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

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

TYPE-II



Ans .

(3) 8


  1. 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





Ans .

(1) 4


  1. Explanation :

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





Ans .

(3) 6


  1. 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





Ans .

(1) 5


  1. 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.





Ans .

(2) 6


  1. 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.





Ans .

(2) 150


  1. 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





Ans .

(2) 6


  1. Explanation :

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





Ans .

(3) 6


  1. Explanation :

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





Ans .

(3)divisible by 9


  1. Explanation :

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





Ans .

divisible by (11 × 13)


  1. Explanation :

    divisible by (11 × 13)





Ans .

(3) 5


  1. 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





Ans .

(4) 5


  1. 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





Ans .

(4) 1


  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





Ans .

(4) 4


  1. 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.





Ans .

(2) 12


  1. 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





Ans .

(4)Smallest 3-digit prime num- ber


  1. 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.





Ans .

(2) 10004


  1. Explanation :

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





Ans .

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


  1. Explanation :

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





Ans .

(4) 48


  1. 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.





Ans .

(2)18


  1. Explanation :

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





Ans .

(3) 24


  1. Explanation :

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





Ans .

(1) 1440


  1. Explanation :

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





Ans .

(2)6


  1. Explanation :

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





Ans .

(2) 10


  1. Explanation :

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





Ans .

(2) 9


  1. 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.





Ans .

(3) 303375


  1. 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.





Ans .

(1) 132


  1. Explanation :

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





Ans .

(1) 15999879


  1. Explanation :

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





Ans .

(2) 2


  1. Explanation :

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





Ans .

(3) 7


  1. Explanation :

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





Ans .

(3) 7387


  1. Explanation :

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





Ans .

(2) 35


  1. 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 \((a^2–b^2)\).





Ans .

(2) 5


  1. Explanation :

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





Ans .

(3) 10 kg


  1. 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.





Ans .

(4) 30


  1. 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





Ans .

(4) 6


  1. Explanation :

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





Ans .

(2) 3


  1. 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





Ans .

(2) 9


  1. 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





Ans .

(4)6, 10, 14, 18


  1. 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





Ans .

(2) 24


  1. Explanation :

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





Ans .

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


  1. Explanation :

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





Ans .

(3) 9


  1. 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





Ans .

(3) 990


  1. Explanation :

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





Ans .

(2) 1250


  1. 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





Ans .

(4) 95


  1. Explanation :

    Required difference = 97 – 2 = 95





Ans .

(4) 10


  1. Explanation :

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





Ans .

(3)Both 3 and 9


  1. 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.





Ans .

(1) 60


  1. 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





Ans .

(3) 5336


  1. 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





Ans .

(3) 5


  1. Explanation :

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





Ans .

(2) 2


  1. Explanation :

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





Ans .

(4) 100008


  1. Explanation :

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





Ans .

(2) 14


  1. Explanation :

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





Ans .

(1) 2


  1. Explanation :

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





Ans .

(2) 1


  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





Ans .

(3) 5


  1. Explanation :

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





Ans .

(2) 7


  1. 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.





Ans .

(2) 2


  1. Explanation :

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





Ans .

paste_right_option


  1. Explanation :

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





Ans .

(1) 6480


  1. 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





Ans .

(2) 11


  1. 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





Ans .

(3) 42


  1. 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





Ans .

(3) 65952


  1. Explanation :

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





Ans .

(1) 21


  1. 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





Ans .

(2) 1683


  1. 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





Ans .

(4) 9


  1. 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.





Ans .

(1) 27


  1. Explanation :

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





Ans .

(1) 27


  1. 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





Ans .

(3) 1


  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





Ans .

paste_right_option


  1. Explanation :

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





Ans .

(4) 69


  1. Explanation :

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





Ans .

(2) 8


  1. 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.





Ans .

(2) 9944


  1. Explanation :

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






Ans .

(1) 57717


  1. Explanation :

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





Ans .

(3) 187


  1. Explanation :

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





Ans .

(3) 38


  1. 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





Ans .

(3) 3


  1. 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





Ans .

(3) 12


  1. Explanation :

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