Tag: maths

Questions Related to maths

If $\displaystyle f\left ( x \right )=\left ( x+2 \right )\left ( x^{2}+8x+15 \right )$ and $\displaystyle g\left ( x \right )=\left ( x+3 \right )\left ( x^{2}+9x+20 \right )$ then find the HCF of $f(x)$ and $g(x)$.

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

  1. $x + 3$

  2. $\displaystyle x^{2}+8x+15$

  3. $x + 4$

  4. $\displaystyle x^{2}+9x+20$

Show answer
Correct Option: B
Explanation:

Prime factorisation of $ (x+2)({x}^{2}+8x+15) = (x+2)  \times [(x+3) \times (x+5)] $
Prime factorisation of $ (x+3)({x}^{2}+9x+20) = (x+3) \times [(x+4) \times (x+5)] $
So,HCF $  =  (x+3) \times (x+5) = ({x}^{2}+8x+15) $

If the HCF of the polynomials $f(x)$ and $g(x)$ is $4x - 6$, then $f(x)$ and $g(x)$ could be :

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

  1. $2, 2x - 3$

  2. $8x - 12, 2$

  3. $\displaystyle 2\left ( 2x-3 \right )^{2},4\left ( 2x-3 \right )$

  4. $\displaystyle 2\left ( 2x+3 \right ),4\left ( 2x+3 \right )$

Show answer
Correct Option: C
Explanation:

Given, HCF $ = 4x-6 = 2(2x-3) $

Since HCF needs to be a factor of both the polynomials, clearly only option C with polynomials $ 2({2x-3)}^{2} , 4(2x-3) $  have both factors $ 2 $ and $ (2x-3) $

HCF of $120, 144$ and $216$ is:

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

  1. $38$

  2. $24$

  3. $120$

  4. $144$

Show answer
Correct Option: B
Explanation:

The HCF of $120,144,216$ is

$120= 2 \times 2 \times 2 \times 3 \times 5 $
$144= 2 \times 2 \times 2 \times 2 \times 3 \times 3 $
$216= 2 \times 2 \times 2\times 3 \times 3 \times 3 $
Common factor is $2\times 2\times 2\times 3=24$ 
Hence, the HCF of $120,144$ and $216$ is $24$.

What is the greatest possible rate at which a man can walk 68 km, 102 km and 51 km in exact number of days?

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

  1. 17

  2. 4

  3. 7

  4. 3

Show answer
Correct Option: A
Explanation:

Required rate = H.C.F. of (68 km, 102 km, 51 km) = 17 km per day

The HCF of $136, 170$ and $255$ is

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

  1. $13$

  2. $15$

  3. $17$

  4. $1$

Show answer
Correct Option: C

Find the greatest number which divides 120, 165 and 210 exactly leaving remainders 5, 4 and 3 respectively

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

  1. 7

  2. 5

  3. 23

  4. None of these

Show answer
Correct Option: D
Explanation:

The required number will be the H.C.F of (120 - 5), (165 - 4) and (210 - 3) i.e. H.C.F. of 115
161 and 207
$\displaystyle \therefore $ Required number = H.C.F. of 115, 161 and 207 = 23

HCF of $24, 36$ 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:

The HCF of $24,36,92$ can be found by factorising all three numbers:

$24= 2 \times 2 \times 2 \times 3 $
$36= 2 \times 2 \times 3 \times 3 $
$92 =2 \times 2 \times 23 $
Now, common factors are $2$ and $2$
So, HCF is $ 2 \times 2=4$
Hence, the answer is $4$.

The HCF of $18$ and $30$ is equal to

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

  1. $6$

  2. $5$

  3. $4$

  4. $3$

Show answer
Correct Option: A
Explanation:

$18 = 2 \times 3 \times 3$

$30 = 2 \times 3 \times 5$
So, H.C.F of $18$ and $30$ is $6$.
So, option A is correct.

The HCF of $75$ and $15$ is equal to

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

  1. $12$

  2. $13$

  3. $14$

  4. $15$

Show answer
Correct Option: D
Explanation:

$75=5^2\times 3$

$15=5\times3$
HCF$=15$

What is the least number by which $825$ must be multiplied in order to produce a multiple of $715 ?$

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

  1. $13$

  2. $15$

  3. $11$

  4. $3$

Show answer
Correct Option: A
Explanation:

$825=3\times5\times5\times 11$

$715=5\times11\times13$
In the factor of both numbers, $13$ is not common. 
Hence, the least number by which $825$ must be multiplied in order to produce a multiple of $715=13.$