Tag: common factors and hcf

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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.

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

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

  1. $6$
  2. $4$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
D Correct answer
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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$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore H.C.F=2\times 2 \times 3=12$