Tag: real numbers on number line

Questions Related to real numbers on number line

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) $

If HCF of numbers $408$ and $1032$ can be expressed in the form of $1032x -408 \times 5$, then find the value of $x$.

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: C
Explanation:
$408=2\times2\times2\times3\times17$

$1032=2\times2\times2\times3\times43$

Hence, $HCF=2\times2\times2\times3=24$

Now, $1032x-408\times5=24\Rightarrow 1032x=2064\Rightarrow x=2$

Find the LCM and HCF of the following integers by the prime factorization mass

  1. 12, 15 and 21

  2. 17, 23, and 29

  3. 8, 9 and 25

  4. 72 and 108

  5. 72 and 108

  6. 306 and 657


Correct Option: A

The ratio of two numbers is 15:11. If their HCF be 13 then these numbers will be

  1. 15, 11

  2. 75, 55

  3. 105, 77

  4. 195, 143


Correct Option: A