Real numbers on number line - class-VIII

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

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 $

Prime factors of numbers are 

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.

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.

$x^3 -1=(x-1)(x^2+x+1)$
$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$
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.

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

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

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

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.

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. 

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

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

For the two polynomials to have an HCF involving x, they must share a common root. x^2 - 2x - 3 factors to (x-3)(x+1). Testing x=3 in the second polynomial: 27 - 18 - 3n - 3 = 0 => 6 - 3n = 0 => n=2.

$ 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) $

The question asks for the LCM and HCF of integers but provides a list of sets. Option A (12, 15, 21) is a valid set of integers for which LCM and HCF can be calculated.

Using Euclidean algorithm: 420 = 130 * 3 + 30; 130 = 30 * 4 + 10; 30 = 10 * 3 + 0. The HCF is 10.

The given number $100$ and $101$ are consecutive numbers and respectively even and odd numbers.

  HCF of 210 and 55 is 5

HCF of $468$ and $222$
$468 = \left(222 \times 2\right) + 24$
$222 = \left(24\times\ 9\right) + 6$
$24 = \left(6\times\ 4\right) + 0$
$\therefore HCF = 6$

Given the H.C.F. of $A$ and $B$ is $24$ and that of $C$ and $D$ is $56.$
Then the H.C.F. of $A, B, C$ and $D$ is the HCF of $24$ and $56$ which is $8.$

136)170(1

Let two expressions $p(x)$ and $q(x)$ then

$480=2^5\times 3\times 5$
$720=2^4\times 3^2\times 5$
$108=2^2\times 3^2$
$HCF=2^2\times 3=12$

$248=2^3 \times 31$

$26=13\times 2$

$1650 = 2\times 3\times 5^{2}\times 11$
$90 = 2\times 3^{2} \times 5^{1}$
$1500 = 2^{2} \times 3^{1} \times 5^{3}$
H.C.F. of $1650, 90$ and $1500$ is $2\times 3\times 5 = 30$

$HCF$ of two consecutive even numbers is always $2$. For example:

The given three ropes are $7$m, $12$ m $95$cm and $3$m$85$cm long. We know that $1$m=$100$cm, therefore,

$26=213$
$91=713$
$HCF=13$
As it is the only common factor

Factors of 6 are $1,2,3,6$
Factors of 72 are $1,2,3,4,6,8,9,12,18,24,36,72$
Factors of 120 are $1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$
The highest factor in all the three is 6 .

Numbers divisible by 7 with remainder 2 or 5 are of form 7k+2 or 7k+5. For two-digit numbers (10-99): 7k+2: 16, 23, ..., 93 (sum = 655); 7k+5: 12, 19, ..., 96 (sum = 701). Total sum = 655 + 701 = 1356.