H.C.F and L.C.M

Subject: Compulsory Maths

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Overview

The highest number that divides exactly into two or more numbers is called highest common factor(HCF) and lowest Common multiples are multiples that two numbers have in common. These can be useful when working with fractions and ratios.

H.C.F and L.C.M

Highest Common Factor (HCF)

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The highest common factor (HCF) of the algebraic expression is the largest number that divides evenly into both numbers. It can be said as largest of all common factors.

For example, HCF of 6x3y2 and 10x5y4 is 2x3y2 since

HCF of 6 and 10 is 2

HCF of x3 and x5 is x3

and HCF of y2 and y4 is y2

To find the HCF of compound expressions, first of all, resolve each expression into factors and then find HCF.

Example:

Find the HCF of 3x2- 6x and x2+ x - 6

Solution:

1st expression = 3x2- 6x

= 3x(x - 2)

2nd expression = x2+ x - 6

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= x2+ 3x - 2x - 6

= x(x + 3) - 2(x + 3)

= (x + 3)(x - 2)

∴ = x - 2

Lowest Common Multiple (LCM)

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The lowest common multiple(LCM) is found by multiplying all the factors which appear on either list. LCM of any number is the smallest whole number which is multiple of both.

For example, LCM of 6x3yand 10x5yis 30 x5ysince

LCM of 6 and 10 is 30, LCM of xand xand LCM of yand yis y4.

To find the LCM of compound expressions, proceed as in the case of HCF and then find LCM.

Example

Find the LCM of 3x2- 6x

1st expression = 3x2- 6x

= 3x(x - 2)

2nd expression = x2+ x - 6

= x2+ 3x - 2x - 6

= x(x + 3) - 2(x + 3)

= (x + 3)(x - 2)

LCM = 3x(x - 2)(x + 3)

Things to remember
  • H.C.F is the largest number that divides every into both numbers.
  • H.C.F is useful when simplifying fraction.
  • L.C.M is the smallest number that is a common multiple of two or more numbers.
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Questions and Answers

Solution:

4x2y and xy2

Here, first expression = 4x2y = 4 × x × x × y

The second expression = xy2= x × y × y

Taking common of both expression = xy.

∴ H.C.F. = xy

Solution:

Here first expression = 9x2y3= 3 × 3 × x × x × y × y × y

Second expression = 15xy2= 3 × 5 × x × y × y

Taking common from both expression

= 3 × x × y × y

∴ H.C.F. = 3xy2

Solution:

Here, first expression = a2bc = a × a × b × c

Second expression=b2ac= b × b × a × c

Third expression= b2ac = b × b ×a× c

Taking common of the three expression=a × b × c

∴ H.C.F = abc

Solution:

Here given x2-4 and 3x+6

First expression = x2-4 = x2-22= (x-2)(x+2)

Second expression = 3x+6 = 3(x+2)

∴ H.C.F = x+2

Solution:

Given, x2-y2 and xy - y2

First expression = x2-y2= (x+y) (x-y)

Second expression = xy - y2= y(x-y)

Taking common from both expression = x-y

∴ H.C.F = x-y

Solution:

Here given,3x2-6x and x2+x-6

1st expression= 3x2-6x

= 3x(x-2)

2nd expression= x2+x-6

=x2+3x-2x-6

=x(x+3)-2(x+3)

=(x+3)(x-2)

∴H.C.F = x-2

Solution:

Here given,3a+b and 15a +5 b

1st expression=3a+b

2nd expression=15a+5b=5(3a+b)

Taking common from both expression =3a+b

∴H.C.F= 3a+b

Solution:

Here given,a3+ 6a2- 4 a-24; a2+5a+6 and a2-4

1st expression=a3+ 6a2- 4 a-24

=a3-4a+6a2-24

=a(a2-4)+6(a2-4)

=(a2-4)(a+6)

=(a-2)(a+2)(a+6)

2nd expression= a2+5a+6

=a2+(2+3)a+6

=a2+2a+3a+6

=a(a+2)+3(a+2)

=(a+2)(a+3)

3rd expression=a2-4

=(a-2)(a+2)

Taking common from three expression=a+2

∴H.C.F= a+2

Solution:

Here given,(x-a), x2-a2 and x2-2ax+a2

1st expression = x-a

2nd expression = x2-a2= (x-a)(x+a)

3rd expression = x2-2ax+a2= (x-a)2= (x-a)(x-a)

Taking common from three expression = x-a

∴ H.C.F = x-a

Solution:

Here given, a2-3a+2; 3a2-2a-8 and 2a2-9a+10

1st expression = a2-3a+2

= a2- 3a + 2

= a2- (1+2) a+2

= a2-a-2a+2

= a(a-1)-2(a-1)

= (a-1)(a-2)

2nd expression=3a2-2a-8

= 3a2-(6-4)a-8

= 3a2-6a + 4a-8

= 3a(a-2)+4(a-2)

= (a-2)(3a+4)

3rd expression = 2a2-9a+10

= 2a2-(4+5)a+10

= 2a2-4a-5a+10

= 2a(a-2)-5(a-2)

= (a-2)(2a-5)

Taking common of three expression = a+2

∴ HCF = a-2

Solution:

Here given, 3x2-6x and x2+x-6

1st expression = 3x2-6x

= 3x(x-2)

2nd expression = x2 + x-6

= x2+ 3x - 2x-6

= x(x+3)-2(x+3)

= (x+3)(x-2)

\(\therefore\) LCM = 3x(x-2)(x+3)

Solution:

Here given, 2x and 4

1st expression = 2x = 2 × x

2nd expression = 4 = 2×2

LCM= 2×2× x = 4x

 

 

Solution:

First expression = a2 – 12a + 35

= a2 – 7a – 5a + 35

= a(a – 7) – 5(a – 7)

= (a – 7) (a – 5)

Second expression = a2 – 8a + 7

= a2 – 7a – a + 7

= a(a – 7) – 1(a – 7)

= (a – 7) (a – 1)

Therefore, the H.C.F. = (a – 7) and L.C.M. = (a – 7) (a – 5) (a – 1)

Solution:

First expression = a2 + 7a – 18

= a2 + 9a – 2a – 18

= a(a + 9) – 2(a + 9)

= (a + 9) (a – 2)

Second expression = a2 + 10a + 9

= a2 + 9a + a + 9

= a(a + 9) + 1(a + 9)

= (a + 9) (a + 1)

Therefore, the H.C.F. = (a + 9)

Therefore, L.C.M. = Product of the two expressions H.C.F.

=\(\frac{(a^2+7a-18)(a^2+10a+9)}{(a+9)}\)

=\(\frac{(a+9)(a-2)(a+9)(a+1)}{(a+9)}\)

= (a – 2) (a + 9) (a + 1)

Solution:

First expression = m3 – 3m2 + 2m

= m(m2 – 3m + 2), by taking common ‘m’

= m(m2 - 2m - m + 2), by splitting the middle term -3m = -2m - m

= m[m(m - 2) - 1(m - 2)]

= m(m - 2) (m - 1)

= m × (m - 2) × (m - 1)

Second expression = m3 + m2 – 6m

= m(m2 + m - 6) by taking common ‘m’

= m(m2 + 3m – 2m - 6), by splitting the middle term m = 3m - 2m

= m[m(m + 3) - 2(m + 3)]

= m(m + 3)(m - 2)

= m × (m + 3) × (m - 2)

In both the expressions, the common factors are ‘m’ and ‘(m - 2)’; the extra common factors are (m - 1) in the first expression and (m + 3) in the 2nd expression.

Therefore, the required L.C.M. = m × (m - 2) × (m - 1) × (m + 3)

= m(m - 1) (m - 2) (m + 3)

Solution:

Factorizing x2 + xy by taking the common factor 'x', we get

x(x + y)

Factorizing xz + yz by taking the common factor 'z', we get

z(x + y)

Factorizing x2 + 2xy + y2 by using the identity (a + b)2, we get

= (x)2 + 2 (x) (y) + (y)2

= (x + y)2

= (x + y) (x + y)

Therefore, L.C.M. of x2 + xy, xz + yz and x2 + 2xy + y2 is xz(x + y) (x + y).

Solution:

Given, 1.20 and 22.5

Converting each of the following decimals into like decimals we get;

1.20 and 22.50

Now, expressing each of the numbers without the decimals as the product of primes we get

120 = 2 × 2 × 2 × 3 × 5 = 23 × 3 × 5

2250 = 2 × 3 × 3 × 5 × 5 × 5 = 2 × 32 × 53

Now, H.C.F. of 120 and 2250 = 2 × 3 × 5 = 30
Therefore, the H.C.F. of 1.20 and 22.5 = 0.30 (taking 2 decimal places)

L.C.M. of 120 and 2250 = 23 × 32 × 53 = 9000
Therefore, L.C.M. of 1.20 and 22.5 = 90.00 (taking 2 decimal places)

Solution:

Given, 0.48, 0.72 and 0.108

Converting each of the following decimals into like decimals we get;

0.480, 0.720 and 0.108

Now, expressing each of the numbers without the decimals as the product of primes we get

480 = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 25 × 3 × 5

720 = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 24 × 32 × 5

108 = 2 × 2 × 3 × 3 × 3 = 22 × 33

Now, H.C.F. of 480, 720 and 108 = 22 × 3 = 12
Therefore, the H.C.F. of 0.48, 0.72 and 0.108 = 0.012 (taking 3 decimal places)
L.C.M. of 480, 720 and 108 = 25 × 33 × 5 = 4320
Therefore, L.C.M. of 0.48, 0.72, 0.108 = 4.32 (taking 3 decimal places)

Solution:

2 18,24
2 9,12
3 3,6
2 1,2
1,1

Lowest common multiple (L.C.M) of 18 and 24 = 2 × 2 × 3 × 2 = 24.

Solution:

3 21
7 7
1

7 49
7 7
1

To find the LCM, multiply all prime factors. But the common factors are included only once.
21 = 3 × 7
49 = 7 × 7 × 1
The required lowest common multiple (L.C.M) of 21 and 49 = 98 = 3 × 7 ×7× 1 = 147

Quiz

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