Tag: common factors and hcf

Questions Related to common factors and hcf

A rectangular veranda is of dimension $18$m $72$cm $\times 13$ m $20$ cm. Square tiles of the same dimensions are used to cover it. Find the least number of such tiles.

  1. $4290$

  2. $4540$

  3. $4620$

  4. $4230$


Correct Option: A
Explanation:

The edge of rectangular veranda are $18\ m\ 72\ cm=1872\ cm$ and $13\ m\ 20\ cm=1320\ cm$.


On taking $HCF$ of $1872$ and $1320$, we get

$HCF=24$

Therefore,
No. of tiles required $=$ $\dfrac{Area\ of\ Veranda}{Area\ of\ tiles}$

                                  $=\dfrac{1872\times 1320}{24\times 24}$

                                  $=4290$

Hence, this is the answer.

What is the H.C.F. of two co-prime numbers ?

  1. $1$

  2. $0$

  3. $2$

  4. none of these


Correct Option: A
Explanation:

The two numbers which have only 1 as their common factor are called co-primes.

For example, Factors of $ 5 $  are $ 1, 5 $
Factors of $ 3 $ are $ 1, 3 $

Common factors is $ 1 $.
$ => HCF = 1 $

The HCF of $256,442$ and $940$ is

  1. $2$

  2. $14$

  3. $142$

  4. none


Correct Option: A
Explanation:

Prime factors of numbers are 

$256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\ 442=2\times 13\times 17\ 940=2\times 2\times 5\times 47\ Hence,\quad HCF=2$
So, correct answer is option A. 

HCF of $x^2 -y^2$ and $x^3-y^3$ is

  1. $x-y$

  2. $x^3-y^3$

  3. $(x^2-y^2)$

  4. $(x+y)(x^2+xy+y^2)$


Correct Option: A
Explanation:

Since, $x^2-y^2=(x+y)(x-y)$
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$H.C.F.=(x-y)$
Option A is correct.

Determine the HCF of $a^2 - 25, a^2 -2a -35$ and $a^2+12a+35$

  1. (a-5)(a+7)

  2. (a+5)(a-7)

  3. (a-7)

  4. (a+5)


Correct Option: D
Explanation:

Since, $a^2 - 25 = (a-5)(a+5) $
$ a^2 -2a -35 = a^2 -7a +5a -35 $
                         $= a(a-7)+5(a-7) $
                         $= (a+5)(a-7) $
and
$a^2+ 12a + 35 =a^2 +7a +5a +35 $
                          $=(a+7)(a+5) $
Clearly HCF of $a^2 - 25, a^2 -2a -35$ and $a^2+ 12a + 35$ i.e $ (a-5)(a+5), (a+5)(a-7)$ and $(a+7)(a+5)$ is $a+5$
Option D is correct.

H.C.F. of $x^3 -1$ and $x^4 + x^2 + 1$ is

  1. $x^2+x+1$

  2. $x-1$

  3. $x^3-1$

  4. $x^4+x^2+1$


Correct Option: A
Explanation:

$x^3 -1=(x-1)(x^2+x+1)$
$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$
Clearly H.C.F of $x^3 -1$ and $x^4 + x^2 + 1$
i.e.H.C.F of $(x-1)(x^2+x+1)$ and $(x^2+x+1)(x^2-x+1)$ is  $(x^2+x+1)$
Option A is correct.

H.C.F. of $x^2-1$ and $x^3-1$ is

  1. $(x^2-1)^2$

  2. $(x-1)$

  3. x+1

  4. $x^2+x+1$


Correct Option: B
Explanation:

$x^2-1 = (x + 1) (x -1)$
$x^3 -1 = (x -1) (x^2 + x + 1)$
$\therefore H.C.F. = (x - 1)$
Option B is correct.

Find the HCF of $x^3y^2, x^2y^3$ and $x^4y^4$

  1. $x^3y^4$

  2. xy

  3. $x^2y^2$

  4. $x^4y^4$


Correct Option: C
Explanation:

$x^3y^2=x^3\times y^2$
$x^2y^3=x^2\times y^3$
and $x^4y^4=x^4\times y^4$
$\therefore$ HCF $=x^2\times y^2=x^2y^2$.
Option C is correct.

The LCM of 54 90 and a third number is 1890 and their HCF is 18 The third number is

  1. 36

  2. 180

  3. 126

  4. 108


Correct Option: C
Explanation:

Given the LCM two numbers 54, 90 and third number is 1890 and HCF is 18

Let the number is 18x because one factor is also 18 the common factor HCF
Then factor 54,90 ,18 =$18\times 3,18\times 5,18\times 18\times x$

$\therefore 18\times 3\times 5\times x=1890\Rightarrow 270x=1890\Rightarrow x=7$
Then third number is $18\times 7=126$ 

HCF of the two numbers =

  1. Product of numbers + their LCM

  2. Product of numbers - their LCM

  3. Product of numbers $\times$ their LCM

  4. Product of numbers $\div$ their LCM

  5. Answer required


Correct Option: D
Explanation:

The product of highest common factor $(H.C.F.)$ and lowest common multiple $(L.C.M.)$ of two numbers is equal to the product of two numbers.

If two numbers are $a$ and $b$ then
$HCF\ \times LCM=a\times b$ 

Hence,
$HCF=\dfrac{ab}{LCM}$

For example: 
Let $a=10\Rightarrow 2\times 5$ and $b=15\Rightarrow 3\times 5$
So, the $LCM$ of both numbers $=2\times 3\times 5\Rightarrow 30$
Then 
$HCF=\dfrac{10\times 15}{30}\Rightarrow 5$

Hence,
$HCF=\dfrac{Product\ of\ numbers}{Their\ LCM}.$