Tag: hcf-lcm

Questions Related to hcf-lcm

HCF of the numbers $24,36$ and $92$ is 

  1. $24$

  2. $36$

  3. $12$

  4. $4$


Correct Option: D
Explanation:

Factors of $ 24 =1,2,3,4,6,8,12,24 $
Factors of $36= 1,2,3,4,6,9,12,18,36$

Factors of $ 92=1,2,4,23,46,92 $

Common factors are $ =1,2,4 $
$\therefore $ HCF $=4.$

If HCF of $m$ and $n$ is $1,$ then what are the HCF of $m + n, m$ and HCF of $m - n, n$ respectively? 

$\displaystyle \left ( m> n \right )$

  1. $1$ and $2$

  2. $2$ and $1$

  3. $1$ and $1$

  4. cannot be determined


Correct Option: C
Explanation:
Let us consider an example.
Let $m =16$ and $n =9$ be relatively prime numbers.
So, $m+n=25$. The HCF of $25$ and $16$ is $1$. 

$m-n=7$. The HCF of $7$ and $9$ is $1$.
Similarly, if we take other values for $m$ and $n,$ we get the same answer. 
Therefore, option $C$ is correct.

The GCD of $\displaystyle \frac{3}{16}$,$\displaystyle \frac{5}{12}$,$\displaystyle \frac{7}{18}$ is 

  1. $\displaystyle \frac{105}{48}$

  2. $\displaystyle \frac{1}{4}$

  3. $\displaystyle \frac{1}{48}$

  4. None


Correct Option: D
Explanation:
The greatest common divisor is same as the highest common factor that is GCD is same as HCF and,

HCF of two or more fractions is given by HCF of Numerators divided by LCM of Denominators 

HCF of the numerators $(3,5,7)=1$
LCM of the denominators $(16,12,18)=2\times 2\times 2\times 2\times 3\times 3=144$

Therefore, 

HCF$\left( \dfrac { 3 }{ 16 } ,\dfrac { 5 }{ 12 } ,\dfrac { 7 }{ 18 }  \right) =\dfrac { 1 }{ 144 }$

Hence, GCD of $\dfrac { 3 }{ 16 } ,\dfrac { 5 }{ 12 } ,\dfrac { 7 }{ 18 }$ is $\dfrac { 1 }{ 144 }$

The HCF of ab, bc, and ca is

  1. a

  2. b

  3. 1

  4. abc


Correct Option: C
Explanation:

$ab = 1.ab$
$bc= 1.bc$
$ca = 1.ca$

If (x + 6) is the HCF of $\displaystyle p\left ( x \right )=x^{2}-a$ and $\displaystyle q\left ( x \right )=x^{2}-bx+6$ then $\displaystyle \frac{p\left ( x \right )}{q\left ( x \right )}$ in its lowest terms is______

  1. $\displaystyle \frac{x-6}{x-2}$

  2. $\displaystyle \frac{x+6}{x+1}$

  3. $\displaystyle \frac{x-6}{x-1}$

  4. $\displaystyle \frac{x-6}{x+1}$


Correct Option: D

The HCF of $(a + b)^2$ and $(a -b)^2$ is

  1. $(a+b)(a-b)$

  2. 1

  3. $a^2+b^2$

  4. $a-b$


Correct Option: B
Explanation:

$(a+b)^2= 1 (a+b)(a+b)$
$(a-b)^2= 1 (a-b)(a-b)$
$HCF= 1$

Two positive numbers have their HCF as $12$ and their sum is $84$. Find the number of pairs possible.

  1. $4$

  2. $3$

  3. $2$

  4. $1$


Correct Option: B
Explanation:

As the HCF is $ 12 $, the numbers can be written as $ 12x $ and $ 12y $, where x and y are co-prime to each other.
So, $ 12x + 12y = 84 => x + y = 7 $

The pair of numbers that are co-prime to each other and sum up to $7$ are $(2, 5), (1,6), (3,4)$.
Hence, only $ 3 $  pairs of such numbers are possible.
 The numbers are $ (24, 60), (12,72) $ and $ (36,48) $

If $\displaystyle f\left ( x \right )=\left ( x+2 \right )\left ( x^{2}+8x+15 \right )$ and $\displaystyle g\left ( x \right )=\left ( x+3 \right )\left ( x^{2}+9x+20 \right )$ then find the HCF of $f(x)$ and $g(x)$.

  1. $x + 3$

  2. $\displaystyle x^{2}+8x+15$

  3. $x + 4$

  4. $\displaystyle x^{2}+9x+20$


Correct Option: B
Explanation:

Prime factorisation of $ (x+2)({x}^{2}+8x+15) = (x+2)  \times [(x+3) \times (x+5)] $
Prime factorisation of $ (x+3)({x}^{2}+9x+20) = (x+3) \times [(x+4) \times (x+5)] $
So,HCF $  =  (x+3) \times (x+5) = ({x}^{2}+8x+15) $

If the HCF of the polynomials $f(x)$ and $g(x)$ is $4x - 6$, then $f(x)$ and $g(x)$ could be :

  1. $2, 2x - 3$

  2. $8x - 12, 2$

  3. $\displaystyle 2\left ( 2x-3 \right )^{2},4\left ( 2x-3 \right )$

  4. $\displaystyle 2\left ( 2x+3 \right ),4\left ( 2x+3 \right )$


Correct Option: C
Explanation:

Given, HCF $ = 4x-6 = 2(2x-3) $

Since HCF needs to be a factor of both the polynomials, clearly only option C with polynomials $ 2({2x-3)}^{2} , 4(2x-3) $  have both factors $ 2 $ and $ (2x-3) $

HCF of $120, 144$ and $216$ is:

  1. $38$

  2. $24$

  3. $120$

  4. $144$


Correct Option: B
Explanation:

The HCF of $120,144,216$ is

$120= 2 \times 2 \times 2 \times 3 \times 5 $
$144= 2 \times 2 \times 2 \times 2 \times 3 \times 3 $
$216= 2 \times 2 \times 2\times 3 \times 3 \times 3 $
Common factor is $2\times 2\times 2\times 3=24$ 
Hence, the HCF of $120,144$ and $216$ is $24$.