FIA 5 Years Past Papers 2018 to 2026 (solved)

There are $n$ cows:
First son gets: $\frac{n}{2}$

Second son gets: $\frac{n}{4}$

Third son gets: $\frac{n}{5}$

Fourth son gets: $7$ cows

Therefore, sum of cows each son gets equals the total number of cows
$\frac{n}{2} + \frac{n}{4} + \frac{n}{5} + 7 = n$
$7 = \frac{n}{2 0}$
$n = 140$

$1 \frac{3}{1 6} = \frac{1 9}{1 6} Its reciprocal= \frac{1 6}{1 9} ∴ Required difference= \frac{1 9}{1 6} - \frac{1 6}{1 9} = \frac{3 6 1 - 2 5 6}{3 0 4} = \frac{1 0 5}{3 0 4}$

There are n cows: First son gets: 2n​Second son gets: 4n​Third son gets: 5n​Fourth son gets: 7 cowsTherefore, sum of cows each son gets equals the total number of cows2n​+4n​+5n​+7=n7=20n​n=140

1163​=1619​ Its reciprocal=1916​∴ Required difference=1619​−1916​=304361−256​=304105​

There are $n$ cows:
First son gets: $\frac{n}{2}$

Second son gets: $\frac{n}{4}$

Third son gets: $\frac{n}{5}$

Fourth son gets: $7$ cows

Therefore, sum of cows each son gets equals the total number of cows
$\frac{n}{2} + \frac{n}{4} + \frac{n}{5} + 7 = n$
$7 = \frac{n}{2 0}$
$n = 140$

$1 \frac{3}{1 6} = \frac{1 9}{1 6} Its reciprocal= \frac{1 6}{1 9} ∴ Required difference= \frac{1 9}{1 6} - \frac{1 6}{1 9} = \frac{3 6 1 - 2 5 6}{3 0 4} = \frac{1 0 5}{3 0 4}$