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LCM and HCF Exercise 2

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LCM and HCF 2

  • This online quiz will test your knowledge of LCM and HCF in Quantitative Aptitude.
  • This Online Test is useful for academic and competitive exams.
  • Multiple answer choices are given for each question in this test. You have to choose the best option.
  • After completing the test, you can see your result.
  • There is no negative marking for wrong answers.
  • There is no specified time to complete this test.

The H.C.F. and L.C.M. of two numbers are 50 and 250 respectively. If the first number is divided by 2, the quotient is 50. The second number is:

First number = (50 × 2) = 100.

Second number lcm-and-hcf-q-40314.png

The sum of two numbers is 2000 and their L.C.M. is 21879. The two numbers are:

Let the numbers be x and (2000 – x).

Then, their L.C.M. = x(2000 – x).

So, x(2000 – x) = 21879

⇒ x² – 2000x + 21879 = 0

⇒ (x –1989)(x – 11) = 0

⇒ x = 1989 or 11.

Hence, the numbers are 1989 and 11.

The H.C.F. of two numbers is 8. Which one of the following can never be their L.C.M.?

H.C.F. of two numbers divides their L.C.M. exactly. 8 is not a factor of 60.

The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?

Since H.C.F. is always a factor of L.C.M., we cannot have three numbers with H.C.F. 35 and L.C.M. 120.

Which of the following fractions is the largest?

L.C.M. of 8, 16, 40 and 80 = 80.

lcm-and-hcf-q-40308.png

lcm-and-hcf-q-40302.png

So, 7/8 is the largest.

Which is the least number that must be subtracted from 1856, so that the remainder when divided by 7, 12 and 16 will leave the same remainder 4?

Suppose least number be x

1856 - x = n(LCM of 7, 12, 16) + 4

⇒ 1856 -x = n (336) + 4

we should take n = 5 so that n(336) is nearest to 1856 and n (336)< 1856

1856 - x = 1680 + 4 = 1684

x = 1856 - 1684 = 172

Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.

Required number

= H.C.F of (148 – 4), (246 – 6) and (623 – 11)

= H.C.F of 144, 240 and 612 = 12

Monica, Veronica and Rachael begin to jog around a circular stadium. They complete their revolutions in 48 seconds, 64 seconds and 72 seconds respectively. After how many seconds will they be together at the starting point?

Required time = LCM of 48, 64 and 72

lcm-and-hcf-q-40284.png

LCM = 2 × 2 × 2 × 2 × 3 × 4 × 3 = 576 seconds.

Three girls start jogging from the same point around a circular track and each one completes one round in 24 seconds, 36 seconds and 48 seconds respectively. After how much time will they meet at one point?

Required time

= L.C.M of 24, 36 and 48

= 144 seconds

= 2 minutes 24 seconds

Three friends A, B and C start running around a circular stadium and complete a single round in 24, 36 and 30 seconds respectively. After how many minutes will they meet again at the starting point?

Required time

= L.C.M of 24, 36 and 30

= 360 seconds

= 6 minutes

Swapnil, Aakash and Vinay begin to jog around a circular stadium. They complete their revolutions in 36 seconds, 48 seconds and 42 seconds respectively. After how many seconds will they be together at the starting point?

They will be together at the starting point after the

L.C.M of 36, 48 and 42 = 1008 seconds

The circumferences of the fore and hind-wheels of a carriage are 2⅖ and 3³⁄₇ respectively. A chalk mark is put on the point of contact of each wheel with the ground at any given moment. How far will the carriage have travelled so that its chalk marks may be again on the ground at the same time?

Required distance

= L.C.M of lcm-and-hcf-q-40278.png

lcm-and-hcf-q-40272.png

Hence, carriage will travelled 24m so that its chalk marks may be again on the ground at the same time.

Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again.

L.C.M of 18, 24 & 32 = 288

Hence they would chime after every 288 min.

⇒ 4 hrs 48 min

HCF of 3240, 3600 and a third number is 36 and their LCM is 2⁴ × 3⁵ × 5² × 7². The third number is:

Let the third number be x.

Product of numbers = LCM × HCF

Therefore,

lcm-and-hcf-q-40266.png

⇒ lcm-and-hcf-q-40260.png

⇒ lcm-and-hcf-q-40254.png

⇒ lcm-and-hcf-q-40248.png

= lcm-and-hcf-q-40242.png

Now check your Result..

Your score is

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