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

The H.C.F. of the numbers $16.5, 0.90$ and $15$ is

  1. $16.5$
  2. $0.90$
  3. $15$
  4. $0.3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$1650 = 2\times 3\times 5^{2}\times 11$
$90 = 2\times 3^{2} \times 5^{1}$
$1500 = 2^{2} \times 3^{1} \times 5^{3}$
H.C.F. of $1650, 90$ and $1500$ is $2\times 3\times 5 = 30$

Therefore, H.C.F. of $16.5, 0.90$ and $15$ is $0.30$.
So, option D is correct.

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

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. $65\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.