The sum of fractions is $3 \frac{13}{24}$ . When the largest fraction is divided by the smallest, the fraction thus obtained is $\frac{7}{8}$ , which is $\frac{1}{4}$ more than the middle one. The fractions are:

(203/180),(29/12), (58/45)

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(203/12),(29/180), (58/45)

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(203/45),(29/12), (58/180)

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(203/12),(29/45), (58/180)

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none of these

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Solution

The correct option is A
(203/180),(29/12), (58/45)

Middle number is $\frac{7}{8} + \frac{1}{4} = \frac{29}{12}$

So we can assume the numbers as 7x and 8x respectively.

So we get x = $\frac{29}{180}$ so the numbers are $\frac{203}{180}$ , $\frac{29}{12}$ , $\frac{58}{45}$

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The sum of fractions is 31324. When the largest fraction is divided by the smallest, the fraction thus obtained is 78, which is 14 more than the middle one. The fractions are:

The correct option is A (203/180),(29/12), (58/45) Middle number is 78+14=2912 So we can assume the numbers as 7x and 8x respectively. So we get x =29180 so the numbers are 203180,2912 ,5845

The sum of fractions is $3 \frac{13}{24}$ . When the largest fraction is divided by the smallest, the fraction thus obtained is $\frac{7}{8}$ , which is $\frac{1}{4}$ more than the middle one. The fractions are:

The correct option is A
(203/180),(29/12), (58/45)

Middle number is $\frac{7}{8} + \frac{1}{4} = \frac{29}{12}$

So we can assume the numbers as 7x and 8x respectively.

So we get x = $\frac{29}{180}$ so the numbers are $\frac{203}{180}$ , $\frac{29}{12}$ , $\frac{58}{45}$