Tag: hcf

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

The greatest common divisor of $878787878787$ and $787878787878$ equals.

  1. $3$
  2. $9$
  3. $27$
  4. $101010101010$
  5. $303030303030$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

$787878787878)878787878787(1\ \quad \quad \quad \quad \quad  -\underline { 787878787878 } \ \quad \quad \quad \quad \quad \quad \quad 90909090909)787878787878(8\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \underline { -727272727272 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 60606060606)90909090909(1\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -60606060606 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 30303030303)60606060606(2\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -60606060606 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0$

$\therefore$ G.D.C = 30303030303

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

Three bells, toll at intervals of $36$ sec, $40$ sec and $48$ sec respectively. They start ringing together at particular time. They next toll together after

  1. $6$ minutes
  2. $12$ minutes
  3. $18$ minutes
  4. $24$ minutes
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

G.C.D of $36,40,48=720\Rightarrow 720sec=12min$
$\therefore$ Next time when three balls toll together is after $12$ mins

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

The G.C.D. of two whole numbers is $5$ and their L.C.M. is $60$. If one of the numbers is $20$, then the other number would be

  1. $23$
  2. $13$
  3. $16$
  4. $15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If we are given two numbers $N _1$ and $N _2$ and their $G.C.D$ and $L.C.M.$.

then by property of numbers $N _1$$\times$$N _2=G.C.D$ $\times$ $L.C.M.$

Here Given:
$N _1=20$
$G.C.D.=5$
and $L.C.M=60$
Let, $N _2=x$

then from  above relation
$20$$\times$$x=5$$\times$$60$

$=>x=\dfrac{300}{20}$

$=>x=15$

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

The HCF of $2{x^2}$ and $12{x^2}$ is

  1. $2{x^2}$
  2. $12{x^2}$
  3. $2x$
  4. $12x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$2x^2=2\times x\times x$


$12x^2=6\times 2\times x\times x$

          $=3\times 2\times 2 \times x\times x$

Common factor between $2x^2$ and $12x^2=2\times x\times x=2x^2$

$\therefore$  H.C.F of $2x^2$ and $12x^2$ is $2x^2$.

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

The solution of: $8\mod x\equiv 6\mod 14$ is,

  1. ${8, 6}$
  2. ${6, 14}$
  3. ${6, 13}$
  4. ${8, 14}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Solution:-
$8x \equiv 6 \left( mod \ 14 \right)$

$\because \; gcd \left( 8, 14 \right) = 2 \text{ divides } 6$

To find solutions, we first solve

$8x − 14y = 6$

By trial and error method, we find a solution

$\left( x, y \right) = \left( 6, 3 \right)$

This means that $x \equiv 6 \left( mod \ 14 \right)$ is a solution

To the congruence $8x \equiv 6 \left( mod \ 14  \right)$

$\therefore$ Incongruent solutions are,

$x = 6 +\left ( k \times \dfrac{14}{2} \right );  k = 0, 1$

$\therefore \; x = 6, 13$

Hence option $C$ is the answer.
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If $G.C.D\ (a , b) = 1$ then $G.C.D\ ( a+b , a-b )$=?

  1. $1$ or $2$
  2. $a$ or $b$
  3. $a+b$ or $a-b$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
It is given that GCD$\left(a,b\right)=1$

Let GCD$\left(a-b,a+b\right)=d$

$\Rightarrow\,d$ divides $a-b$ and $a+b$

there exists integers $m$ and $n$ such that 

$a+b=m\times d$        ..........$(1)$

and $a-b=n\times d$        ..........$(2)$

Upon adding and subtracting equation $(1)$ and $(2)$ we get

$2a=\left(m+n\right)\times d$         ..........$(3)$

and $2b=\left(m-n\right)\times d$         ..........$(4)$

Since, GCD$\left(a,b\right)=1$(given)

$\therefore\,2\times GCD\left(a,b\right)=2$

$\therefore\,GCD\left(2a,2b\right)=2$ since $GCD\left(ka,kb\right)=kGCD\left(a,b\right)$

Upon substituting  value of $2a$ and $2b$ from equations $(3)$ and $(4)$ we get

$\therefore\,gcd\left(\left(m+n\right)\times d,\left(m-n\right)\times d\right)=2$

$\therefore\,d\times gcd\left(\left(m+n\right),\left(m-n\right)\right)=2$

$\therefore\,d\times$ some integer$=2$

$\therefore\,d$ divides $2$

$\therefore\,d\le 2$ if $x$ divides $y,$ then $\left|x\right|\le \left|y\right|$

$\therefore\,d=1$ or $2$ since, gcd is always a positive integer.
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

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

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $xy$
  2. $xy^2$
  3. $x^3y^3$
  4. $x^2y^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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  } $