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Application problems on HCF and LCM

Application problems on HCF and LCM. Co-relation between HCF and LCM. HCF and LCM relationship states that the product of two numbers is equal to the product of their LCM and HCF. 1st number x 2nd number = LCM x HCF. Proof : Consider the numbers 8 and 12.

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Application problems on HCF and LCM

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  1. Application problems on HCF and LCM

  2. Co-relation between HCF and LCM. HCF and LCM relationship states that the product of two numbers is equal to the product of their LCM and HCF. 1st number x 2nd number = LCM x HCF

  3. Proof : Consider the numbers 8 and 12. Let us find the L.C.M And H.C.F of 8 and 12. To do that we need to first find prime factorization of 8 and 12 Prime factorization is 8 = 2 x 2 x 2 Prime factorization is 12 = 2 x 2 x 3 HCF = 2 x 2 = 4 LCM = 2 x 2 x 2 x 3 = 24

  4. 1st number x 2nd number = LCM x HCF 8 x 12 24 x 4 = 96 96 ∴ It is proved product of two number is product of its HCF and LCM

  5. We know that 1st number x 2nd number = LCM x HCF • On the basis of this relation, we find that: • 1st number = (LCM x HCF) / 2nd number • 2nd number = (LCM x HCF) / 1st number • LCM = (1st number x 2nd number) / HCF • HCF = (1st number x 2nd number) / LCM

  6. Example 1: H.C.F and L.C.M of two numbers are 18 and 1782 respectively. One number is 162, find the other. Solution:- We know, H.C.F. × L.C.M. = First number × Second number then we get, 18 × 1782 = 162 × Second number Second number = = 198 Therefore, the second number = 198

  7. Applications of LCM and HCF Now there are various real-life applications of LCM and HCF. The best way to understand these and grasp the concept of LCM and HCF is to learn via examples. So let us take a look at a few examples which will help you understand LCM and HCF. • H.C.F. : We can use the H.C.F. method where we need: • To split things into smaller sections? • To equally distribute 2 or more sets of items into their largest grouping? • To figure out how many people we can invite? • To arrange something into rows or groups? • L.C.M. : we use find L.C.M. where , • About an event that is or will be repeating over and over. • To purchase or get multiple items in order to have enough. • To figure out when something will happen again at the same time

  8. Example2 :Maya has two pieces of cloth. One piece is 36 inches wide and the other piece is 24 inches wide. She wants to cut both pieces into strips of equal width that are as wide as possible. How wide should she cut the strips? Solution:- This problem can be solved using H.C.F. because we are cutting or “dividing” the strips of cloth into smaller pieces and also we are looking for the widest possible strips. 36 = 2 x 2 x 3 x 3 24 = 2 x 2 x 2 x 3 H.C.F. of 36 and 24 is 2 x 2 x 3 = 12 Maya should cut each piece to be 12 inches wide

  9. Example3:A, B and C start to jog around a circular stadium. They complete their rounds in 36 seconds, 48 seconds and 42 seconds respectively. After how many seconds will they be together at the starting point? Solution:- The required time is the LCM of all their lap times. This is the earliest when all three will intersect at the same point. Required time is the LCM (36,48,42) LCM (36,48,42) = 2 x 3 x 2 x 2 x 3 x 7 x 2 = 1008 Therefore the required time is 1008 seconds

  10. Example4:Find the largest number of 4-digits divisible by 12, 15 and 18. Solution:- Largest 4-digit number is 9999. Remember: The question may be asked in a tricky way. Here, largest number does not mean H.C.F.. We have to find a number which is divisible by 12, 15 and 18 Required largest number must be divisible by the L.C.M. of 12, 15 and 18 L.C.M. of 12, 15 and 18 L.C.M. 3 x 2 x 2 x 5 x 3 = 180 Now divide 9999 by 180, we get remainder as 99 The required largest number = (9999 – 99) =9900 Number 9900 is exactly divisible by 180.

  11. Example5:Find the least number, which when divided by 12, 15, 20 and 54 leaves a remainder of 8 in each case. Solution:- We are given that, the least number, when divided by 12, 15, 20 and 54 leaves a remainder of 8 in each case. Therefore, add remainder 8 to the L.C.M. of divisors. The required least number = (L.C.M. of 12, 15, 20 and 54) + remainder (8) L.C.M. of 12, 15, 20 and 54 L.C.M. = 2 × 3 × 2 × 3 × 1 x 5 x 5 x 3 = 540 The required least number = (540) + (8) = 548 Least number = 548

  12. Example6:What is the greatest number that will divide 62, 78 and 109 leaving remainders 2,3 and 4 respectively? Solution:- Step 1: Subtract the remainders from their respective divisor. 62-2 = 60 78-3 = 75 109-4 = 105 Step 2: Prime factorize 60, 75 and 105 60 = 2 x 2 x 3 x 5 75 = 3 x 5 x 5 105 = 3 x 5 x 7 Step 3: Identify the greatest number that divides all three numbers. Here it is 3 x 5 = 15 i.e., the HCF of 60, 75 and 105 is 15. Ans: 15 is the greatest number that will divide 62, 78 and 109 leaving 2, 3 and 4 as remainders respectively.

  13. Try these Find the least number which is exactly divisible by 12, 15, and 20. Find the least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18? The HCF and LCM of two numbers are 25 and 1750 respectively, if one of the number is 175, find the other.

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