Tag: common factors and hcf

Questions Related to common factors and hcf

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

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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

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

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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. $a-b$

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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. $1$

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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) $

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

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

  1. $38$

  2. $24$

  3. $120$

  4. $144$

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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 $24$.

What is the greatest possible rate at which a man can walk 68 km, 102 km and 51 km in exact number of days?

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

  1. 17

  2. 4

  3. 7

  4. 3

Show answer
Correct Option: A
Explanation:

Required rate = H.C.F. of (68 km, 102 km, 51 km) = 17 km per day

The HCF of $136, 170$ and $255$ is

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

  1. $13$

  2. $15$

  3. $17$

  4. $1$

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Correct Option: C