Solved Examples

Problems on H.C.F and L.C.M

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Problems on H.C.F and L.C.M

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Solved Examples

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Solved Examples – Problems on H.C.F and L.C.M

Solved examples help students understand the practical application of H.C.F and L.C.M concepts in competitive examinations. These examples are designed from basic to advanced level and cover important aptitude questions frequently asked in SSC, Banking, Railway, CDS, NDA, CAT, UPSC, and placement exams.

Topics Covered in Solved Examples

  • Factors and Multiples
  • HCF Calculation
  • LCM Calculation
  • Prime Factorization Method
  • Division Method
  • HCF and LCM of Fractions
  • Repeated Event Problems
  • Remainder-Based Questions
  • Circular Track Problems
  • Greatest and Smallest Number Problems

Example 1: Factors and Multiples

Question: Find the factors and first five multiples of 12.

Solution:

Factors of 12:

1, 2, 3, 4, 6, 12

First five multiples of 12:

12, 24, 36, 48, 60

Example 2: Finding H.C.F Using Factors

Question: Find the HCF of 18 and 24.

Solution:

Factors of 18:

1, 2, 3, 6, 9, 18

Factors of 24:

1, 2, 3, 4, 6, 8, 12, 24

Highest common factor = 6

Example 3: Finding L.C.M Using Multiples

Question: Find the LCM of 4 and 6.

Solution:

Multiples of 4:

4, 8, 12, 16...

Multiples of 6:

6, 12, 18...

Least common multiple = 12

Example 4: Prime Factorization Method

Question: Find the HCF and LCM of 15, 25, and 27.

Solution:

15 = 3 × 5

25 = 5²

27 = 3³

No common factor exists in all three numbers.

Therefore:

HCF = 1

LCM = 3³ × 5²

= 27 × 25

= 675

Example 5: Product Formula

Question: Two numbers have HCF = 6 and LCM = 180. Find the product of the numbers.

Solution:

Using formula:

Product of numbers = HCF × LCM

= 6 × 180

= 1080

Example 6: Ratio-Based HCF Problem

Question: The ratio of two numbers is 3 : 4 and their HCF is 4. Find their LCM.

Solution:

Let the numbers be:

3x and 4x

Since HCF = x

x = 4

Numbers:

12 and 16

LCM of 12 and 16:

= 48

Example 7: Greatest 4-Digit Number Divisible by Given Numbers

Question: Find the greatest 4-digit number divisible by 15, 25, 40, and 75.

Solution:

LCM(15, 25, 40, 75)

= 600

Largest 4-digit number = 9999

9999 ÷ 600 leaves remainder 399

Required number:

9999 − 399

= 9600

Example 8: Circular Track Problem

Question: A, B, and C complete one round of a stadium in 252 sec, 308 sec, and 198 sec respectively. After how much time will they meet again at the starting point?

Solution:

Required time = LCM(252, 308, 198)

= 2772 seconds

Convert into minutes:

2772 ÷ 60

= 46 min 12 sec

Therefore:

Required time = 46 min 12 sec

Example 9: Bells Ringing Together

Question: Six bells toll at intervals of 2, 4, 6, 8, 10, and 12 seconds. How many times will they toll together in 30 minutes?

Solution:

LCM(2, 4, 6, 8, 10, 12)

= 120 seconds

= 2 minutes

In 30 minutes:

30 ÷ 2 = 15

Including starting time:

15 + 1 = 16 times

Example 10: Same Remainder Problem

Question: Find the greatest number that divides 183, 91, and 43 leaving the same remainder in each case.

Solution:

Differences:

183 − 91 = 92

91 − 43 = 48

183 − 43 = 140

Required number:

HCF(92, 48, 140)

= 4

Example 11: Same Remainder and Divisibility

Question: Find the smallest number which leaves remainder 3 when divided by 5, 6, 7, and 8, but is divisible by 9.

Solution:

LCM(5, 6, 7, 8)

= 840

Required number:

= 840t + 3

Try smallest value of t such that number becomes divisible by 9.

For t = 2:

840 × 2 + 3

= 1683

1683 is divisible by 9.

Therefore:

Required number = 1683

Example 12: HCF of Fractions

Question: Find the HCF of 2/3 and 4/9.

Solution:

Using formula:

HCF of fractions =

HCF of numerators / LCM of denominators

HCF(2, 4) = 2

LCM(3, 9) = 9

Therefore:

HCF = 2/9

Example 13: LCM of Fractions

Question: Find the LCM of 2/3 and 4/9.

Solution:

Using formula:

LCM of fractions =

LCM of numerators / HCF of denominators

LCM(2, 4) = 4

HCF(3, 9) = 3

Therefore:

LCM = 4/3

Example 14: HCF Using Division Method

Question: Find the HCF of 96 and 404 using division method.

Solution:

404 ÷ 96 = 4 remainder 20

96 ÷ 20 = 4 remainder 16

20 ÷ 16 = 1 remainder 4

16 ÷ 4 = 0 remainder

Therefore:

HCF = 4

Example 15: Repeated Event Problem

Question: Three traffic lights change after every 30 sec, 45 sec, and 60 sec. If they change together now, after how much time will they change together again?

Solution:

Required time:

LCM(30, 45, 60)

= 180 seconds

= 3 minutes

Therefore:

They will change together again after 3 minutes.

Important Exam Tips

  • Use prime factorization for faster calculations.
  • Use HCF for greatest divisor problems.
  • Use LCM for repeated event problems.
  • Memorize divisibility rules properly.
  • Practice remainder-based questions regularly.
  • Use product relation for two-number problems.
  • Always check whether HCF or LCM is actually required.

Practicing solved examples regularly improves speed, conceptual clarity, and accuracy in solving H.C.F and L.C.M aptitude questions in competitive examinations.

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