What is the greatest length x such that $3\frac{1}{2}$m and $8\frac{3}{4}$m are integral multiple of x?

Correct Option

To determine the greatest length x such that 3½m and 8¾m are integral multiples of x, it is necessary to find the Highest Common Factor (HCF) of these two lengths.

First, convert the given mixed fractions into improper fractions:

• 3½m = 7/2m
• 8¾m = 35/4m

The HCF of two fractions a/b and c/d is calculated using the formula: H.C.F of (a, c) / L.C.M of (b, d).

• The HCF of the numerators (7 and 35) is 7.
• The LCM of the denominators (2 and 4) is 4.

Therefore, the greatest length x is 7/4m.

Converting this improper fraction back to a mixed fraction yields 1¾m.

Incorrect Options

For a length x to be a common divisor of 3½m and 8¾m, both given lengths must be integral multiples of x. This implies that the division of each given length by x must result in an integer.

• Option (1) 1½m = 3/2m: 7/2m ÷ 3/2m = 7/3, which is not an integer. Thus, 3/2m is not a common divisor.
• Option (2) 1⅓m = 4/3m: 7/2m ÷ 4/3m = 21/8, which is not an integer. Thus, 4/3m is not a common divisor.
• Option (3) 1¼m = 5/4m: 7/2m ÷ 5/4m = 14/5, which is not an integer. Thus, 5/4m is not a common divisor.

As these options do not satisfy the condition of being a common divisor, they cannot be the greatest common divisor. The calculated HCF is 1¾m.