Tag: maths

Questions Related to maths

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

What is the HCF of $4x^{3} + 3x^{2}y - 9xy^{2} + 2y^{3}$ and $x^{2} + xy - 2y^{2}$?

  1. $x - 2y$

  2. $x - y$

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

  4. $(x - 2y)(x - y)$


Correct Option: C
Explanation:

Let $f(x, y) = 4x^{2} + 3x^{2}y - 9xy^{2} + 2y^{3}$
and $g(x, y) = x^{2} + xy - 2y^{2}$
Clearly $(x - y) = 0$ or $x = y$ satisfy both equation, since for $x = y$ makes both equations zero. 

Hence, $(x - y)$ is the H.C.F. 
Similarly $(x + 2y) = 0$ i.e., $x = -2y$ also makes both equations zero. 
Thus, $(x + 2y)(x - y)$ is the H.C.F.

H.C.F. of $(10224, 1608)$ is _________.

  1. $12$

  2. $24$

  3. $48$

  4. $96$


Correct Option: B
Explanation:

$10224 = 2^{4}\times3^{2}\times71$

$1608 = 2^{3}\times3\times67$

$\therefore$ H.C.F$(10224,1608) = 2^{3}\times3=24$

The greatest number that will divided $398, 436$ and $542$ leaving $7,11$ and $14$ remainders, respectively, is

  1. $16$

  2. $17$

  3. $18$

  4. $19$


Correct Option: B
Explanation:

Out of tour choices only $17$ is the number that divides $398,436$ and $542$ leaving $7,11$ and $15$ as remainder

The smaller value of n for which $x^{2} - 2x - 3$ and $x^{3} - 2x^{2} - nx - 3$ have an H.C.F. involving $x$ is

  1. 0

  2. 1

  3. 2

  4. 3


Correct Option: C

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: B
Explanation:

$ Given\quad that\quad product\quad of\quad the\quad number\quad is\quad 5400=30\times 3\times 2\times 30.\ \therefore \quad Possible\quad pairs\quad as\quad per\quad the\quad requirment\quad are-\ (1)\quad 30\times (3\times 2\times 30)=30\times 180\ (2)\quad (30\times 3)\times (2\times 30)=90\times 60\ \therefore \quad Total\quad number\quad of\quad pairs=2\quad \quad (Ans) $