Explanation / Important formulas:
 (a + b)(a – b) = (a^{2} – b^{2})
 (a + b)^{2} = (a^{2} + b^{2} + 2ab)
 (a – b)^{2} = (a^{2} + b^{2} – 2ab)
 (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)
 (a^{3} + b^{3}) = (a + b)(a^{2} – ab + b^{2})
 (a^{3} – b^{3}) = (a – b)(a^{2} + ab + b^{2})
 (a^{3} + b^{3} + c^{3} – 3abc) = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ac)
 When a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc
For more information: https://en.wikipedia.org/wiki/Number
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Number system and number property  Test
Number system and number property  Question and Answers
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Question 1

A

7

B

8

C

12

D

13

=> 124/2 = 62 & 62  2 = 60
=> 60/2=30 & 30  2 =28
=> 28/2=14 & 14  2 =12
Question 2

A

2

B

3

C

4

D

Cannot be determined

(Here a = First term, r = ratio)
=1 / (1  1/2)
=2
Question 3

A

1123346

B

10224

C

100234

D

123458

Question 4

A

8

B

9

C

10

D

11

1 yard = 36 inches
4 yards = 4*36=144 inches
No.of maximum bows that can be made = 144/15 = 9.6
So maximum 9 bows can be made
Question 5

A

145

B

253

C

370

D

352

Hence , x + y + z = 10………(1)
x + z = y………(2)
As per the condition
(100z + 10y + x)  (100x + 10y + z) = 99
Solving, the value of (z  x) =1
From equation(1) and (2)
2y = 10, y = 5
So 253 satisfies all the condition
Question 6

A

1

B

2

C

4

D

6

Unit digit in 365 = Unit digit in [ (34)16 x 3 ] = (1 x 3) = 3
Unit digit in 659 = 6
Unit digit in 74 Unit digit in (74)17 is 1.
Unit digit in 771 = Unit digit in [(74)17 x 73] = (1 x 3) = 3
Required digit = Unit digit in (3 x 6 x 3) = 4.
Question 7

A

1

B

5

C

7

D

9

But, unit digit in (7^{4})^{26} = 1
Unit digit in 7^{105} = (1 x 7) = 7
Question 8

A

1250

B

1300

C

1375

D

1200

+ 587
369

4207

Let 4207  x = 3007
Then, x = 4207  3007 = 1200
Question 9

A

4242

B

4155

C

1123

D

11023

Then, x = 7589  3434 = 4155
Question 10

A

79698

B

80578

C

80698

D

81268

= 2 x [(200)^{2} + (17)^{2}] [Ref: (a + b)^{2} + (a  b)^{2} = 2(a^{2} + b^{2})]
= 2[40000 + 289]
= 2 x 40289
= 80578
Question 11

A

2

B

3

C

4

D

5

Question 12

A

3883203

B

3893103

C

3639403

D

3791203

= 3897 x 1000  3897 x 1
= 3897000  3897
= 3893103
Question 13

A

3

B

10

C

11

D

13

= 460 x (4 x 85)
= (460 x 340), which is divisible by 10
Question 14

A

2400

B

2000

C

1904

D

1906

= (106 + 94)(106  94) [Ref: (a^{2}  b^{2}) = (a + b)(a  b)]
= (200 x 12)
= 2400
Question 15

A

4494

B

561.75

C

2247

D

280.875

Question 16

A

362.3

B

372.33

C

702.33

D

.702

+ 33
+ 333
+ 3.33

372.33

Question 17

A

10000

B

1000

C

100

D

10

Question 18

A

2906

B

3116

C

2704

D

2904

Consider a = 476 and b = 424
= [(476 + 424)^{2}  4 x 476 x 424]
= [(900)^{2}  807296]
= 810000  807296
= 2704
Question 19

A

1, 2

B

2, 3

C

3, 2

D

4, 1


5  y 1

 1 4
y = (5 x 1 + 4) = 9
x = (4 x y + 1) = (4 x 9 + 1) = 37
Now, 37 when divided successively by 5 and 4, we get
5  37

4  7  2

 1  3
Respective remainders are 2 and 3
Question 20

A

2736900

B

2638800

C

2658560

D

2716740

= (8796 x 300)
= 2638800
Question 21

A

534

B

446

C

354

D

324

= (a  b)^{2} = (287  269)^{2}
= (18^{2})
= 324
Question 22

A

30976

B

75625

C

28561

D

143642

143642 is not the square of natural number.
Question 23

A

6

B

28

C

240

D

512

Let the number of terms be n. Then,
a + (n  1)d = 30
2 + (n  1) x 2 = 30
n = 15
Therefore, S_{n = }n/2(a + l) = 15/2 x (2 + 30) = (15 x 16) = 240
Question 24

A

80

B

100

C

75

D

90

60% of 3/5 of x = 36
=> 60/100 * 3/5 * x = 36
=> x = (36 * 25/9) = 100
Required number = 100
Question 25

A

2 or 6

B

4

C

4 or 8

D

8

Since 653xy is divisible by 2 and 5 both, so y = 0
Now, 653x is divisible by 8, so 13x should be divisible by 8
This happens when x = 6
x + y = (6 + 0) = 6
Question 26

A

3

B

6

C

7

D

8

(2n + 3)^{2}  (2n + 1)^{2} = (2n + 3 + 2n + 1) (2n + 3  2n  1)
= (4n + 4) x 2
= 8(n + 1), which is divisible by 8
Question 27

A

1

B

3

C

7

D

9

=Unit digit in { 292915317923361 x 17114769 }
= (1 x 9) = 9
Question 28

A

214

B

476

C

954

D

1908


5  y 2

6  z  3

 1  4
z = 6 x 1 + 4 = 10
y = 5 x z + 3 = 5 x 10 + 3 = 53
x = 4 x y + 2 = 4 x 53 + 2 = 214
Hence, required number = 214.
Question 29

A

111111

B

110011

C

100011

D

110101

111) 100000 (900
999

100

Required number = 100000 + (111  100)
= 100011
Question 30

A

99921

B

99918

C

99981

D

99971

91) 99999 (1098
91

899
819

809
728

81

Required number = (99999  81)
= 99918
Question 31

A

8

B

9

C

10

D

11

Then, t_{n} = 384 ar^{n}^{1} = 384
=>3 x 2^{n}^{  1}= 384
=>2^{n}^{1} = 128 = 2^{7}
=>n  1 = 7
=> n = 8
Number of terms = 8
Question 32

A

4x+6y

B

x+y+4

C

9x+4y

D

4x9y

(4x + 6y) = ( 4 x 5 + 6 x 1) = 26, which is not divisible by 11
(x + y + 4 ) = (5 + 1 + 4) = 10, which is not divisible by 11
(9x + 4y) = (9 x 5 + 4 x 1) = 49, which is not divisible by 11
(4x  9y) = (4 x 5  9 x 1) = 11, which is divisible by 11
Question 33

A

8230

B

8410

C

8500

D

8600

+ 7314

16862

16862 = 8362 + x
x = 16862  8362
x = 8500
Question 34

A

0

B

12

C

13

D

20

Correct Quotient = 420 ÷ 21 = 20
Question 35

A

2044

B

1022

C

1056

D

None of these

S_{n}= a(r^{n}1) / (r1) = 2 *(2^{9} 1) / (21) = 2 *(512  1) = 2*511 = 1022
Question 36

A

3

B

4

C

6

D

7

(2n + 2)2 = (2n + 2 + 2n)(2n + 2  2n)
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4
Question 37

A

11

B

16

C

25

D

30

= 3^{24} x 3 x 4 x 10
= (3^{24} x 4 x 30), which is divisible by 30
Question 38

A

0

B

1

C

2

D

3

Then, x^{2} = (6q + 3)^{2}
= 36q^{2} + 36q + 9
= 6(6q^{2} + 6q + 1) + 3
Thus, when x^{2} is divided by 6, then remainder = 3
Question 39

A

17

B

16

C

1

D

2

(17^{200}  1^{200}) is completely divisible by (17 + 1), i.e., 18
(17^{200}  1) is completely divisible by 18
On dividing 17^{200} by 18, we get 1 as remainder
Question 40

A

0

B

1

C

2

D

3

Then, x^{2} = (6q + 3)^{2}
= 36q^{2} + 36q + 9
= 6(6q^{2} + 6q + 1) + 3
Thus, when x^{2} is divided by 6, then remainder = 3
Question 41

A

339

B

349

C

369

D

Data inadequate

13p + 11 = 17q + 9
17q  13p = 2
q= 2+13p / 17
The least value of p for which q = 2+13p / 17 is a whole number is p = 26
Therefore,x = (13 x 26 + 11)
= (338 + 11)
= 349
Question 42

A

2654

B

2975

C

3225

D

3775

= 100/2 * (1 + 100) – 50/2 * (1 + 50)
= (50 x 101)  (25 x 51)
= (5050  1275)
= 3775
Question 43

A

3

B

2

C

1

D

0

Thus, when 2n is divided by 4, the remainder is 2
Question 44

A

144

B

864

C

2

D

4

Question 45

A

7 and 4

B

7 and 5

C

8 and 5

D

None of these

Then (4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3
And, (0 + x + 7)  (y + 6 + 4) = (x  y 3) must be either 0 or 11
x  y  3 = 0 y = x  3
(17 + x + y) = (17 + x + x  3) = (2x + 14)
x= 2 or x = 8
x = 8 and y = 5
Question 46

A

10

B

11

C

12

D

13

Now, 1397 = 11 x 127
The required 3digit number is 127, the sum of whose digits is 10
Question 47

A

553681

B

555181

C

555681

D

556581

So, the required number must be divisible by each one of 3, 7, 47
553681 (Sum of digits = 28, not divisible by 3)
555181 (Sum of digits = 25, not divisible by 3)
555681 is divisible by 3, 7, 47
Question 48

A

0

B

3

C

5

D

11

x = 357 x y + 39
= (17 * 21 x y) + (17 * 2) + 5
= 17 * (21y + 2) + 5)
Required remainder = 5
Question 49

A

1

B

5

C

6

D

8

So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3
P = 1.
Question 50

A

(47  43)

B

(47 + 43)

C

(4743 + 4343)

D

None of these

Each one of (47^{43} + 43^{43}) and (47^{47} + 43^{47}) is divisible by (47 + 43).
Question 51

A

0

B

1

C

2

D

4

x = 5k + 3
x^{2} = (5k + 3)^{2}
= (25k^{2} + 30k + 9)
= 5(5k^{2} + 6k + 1) + 4
On dividing x^{2} by 5, we get 4 as remainder
Question 52

A

1

B

63

C

66

D

67

(67^{67} + 1) will be divisible by (67 + 1)
(67^{67} + 1) + 66, when divided by 68 will give 66 as remainder
Question 53

A

1035

B

1280

C

2070

D

2140

S_{n}= n/2[2a+(n1)d] = 45/2 * [2*1+(45  1)*1)] = (45/2*46) =(45 x 23)
= 45 x (20 + 3)
= 45 x 20 + 45 x 3
= 900 + 135
= 1035
Question 54

A

31

B

61

C

71

D

91

Question 55

A

9

B

11

C

13

D

15

Then, 3x = 2(x + 4) + 3 x = 11
Third integer = x + 4 = 15
Question 56

A

69

B

78

C

96

D

Cannot be determined

Then, x + y = 15 and x  y = 3 or y  x = 3
Solving x + y = 15 and x  y = 3, we get: x = 9, y = 6
Solving x + y = 15 and y  x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69
Hence, the number cannot be determined
Question 57

A

20

B

30

C

40

D

None of these

Then, a^{2} + b^{2} + c^{2} = 138 and (ab + bc + ca) = 131
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = 138 + 2 x 131 = 400
(a + b + c) = 400 = 20
Question 58

A

20

B

23

C

169

D

None of these

Then, xy = 120 and x^{2} + y^{2} = 289
(x + y)^{2} = x^{2} + y^{2} + 2xy = 289 + (2 x 120) = 529
x + y = 529 = 23
Question 59

A

145

B

253

C

370

D

352

Then, 2x = 10 or x = 5. So, the number is either 253 or 352
Since the number increases on reversing the digits,so the hundred's digits is smaller than the unit's digit.
Hence, required number = 253
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