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** 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

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