Tag: maths

Questions Related to maths

The GCD of two numbers is $17$ and their LCM is $765$. How many pairs of values can the numbers assume?

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

Since the GOD of numbers is $17$. So, the numbers are $17a$ and $17b$, where a and b are relatively prime.
LCM$=765$

$\Rightarrow 17a\times 17b=765$
$\Rightarrow ab=45$
$\Rightarrow a=15, b=9$ or $a=9$, $b=5$.
So, the numbers are $17\times 5=85$ and $17\times 9=153$.
The numbers can be $17\times 1=17$ and $765$. So, two pairs are possible.

If two positive integers $a$ and $b$ are written as $a=x^3y^2$ and $b=xy^3$; $x, y$ are prime numbers, then HCF of $a$ and $b$ is

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

  1. $xy$

  2. $xy^2$

  3. $x^3y^3$

  4. $x^2y^2$

Show answer
Correct Option: B
Explanation:

Given,

$a={  x}^{3  }{ y }^{2  } = x\times x\times x\times y\times y$

$b={  x}{ y }^{3  }         =x\times y\times y\times y$

H.C.F of $a,b$  = ${  x}{ y }^{2  } $

When teams of same size are formed from three groups of $512, 430$ and $489$ students separately $8, 10$ and $9$ students respectively are left out What could be the largest size of the team?

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

  1. $6$

  2. $12$

  3. $18$

  4. $20$

Show answer
Correct Option: B
Explanation:

It is given that $8,10$ and $9$ students are respectively left out from the three separate groups $512,430$ and $489$ when the teams of same size are formed.


Number of students taken from first group are $512-8=504$
Number of students taken from second group are $430-10=420$
Number of students taken from third group are $489-9=480$

Now, we factorize $504,420$ and $480$ as follows:

$504=2\times 2\times 2\times 3\times 3\times 7\ 430=2\times 2\times 3\times 5\times 7\ 480=2\times 2\times 2\times 2\times 2\times 3\times 5$

Therefore, the HCF of $504,420$ and $480$ is:

HCF$\left( 504,430,480 \right) =2\times 2\times 3=12$

Hence, the largest size of the team is of $12$ students.

HCF of the numbers $24,36$ and $92$ is 

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

  1. $24$

  2. $36$

  3. $12$

  4. $4$

Show answer
Correct Option: D
Explanation:

Factors of $ 24 =1,2,3,4,6,8,12,24 $
Factors of $36= 1,2,3,4,6,9,12,18,36$

Factors of $ 92=1,2,4,23,46,92 $

Common factors are $ =1,2,4 $
$\therefore $ HCF $=4.$

If HCF of $m$ and $n$ is $1,$ then what are the HCF of $m + n, m$ and HCF of $m - n, n$ respectively? 

$\displaystyle \left ( m> n \right )$

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

  1. $1$ and $2$

  2. $2$ and $1$

  3. $1$ and $1$

  4. cannot be determined

Show answer
Correct Option: C
Explanation:
Let us consider an example.
Let $m =16$ and $n =9$ be relatively prime numbers.
So, $m+n=25$. The HCF of $25$ and $16$ is $1$. 

$m-n=7$. The HCF of $7$ and $9$ is $1$.
Similarly, if we take other values for $m$ and $n,$ we get the same answer. 
Therefore, option $C$ is correct.

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

Show answer
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

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

Show answer
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. $a-b$

Show answer
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. $1$

Show answer
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) $