- Staff Selection Commission Mathematics 1999 to 2017 - LCM and HCF

Staff Selection Commission Mathematics - LCM and HCF



Ans .

1


  1. Explanation :


    2   28, 42
    2   14, 21
    7   7, 21
         1, 3
    = 2 × 2 × 7 × 3 = 84 H.C. F. of 28 and 42
                 28 ) 42 ( 1
                        28
                     14 ) 28 ( 2
                          28
                     0
    H.C. F = 14
    Required ratio = 84 = 6:1
        14





Ans .

3


  1. Explanation :

    Let the two numbers are 2x and 3x respectively. According to question, LCM = 54 x ( ) 3 2 54 ´ = Þ x = 9 Numbers = 2x = 2 × 9 = 18 and, 3x = 3 × 9 = 27 Sum of the two numbers = 18 + 27 = 45





Ans .

3


  1. Explanation :

    Suppose the numbers are 4x and 5x respectively According to question x × 4 × 5 = 120 Þ x = 6 Required numbers = 4 × 6 = 24 and = 5 × 6 = 30





Ans .

1


  1. Explanation :

    Let the numbers be 2x, 3x and 4x respectively. HCF = x = 12 Numbers are : 2 ×12 = 24 3 ×12 = 36, 4 ×12 = 48 LCM of 24, 36, 48 = 2 × 2 × 2 × 3 × 3 × 2 = 144





Ans .

3


  1. Explanation :

    Let the number be 3x and 4x. Their LCM = 12x According to the question, 12x = 240
    x = 240 = 20
            12
    Smaller number = 3x = 3 × 20 = 60





Ans .

3


  1. Explanation :

    Let the numbers be 3x and 4x. Their LCM = 12x 12x = 84
    x = 84 = 7
        12
    Larger number = 4x = 4 × 7 = 28





Ans .

1


  1. Explanation :

    Numbers = 3x and 4x HCF = x = 4
    LCM = 12x = 12 × 4 = 48





Ans .

3


  1. Explanation :

    Let the numbers be 4x and 4y where x and y are prime to each other. LCM = 4xy
    (4x + 4y) = 7
    4xy    12
    12 (x + y) = 7 xy
    x = 3, y = 4
    Smaller number = 4 × 3 = 12





Ans .

4


  1. Explanation :

    Using Rule 1, Let the numbers be 3x and 4x respectively First number × second number
    = HCF × LCM
    3x × 4x = 2028
    x2 = 2028 = 169
         3 x 4
    x = √169 = 13 Sum of the numbers = 3x + 4x = 7x = 7 × 13 = 91





Ans .

3


  1. Explanation :

    If the numbers be 2x and 3x, then LCM = 6x 6x = 48
    x = 8 Required sum = 2x + 3x = 5x = 5 × 8 = 40





Ans .

4


  1. Explanation :

    Let the numbers be 4x and 5x. H.C.F. = x = 8 Numbers = 32 and 40 Their LCM = 160





Ans .

2


  1. Explanation :

    If the numbers be 3x and 4x, then HCF = x = 5 Numbers = 15 and 20 LCM = 60





Ans .

1


  1. Explanation :

    Numbers = x , 2 x and 3 x (let) Their H.C.F. = x = 12 Numbers = 12, 24 and 36





Ans .

4


  1. Explanation :

    Using Rule 1, Product of two numbers = HCF × LCM
    Numbers = zx and zy
    zx × zy = z × LCM
    LCM = xyz





Ans .

4


  1. Explanation :

    HCF of numbers = 21 Numbers = 21x and 21y Where x and y are prime to each other. Ratio of numbers = 1 : 4
    Larger number = 21 × 4 = 84





Ans .

4


  1. Explanation :

    Using Rule 1, Let the numbers be x and (x + 2). Product of numbers = HCF × LCM x (x + 2) = 24
    x2 + 2x – 24 = 0
    x2 + 6x – 4x – 24 = 0
    x (x + 6) – 4 (x + 6) = 0
    (x – 4) (x + 6) = 0
    x = 4, as x ¹ – 6 = 0
    Numbers are 4 and 6.





Ans .

1


  1. Explanation :

    Using Rule 1, Suppose 1st number is x then, 2nd number = 100 – x
    LCM × HCF = 1st number × 2nd number
    495 × 5 = x × (100 – x)
    495 × 5 = 100x – x2
    x2 – 55x – 45x – 2475 = 0
    (x – 45) (x – 55) = 0
    x = 45 or x = 55

    Then, difference = 55 – 45 = 10





Ans .

2


  1. Explanation :

    Let the number be 29x and 29y respectively where x and y are prime to each other.
    LCM of 29x and 29y = 29xy
    Now, 29xy = 4147
    xy = 4147 = 143
        29
    Thus xy = 11 × 13
    Numbers are 29 × 11
    = 319 and 29 × 13 = 377
    Required sum
    = 377 + 319 = 696





Ans .

4


  1. Explanation :

    Let HCF be h and LCM be l. Then, l = 84h and l + h = 680 84h + h = 680 h = 680 = 8
    85 l = 680 – 8 = 672
    Other number
    = 672 x 8
    56
    = 96





Ans .

2


  1. Explanation :

    HCF = 12 Numbers = 12x and 12y where x and y are prime to each other.
    12x + 12y = 84
    12 (x + y) = 84
    x + y = 84 = 7
    12
    Possible pairs of numbers satisfying this condition = (1,6), (2,5) and (3,4). Hence three pairs are of required numbers





Ans .

3


  1. Explanation :

    Let the numbers be 21x and 21y where x and y are prime to each other. 21x + 21y = 336
    21 (x + y) = 336
    x + y = 336 =16 21 Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)





Ans .

3


  1. Explanation :

    Let the number be x and y. According to the question, x + y = 45 ......... (i)
    Again, x – y = 1( x + y)
    9
    or x – y = 145
    9 or x – y = 5 ..... (ii)
    By (i) + (ii) we have,
    x + y = 45
    x – y = 5
    = 50 2x
    or, x = 25
    y = 45 – 25 = 20.
    Now, LCM of 25 and 20 = 100





Ans .

3


  1. Explanation :

    Let the numbers be 17x and 17y where x and y are co-prime.
    LCM of 17x and 17y = 17 xy
    According to the question,
    17xy = 714
    xy = 714 = 42 = 6 x 7
         17
    x = 6 and y = 7
    or, x = 7 and y =6.
    First number = 17x
    = 17 × 6 = 102
    Second number = 17y
    =17 × 7 = 119
    Sum of the numbers
    = 102 + 119 = 221





Ans .

3


  1. Explanation :

    Using Rule 1, Let the larger number be x. Smaller number = x – 2
    First number × Second number = HCF × LCM x (x – 2) = 24
    x2 – 2x – 24 = 0
    x2 – 6x + 4x – 24 = 0
    x (x – 6) + 4 (x – 6) = 0
    (x – 6) (x + 4) = 0
    x = 6 because x not equal to 4





Ans .

2


  1. Explanation :

    HCF of two numbers = 27 Let the numbers be 27x and 27y
    where x and y are prime to each other.
    According to the question,
    27x + 27y = 216
    27 (x + y) = 216
    x + y = 216 = 8
        27
    Possible pairs of x and y = (1, 7) and (3, 5)
    Numbers =(27, 189) and (81, 135)





Ans .

1


  1. Explanation :

    Using Rule 1, Let the HCF of numbers = H
    Their LCM = 12H
    According to the question,
    12H +H = 403
    13H = 403
    H =403 = 31
         13
    LCM = 12 × 31
    Now,
    First number × second number
    = HCF × LCM
    = 93 × Second Number
    = 31 × 31 × 12
    Second number =
    31 x 31 x 12
          93
    = 124





Ans .

3


  1. Explanation :

    Let the numbers be 48x and 48y where x and y are co-primes.
    48x + 48y = 384
    48 ( x + y) = 384
    x + y =384 =8
       48
    Possible and acceptable pairs of x and y satisfying this condition are : (1, 7) and (3, 5).
    Numbers are : 48 × 1 = 48
    and 48 × 7 = 336
    and 48 × 3 = 144 and 48 × 5 = 240
    Required difference
    = 336 – 48 = 288





Ans .

2


  1. Explanation :

    Let the numbers be 3x and 3y.
    3x + 3y = 36
    x + y = 12 ... (i)
    and 3xy = 105 ... (ii)
    Dividing equation (i) by (ii), we have
    x + y =12
    3xy    3xy    105
    1 + 1 =4
    3xy    3xy    35





Ans .

4


  1. Explanation :

    Let the numbers be 10x and 10y where x and y are prime to each other.
    LCM = 10 xy
    10xy = 120
    xy = 12
    Possible pairs = (3, 4) or (1, 12)
    Sum of the numbers
    = 30 + 40 = 70





Ans .

3


  1. Explanation :

    Let the numbers be x, y and z which are prime to one another. Now, xy = 551
    yz = 1073
    y = HCF of 551 and 1073
    y = 29
    x = 551 =19
      29
    and
    z = 1073 =37
      29
    Sum = 19 + 29 + 37 = 85





Ans .

3


  1. Explanation :

    HCF of two numbers = 4.
    Hence, the numbers can be given by 4x and 4y where x and y are co-prime. Then,
    4x + 4y = 36
    4 (x + y) = 36
    x + y = 9
    Possible pairs satisfying this condition are : (1, 8), (4, 5), (2, 7)





Ans .

2


  1. Explanation :

    Let the numbers be 2x and 2y where x and y are prime to each other.
    LCM = 2xy
    2xy = 84
    xy = 42 = 6 × 7
    Numbers are 12 and 14.
    Hence Sum = 12 + 14 = 26





Ans .

3


  1. Explanation :

    Let the numbers be x H and yH where H is the HCF and yH x H.
    LCM = xy H
    xyH = 2yH
    x = 2 Again, x H – H = 4
    2H – H = 4
    H = 4
    Smaller number = x H = 8





Ans .

4


  1. Explanation :

    Using Rule 1, Let the H.C.F. be H.
    L.C.M. = 20H
    Then, H + 20H = 2520
    21 H = 2520
    H = 2520 = 120
         21
    L.C.M. = 20H = 20×120= 2400
    As,
    First number × Second number
    = L.C.M. × H.C.F.
    480 × Second number = 2400 × 120
    Second number = 2400 x 120 = 600
         480





Ans .

1


  1. Explanation :

    Using Rule 1, If the HCF = H, then
    LCM = 44 H
    44 H + H = 1125
    45 H = 1125
    H = 1125 = 25
       45
    LCM = 44 × 25 = 1100
    Now
    First number × Second number
    = LCM × HCF
    25 × Second number
    = 1100 × 25
    Second number
    = 1100 x 25 = 1100
       25





Ans .

4


  1. Explanation :

    Let no. are x and y and HCF = A, LCM = B
    Using Rule, we have
    xy = AB
    x + y = A + B (given) ...(i)
    (x–y)2 = (x + y)2 – 4xy
    or, (x–y)2 = (A + B)2 – 4 AB
    or, (x–y)2 = (A – B)2
    or, (x–y) = A – B ...(ii)
    Using (i) and (ii), we get
    x = A and y = B
    A3 + B3 = x3 + y3





Ans .

3


  1. Explanation :

    Let the numbers be 7x and 7y where x and y are co-prime.
    Now, LCM of 7x and 7y = 7xy
    7xy = 140
    xy = 140 = 20
       7
    Now, required values of x and y whose product is 50 and are coprime, will be 4 and 5.
    Numbers are 28 and 35 which lie between 20 and 45.
    Required sum = 28 + 35 = 63





Ans .

4


  1. Explanation :


    Firstly, we find the LCM of 30, 36 and 80.
    2   30, 36, 80
    2   15, 18, 40
    3   15, 9, 20
    5    5, 3, 20
         1, 3, 4
    LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720
    Required number = Multiple of 720 = 720 × 5 = 3600; because 3000 < 3600 < 4000





Ans .

3


  1. Explanation :

    LCM of 5, 6, 7 and 8 = 840
    2   5, 6, 7, 8
         5, 3, 7, 4
    LCM = 2 × 5 × 3 × 7 × 4 = 840
    Required number = 840x + 3
    which is divisible by 9 for a certain least value of x.
    Now,
    840x + 3 = 93x × 9 + 3x + 3
    3x + 3, is divisible by 9 for x = 2
    Required number = 840 × 2 + 3
    = 1680 + 3 = 1683
    Sum of digits = 1 + 6 + 8 + 3
    = 18





Ans .

1


  1. Explanation :


    2   12, 18, 21, 28
    2   6, 9, 21, 14
    3   3, 9, 21, 7
    7   1, 3, 7, 7
         1, 3, 1, 1
    LCM = 2 × 2 × 3 × 3 × 7= 252 The largest 4-digit number = 9999
                     252 ) 9999 ( 39
                      756
                     2439
                     2268
                     171
    Required number = 9999 – 171 = 9828





Ans .

4


  1. Explanation :

    LCM of 8, 12 and 16 = 48 Required number = 48a + 3 which is divisible by 7.
    x = 48a + 3
    = (7 × 6a) + (6a + 3) which is divisible by 7.
    i.e. 6a + 3 is divisible by 7.
    When a = 3, 6a + 3 = 18 + 3
    = 21 which is divisible by 7.
    x = 48 × 3 + 3 = 144 + 3 = 147





Ans .

1


  1. Explanation :


    2   12, 16, 18, 21
    2   6, 8, 9, 21
    3   3, 4, 9, 21
         1, 4, 3, 7
    LCM = 2 × 2 × 3 × 4 × 3 × 7 = 1008
    Multiple of 1008 = 2016
    Required number = 2016 – 2000 = 16 = x
    Sum of digits of x = 1 + 6 = 7





Ans .

3


  1. Explanation :


    2   12, 18, 21
    3   6, 9, 21
         2, 3, 7
    LCM of 12, 18 and 21
    = 2 × 3 × 2 × 3 × 7 = 252
    Of the options, 10080 ÷ 252 = 40





Ans .

1


  1. Explanation :

    2   30, 36, 80
    2   15, 18, 40
    3   15, 9, 20
    5   5, 3, 20
         1, 3, 4

    LCM = 2 × 2 × 3 × 3 × 4 × 5 = 720 Required number = 2 × 720 + 11 = 1440 + 11 = 1451





Ans .

2


  1. Explanation :

    2   12, 18, 21, 32
    2   6, 9, 21, 16
    3   3, 9, 21, 8
         1, 3, 7, 8
    LCM = 2 × 2 × 3 × 3 × 7 × 8 = 2016 Required number = 2016 × 2 = 4032





Ans .

4


  1. Explanation :

    2   210
    3   105
    5   35
         7
    210 = 2 × 3 × 5 × 7 = 5 × 6 × 7 Required answer = 5 + 6 = 11





Ans .

4


  1. Explanation :

    Let the numbers be 12x and 12y.
    Their LCM = 12xy when x and y are prime to each other.
    y = 1056 = 8
       132 Other number = 12y
    = 12 × 8 = 96





Ans .

2


  1. Explanation :

    When 36798 is divided by 78, remainder = 60
    The least number to be subtracted = 60





Ans .

1


  1. Explanation :

    LCM of 18, 21 and 24
    2   18, 21, 24
    3   9, 21, 12
         3, 7, 4
    LCM = 2 × 3 × 3 × 7 × 4 = 504 Now compare the divisors with their respective remainders. We observe that in all the cases the remainder is just 11 less than their respective divisor. So the number can be given by 504 K – 11. Where K is a positive integer Since 23 × 21 = 483 We can write 504 K – 11 = (483 + 21) K – 11 = 483 K + (21K – 11) 483 K is multiple of 23, since 483 is divisible by 23.
    So, for (504K – 11) to be multiple of 23, the remainder (21K – 11) must be divisible by 23.
    Put the value of K = 1, 2, 3, 4, 5, 6, ..... and so on successively.
    We find that the minimum value of K for which (21K – 11) is divisible by 23. is 6, (21 × 6 – 11) = 115 which is divisible by 23.
    Therefore, the required least number = 504 × 6 – 11 = 3013





Ans .

4


  1. Explanation :

    Using Rule 7, Clearly, 122 – 2 = 120 and 243 – 3 = 240 are exactly divisible by the required number.
    Required number = HCF of 120 and 240 = 120





Ans .

2


  1. Explanation :

    P = 23 × 310 × 5 Q = 25 × 3 × 7 HCF = 23 × 3





Ans .

4


  1. Explanation :

    Let the original fraction be x.
    y
    x+2 = 1
    y+1 6
    6x – 24 = y + 1
    6x – y = 25 .......(i) Again, x+2 = 1
    y+1 3
    3x + 6 = y + 1
    3x – y = –5 .......(ii) By equation (i) – (ii),
    6x – y – 3x + y = 25 + 5
    = 3x = 30
    x = 10 From equation (i), 60 – y = 25
    = y = 35 LCM of 10 and 35 = 70





Ans .

4


  1. Explanation :

    HCF of a and b = 12
    Numbers = 12x and 12y where x and y are prime to each other. Q a > b > 12
    a = 36; b = 24





Ans .

4


  1. Explanation :

    Let the numbers be 9x and 9y where x and y are prime to each other. According to the question, 9x + 9y = 99 Þ 9(x + y) = 99
    x + y = 11 Possible pairs = (1, 10) (2, 9), (3, 8), (4, 7), (5, 6