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 54
= 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
= 16072
= (1600+7)2
= 16002 + 72 + 2 x 1600 x 7
= 2560000 + 49 +22400
= 2582449
Ques 7: 1396×1396 = ?
- 1848946
- 1547966
- 1948816
- 1948836
1948816
Sol. 1396 x 1396
= 13962
= (1400-4)2
= 14002 + 42 – 2 x 1400 x 4
= 1960000 + 16 – 11200
= 1948816
Ques 8: (475×475+125×125) = ?
- 356450
- 241250
- 124550
- 521250
241250
Sol.We have (a2 + b2) =
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=a2+b2+2ab]
= 5002
=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
(a2+b2−2ab) 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)
= (a3+b3)/(a2-ab+b2), 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)
= (a3-b3)/(a2+ab+b2), 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)153x(251)72}?
- 7
- 5
- 4
- 9
7
Sol. Required digit = unit digit in (7153x172)
Now, 74 gives unit digit 1 and 172 = 1
(7153x172) = [(74)38x7x1] = 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)102x265
Required unit digit = unit digit in [(4)102x5]
=unit digit in [(44)25x42x5]
=unit digit in (6x6x5) = 0
Ques 25: Find the total number of prime factors in the product {(4)11x75x(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
= 211 x 211 x 75 x 112
= 222 x 75 x 112
Total number of prime factors = (22 + 5 + 2) = 29
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