Fill in the blanks to make the statement true.
Narain bought $120$ oranges $R s .4$ each. He sold $60 %$ of the oranges at $R s .5$ each and the remaining at $R s .3 .50$ each. His ......... is ........... $% .$

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Solution

Total oranges $= 120$

$60 %$ of total oranges $= \frac{60}{100} \times 120 = 72$

Remaining oranges $= 120 - 72 = 48$

Now,
Cost price of $120$ oranges $= 120 \times 4 = R s .480$
Selling price of $72$ oranges $= 72 \times 5 = R s .360$
Selling price of $48$ oranges $= 48 \times 3.50 = R s .168$
Selling price of $120$ oranges $= 360 + 168 = R s .528$

Here, $S . P . > C . P .$

therefore, profit that Narain gets
$\Rightarrow 528 - 480 = R s .48$

$P r o f i t % = \frac{48}{480} \times 100$
$= 10 %$

His Profit is 10 $% .$

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Fill in the blanks to make the statement true.Narain bought 120 oranges Rs.4 each. He sold 60% of the oranges at Rs.5 each and the remaining at Rs.3.50 each. His ......... is ........... %.

Fill in the blanks to make the statement true.
Narain bought $120$ oranges $R s .4$ each. He sold $60 %$ of the oranges at $R s .5$ each and the remaining at $R s .3 .50$ each. His ......... is ........... $% .$