LCM Questions

LCM Questions are given here, along with detailed solutions and proper explanations to help out students regarding the concept of LCM. These questions on LCM will help the students to be able to solve the problems efficiently. Learn more about What is LCM?

The LCM or Lowest Common Multiple of two or more numbers is the least among all the common multiples of given numbers. For example, LCM of 2, 4 and 5 is 20, which is the lowest common multiple of 2, 4 and 5, or we can say 20 is the lowest number which 2, 4 and 5 can divide.

LCM Questions with Solutions

1. If HCF(252, 594) = 18, find LCM(252, 594).

Solution: We have LCM of two numbers = (Product of two numbers)/ their HCF

= (252 × 594)/18 = 8316.

Hence, LCM(252, 594) = 8316

2. Player 1 and player 2 are running around a circular field. Player 1 takes 16 minutes to take one round, while Player 2 completes the round in 20 minutes. If both start simultaneously and go in the same direction, after how much time will they meet at the starting point?

Solution: Time taken by the players to meet again = LCM(16, 20)

Now 16 = 2⁴ and 20 = 2² × 5

Therefore, LCM(16, 20) = 2⁴ × 4 = 80

Hence, both will meet at the starting point after 80 minutes.

3. Is it possible to have two numbers whose HCF is 18 and LCM is 540?

Solution: Since HCF always divides LCM, we see that 540 is divisible by 18, 540/18 = 30.

Thus, it is possible to have two numbers.

4. Find the least number divided by 28 and 32, leaving the remainder 8 and 12, respectively.

Solution: Since 28 – 8 = 20 and 32 – 12 = 20

So we need LCM{(28, 32) – 20} = 224 – 20 = 204

Thus, 204 is required number.

5. Find the least number, which, when divided by 35, 56 and 91, leaves the same remainder of 7, respectively.

Solution: Let us find LCM(35, 56, 91) + 7

35 = 5 × 7

56 = 2 × 2 × 2 × 7

91 = 7 × 13

Thus, LCM(35, 56, 91) = 2³ × 5 × 7 × 13 = 3640

The required number = LCM(35, 56, 91) + 7 = 3640 + 7 = 3647.

6. Two alarm clocks ring their alarms at regular intervals of 72 seconds and 50 seconds. If they beep together at noon, at what time will they beep again for the first time?

Solution: We find the LCM of 72 and 50.

Prime factorisation of 72 and 50,

72 = 2 × 2 × 2 × 3 × 3

50 = 2 × 5 × 5

Therefore, the LCM of 72 and 50 = 2³ × 3² × 5² = 1800

1800 seconds = 1800/60 min = 30 min

Hence, the clocks will beep again for the first time at 12:30 pm.

7. There are 56 students in section A and 58 students in section B of a class in a school. Find the minimum number of books required for their class library so that they can be distributed equally among the students of section A or section B.

Solution: Clearly, the number of books that are to be equally distributed should be multiple of 56 and of 58. Thus, we have to find LCM of 56 and 58.

Now, 56 = 2 × 2 × 2 × 7

58 = 2 × 29

LCM (56, 58) = 2³ × 7 × 29 = 1624.

Hence, atleast 1624 books are required in the library.

Also Read:

• Euclid’s Division Lemma
• Real Numbers
• Rational and Irrational Numbers
• Polynomials

8. Find the LCM of 96(x – 1)(x + 1)²(x + 3)³ and 64(x² – 1)(x + 3)(x + 2)².

Solution: Let f(x) = 96(x – 1)(x + 1)²(x + 3)³

And g(x) = 64(x² – 1)(x + 3)(x + 2)²

Now, factorising the polynomials into irreducible factors.

f(x) = 2 × 2 × 2 × 2 × 2 × 3 × (x – 1)(x + 1)²(x + 3)³

g(x) = 2 × 2 × 2 × 2 × 2 × 2 × (x + 1)(x – 1)(x + 3)(x + 2)²

Taking all the factors raised to their highest exponents: 2⁶, 3, (x – 1), (x + 1)², (x + 3)³, (x + 2)²

⇒ The LCM of the given polynomials = 192(x – 1)(x + 1)²(x + 2)²(x + 3)³.

9. Find the LCM of ⅔, ¾ and 7/2.

Solution: LCM of ⅔, ¾ and 7/2 = [LCM of (2, 3, 7)]/[HCF of (3, 4, 2)]

LCM of (2, 3, 7) = 2 × 3 × 7 = 42

HCF of (3, 4, 2) = 1

Therefore, LCM of ⅔, ¾ and 7/2 = 42.

10. Find the LCM of 22.5, 3.5 and 0.55.

Solution: Converting the decimals into integers,

22.5 = 22.5 × 100 = 2250

3.5 = 3.5 × 100 = 350

0.55 × 100 = 55

Now, 2250 = 2 × 3 × 3 × 5 × 5 × 5

350 = 2 × 5 × 5 × 7

55 = 5 × 11

LCM (225, 350, 55) = 2 × 3² × 5³ × 7 × 11 = 173250

Place a decimal point after place from right

Then, LCM(22.5, 3.5, 0.55) = 1732.5.

11. Find the smallest number, which is, when reduced by 7, is divisible by 12, 16, 18, 21 and 28.

Solution: Let x be a number when reduced by 7, is divisible by 12, 16, 18, 21 and 28.

Then, x – 7 = m × 12 ⇒ x = m × 12 + 7.

Thus, when x is divided by 12, it leaves a remainder of 7, which is the same for each given number.

∴ x = LCM (12, 16, 18, 21, 28) + 7 = 1008 + 7 = 1015.

12. Find the largest four-digit number, which is divided by 4, 7 and 13, leaving a remainder of 3, respectively.

Solution: Largest 4-digit number = 9999

LCM (4, 7, 13) = 364

Now 9999 = 27 × 364 + 171

Thus, the 4-digit number divisible by 4, 7 and 13 = 9999 – 171 = 9828.

Since the number we have to find leaves a remainder of three when divided by 4, 7 and 13,

∴ 9828 + 3 = 9831 is the required number.

13. Six bells commence tolling together. After that, they toll at a time interval of 2, 4, 6, 8, 10 and 12 seconds, respectively. In 60 minutes, how many times do they toll together?

Solution: Taking LCM (2, 4, 6, 8, 10, 12) = 120

So, the bells will toll together after each 120 seconds = 2 min

∴ In 60 minutes number of times the bells will toll = 60/2 + 1 = 31 times.

Video Lesson on Application of LCM

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Practice Questions:

1. If the LCM of 12x³y² and 18x^py³ is 36x⁴y³. Find the value of p.

2. The sum of LCM and HCF of the two numbers is 1260. If their LCM is 900 more than their HCF, find the product of two numbers.

3. The LCM and HCF of two numbers are equal; the numbers must be ______.

4. Two bells toll at an interval of 24 minutes and 36 minutes, respectively. If they tolled together at 9 am, after how many minutes do they toll together again, at the earliest?

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