** Number Series Reasoning Questions** So, in this article we are providing you with some questions on Number system, we hope you will practice them and make your concepts of the chapter much stronger. These questions on **Number System** are provided to you in objective type format, so that it will help you in solving your objective type questions during your examination.

Solve these **Number series reasoning questions**, All the questions provided in this online quiz are important from your examination point of view. So, practice these questions on **Number System** with full dedication and gain good marks in this subject. number series questions and answers given.

## Number Series Reasoning Questions part 1

Ques 1: 9587-? = 7429-4358

- 6516
- 7421
- 6561
- 6428

6516

Sol. Let 9587-x = 7429-4358. Then,

9587-x = 3071

x = 9587-3071 = 6516

Ques 2: 5793405×9999 = ?

- 57928256586
- 65928256545
- 75928256595
- 57928256595

57928256595

Sol. = 5793405x(10000-1)

= 57934050000-5793405

= 57928256595

Ques 3: 839478×625 = ?

- 524673710
- 524673750
- 534673750
- 598673750

524673750

Sol. 839478 x 5^{4}

= 8394780000/16

= 524673750

Ques 4: 976×237+976×763 = ?

- 824000
- 769000
- 976000
- 912000

976000

Sol. Using distributive law, we get :

= 976×237+976+763 = 976x(237+763)

= 976×1000

=976000

Ques 5: 986×307-986×207 = ?

- 72600
- 98600
- 98621
- 45870

98600

Sol. By using distributive law, we get

= 986×307-986×207 = 986x(307-207)

= 986×100

= 98600

Ques 6: 1607×1607 =?

- 3582789
- 2582449
- 4587659
- 6548449

2582449

Sol. 1607 x 1607

= 1607^{2}

= (1600+7)^{2}

= 16002 + 7^{2} + 2 x 1600 x 7

= 2560000 + 49 +22400

= 2582449

Ques 7: 1396×1396 = ?

- 1848946
- 1547966
- 1948816
- 1948836

1948816

Sol. 1396 x 1396

= 1396^{2}

= (1400-4)^{2}

= 1400^{2} + 4^{2} – 2 x 1400 x 4

= 1960000 + 16 – 11200

= 1948816

Ques 8: (475×475+125×125) = ?

- 356450
- 241250
- 124550
- 521250

241250

Sol.We have (a^{2} + b^{2}) =

1/2 [(a + b)^{2} + (a – b)^{2}]

=(475)^{2} + (125)^{2} =

1/2 [(475 + 125)^{2} + (475 – 125)^{2}]

= 1/2 [360000 + 122500]

= 241250

Ques 9: (796×796-204×204) = ?

- 632000
- 592000
- 548700
- 658900

592000

Sol. 796×796-204×204

=(796)^{2}−(204)^{2}

= (796+204)(796-204)

= 1000×592

= 592000

Ques 10: (387×387+113×113+2x387x113) = ?

- 245600
- 458700
- 250000
- 256400

250000

Sol. 387×387+114×114+2×387×114

= (387)^{2}+(113)^{2} + 2×387×113

= (387+113)^{2}

[ Since , (a+b)^{2}=a^{2}+b^{2}+2ab]

= 500^{2}

=250000

### Number Series Reasoning Questions part – 2

Ques 11: (87×87+61×61-2x87x61) = ?

- 476
- 676
- 872
- 576

676

Sol. (87)^{2}+(61)^{2}-2x87x61

(a^{2}+b^{2}−2ab) where a = 87 and b = 61.

=(a−b)^{2}=(87−61)^{2}

=(26)^{2}

=(20+6)^{2}

=(20)^{2}+(6)^{2}+2x20x6

= 400 + 36 + 240)

= 676

Ques 12: Find the least value of * for which 5967*13 is divisible by 3.

- 2
- 1
- 5
- 10

2

Sol. Let the required value be x. Then,

(5+9+6+7+x+1+3 = (x+31) is divible by 3

Least value of x is 2

Ques 13: Find the least value of * for which 7*5462 is divisible by 9.

- 4
- 9
- 8
- 3

3

Sol. Let the required value of x . Then

(7+x+5+4+6+2)=(24+x) is divisible by 9

Least value of * is 3

Ques 14: Find the least value of * for which 4832*18 is divisible by 11.

- 5
- 11
- 7
- 4

7

Sol. (Sum of digits at odd places) – (Sum of digits at even places)

= (8+x+3+4) – (1+2+8)

= (4+x), which should be divisible by 11

x = 7

Ques 15: What least number must be subtracted from 1672 to obtain a number which is completely divisible by 17?

- 6
- 5
- 4
- 7

6

Sol. When 1672 is divided by 17, then 6 is the reminder.

Hence, 6 must be subtracted.

Ques 16: What is the least number to be added to 2010 to obtain a number which is completely divisible by 19?

- 5
- 9
- 4
- 7

4

Sol. When 2010 is divided by 19, 15 is the reminder.

So, 19-15 = 4

Hence, 4 must be added.

Ques 17: On dividing 12401 by a certain number, we get 76 as quotient and 13 as reminder. What is the divisor?

- 173
- 185
- 163
- 142

163

Sol. Dividend = Divisor × Quotient + Remainder

12401 = 76*N + 13

N = (12401 – 13)/76

= 12388/76

= 3097/19

= 163

Ques 18: On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder?

- 21
- 13
- 10
- 11

11

Sol. Suppose that on dividing the given number by 342, we get quotient = k and remainder = 47. Then,

Number = 342k + 47

= (18x19k) + (18×2) = 11

=18x(19k+2)+11

So, the number when divided by 18 gives remainder = 11.

Ques 19: Simplify : (789x789x789 + 211x211x211)/(789×789-789×211+211×211) =?

- 1500
- 500
- 100
- 1000

1000

Sol. Given exp. = ((789)^{3}+(211)^{3})/(789)^{2}-(789×211)+(211)^{2})

= (a^{3}+b^{3})/(a^{2}-ab+b^{2}), where a = 789 and b = 211

=(a+b) = (789+211) = 1000

Ques 20: Simplify : (658x658x658 – 328x328x328)/(658×658+658×328+328×328) =?

- 315
- 330
- 320
- 350

330

Sol. Given exp. = ((658)^{3}-(328)^{3})/(658)^{2}+(658×328)+(328)^{2})

= (a^{3}-b^{3})/(a^{2}+ab+b^{2}), where a = 658 and b = 328

=(a-b) = (658-328) = 330

### Number Series Reasoning Questions part 3

Ques 21: Simplify :

((893+786)^{2}-(893-786)^{2})/(893×786)

- 8
- 5
- 4
- 3

4

Sol. Given exp. = ((a+b)^{2}-(a-b)^{2})/ab, where a = 893 and b = 786

= 4ab/ab

= 4

Ques 22: What is the unit digit in the product (684x759x413x676)?

- 4
- 8
- 10
- 9

8

Sol. Unit digit in the given product

= Unit digit in the product (4x9x3x6) = 8

Ques 23: What is the unit digit in the product {(3547)^{153}x(251)^{72}}?

- 7
- 5
- 4
- 9

7

Sol. Required digit = unit digit in (7^{153}x1^{72})

Now, 7^{4} gives unit digit 1 and 1^{72} = 1

(7^{153}x1^{72}) = [(7^{4})^{38}x7x1] = 7

Required unit digit = (1x7x1) = 7

Ques 24: What is the unit digit in {(264)^{102}+(264)^{103}}?

- 0
- 1
- 7
- 5

0

Sol. (264)^{102}+(264)^{103} = (264)^{102}{1+264} = (264)^{102}x265

Required unit digit = unit digit in [(4)^{102}x5]

=unit digit in [(4^{4})^{25}x4^{2}x5]

=unit digit in (6x6x5) = 0

Ques 25: Find the total number of prime factors in the product {(4)^{11}x7^{5}x(11)^{2}}

- 29
- 20
- 16
- 10

29

Sol. (4)^{11} x (7)^{5} x (11)^{2}

= (2 x 2)^{11} x (7)^{5} x (11)^{2}

= 2^{11} x 2^{11} x 7^{5} x 11^{2}

= 2^{22} x 7^{5} x 11^{2}

Total number of prime factors = (22 + 5 + 2) = 29

#### number series reasoning questions pdf download, number series online quiz, number series mcq questions and

#### answers.

We hope you have practiced all the questions provided to you above in the article and now you are feeling confident about the chapter. To learn more, stay in touch with the page and don’t forget to share your feedback with us.

## Recent Comments