QUESTION

CSAT

Hard

Maths

Prelims 2020

What is the greatest length xx such that 3123 \frac{1}{2} m and 8348 \frac{3}{4} m are integral multiples of xx?

Select an option to attempt

Explanation

First, convert the mixed fractions into improper fractions: 312=723 \frac{1}{2} = \frac{7}{2} 834=3548 \frac{3}{4} = \frac{35}{4}

Now, find the greatest common divisor (GCD) of 7/27/2 and 35/435/4. The GCD of two fractions a/ba/b and c/dc/d is given by:

GCD(a/b,c/d)=GCD(a,c)LCM(b,d)\text{GCD}(a/b, c/d) = \frac{\text{GCD}(a, c)}{\text{LCM}(b, d)}

  1. Find the GCD of the numerators (7 and 35): GCD(7,35)=7\text{GCD}(7, 35) = 7
  2. Find the LCM of the denominators (2 and 4): LCM(2,4)=4\text{LCM}(2, 4) = 4

Now, the GCD of the two fractions is: 7/47/4

Convert 7/47/4 back to a mixed fraction: 7/4=1347/4 = 1 \frac{3}{4}

The greatest length xx is 1341 \frac{3}{4} m, which corresponds to Option D.

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