Staff Selection Commission Mathematics - Number system (Type I and Type II)

TYPE-II

81.If a and b are two odd positive integers, by which of the follow- ing integers is \((a^4 – b^4)\) always divisible ?

(1) 3

(2) 6

(3) 8

(4) 12

Ans .

(3) 8

Explanation :
\(a^2-b^2 = (a^2+b^2)(a+b)(a-b)\)
Let a = 3, b = 1
 Required number
= (3 + 1) (3 – 1) = 8

82.If m and n are positive integers and (m – n ) is an even number, then \((m^2 – n^2)\) will be always di- visible by

(1) 4

(2) 6

(3) 8

(4) 12

Ans .

(1) 4

Explanation :
82. Let m = n = p and m – n = 2p m + n = 2p
 (m – n) (m + n) = 4p2

83. If 5432*7 is divisible by 9, then the digit in place of * is :

(1) 0

(2) 1

(3) 6

(4) 9

Ans .

(3) 6

Explanation :
83. A number is divisible by 9, if sum of its digits is divisible by 9.
Let the number be x.
 5 + 4 + 3 + 2 + x + 7 = 21 + x
 x = 6

84.The least number, which must be added to 6709 to make it ex- actly divisible by 9, is

(1) 5

(4) 4

(3) 7

(4) 2

Ans .

(1) 5

Explanation :
84.A number is divisible by 9 if the sum of its digits is divisible by 9.
Here, 6 + 7 + 0 + 9 = 22
Now, 22 + 5 = 27, which is divis- ible by 9. Hence 5 must be add- ed to 6709.

85. The total number of integers be- tween 100 and 200, which are divisible by both 9 and 6, is :

(1) 5

(2) 6

(3) 7

(4) 8

Ans .

(2) 6

Explanation :
A number is divisible by 9 and 6 both, if it is divisible by LCM of 9 and 6 i.e., 18. Hence, the numbers are 108, 126, 144, 162, 180, 198.

86. How many 3-digit numbers, in all, are divisible by 6 ?

(1) 140

(2) 150

(3) 160

(4) 170

Ans .

(2) 150

Explanation :
86.First 3–digit number divisible by 6 = 102
Last such 3-digit number =996
 996 = 102 + (n –1) 6
 (n – 1)6 = 996 – 102 = 894
 n – 1 = \(\frac{894}{6}=149\)
 n = 150

87. If ‘n’ be any natural number, then by which largest number \((n^3 – n)\) is always divisible ?

(1) 3

(2) 6

(3) 12

(4) 18

Ans .

(2) 6

Explanation :
\( n^3 – n = n (n^2 – 1)\)
= n (n + 1) (n – 1)
For n = 2, n3 – n = 6

If n is an integer, then \((n^3 – n)\) is always divisible by :

(1) 4

(2)5

(3) 6

(4)7

Ans .

(3) 6

Explanation :
\( n^3 – n = n (n + 1) (n – 1)\)
\(n = 4, n^3 – n = 4 × 5 × 3 = 60\)
60 ÷ 6 = 10

89.If the sum of the digits of any integer lying between 100 and 1000 is subtracted from the number, the result always is

(1)divisible by 6

(2)divisible by 2

(3)divisible by 9

(4)divisible by 5

Ans .

(3)divisible by 9

Explanation :
Number = 100x + 10y + z
Sum of digits = x + y + z
Difference = 100x + 10y + z – x – y – z
= 99x + 9y = 9 (11x + y)

90.If a number is divisible by both 11 and 13, then it must be necessarily :

(1)divisible by (11 + 13)

(2)divisible by (13 – 11)

divisible by (11 × 13)

(4) 429

Ans .

divisible by (11 × 13)

Explanation :
divisible by (11 × 13)

91. If * is a digit such that 5824* is divisible by 11, then * equals :

(1) 2

(2) 3

(3) 5

(4) 6

Ans .

(3) 5

Explanation :
91.Any number is divisible by 11 when the differences of alterna- tive digits is 0 or multiple of 0, 11 etc. Here,
5 + 2 + * = 7 + *
8 + 4 = 12
 * = 12 – 7 = 5

If 78*3945 is divisible by 11, where * is a digit, then * is equal to

(1) 1

(2) 0

(3) 3

(4) 5

Ans .

(4) 5

Explanation :
92.A number is divisible by 11, if the difference of the sum of its dig- its at odd places and the sum of its digits of even places, is either 0 or a number divisible by 11.
 (5 + 9 + * + 7) – (4 + 3 + 8) = 0
or multiple of 11
 21 + * – 15
 * + 6 = a multiple of 11
 * = 5

93.If the number 4 8 3 2 7 * 8 is divisible by 11, then the miss- ing digit (*) is

(1) 5

(2) 3

(3) 2

(4) 1

Ans .

(4) 1

Explanation :
93. A number is divisible by 11, if the difference of sum of its dig- its at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.
Difference = (4 + 3 + 7 + 8) – (2 + 8 + ) = 22 – 10 – * = 12 – *
Clearly, * = 1

94.Both the end digits of a 99 digit number N are 2. N is divisible by 11, then all the middle digits are :

(1) 1

(2) 2

(3) 3

(4) 4

Ans .

(4) 4

Explanation :
94. A number is divisible by 11 if the difference of the sum of digits at odd and even places be either zero or multiple of 11.
If the middle digit be 4, then 24442 or 244442 etc are divisi- ble by 11.

95.If n is a whole number greater than 1, then n2(n2 – 1) is always divisible by :

(1) 16

(2) 12

(3) 10

(4) 8

Ans .

(2) 12

Explanation :
\(n^2(n^2–1) = n^2 (n + 1) (n – 1)\)
Now, we put values n = 2, 3..... When n = 2
 \(n^2(n^2 –1)\) = 4 × 3 × 1 = 12, which is a multiple of 12
When n = 3. n\(n^2(n^2 –1)\) = 9 × 4 × 2 = 72,
which is also a multiple of 12. etc

96.A 4-digit number is formed by repeating a 2-digit number such as 2525, 3232, etc. Any number of this form is always exactly di- visible by :

(1) 7

(2) 11

(3) 13

(4)Smallest 3-digit prime num- ber

Ans .

(4)Smallest 3-digit prime num- ber

Explanation :
Let the unit digit be x and ten’s digit be y.
 Number = 1000y + 100x + 10y + x = 1010y + 101x = 101(10y + x)
Clearly, this number is divisible by 101, which is the smallest three-digit prime number.

What least number, of 5 digits is divisible by 41?

(1) 10045

(2) 10004

(3) 10041

(4) 41000

Ans .

(2) 10004

Explanation :
The least number of 5 digits = 10000
 Required number = 10000 + (41–37)
= 10004

98.It is given that \((2^{32} + 1)\) is exactly divisible by a certain number. which one of the following is also definitely divisible by the same number ?

\(2^{96} + 1\)

\(2^{66} + 1\)

\(2^{86} + 1\)

\(2^{76} + 1\)

Ans .

(1) \(2^{96} + 1\)<

Explanation :
\(2^{96} + 1 = (2^{32})^3 + 1^3\)
Clearly, \(2^{32}+1\) is a factor of \(2^{96} +1\)

99.The greatest whole number, by which the expression \(n^4 + 6n^3 + 11n^2 + 6n + 24\) is divisible for every natural num- ber n, is

(1) 6

(2) 24

(3) 12

(4) 48

Ans .

(4) 48

Explanation :
For n = 2
\(n^4 + 6n^3 + 11n^2 + 6n + 24\)= 16 + 48 + 44 + 12 + 24 = 144
which is divisible by 48. Clearly, 48 is the required num- ber. = 144
which is divisible by 48. Clearly, 48 is the required num- ber.

100.How many numbers between 1000 and 5000 are exactly di- visible by 225 ?

(1) 16

(2)18

(3) 19

(4)12

Ans .

(2)18

Explanation :
When we divide 1000 by 225, quotient = 4
When we divide 5000 by 225, quotient = 22
 Required answer = 22 – 4 = 18

101.Find the largest number, which exactly divides every number of the form \((n^3 – n) * (n –2)\) where n is a natural number greater than 2.

(1) 6

(2) 12

(3) 24

(4) 48

Ans .

(3) 24

Explanation :
\((n^3 – n) (n – 2) = n (n – 1) (n + 1) (n – 2)\)
When n = 3, Number = 3 × 2 × 4 = 24

102.The greatest number less than 1500, which is divisible by both 16 and 18, is

(1) 1440

(2) 1404

(3) 1386

(4) 1368

Ans .

(1) 1440

Explanation :
LCM of 16 and 18 = 144 Multiple of 144 that is less than 1500 = 1440

103.The least number, which is to be added to the greatest number of 4 digits so that the sum may be divisible by 345, is

(1) 50

(2)6

(3) 60

(4)5

Ans .

(2)6

Explanation :
The largest 4-digit number = 9999
 Required number = 345 – 339 = 6

104. \(4^{61} + 4^{62} + 4^{63} + 4^{64}\) is divisible by

(1) 3

(2) 10

(3) 11

(4) 13

Ans .

(2) 10

Explanation :
\(4^{61} + 4^{62} + 4^{63} + 4^{64} = 4^{61}*(1 + 4 + 16 + 64) = 4^{61} * 85\)
Which is a multiple of 10.

105.The difference of a number con- sisting of two digits from the number formed by interchanging the digits is always divisible by

(1) 10

(2) 9

(3) 11

(4) 6

Ans .

(2) 9

Explanation :
Let the number be 10x + y
where y < x. Number obtained by interchang- ing the digits = 10y + x
 Difference = 10x + y – 10y – x
= 9x – 9y = 9 (x – y)
Hence, the difference is always exactly divisible by 9.

106.Which one of the numbers is di- visible by 25 ?

(1) 303310

(2) 373355

(3) 303375

(4) 22040

Ans .

(3) 303375

Explanation :
Check through option
\(\frac{303375}{25} = \frac{303375*4}{25*4} = 12135\)
A number is divisible by 25 if the last two digits are divisible by 25 or zero.

The least number which must be added to the greatest number of 4 digits in order that the sum may be exactly divisible by 307 is

(1) 132

(2) 32

(3) 43

(4) 75

Ans .

(1) 132

Explanation :
307 × 32 = 9824
307 × 33 = 10131
 Required number = 10131 – 9999 = 132

108.If a = 4011 and b = 3989 then value of ab = ?

(1) 15999879

(2) 15899879

(3) 15989979

(4) 15998879

Ans .

(1) 15999879

Explanation :
a = 4011, b = 3989
 ab = 4011 × 3989
= (4000 + 11) (4000 – 11) = (4000)2 – (11)2
= 16000000 – 121 = 15999879

109.For any integral value of n,\( 3^{2n} + 9^n + 5 \)when divided by 3 will leave the remainder

(1) 1

(2) 2

(3) 0

(4) 5

Ans .

(2) 2

Explanation :
Expression =\( 3^{2n} + 9^n + 5 \) = \( 3^{2n} + 9^n + 3 \) + 2
Clearly, remainder = 2

111.5349 is added to 3957. Then 7062 is subtracted from the sum. The result is not divisible by

(1) 4

(2) 3

(3) 7

(4) 11

Ans .

(3) 7

Explanation :
Resulting number = 3957 + 5349 – 7062 = 2244 which is divisible by 4, 3 and 11.
2244 ÷ 4 = 561
2244 ÷ 3 = 748
2244 ÷ 11 = 204

The product of all the prime numbers between 80 and 90 is

(1) 83

(2) 89

(3) 7387

(4) 598347

Ans .

(3) 7387

Explanation :
Prime numbers between 80 and 90.
= 83 and 89
 Required product = 83 × 89 = 7387

If n is even, \((6^n – 1)\) is divisible by

(1) 37

(2) 35

(3) 30

(4) 6

Ans .

(2) 35

Explanation :
When n = 2, \(6^n – 1 = 6^2 – 1 = 36 – 1 = 35\)
When, n = an even number,
\(a^n–b^n\) is always divisible by \((a^2–b^2)\).

114.I have x marbles. My elder broth- er has 3 more than mine, while my younger brother has 3 less than mine. If the total number of marbles is 15, the number of marbles that I have is

(1) 3

(2) 5

(3) 8

(4) 7

Ans .

(2) 5

Explanation :
Total number of marbles = x
+ x + 3 + x – 3 = 3x
 3x = 15  x = 5

115. Weight of a bucket when filled fully with water is 17 kg. If the weight of the bucket when half filled with water is 13.5 kg, what is the weight of empty bucket ?

(1) 12 kg

(2) 8 kg

(3) 10 kg

(4) 7 kg

Ans .

(3) 10 kg

Explanation :
Bucket + full water = 17 kg.
Bucket + 1/2 water = 13.5 kg.
 Water = 2 × 3.5 = 7 kg.
 Weight of empty bucket 122.
= 17 – 7 = 10 kg.

116.In a farm there are cows and hens. If heads are counted they are 180, if legs are counted they are 420. The number of cows in the farm is

(1) 130

(2) 150

(3) 50

(4) 30

Ans .

(4) 30

Explanation :
A cow and a hen each has a head.
If the total number of cows be x,
then Number of hens = 180 – x
A cow has four legs and a hen has two legs.
 (180 – x) × 2 + 4x = 420
 360 – 2x + 4x = 420
 2x = 420 – 360 = 60

The number which can be writ- ten in the form of n (n + 1) (n + 2), where n is a natural number, is

(1) 7

(2) 3

(3) 5

(4) 6

Ans .

(4) 6

Explanation :
On putting n = 1
n(n +1) (n + 2) = 1 × 2 × 3 = 6

118.A number when divided by 2736 leaves the remainder 75. If the same number is divided by 24, then the remainder is

(1) 12

(2) 3

(3) 0

(4) 23

Ans .

(2) 3

Explanation :
2736 ÷ 24 = 114
Hence, first divisor (2736) is a multiple of second divisor (24).
 Required remainder = Remainder obtained on dividing 75 by 24 = 3

119.The maximum value of F in the following equation 5E9 + 2F8 + 3G7 = 1114 is where E, F, G each stands for any digit.

(1) 8

(2) 9

(3) 7

(4) 5

Ans .

(2) 9

Explanation :
5 E9 + 2 F8 + 3 G7 = 1114
Value of ‘F’ will be maximum if the values of E and G are mini- mum.
 509 + 2 F8 + 307 = 1114
 2 F8 = 1114 – 509 – 307= 298
 F = 9

120.The sum of four numbers is 48. When 5 and 1 are added to the first two; and 3 and 7 are sub- tracted from the 3rd and 4th, the numbers will be equal. The numbers are

(1) 9, 7, 15, 17

(2)4, 12, 12, 20

(3) 5, 11, 13, 19

(4)6, 10, 14, 18

Ans .

(4)6, 10, 14, 18

Explanation :
Let four numbers be a, b, c and d respectively.
 a + b + c + d = 48 (i) and,
a + 5 = b + 1 = c – 3 = d – 7 = x (let)
 a = x – 5; b = x – 1, c = x + 3, d = x + 7 From equation (i),
x – 5 + x – 1 + x + 3 + x + 7 = 48  4x + 4 = 48
 4x = 48 – 4 = 44
 x = 44 = 11
 a = x – 5 = 11 – 5 = 6
b = x – 1 = 11 – 1 = 10
c = x + 3 = 11 + 3 = 14
d = x + 7 = 11 + 7 = 18

121.The least number that should be added to 2055, so that the sum is exactly divisible by 27 is

(1) 28

(2) 24

(3) 27

(4) 31

Ans .

(2) 24

Explanation :
\(\frac{2055}{27}\)
Remainder is 3
 Required number = 27 – 3 = 24

122.What is the Arithmetic mean of the first ‘n’ natural numbers ?

\(\frac{n*(n+1)}{2}\)

\(\frac{n+1}{2}\)

\(\frac{n^2*(n+1)}{2}\)

2(n + 1)

Ans .

\(\frac{n+1}{2}\)

Explanation :
122.Sum of first n bers = \(\frac{n*(n+1)}{2}\)
Required avg. = \(\frac{n*(n+1)}{2*n}\) = \(\frac{(n+1)}{2}\)

123.A number when divided by 361 gives a remainder 47. If the same number is divided by 19, the re- mainder obtained is

(1) 3

(2) 8

(3) 9

(4) 1

Ans .

(3) 9

Explanation :
123.Here, the first divisor (361) is a multiple of second divisor (19).
 Required remainder = Re- mainder obtained on dividing 47 by 19 = 9

124.The difference between the great- est and the least four digit num- bers that begin with 3 and ends with 5 is

(1) 999

(2) 900

(3) 990

(4) 909

Ans .

(3) 990

Explanation :
Largest number = 3995
Smallest number = 3005
Difference = 3995 – 3005 = 990

125.The sum of two numbers is 75 and their difference is 25. The product of the two numbers is :

(1) 1350

(2) 1250

(3) 125

(4) 1000

Ans .

(2) 1250

Explanation :
Let the numbers be x and y. According to the question,
x + y = 75
x – y = 25
Q (x + y)2 – (x – y)2 = 4xy
 752 – 252 = 4xy
 4xy = (75 + 25) (75 – 25)
4xy = 100 × 50
xy = 1250

126.The difference between the greatest and least prime numbers which are less than 100 is

(1) 96

(2) 97

(3) 94

(4) 95

Ans .

(4) 95

Explanation :
Required difference = 97 – 2 = 95

127.Which one of the following is the minimum value of the sum of two integers whose product is 24?

(1) 25

(2) 11

(3) 8

(4) 10

Ans .

(4) 10

Explanation :
) xy = 24
 (x, y)
= (1 × 24), (2 ×12), (3 × 8), (4 × 6)
 Minimum value of (x + y) = 4 + 6 = 10.

128. If the sum of the digits of a three digit number is subtracted from that number, then it will always be divisible by

(1)3 only

(2)9 only

(3)Both 3 and 9

(4)All of 3, 6 and 9

Ans .

(3)Both 3 and 9

Explanation :
128. Let the 3–digit number be 100x + 10y + z.
Sum of the digits = x + y + z According to the question, Difference
= 100x + 10y + z – (x + y + z)
= 99x + 9y = 9 (11x + y)
Clearly, it is a multiple of 3 and 9.

. The greater of the two numbers whose product is 900 and sum exceeds their difference by 30 is

(1) 60

(2) 75

(3) 90

(4) 100

Ans .

(1) 60

Explanation :
129. Let the numbers be x and y
where x > y.
According to the question, (x + y) – (x – y) = 30
 x + y – x + y = 30
 2y = 30
 y = 30/2 = 15
 xy = 900
 15x = 900
x = 60

130.In a division sum, the divisor ‘d’ is 10 times the quotient ‘q’ and 5 times the remainder ‘r’. If r = 46, the dividend will be

(1) 5042

(2) 5328

(3) 5336

(4) 4276

Ans .

(3) 5336

Explanation :
130. According to the question, Divisor (d) = 5r = 5 × 46 = 230 Again, Divisor (d) = 10 × Quo- tient (q)
 230 = q × 10
 q = 230/10 = 23
 Dividend = Divisor × Quotient + Remainder
= 230 × 23 + 46
= 5290 + 46 = 5336

131.A number when divided by 44, gives 432 as quotient and 0 as remainder. What will be the re- mainder when dividing the same number by 31?

(1) 3

(2) 4

(3) 5

(4) 6

Ans .

(3) 5

Explanation :
Divided = 44 × 432 = 19008
19008/31 Remainder = 5

A number when divided by 729 gives a remainder of 56. What will we get as remainder if the same number is divided by 27?

(1) 4

(2) 2

(3) 0

(4) 1

Ans .

(2) 2

Explanation :
Here, first divisor (729) is a multiple of second divisor (27).
 Required remainder = Remainder got on dividing 56 by 27 = 2.

133.What is the smallest 6-digit number that is completely divis- ible by 108 ?

(1) 100003

(2) 100004

(3) 100006

(4) 100008

Ans .

(4) 100008

Explanation :
Smallest number of six dig- its = 100000
108/100000 remainder = 100
 Required number = 100000 + (108 – 100) = 100008

134.If 25 is added to a number it becomes 3 less than thrice of the number. Then number is :

(1) 15

(2) 14

(3) 19

(4) 20

Ans .

(2) 14

Explanation :
Let the number be x. According to the question, x + 25 = 3x – 3
 3x – x = 25 + 3
 2x = 28
 x = 14

The number 334 × 545 × 7p is dividible by 3340 if p is at least.

(1) 2

(2) 4

(3) 3

(4) 1

Ans .

(1) 2

Explanation :
334 × 545 × 7p is divisible by 3340.
 334 × 5 × 109 × 7 × p, is
divisible by 334 × 2 × 5
Clearly, p = 2

136. If the sum of a number and its reciprocal be 2, then the num- ber is

(1) 0

(2) 1

(3) –1

(4) 2

Ans .

(2) 1

Explanation :
Let the number be a. According to the question,
a + 1/a = 2
 a2 + 1 = 2a  a2 – 2a + 1 = 0
 (a – 1)2 = 0  a – 1 = 0
 a = 1

137.When a number is divided by 56, the remainder will be 29. If the same number is divided by 8, then the remainder will be

(1) 6

(2) 7

(3) 5

(4) 3

Ans .

(3) 5

Explanation :
First divisor (56) is a mul- tiple of second divisor (8).
 Required remainder = Remainder obtained after divid- ing 29 by 8 = 5

138.A positive number when de-creased by 4, is equal to 21 times the reciprocal of this number. The number is :

(1) 3

(2) 7

(3) 5

(4) 9

Ans .

(2) 7

Explanation :
Let the number be x. According to the question,
x – 4 = 21/x
 x2 – 4x = 21
 x2 – 4x – 21 = 0
 (x + 3) (x – 7) = 0
 x = 7 because x is not equal to – 3.

139.When n is divided by 4, the re- mainder is 3. The remainder when 2n is divided by 4 is :

(1) 1

(2) 2

(3) 3

(4) 6

Ans .

(2) 2

Explanation :
Let quotient be 1.
 n = 4 × 1 + 3 = 7
 2n = 2 × 7 = 14,
On dividing 14 by 4, remainder = 2

140.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is

(1) 220030

(2) 22030

(3) 1220

(4) 1250

Ans .

paste_right_option

Explanation :
Divisor = 555 + 445 = 1000
Quotient = (555 – 445) × 2 = 110 × 2 = 220
Remainder = 30
 Dividend = Divisor × Quotient + Remainder
= 1000 × 220 + 30 = 220030

When a number x is divided by a divisor it is seen that the divi- sor = 4 times the quotient = dou- ble the remainder. If the remain- der is 80 then the value of x is

(1) 6480

(2) 9680

(3) 8460

(4) 4680

Ans .

(1) 6480

Explanation :
According to the question, Divisor = 2 × remainder = 2 × 80 = 160
Again, 4 × quotient = 160
 Quotient = 160/4 = 40
 x = Divisor × Quotient + re- mainder = 160 × 40 + 80 = 6480

142. 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 ?

(1) 15

(2) 11

(3) 17

(4) 13

Ans .

(2) 11

Explanation :
Here, first divisor (342) is a multiple of second divisor (18). i.e. 342 ÷ 18 = 19
 Required remainder = Remainder on dividing 47 by 18 = 11

143.The sum of three numbers is 252. If the first number is thrice the second and third number is two–third of the first, then the second number is

(1) 41

(2) 21

(3) 42

(4) 84

Ans .

(3) 42

Explanation :
Let second number = x.  First number = 3x
Third number = 2/3 × 3x = 2x
According to the question, 3x + x + 2x = 252
 6x = 252
 x = 252/6 = 42

144. The difference between the great- est and the least five-digit num- bers formed by the digits 2, 5, 0, 6 and 8 is (repetition of digits is not allowed)

(1) 69552

(2) 65925

(3) 65952

(4) 63952

Ans .

(3) 65952

Explanation :
Five-digit numbers formed by 2, 5, 0, 6 and 8 :
Largest number = 86520
Smallest number = 20568
Required difference = 86520 – 20568 = 65952

145. A man has some hens and some cows. If the total number of heads of hens and cows together is 50 and the number of feet of hens and cows together is 142, then the number of cows is

(1) 21

(2) 25

(3) 27

(4) 29

Ans .

(1) 21

Explanation :
Let the number of cows be x. Q A hen or a cow has only one head.
 Number of hens = 50 – x
A hen has two feet. A cow has four feet.
According to the question, 4x + 2 (50 – x) = 142
 4x + 100 – 2x = 142
 2x = 142 – 100 = 42
x = 21

The least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case, but when divided by 9 leaves no re- mainder, is :

(1) 1677

(2) 1683

(3) 2523

(4) 3363

Ans .

(2) 1683

Explanation :
Firstly, we find LCM of 5, 6, 7 and 8.
 LCM = 2 × 5 × 4 × 3 × 7= 840
Required number = 840x + 3 which is exactly divis- ible by 9.
Now, 840x + 3 = 93x × 9 + 3x + 3
When x = 2 then 840x + 3, is di- visible by 9.
 Required number = 840 × 2 + 3 = 1683

147.If the sum of the digits of any integer between 100 and 1000 is subtracted from the same inte- ger, the resulting number is al- ways divisible by

(1) 2

(2) 5

(3) 6

(4) 9

Ans .

(4) 9

Explanation :
A 3–digit number = 100x + 10y + z
Sum of digits = x + y + z
Difference = 100x + 10y + z – x – y – z
= 99x + 9y = 9 (11x + y)
i.e., multiple of 9.

148.The least number that must be added to 8961 to make it exact- ly divisible by 84 is :

(1) 27

(2) 57

(3) 141

(4) 107

Ans .

(1) 27

Explanation :
8961/84 Remainder is 57
 Required number = 84 – 57 = 27

Number of composite numbers lying between 67 and 101 is :

(1) 27

(2) 24

(3) 26

(4) 23

Ans .

(1) 27

Explanation :
Number of numbers lying be- tween 67 and 101
 101 – 67 – 1 = 33
Prime numbers  71, 73, 79, 83, 89 and 97 = 6
 Composite numbers = 33 – 6 = 27

150.The least number that must be subtracted from 1294 so that the remainder when divided by 9, 11 and 13 will leave in each case the same remainder 6, is :

(1) 2

(2) 3

(3) 1

(4) 4

Ans .

(3) 1

Explanation :
LCM of 9, 11 and 13 = 9 × 11 × 13 = 1287
 Required lowest number that leaves 6 as remainder = 1287 + 6 = 1293
 Required answer = 1294 – 1293 = 1

151. What least value must be as- signed to ‘*’ so that the number 63576*2 is divisible by 8 ?

(1) 1

(2) 2

(3) 3

(4) 4

Ans .

paste_right_option

Explanation :
A number is divisible by 8 if number formed by the last three
 If * is replaced by 3, then 632

152. The least number to be added to 13851 to get a number which is divisible by 87 is :

(1) 18

(2) 43

(3) 54

(4) 69

Ans .

(4) 69

Explanation :
13851/87 Remainder = 18
Required no. = 87 - 18 = 69

153. What least value must be as- signed to ‘*’ so that the number 451 * 603 is exactly divisible by

(1) 7

(2) 8

(3) 5

(4) 9

Ans .

(2) 8

Explanation :
If the sum of the digits of a number be divisible by 9, the number is divisible by 9.
Sum of the digits of 451 * 603
= 4 + 5 + 1 + * + 6 + 0 + 3 = 19 + * If * = 8, then 19 + 8 = 27 which is divisible by 9.

154. The largest number of four dig-its exactly divisible by 88 is :

(1) 9988

(2) 9944

(3) 8888

(4) 9768

Ans .

(2) 9944

Explanation :
The largest 4-digit number = 9999
9999/88 Remainder is 55  Required number = 9999 – 55 = 9944

155. Which of the following numbers is completely divisible by 99?

(1) 57717

(2) 57627

(3) 55162

(4) 56982

Ans .

(1) 57717

Explanation :
= 5 + 7 + 7 + 1 + 7 = 27 which is divisible by 9
 Required number = 57717

156.The sum of all prime numbers between 58 and 68 is

(1) 179

(2) 178

(3) 187

(4) 183

Ans .

(3) 187

Explanation :
Prime numbers between 58 and 68  59, 61 and 67
 Required sum = 59 + 61 + 67 = 187

157.The product of digits of a 2-digit number is 24. If we add 45 to the number, the new number ob- tained is a number formed by interchanging the digits.What is the original number?

(1) 54

(2) 83

(3) 38

(4) 45

Ans .

(3) 38

Explanation :
Let the two digit number be 10x + y.
According to the question,
xy = 24 (i)
and, 10x + y + 45 = 10y + x
y - x = 5 ....(ii)
x + y = 11 ....(iii)
On adding equations (ii) and (iii),
y – x + x + y = 5 + 11
 2y = 16  y = 8
 xy = 24  8x = 24
x = 3
 Required number = 10x + y = 10 × 3 + 8 = 38

The smallest number, which should be added to 756896 so as to obtain a multiple of 11, is

(1) 1

(2) 2

(3) 3

(4) 5

Ans .

(3) 3

Explanation :
A number is divisible by 11 if the difference between the sum of digits at odd places and that at even places is either zero or a multiple of 11.
Sum of the digits at odd places = 6 + 8 + 5 = 19
Sum of the digits at even places = 9 + 6 + 7 = 22
 Required number=22–19 = 3

159. The product of two numbers is 48. If one number equals “The number of wings of a bird plus 2 times the number of fingers on your hand divided by the num- ber of wheels of a Tricycle”. Then the other number is

(1) 9

(2) 10

(3) 12

(4) 18

Ans .

(3) 12

Explanation :
According to the question
\(\frac{2+2*5}{3} = 4\)
Second No. \(\frac{48}{4} = 12\)

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