Tag: real numbers on number line

Questions Related to real numbers on number line

The HCF of two consecutive even numbers is

  1. $6$

  2. $3$

  3. $4$

  4. $2$


Correct Option: D
Explanation:

$HCF$ of two consecutive even numbers is always $2$. For example:


$HCF$ of $2$ and $4$ is $2$ and similarly,

$HCF$ of $22$ and $24$ is $2$ and we can do so on..

The HCF of two consecutive odd numbers is

  1. $28$

  2. $30$

  3. $1$

  4. $5$


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?

  1. $2$

  2. $1$

  3. $6$

  4. $5$


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. 

  1. $35$

  2. $30$

  3. $15$

  4. $25$


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?

  1. $35\ cm$

  2. $55\ cm$

  3. $1\ m$

  4. $65\ cm$


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:

  1. $13$

  2. $2366$

  3. $91$

  4. $182$


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:

  1. $6$

  2. $2$

  3. $3$

  4. $1$


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 

  1. 1300

  2. 1345

  3. 1465

  4. 1356


Correct Option: D