model 6 operations of consecutive numbers (odd, even, square, etc.) Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

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The following question based on number system topic of quantitative aptitude

Questions : The sum of the squares of three consecutive natural numbers is 2030. Then, what is the middle number?

(a) 27

(b) 25

(c) 26

(d) 28

The correct answers to the above question in:

Answer: (c)

Let the three consecutive natural numbers be x, x + 1 and x + 2.

According to question,

$x^2 + (x + 1)^2 + (x + 2)^2$ = 2030

or $x^2 + x^2 + 2x + 1 + x^2 + 4x + 4$ = 2030

or $3x^2 + 6x + 5 = 2030 $

or $3x^2 + 6x – 2025$ = 0

or $x^2 + 2x – 675$ = 0

or $x^2 + 27x – 25x – 675$ = 0

$x (x + 27) – 25 (x + 27)$ = 0

or $(x – 25) (x + 27)$ = 0

$x = 25 and – 27$

∴ Required number = $x$ + 1 = 25 + 1 = 26

Practice number system (model 6 operations of consecutive numbers (odd, even, square, etc.)) Online Quiz

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Read more operations on consecutive numbers Based Quantitative Aptitude Questions and Answers

Question : 1

The sum of all those prime numbers which are not greater than17 is

a) 41

b) 59

c) 58

d) 42

Answer: (c)

Prime numbers upto 17⇒ 2, 3, 5, 7, 11, 13, 17

Required sum = 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58

Question : 2

Find the sum of all positive multiples of 3 less than 50

a) 408

b) 400

c) 404

d) 412

Answer: (a)

Sum of all multiples of 3 upto 50

= 3 + 6 + ..... + 48

= 3 (1 + 2 + 3 + .... + 16)

= 3×${16(16+1)}/2$= 3 × ${272}/2$ = 408

[Since 1+2+3+…+n = ${n(n+1)}/2$]

Question : 3

What is the sum of two consecutive even numbers, the difference of whose square is 84?

a) 42

b) 38

c) 34

d) 46

Answer: (a)

$(x + 2)^2 – x^2 = 84$

or $x^2 + 4x + 4 – x^2$ = 84

4x = 84 – 4 = 80

x = $80/4$ = 20

x + 2 = 20 + 2 = 22

∴ The required sum = 20 + 22 = 42

Question : 4

The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is

a) 36

b) 30

c) 33

d) 45

Answer: (c)

$10^2 + 11^2 + 12^2$ = 100 + 121 + 144 = 365

Required sum =10 + 11 + 12 = 33

Question : 5

The sum of all natural numbers from 75 to 97 is :

a) 1958

b) 1598

c) 1798

d) 1978

Answer: (d)

Series of all natural numbers from 75 to 97 is in A.P. whose first term,

a = 75, last term, l = 97

If number of terms be n, then $a_n$ = a + (n–1)d

97 = 75 + (n – 1)

n = 97 – 74 = 23

$S_n = n/2(a+l)$

$S_23 = 23/2(75+97)$

= $23/2$× 172=1978

Question : 6

The sum of first 20 odd natural numbers is equal to :

a) 400

b) 210

c) 300

d) 420

Answer: (a)

Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2.

Sum of n terms in arithmetic progression is given by.

$S_n =1/2n[2a + (n – 1)d]$

Where a : First term; d : common difference

$S_{20}= 1/2×20[(2×1) + (20 -1) × 2]$

= 10 [2 + 38]=10 × 40 = 400

Note :Sum of first n consecutive odd numbers = $n^2$

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