Tricks to solve questions on LCM and HCF in SSC CGL Exam

Questions based on the concept of HCF and LCM always appear in SSC CGL Exam. Also, these two concepts make calculations of other complex problems easier. So, here are some tips which aspirants can use while solving questions on LCM and HCF.

Basics of LCM:

LCM of two are number is basically a lowest possible number which is a multiple of given numbers.

For example: Multiples of 12 are 12, 24,36, …. & Multiples of 8 are 8,16, 24,….

Now the least common multiple of both numbers is 24. Therefore 24 is an LCM of 12 and 8

Shortcuts which come in handy while Solving LCM

  1. a) Let’s try to find LCM of 108, 125 and 225

Sol: Step 1: Write down the prime factorization of all numbers

108 =

125 =

225 =

Step 2: Check the highest power of each factor. In above example factors are 2, 3 and 5

Highest power of 2 = 2 (in 108)

The highest power of 3= 3 (in 108)

Highest power of 5 = 3 (in 125)

Therefore,LCM of 108 ,125 and 225 = 13500

(b) LCM of prime numbers: If a and b are two prime numbers then their LCM will be equal to the Product of a and b.

For example: The LCM of 13, 3 and 2 = 13×3×2 = 78

(c)LCM of Co – prime* numbers: If a and b are two co-prime numbers then their LCM will be equal to the product of a and b.

For example: The LCM of 4 and 9 = 4×9 = 36

*Co-prime numbers: Numbers which do not have any common factor other than 1 are called Co-Prime numbers.

(d) LCM of two same numbers: If there are two or more same numbers then their LCM will be the number itself.

For example: LCM of 10 and 10 is 10.

(e) LCM of 1 and number: LCM of 1 and any other number will be number itself.

For example LCM of 1 and 10 is 10.

(f) LCM of a fraction:

would be

For example LCM of

(g) LCM of two numbers in which one is the factor of another is the larger number.

For example: LCM of 4 and 28 is 28 as 4 is the factor of 28.

Basics of HCF

HCF of two or more numbers is basically the highest number which completely divides the given numbers.

For example: Factors of 9 are 1, 3 &9 and Factors of 6 are 1, 2,3 &6.Since, the highest common factor is 3.

Therefore, HCF of 9 and 6 is 3.

Shortcuts which come in handy while Solving HCF

(a) Let’s try to find HCF of 125 and 225

 Step 1: Write down the prime factorisation of all numbers

125 =

225 =

Step 2: Checkwhich factor is common in both numbers. Here it is 5.

Step 3:Find the lowest power of 5 which is 2.

Therefore,HCFof 125 and 225 =

(b) HCF of prime numbers: If a and b are two prime numbers then their HCF will be 1.

For example: The HCF of 13, 3 and 2 = 1

(c) HCF of Co – primenumbers: If a and bare two co prime numbers then their HCF will be equal to 1

For example: The LCM of 4 and 9 = 1

(d) HCF of two same numbers: If there are two or more same numbers then their HCFwill be number itself.

(e) HCF of 1 and number: HCF of 1 and any other number will be 1

(f) HCF of fraction:

would be

For example: HCF of

(g) HCF of two numbers in which one is the factor of another is the smallest number.

For example: HCF of 4 and 28 is 4as 4 is the factor of 28.

(h) The HCF of two numbers is generally the difference between them or the factors of difference.

For example:

  1. HCF of 36 and 72 is 36 which also the difference between 72 and 36.
  2. HCF of 36 and 108 is 36 which is the factor of difference between 108 and 36.

Important Formulae and Results based on Result of HCF and LCM

  • Product of Numbers = HCF × LCM of those numbers.
  • If H is the HCF of two numbers and L is the LCM of these numbers, then there must exist pairs of co-prime numbers say a and b such that:

    abH = L, where aH and bHare possiblepairs ofnumbers whose HCF and LCM has been given.

    For example:

    Let HCF of two numbers be 6 and LCM be 36

    Therefore ,6ab = 36, ab = 6

    Now co-primepairs of a and b are: 2 & 3and 1 & 6.

    Therefore, Possible pairs of numbers whose HCF and LCM has been given are 12 & 18 and 6&36.

  • Least number which when divided by a, b and c and leaves same remainder k in each case = (LCM of a, b and c)+ K.
  • Largest number which divides a,b,c to leave remainder.same remainder k in each case = HCF(a-k,b-k,c-k).
  • Greatest number which divides a,b,c toleave same remainder = (HCF(b-a,c-b,c-a).
  • Largest number which divides a, b, c to leave same remainder p,q,r = H.C.F (a-p,b-q,c-r).
  • Least number which when divided by a, b and c leaves remainders p, q & r respectively. In this case follow these steps:

    Let a-p = b-q = c-r = k

    Now find (LCM of a, b, c) -k to solve the given problem.

Previous year questions asked in SSC CGLbased on HCF and LCM

Q-1If product of two co-prime numbers is 117,then find their L.C.M.

SolWe Know that HCF of co-prime numbers is 1

Also, HCF × LCM = Product of numbers

Therefore LCM =

Q2Find the greatest number which divides 110 and 128 leaving a remainder 2 in each case.

SolHCF of 110-2 and 128-2

This implies HCF of 108 and 126 = 18,which is required answer.

Q3Find the smallest number which when divided by 15,20 or 35, each time the remainder is 8.

Sol LCM of 15, 20 and 35 is 420.

Therefore, number would be 420+ 8 = 428

Q4 Find the smallest number, which when divided by 12 and 16 leaves remainder 5 and 9 respectively.

Sol We have 12-5 = 16-9 =7.

Also, LCM of 12 and 16 = 48.

Therefore, required number = 48-7 = 41

Q5 Four bells ring at intervals of 4,6, 8 and 14 seconds start ringing simultaneously at 12.00 P.M. At what time will they again simultaneously?

Sol In such questions we have to find least common multiple.

Now LCM of 4, 6,8 and 14 = 168 = 2min and 48 sec.

Therefore, bells will ring again simultaneously at 12:00 hrs, 2 min and 48 sec.

Q6Find the maximum number of students among whom 1001 chocolates and 910 toys can be distributed in such a way that each student gets same number of chocolates and toys.

Sol In such questions, we have to determine HCF.

Now HCF of 1001 and 910 = 91.

Therefore, required answer is 91.

Q7The LCM of two numbers is 48.The numbers are in the ratio 2 : 3.Determine the sum of numbers.

SolLet the HCF be x.

Therefore, first number be 2x and another be 3x

Now 6x = 48, or x=8

Sum of numbers = 5x=5×8= 40.

Q8The sum of two numbers is 84 and their HCF is 12. Find the total number of such pairs of number.

Sol: Let a and b be two coprime numbers such that

12a + 12b = 84

Or a+b= 7

Now possible pairs of such numbers are (1,6);(2,5),(3,4)

Therefore, there are only 3 possible number of such pairs.

Q9The LCM of two multiples of 12 is 1056. If one of the number is132. Determine the other number.

Sol Let number be 12a and 12b where a and b are coprime. Also say 12a = 132

Therefore 12ab = 1056

b =

or, 12b = 96, which is required answer.

Q10 Let a be the smallest number, which when added to 2000 makes the resulting number di-

visible by 12, 16, 18 and 21. Find the sum of the digits of a.

Sol LCM of 12 ,16 ,18 and 21 = 1008

Since it is less than 2000. Therefore multiply 1008 by 2, as it will also be divisible by given numbers.

1008×2 = 2020

Therefore, number added = 2020-2000= 16

Now sum of digits of 16= 1+6= 7,which is required answer.

Always remember that Practice is the Key to Success. So, keep practicing to score better in the upcoming SSC CGL exam!

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