Tag: common factors and hcf

Questions Related to common factors and hcf

The HCF of two consecutive odd numbers is

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. $28$

  2. $30$

  3. $1$

  4. $5$

Show answer
Correct Option: C
Explanation:


Let's take two odd numbers $15$ and $17$.
Factors of $15 = 1, 3, 5$
Factors of $17 = 1, 17$
HCF is $1$
Therefore, $C$ is the correct answer.
OR
HCF of $25$ and $27$ is also $1$

There are five odd numbers $1, 3, 5, 7, 9$. What is the HCF of these odd numbers?

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. $2$

  2. $1$

  3. $6$

  4. $5$

Show answer
Correct Option: B
Explanation:
$3=3 \times 1$
$5=5 \times 1$
$7=7 \times 1$
$9=3 \times 3$
HCF is $1$ for the five odd numbers.
Therefore, $B$ is the correct answer.

Find HCF of $70$ and $245$ using Fundamental Theorem of Arithmetic. 

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. $35$

  2. $30$

  3. $15$

  4. $25$

Show answer
Correct Option: A
Explanation:
using Fundamental Theorem of Arithmetic
$70 =2\times 5\times 7$
$245=5\times 7 \times 7 $
Common factors of 70 and 245 are 5 and 7.
$\therefore HCF = 5\times 7=35$
hence, option $A$ is correct.

Three ropes are $7\ m, 12\ m\ 95\ cm$ and $3\ m\ 85\ cm$ long. What is the greatest possible length that can be used to measure these ropes?

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. $35\ cm$

  2. $55\ cm$

  3. $1\ m$

  4. $65\ cm$

Show answer
Correct Option: A
Explanation:

The given three ropes are $7$m, $12$ m $95$cm and $3$m$85$cm long. We know that $1$m=$100$cm, therefore,


The length of the respective ropes will be:

1st rope $=7\times 100=700$cm
2nd rope $=(12\times 100)+95=1200+95=1295$cm
3rd rope $=(3\times 100)+85=300+85=385$cm

Now, let us factorize the length of the ropes as follows:

$700=2\times 2\times 5\times 5\times 7\ 1295=5\times 7\times 37\ 385=5\times 7\times 11$

The highest common factor (HCF) is $5\times 7=35$

Hence, the greatest possible length that can be used to measure these ropes is $35$cm.

H.C.F. of $26$ and $91$ is:

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. $13$

  2. $2366$

  3. $91$

  4. $182$

Show answer
Correct Option: A
Explanation:

$26=2*13$
$91=7*13$
$HCF=13$
As it is the only common factor

H.C.F. of $6, 72$ and $120$ is:

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. $6$

  2. $2$

  3. $3$

  4. $1$

Show answer
Correct Option: A
Explanation:

Factors of 6 are $1,2,3,6$
Factors of 72 are $1,2,3,4,6,8,9,12,18,24,36,72$
Factors of 120 are $1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$
The highest factor in all the three is 6 .

Sum of all two digit numbers divisible by $7$ leaves remainder $2$ or $5$ is 

maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

  1. 1300

  2. 1345

  3. 1465

  4. 1356

Show answer
Correct Option: D

What is the H.C.F . of $348$ and $319$

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

  1. $348$

  2. $319$

  3. $56$

  4. $29$

Show answer
Correct Option: D
Explanation:

$348=$$2\times$$2\times$$3\times$$29$

$319=11$$\times29$
thus $HCF=29$

HCF of $144$ and $198$ is

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

  1. $9$

  2. $18$

  3. $6$

  4. $12$

Show answer
Correct Option: A
Explanation:

HCF of 144 and 198 is:

$ 198 = 2 \times 3^2 \times 11$
$ 144 = 2^4 \times 3^2$

Highest Common factor is 9.
Option A is correct

HCF of $24 $ and $36$ is ..............

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

  1. $6$

  2. $4$

  3. $9$

  4. $12$

Show answer
Correct Option: D
Explanation:
$2\underline {|24} {\;} 2\underline {|36}$
$2\underline {|12} {\;} 2\underline {|18}$
$2\underline {|6} {\;} 3\underline {|9}$
$3\underline {|3} {\;} 3\underline {|3}$
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$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore H.C.F=2\times 2 \times 3=12$