Tag: highest common factor (h.c.f.)

Questions Related to highest common factor (h.c.f.)

When teams of same size are formed from three groups of $512, 430$ and $489$ students separately $8, 10$ and $9$ students respectively are left out What could be the largest size of the team?

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $6$

  2. $12$

  3. $18$

  4. $20$

Show answer
Correct Option: B
Explanation:

It is given that $8,10$ and $9$ students are respectively left out from the three separate groups $512,430$ and $489$ when the teams of same size are formed.


Number of students taken from first group are $512-8=504$
Number of students taken from second group are $430-10=420$
Number of students taken from third group are $489-9=480$

Now, we factorize $504,420$ and $480$ as follows:

$504=2\times 2\times 2\times 3\times 3\times 7\ 430=2\times 2\times 3\times 5\times 7\ 480=2\times 2\times 2\times 2\times 2\times 3\times 5$

Therefore, the HCF of $504,420$ and $480$ is:

HCF$\left( 504,430,480 \right) =2\times 2\times 3=12$

Hence, the largest size of the team is of $12$ students.

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

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $24$

  2. $36$

  3. $12$

  4. $4$

Show answer
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? 

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

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $1$ and $2$

  2. $2$ and $1$

  3. $1$ and $1$

  4. cannot be determined

Show answer
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 

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

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

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

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

  4. None

Show answer
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

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. a

  2. b

  3. 1

  4. abc

Show answer
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______

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

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

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

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

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

Show answer
Correct Option: D

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

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

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

  2. 1

  3. $a^2+b^2$

  4. $a-b$

Show answer
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.

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $4$

  2. $3$

  3. $2$

  4. $1$

Show answer
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 $g(x)$.

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $x + 3$

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

  3. $x + 4$

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

Show answer
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 :

maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

  1. $2, 2x - 3$

  2. $8x - 12, 2$

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

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

Show answer
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) $