LCM and HCF


  1. If A and B are the H.C.F. and L.C.M. respectively of two algebraic expressions p and q, and A + B = p + q, then the value of A3 + B3 is









  1. View Hint View Answer Discuss in Forum

    Let the numbers are p and q and HCF = A, LCM = B
    Using Rule, we have
    Product of numbers = HCF × LCM
    pq = AB
    ⇒ p + q = A + B (given) ...(i)
    (p – q)2 = (p + q)2 – 4pq
    or, (p – q)2 = (A + B)2 – 4AB
    or, (p – q)2 = (A – B)2
    or, (p – q) = A – B ...(ii)
    Using (i) and (ii), we get
    p = A and q = B

    Correct Option: D

    Let the numbers are p and q and HCF = A, LCM = B
    Using Rule, we have
    Product of numbers = HCF × LCM
    pq = AB
    ⇒ p + q = A + B (given) ...(i)
    (p – q)2 = (p + q)2 – 4pq
    or, (p – q)2 = (A + B)2 – 4AB
    or, (p – q)2 = (A – B)2
    or, (p – q) = A – B ...(ii)
    Using (i) and (ii), we get
    p = A and q = B
    ∴ A3 + B3
    = p3 + q3


  1. The LCM of two numbers is 44 times of their HCF. The sum of the LCM and HCF is 1125. If one number is 25, then the other number is









  1. View Hint View Answer Discuss in Forum

    Let the HCF of two numbers be H . then
    According to question ,
    LCM = 44H
    ∴ 44H + H = 1125
    ⇒ 45H = 1125

    ∴ H =
    1125
    = 25
    45

    ∴ LCM = 44 × 25 = 1100
    As we know that ,
    First number × Second number = LCM × HCF
    ⇒ 25 × Second number = 1100 × 25

    Correct Option: A

    Let the HCF of two numbers be H . then
    According to question ,
    LCM = 44H
    ∴ 44H + H = 1125
    ⇒ 45H = 1125

    ∴ H =
    1125
    = 25
    45

    ∴ LCM = 44 × 25 = 1100
    As we know that ,
    First number × Second number = LCM × HCF
    ⇒ 25 × Second number = 1100 × 25
    ∴ Second number =
    1100 × 25
    = 1100
    45



  1. The L.C.M. of two numbers is 20 times their H.C.F. The sum of H.C.F. and L.C.M. is 2520. If one of the number is 480, the other number is :









  1. View Hint View Answer Discuss in Forum

    Let the H.C.F. be H.
    According to question ,
    ∴ L.C.M. = 20H
    and H + 20H = 2520
    ⇒ 21 H = 2520

    ⇒ H =
    2520
    = 120
    21

    ∴ L.C.M. = 20H = 20 × 120 = 2400
    We can find the Second number with the help of given formula ,
    First number × Second number = L.C.M. × H.C.F.
    ⇒ 480 × Second number = 2400 × 120

    Correct Option: D

    Let the H.C.F. be H.
    According to question ,
    ∴ L.C.M. = 20H
    and H + 20H = 2520
    ⇒ 21 H = 2520

    ⇒ H =
    2520
    = 120
    21

    ∴ L.C.M. = 20H = 20 × 120 = 2400
    We can find the Second number with the help of given formula ,
    First number × Second number = L.C.M. × H.C.F.
    ⇒ 480 × Second number = 2400 × 120
    ⇒ Second number =
    2400 × 120
    = 600
    480


  1. The LCM of two positive integers is twice the larger number. The difference of the smaller number and the GCD of the two numbers is 4. The smaller number is :









  1. View Hint View Answer Discuss in Forum

    Let the numbers be pH and qH where H is the HCF and qH > pH.
    ∴ LCM = pqH
    ∴ pqH = 2qH ⇒ p = 2
    Again, pH – H = 4
    ⇒ 2H – H = 4 ⇒ H = 4

    Correct Option: C

    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 = pH = 2 × 4 = 8



  1. If the HCF and LCM of two consecutive (positive) even numbers be 2 and 84 respectively, then the sum of the numbers is









  1. View Hint View Answer Discuss in Forum

    Let the numbers be 2p and 2q , where p and q are prime to each other.
    ∴ LCM = 2pq
    ⇒ 2pq = 84
    ⇒ pq = 42 = 6 × 7
    Here p = 6 and q = 7

    Correct Option: B

    Let the numbers be 2p and 2q , where p and q are prime to each other.
    ∴ LCM = 2pq
    ⇒ 2pq = 84
    ⇒ pq = 42 = 6 × 7
    Here p = 6 and q = 7
    ∴ Numbers are 2p = 2 × 6 = 12 and 2q = 2 × 7 = 14.
    Hence , Sum of numbers = 12 + 14 = 26