Hello Friends, Under " HCF and LCM : Types of Questions Asked in Competitive Exams ", we will discuss various varieties of question asked under section HCF and LCM in various competitive exams.

Type 1 : Find the HCF of Composite Number

Qusetion 1 : Find the HCF of 224, 448 and 550

Solution : By Prime Factorisation Method,
224 = 2 x 2 x 2 x 2 x 2 x 7
448 = 2 x 2 x 2 x 2 x 2 x 2 x 7
550 = 2 x 5 x 5 x 11
So, HCF of 224, 448 and 550 is 2

Type 2 : Find the LCM of Composite Number

Question 2 : Find the LCM of 25, 55 and 75

Solution : By Prime Factorisation Method,
25 = 5 x 5
55 = 5 x 11
75 = 3 x 5 x 5
So, LCM of 25, 55 and 75 = 5 x 5 x 11 = 275

Type 3 : Find the Least Number which is exactly divisible by given numbers.

Question 3 : Find the least number which is exactly divisible by 25, 55 and 75

Note : Same question as Type 2 only it is asked in a tricky way

Type 4 : Find the HCF and LCM of Fractions and Decimals

Question 4 : Find L.C.M. of 1.05 and 2.1

Solution : First, take LCM of 105 and 210 i.e 210
Now, write in decimal i.e 2.1 hence, LCM of 1.05 and 2.1 is 2.10

Type 5 : Find the numbers if their HCF and ratio are given.

Question 5 : H.C.F. of two numbers is 11. If these two numbers are in the ratio of 12 : 11, then find the numbers.

Solution : The required two numbers are 12 x 11 = 132 and 11 x 11 =121

Type 6 : Find the other number if one number and their HCF and LCM is given

Question 6 : The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then find the other number

Solution : Product of two numbers = Product of their H.C.F. and L.C.M.
So, 275 x Y = 11 x 7700
=> Y = 308

Type 7 : Find the least number, which when divided by another number leaves a remainder

Question 7 : Find the least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8

Solution : Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8 = 548