Question

answer with full processs



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Answer 1

→ Total weight -

$19 \frac{1}{3} \: = \: \frac{58}{3} kg$

→ Weight of apples -

$8 \frac{1}{9} \: = \frac{73}{9} kg$

→ Weight of oranges -

$3 \frac{1}{6} kg = \frac{19}{6} kg$

→ Let, weight of pears be 'x'

So, we got an simple equation to compute, i.e -

weight of all the fruits = total weight

$\frac{73}{9} + \frac{19}{6} + x \: = \frac{58}{3} \\ \\ \frac{146 + 57}{18} + x = \frac{58}{3} \\ \\ \frac{203}{18} + x = \frac{58}{3} \\ \\ x = \frac{58}{3} - \frac{203}{18} \\ \\ x = \frac{348 - 203}{18} \\ \\ x = \frac{145}{18} = 8 \frac{1}{18} kg$

∴ weight of pears = x = 8 ⅛ kg

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Answer 2

$\huge\tt{\bold{\underline{\underline{Given᎓}}}}$

Total weights of the fruits =$19 \frac{1}{3} kg$

weights of apples=$8 \frac{1}{9} kg$

weights of oranges =$3 \frac{1}{6}$kg

To find :

weight of pears ..??

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

Step by step explanation:

Total weight of the fruits =$19 \frac{1}{3} kg = \frac{58}{3} kg$

Weight of apples=$8 \frac{1}{9} kg = \frac{73}{9} kg$

weights of oranges =$3 \frac{1}{6} = \frac{19}{6} kg$

let the weight of the pears be 'x'

According to the question:-

$= > \frac{73}{9} + \frac{19}{6} + x = \frac{58}{3}$

L.C.M OF 9 & 6 IS 18

$= > \frac{73 \times 2 + 19 \times 3}{18} + x = \frac{58}{3}$

$= > \frac{146 + 57}{18} + x = \frac{58}{3}$

$= > \frac{203}{18} + x = \frac{58}{3}$

$= > x = \frac{58}{3} - \frac{203}{18}$

L.C.M OF 3&18 IS 18

$= > \frac{58 \times 6 - 203}{18} = \frac{348 - 203}{18}$

$= \frac{145}{18} = 18 \frac{1}{8} kg$

Therefore,$\bold{\red{18 \frac{1}{8} kg}}$ is the weight of the pears