ (3x + 6) – (x + 2) = 28

⇒ 3x + 6 – x – 2 = 28

⇒ 2x + 4 = 28

⇒ 2x = 28 – 4 = 24

⇒ x = $24/2$ = 12

Question : 4

The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m + 2) ?

a) ${2n+9}/2$

b) (n+3)

c) (n+2)

d) ${2n+5}/2$

Answer: (d)

m + m + 1 + m + 2 + m + 3 + m + 4 = 5n

⇒ 5m + 10 = 5n

⇒ m + 2 = n ....(i)

Required average = ${m+2+m+3+m+4+m+5+m+6+m+7}/6$

= ${6m+27}/6$

= ${2m+9}/2$ = ${2(n-2)+9}/2$= ${2n+5}/2$ By (i) [m = n – 2]

Question : 5

If a, b, c, d, e are five consecutive odd numbers, their average is

a) a+4

b) 5(a+b+c +d+e)

c) ${abcde}/5$

d) 5(a+4)

Answer: (a)

b = a + 2

c = b + 2 = a + 4

d = c + 2 = a + 6

e = d + 2 = a + 8

∴ Required average

= ${a+a+2+a+4+a+6+a+8}/5$

= ${5a+20}/5$ = a+4

Question : 6

What is the average of the first six (positive) odd numbers each of which is divisible by 7?

a) 49

b) 47

c) 43

d) 42

Answer: (d)

Required average

= ${7+21+35+49+63+77}/6$

= ${7(1+3+5+7+9+11)}/6$

= ${7×36}/6$ = 42

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