LCM stands for Lowest Common Multiple, and HCF stands for Highest Common Factor.
The key to telling the difference between these two things is knowing the difference between a multiple and a factor.
A multiple of an integer (whole number) is any integer that appears in its times table. For example, the multiples of 3 are 3, 6, 9, 12, and so on.
A factor of an integer is any integer that divides the integer with no remainder. For example, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
We use LCM and HCF to compare two (or more) integers.
The following diagram shows how to find the HCF and LCM of 24 and 36 using Repeated Division.
Question / Answer
Question #1 | LCM of two prime numbers x and y, (x > y), is 161. The value of 3y – x is
[Mathematics] [CTET-2014-09]
|
|||||||||||||||||||||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | As these are two prime numbers so HCF is 1
And as (x) * (y) = HCF * LCM x * y = 161 Now as we know these are prime numbers so check the divisors of 161 It is clearly not divisible by 2, 3 & 5 But 7 divides it 161/7 = 23 So two no.s are 7 & 23 As x > y so x =23 & y = 7 so 3y – x = 3*7 – 23 = 21 – 23 = -2 |
|||||||||||||||||||||||||||||||||||
Question #2 | HCF of two numbers is 28 and their LCM is 336. If one number is 112, then the other number is
[Mathematics] [CTET-2013-07]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | Using the formula HCF X LCM = number1 x number2 => 28×336 = 112 x number2 => number2 = 28×336/112 = 28×3 = 84 |
|||||||||||||||||||||||||||||||||||
Question #3 | LCM of 22, 54, 135 and 198 is
[Mathematics] [CTET-2015-02]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation |
|
|||||||||||||||||||||||||||||||||||
Question #4 | The HCF and LCM of two numbers are 12 and 72 respectively. If the sum of these numbers is 60,then one of the numbers will be
[Mathematics] [UPTET-2017-10]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | Let two numbers be a & bAs product of two numbers = HCF x LCM
So, a x b = 12 x 72 And as a + b = 60 So, a x (60-a) = 12 x 72 ==> (60a) – (a^2) = 12 x 72 Breaking 60a into 24 & 36, and 72 into 2 & 36 So a can have value either 24 or 36 |
|||||||||||||||||||||||||||||||||||
Question #5 | The HCF and LCM of two number are 13 and 1989 respectively. if one of the number is 117 , then other number is
[Mathematics] [UPTET-2017-10]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | Let the two numbers are a and b
a=117, b=? product of the given numbers = lcm * hcf a*b =lcm *hcf |
|||||||||||||||||||||||||||||||||||
Question #6 | The ratio of two numbers is 5:6 and their HCF is 12. Their LCM is
[Mathematics] [UPTET-2016-12]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | LCM = (The part of first no. in ratio) x (The part of second no. in ratio) x HCFSo, LCM = 5 x 6 x 12 = 360 | |||||||||||||||||||||||||||||||||||
Question #7 | What is the LCM of two consecutive odd numbers?
[Mathematics] [OTET-2018-12]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Question #8 | Three numbers are in the ratio 1:2:3. If their HCF is 13, then the LCM of the three numbers is
[Mathematics] [SCERT-Odisha-BEd-Science-09Aug18-Batch3]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | LCM = (The part of first no. in ratio) x (The part of second no. in ratio) x (The part of third no. in ratio) x HCF
LCM = 1 x 2 x 3 x 13 = 78 |
|||||||||||||||||||||||||||||||||||
Question #9 | Three numbers are in the ratio 1:2:3. If their HCF is 13. then the LCM of the three numbers is
[] [Mathematics] [SCERT-Odisha-BEd-Science-08Aug18-Batch3]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | LCM = (The part of first no. in ratio) x (The part of second no. in ratio) x (The part of third no. in ratio) x HCF
LCM = 1 x 2 x 3 x 13 = 78 |
|||||||||||||||||||||||||||||||||||
Question #10 | If two positive integers a, b expressed as a = pq2 and b = p3q where p, q are prime numbers, then LCM ( a, b) is
[Mathematics] [SCERT-Odisha-DElEd-Odia-09Aug18-Batch3]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | As p & q are prime numbersDivisors Number 1 Number 2 p pq2 p3q p q2 p2q p q2 pq q q2 q q q 1 1 1 Multiply all divisors = p * p * p * q * q = p3q2 |
|||||||||||||||||||||||||||||||||||
Question #11 | The LCM of 90 and 144 is
[Mathematics] [SCERT-Odisha-DElEd-Odia-14Aug18-Batch4]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | Divisor Number 1 Number 2 2 90 144 2 45 72 2 45 36 2 45 18 3 45 9 3 15 3 5 5 1 1 1 So, LCM = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720 |
|||||||||||||||||||||||||||||||||||
Question #12 | The product of LCM and HCF of two numbers is 7605. If one number is 117 then other number is
[Mathematics] [SCERT-Odisha-DElEd-Odia-13Aug18-Batch1]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | As Number 1 x Number 2 = HCF x LCM
So, 117 x Number 2 = 7605 Number 2 = 7605/117 = 65 |
|||||||||||||||||||||||||||||||||||
Question #13 | If a = 8×3, b = 2×3×5, c = 3n×5 and LCM (abc) is 8×32×5 then what is the value of ‘n’?
[Mathematics] [SCERT-Odisha-DElEd-Odia-13Aug18-Batch2]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | LCM would be formed by taking highest power of each factor from each no.
As, Highest power of 2 = 3 So, LCM = 23 x 3n×5 Comparing to LCM = 8 x 32 x 5 = 23 x 32 x 5 So, |
|||||||||||||||||||||||||||||||||||
Question #14 | Two numbers are in the ratio 3:4. The product of their HCF and LCM is 2028. The sum of the numbers will be
[Mathematics] [UPTET-2018-11]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | LCM = (The part of first no. in ratio) x (The part of second no. in ratio) x HCF = 3 x 4 x HCF
Product of two no.s = HCF x LCM = 2028 So, HCF2 = 2028/12 = 169 HCF = 13 Number 1 = 3 x HCF = 3 x 13 = 39 Sum of Numbers = 39 + 52 = 91 |
|||||||||||||||||||||||||||||||||||
Question #15 | The HCF of 14, 21 and another number is 7 and their LCM is 210. Then what is the third number?
[Mathematics] [OTET-2017-09]
|
|||||||||||||||||||||||||||||||||||
Options |
|
|||||||||||||||||||||||||||||||||||
Answer Explanation | Let third number be yIf we divide all numbers and their LCM by HCF then ==> 14/7, 21/7, y/7 & 210/7
Now, (14/7) x (21/7) x (y/7) = (210/7) ==> 2 x 3 x (y/7) = 30 ==> (y/7) = 5 ==> y = 35 |