Tag: maths

Questions Related to maths

If the HCF of 85 and 153 is expressible in the form 85n $-$ 153, then value of n is :

  1. 3

  2. 2

  3. 4

  4. 1


Correct Option: B
Explanation:

HCF of $85\  and\  153 = 17$

Now given HCf can be expressed in the gorm of $85n-153$
So $17=85n-173$
On solving the above equation we get $n=2$
So correct answer will be option B

Choose the correct answer form the alternatives given.
What is the HCF of $(x^4 \, - \, x^2 \, - \, 6) \, and \, (x^4 \, - \, 4x^2 \, + \, 3)$? 

  1. $x^2$ - $3$

  2. $x + 2$

  3. $x + 3$

  4. $x^2$ + $3$


Correct Option: A
Explanation:

$\displaystyle (x^4 \, - \, x^2 \, - \, 6) \, = \, (x^2 \, - \, 3) (x^2 \, + \, 2)$
$\displaystyle (x^4 \, - \, 4x^2 \, + \, 3) \, = \, (x^2 \, - \, 3) (x^2 \, - \, 1)$
HCF is = $x^2$ - $3$

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

  1. $3$

  2. $9$

  3. $27$

  4. $101010101010$

  5. $303030303030$


Correct Option: E
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

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


Correct Option: B
Explanation:

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

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. $15$


Correct Option: D
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$

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

$=>x=15$

The HCF of $2{x^2}$ and $12{x^2}$ is

  1. $2{x^2}$

  2. $12{x^2}$

  3. $2x$

  4. $12x$


Correct Option: A
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 $2x^2$.

Find the HCF of $25$ and $30$.

  1. $25$

  2. $6$

  3. $1$

  4. $5$


Correct Option: D
Explanation:

$\Rightarrow$$25=5\times5$

$\Rightarrow$$30=5\times6$

Hence the Highest common factor is 5

Therefore HCF is $5$

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

  1. ${8, 6}$

  2. ${6, 14}$

  3. ${6, 13}$

  4. ${8, 14}$


Correct Option: C
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.

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. $4$


Correct Option: A
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.

ILLUSTRATION  2 The total number of factors (exculding 1) of 2160 is 

  1. 40

  2. 39

  3. 41

  4. 38


Correct Option: A