Question

A basket contains fruits weighing 58/3 in all. If 73/9 kg of these be apples 19/6 kg be oranges and the rest years what is the weight of the pears in the basket

Answer 1

Answer:

$\large{\underline{\boxed{\sf Weight \: of \: pears = \dfrac{145}{18}}}}$

Step-by-step explanation:

Given that ;

Total weight of fruits = $\sf \dfrac{58}{3 }$

Weight of apples = $\sf \dfrac{73}{9}$

Weight of oranges = $\sf \dfrac{19}{6}$

Let the weight of pears be x.

According to the question ;

$\implies \sf \frac{73}{9} + \frac{19}{6} + x = \frac{58}{3} \\ \\ \implies \sf \frac{146 + 57 + 18x}{18} = \frac{58}{3} \\ \\ \implies \sf \frac{203 + 18x}{18} = \frac{58}{3} \\ \\ \bf on \: cross - multiplying : \\ \\ \implies \sf 3(203 + 18x) = 58 \times 18 \\ \\ \implies \sf 203 + 18x = \frac{58 \times 18}{3} \\ \\ \implies \sf 203 + 18x = 348 \\ \\ \implies \sf 18x = 348 - 203 \\ \\\implies \sf 18x = 145\\ \\ \implies \sf x = \frac{145}{18}$

Hence, the weight of pears is $\bf \dfrac{145}{18}$ .

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Verification ;

$\implies \sf \frac{73}{9} + \frac{19}{6} + \frac{145}{18} = \frac{58}{3} \\ \\ \implies \sf \frac{146 + 57 + 145}{18} = \frac{58}{3} \\ \\ \implies \sf \frac{348}{18} = \frac{58}{3} \\ \\ \implies \sf \frac{58}{3} = \frac{58}{3}$

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Hence, the weight of pears is $\bf \dfrac{145}{18}$ .