Question

please help help....... Q no. 19.

Answer 1

hey mate
here's ur answer

given
$total \: weight = 60 \times \frac{1}{2} = \frac{121}{2} kg$
$box \: b = 3 \frac{1}{2} = \frac{7}{2} kg \: more \: than \: box \: a$
$box \: c = 5 \frac{1}{3} kg \: more \: than \: box \: b$
consider weight of box à as x

total weight = box à+box b + box ç
$60 \frac{1}{2} = x + (x + \frac{7}{2} ) +( x + \frac{7}{2} + \frac{16}{3} ) \\ \frac{121}{2} = 3x + 7 + \frac{16}{3}$
$121 \times 3 = 2(3x + 7 + 16) \\ 363 = 6x + 14 + 32 \\ 363 - 46 = 6x \\ 317 = 6x \\ \frac{317}{6} = x \\ 52 \frac{5}{6} = x$
weight of box a is 52 integer 5 /6 kg

hope it helps