HCF and LCM word Problems, tricks & Solved examples

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  • The product of the two numbers is always equal to the product of their HCF and LCM.
  • In case of HCF, if some remainders are given, then firstly those remainders are subtracted from the numbers given and then their HCF is calculated.
  • In case of LCM, if a single remainder is given, then firstly the LCM is calculated and then that single reminder is added in that.
  • In case of LCM, if for different numbers different remainders are given, then the difference between the number and its respective remainder will be equal. In that case, firstly the LCM is calculated, then that common difference between the number and its respective remainder is subtracted from that.
  • Sometimes in case of HCF questions, the required remainder is given and when the remainder is not given, in those cases you will generally have three numbers given. For answering the question, you need to take the difference of the three pairs of numbers, now the HCF of these differences will become the answer e.g. if you have to find the greatest number, which when divides 83, 93 and 113 and leaves the same remainder. Here you will take the three differences i.e. 93 – 83 = 10 ; 113 – 93 = 20 ; 113 – 83 = 30, after that find the HCF of these differences, which comes out to be 10. Now you can check for yourself- when 10 divides these three numbers, the reminder obtained is 3 in each case and that is what the question was asking for.
  • Whenever the question talks about the greatest or maximum, then in most of these cases it will be a question of HCF. Secondly, whenever the question is related to classification or distribution into groups, then in all the cases it is HCF only.
  • Whenever the question talks about the smallest or minimum, then in most of the cases it will be a question of LCM. Secondly, whenever the word ‘together’ or ‘simultaneous’ is used in the question, then in all the cases it is LCM.
  • Before solving the problems on HCF and LCM in the real exam, you must practice some HCF and LCM worksheets.
HCF and LCM Questions (Solved)
 
Example 1: Find the greatest number which when divides 259 and 465 leaves remainders 4 and 6 respectively.
Sol:  Here, the numbers 259 and 465 leave the remainders 4 and 6 respectively. So, the required number will be obtained by finding the H.C.F. of 259 – 4 = 255 and 465 – 6 = 459. The HCF of 255 and 459 is 51 which is the number.
Example 2:  Find the least number which when divided by 6, 14, 18 and 22 leaves remainder 4 in each case.
Sol: The LCM of 6, 14, 18 and 22 is 1386. In order to get remainder 4 in each case, we will add 4 to the LCM. So, the number is 1386 + 4 = 1390.
Example 3: Find the least number which when divided by 8, 12, 20 and 36 leaves remainders 6, 10, 18 and 34 respectively.
Sol: Here, the numbers are 8, 12, 20 and 36 and the respective remainders are 6, 10, 18 and 34. The difference between numbers and the respective remainders is equal to 2. So, first of all find the LCM of 8, 12, 20 and 36 which is 360. The required number is 360 – 2 = 358.
Example 4: Find the greatest number, which when divides 41, 71 and 91, leaves the same remainder in each case.
Sol: Take the difference between all the three pairs of numbers and their HCF will be the answer i.e. 91 – 41 = 50, 71 – 41 = 30 and 91 – 71 = 20. Now, the HCF of 50, 30 and 20 is 10.
Sometimes in such questions, the common remainder can also be asked. You can divide any of the numbers given by HCF (41 ¸ 10) and find the remainder to be equal to 1.
Example 5: Find the L.C.M of 2/5, 3/10 and 6/25
L.C.M. of 2/5, 3/10 and 6/25 = L.C.M of 2,3 and 6 / H.C.F of 5,10 and 25
L.C.M. of 2, 3 and 6 = 6 ; H.C.F. of 5, 10 and 25 = 5. Thus, the LCM of these three fractions will be 6/5.
Example 6: How often will five bells toll together in one hour if they start together and toll at intervals of 5, 6, 8, 12, 20 seconds, respectively?
Sol: The time after which the bells will ring together is the L.C.M. of 5, 6, 8, 12 and 20 seconds, i.e. 120 seconds. The number of times they will toll together in one hour = (3600 ¸ 120) = 30. Thus, they will toll together 30 times in an hour.
⇒ Sometimes the question is how many times they toll together in the first hour, in that case after finding the answer like above, you need to add 1 for the start together as well i.e. in the first hour it is 1 more than the usual number of times. 
Example 7: Find the greatest number of four digits which when divided by 11, 21, 15, and 28 leaves 5, 15, 9 and 22 as remainders, respectively?
Sol:  The LCM of 11, 21, 15 and 28 is 4620. The greatest 4 digit multiple of 4620 is 9240. Now the difference between the number and the respective remainder is same here. So to find the greatest 4 digit number which satisfies the given condition, subtract the common difference from 9240. Hence the required number is 9240 – 6 = 9234.
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