1.

What is the greatest length x such that \(3 \dfrac{1}{2}\) m and \(8 \dfrac{3}{4}\) m are integral multiplies of x?1. \(1 \dfrac{1}{2} \ m\)2. \(1 \dfrac{1}{3} \ m\)3. \(1 \dfrac{1}{4} \ m\)4. \(1 \dfrac{3}{4} \ m\)

Answer» Correct Answer - Option 4 : \(1 \dfrac{3}{4} \ m\)

Formula used :

HCF = Highest Common Factor 

HCF of fraction = (HCF of numerator/LCM of denominator)

Calculations :

For finding the HCF first convert mixed fraction to simple fraction 

⇒ \(3\frac{1}{2}\;=\;\frac{7}{2}\)  and \(8\frac{3}{4}\;=\;\frac{35}{4}\) 

Now HCF of 7/2 and 35/4 

For numerator,

HCF of 7 and 35 = 7 

For denominator, 

LCM of 2 and 4 = 4 

Fraction = 7/4 = \(1\dfrac{3}{4}\)

So, 

x = \(1\dfrac{3}{4}\)      (x is the greatest such length or HCF)

∴ The value of x will be \(1\dfrac{3}{4}\)m.


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