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LCM & HCF Test 284
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LCM & HCF Test 284

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  • Question 1/10
    1 / -0.25
    In a school, 391 boys and 323 girls have been divided into the largest possible equal classes, so that the number of boys and girls in each class are same. What is the number of classes of students?
    Solutions
    Short Approach:
    The largest possible number of students in classes is given by the H.C.F. of 391 and 323 i,e. 17
    Thus, No. of classes for boys = 391/17 = 23
    Alternate Approach:
    total number of boys = 391
    factors of 391 = 17 × 23
    total number of girls = 323
    factors of 323 = 17 × 19
    therefore for no. of students in classes to be equal and we have to select the common factor from both
    17 is a common factor.
    Thus, No. of classes for boys = 391/17 = 23
  • Question 2/10
    1 / -0.25
    The product of two numbers is 4107. If the HCF of the number is 37. The greatest number is
    Solutions

    Let 37X and 37Y be two numbers 
    Given 37*37 XY = 4107 
    XY = 3

    Only possible integer pair of (X,Y) is (1,3) or (3,1)
    Hence the greater number is 3*37 = 111

  • Question 3/10
    1 / -0.25
    The ratio of two numbers is 3:4 and their HCF is 5. Their LCM Is:
    Solutions
    If the number be 3x and 4x, then
    Their HCF = x
    According to the question x=5
    so the numbers are = 15 and 20
    and their LCM,
    15 = 3×5
    20 = 2×2×5
    so, LCM = 2×2×3×5 = 60
  • Question 4/10
    1 / -0.25
    What is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
    Solutions
    Length of the floor = 15 m 17 cm = 1517 cm
    Breadth of the floor = 9 m 2 cm = 902 cm
    Area of the floor = 1517 × 902 cm2
    The number of square tiles will be least, when the size of each tile is maximum.
    ∴Size of each tile = HCF of 1517 and 902 = 41
    ∴Required number of tiles =
  • Question 5/10
    1 / -0.25
    A wine seller has 3 varieties of wines measuring 481 litres, 629 litres and 703 litres what is the minimum number of containers of equal size so that whole of the wine is filled in them without mixing?
    Solutions
    A.T.Q
    HCF of 481, 629 and 703 = 37
    ⇒ Each container contain 37 litres of wine.
    ⇒ Minimum number of containers required =
  • Question 6/10
    1 / -0.25
    Find the HCF of 36, 56 and 84.
    Solutions
    Factors of 36 = 2 x 2 x 3 x 3
    Factors of 56 = 2 x 2 x 2 x 7
    Factors of 84 = 2 x 2 x 3 x 7
    HCF = 2 x 2 = 4
  • Question 7/10
    1 / -0.25
    There are 2 large containers, 1 contains 1638 Green balls, and other contains 1485 Blue balls. A small container has ‘x’ compartments to accommodate all balls so that no compartment is empty. Also, all balls in 1 small container should be of same colour. What is ‘x’?
    Solutions
    In the above question we basically need to find the Highest Common Factor (HCF) of 1638 and 1485.
    ⇒ 1638 = 2 x 7 x 9 x 13
    ⇒ 1485 = 3 x 5 x 9 x 11
    From the above prime factorisation we can say that ‘x’ is 9.
  • Question 8/10
    1 / -0.25
    Three natural number whose LCM is 360, are in the ratio 2:3:4, the largest of them is:
    Solutions
    Let the natural numbers are
    Then, their LCM and HCF

    Largest number
  • Question 9/10
    1 / -0.25
    If least common multiple of two numbers is 175 and the highest common factor is 5 then find the numbers when one of the numbers is 25.
    Solutions

    Let the 2nd number be x


    HCF×LCM = Product of two numbers


    175×5 = 25x


    x = 35

  • Question 10/10
    1 / -0.25
    The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is
    Solutions
    Let the no. be 3x and 3y
    3x + 3y = 36
    x + y =12 ……(1)
    3xy =105     
    xy =35 ……(2)
    Dividing eq1 by 2
    We have,
    (x+y)/xy = 12/35 
    1/y + 1/x = 12/35
    now, since the numbers were 3x and 3y, divide whole equation by 3.
    1/3y + 1/3x = 4/35
    Hence option 3 is correct.
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