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Number System LCM & HCF Test 50
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Number System LCM & HCF Test 50

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  • Question 1/10
    1 / -0
    If the number 23583ab ,is divisible by 80 then find the value of (a-b).
    Solutions

    if a number is divisible by 80 then it also be divisible by their factors . so that 80 is also divisible by 8 and 10 .

    To be divisible by 10 the unit digit number should be 0.

    Therefore , b=0

    To be divisible by 8 the last 3 digit of that number is divisible by 8.

    Hence 3ab =3a0 is also divisible by 8 .

    If 3a0 is divisible by 8 then value of a must be 2 .

    a-b=2-0=2

  • Question 2/10
    1 / -0
    The value of is
    Solutions

    Given, series is


    = 1 + 1.1 + 0.11 + 0.111 + 0.0111


    = 2.3321


    Hence, option (B) is correct

  • Question 3/10
    1 / -0
    3 × 105 + 4 × 103 + 7 × 102 + 5 is equal to
    Solutions

    Hence, option C is correct.
  • Question 4/10
    1 / -0
    If a, b and c are three natural numbers in ascending order, then
    Solutions
    a, b and c are three natural numbers in ascending order
    Given; a<b<c
    ⇒ (c + a) > b & (c – a) > 0
    Now, 
    ⇒(c + a) (c – a) > b
    ⇒ c2 - a2 > b
  • Question 5/10
    1 / -0
    How many numbers exist between 225 and 550 which are completely divisible by all of 4, 5 and 6?
    Solutions

    First number after which is divisible by 60 after 225 is 240 and last number below 550 is 540
    Now by
    An = a + (n − 1)d
    540 = 240 + (n − 1)60
    ⇒ 300 = (n − 1)60
    ⇒ n – 1 = 5
    ⇒ n = 6

  • Question 6/10
    1 / -0
    Traffic light at a road crossing changes in every 25 seconds. On the next crossing the traffic lights change in every 30 seconds. If they both change simultaneously at the same time, then at what time interval they again change together?
    Solutions

    The LCM of 25 and 30 sec = 150 sec = 2 min 30 sec.

  • Question 7/10
    1 / -0
    If each among 63, 87 and 123 is divided by a number the remainders are same in every condition. What is the maximum possible value of the divisor?
    Solutions

    The required number will be GCD of 87 – 63, 123 – 63, and 123 – 87 means 24, 60 and 36.

    GCD of 24, 60 and 36 is 12

  • Question 8/10
    1 / -0
    The sepoy of a regiment have to make a perfect square in the lines of 10, 15 and 20. What will be the lowest number of army men?
    Solutions

    LCM of 10, 15, 20 = 60 = 4 × 15, to make it a perfect square it will have to multiplied with 15.

    So, required number = 60 × 15 = 900

  • Question 9/10
    1 / -0
    Find the least number of 4 digit exactly divisible by 5, 6, 7, 8.
    Solutions

    Smallest number of 4 digit = 1000

    LCM of 5, 6, 7, 8 = 840

    On dividing 1000 by 840 we get 160 as remainder.

    Required number = 1000 + (840 – 160)

    = 1680

  • Question 10/10
    1 / -0
    LCM of two numbers is 340 times their HCF. 1f HCF is 26 and one of the numbers is 442, then the other number is:
    Solutions

    LCM = 340 × HCF


    LCM = 340 × 26


    Now we have


    P × Q = HCF × LCM


    442 × Q = 26 × 340 × 26


    Q = 520

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