Question

A trader had 22 quintals of wheat. He sold a part of it at 23% profit and the rest at 33% profit, so that he made a total profit of 27%. How much wheat did he sell at 33% profit?

Answer 1

Answer:

8.8 kg

Given:

A trader has 22 quintal of wheat. He sold a part for 23% and rest at 33%

He made total of profit of 27%

To find:

Wheat he selled at 33% profit

Solution:-

Let the wheat sold at 23% be x kg and wheat at 33% be 22-x

Now

According to the question

$x + (1 + \frac{23}{100}) + (22 - x)(1 + \frac{33}{100}) = 22(1 + \frac{27}{100})$

$x (\frac{123}{100}) - x( \frac{133}{100}) + (\frac{22 \times 133}{100}) = (\frac{22 \times 127}{100})$

$\frac{ - 10x}{100} + \frac{2926}{100} = \frac{2794}{100}$

$\frac{ - 10x}{100} = \frac{2794}{100} - \frac{2926}{100}$

$\frac{ - 10x}{100} = \frac{ - 132}{100}$

$x = \frac{132}{10}$

$= 13.2 \: kg$

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Trader sold (22-x) at 33%

=22-13.2

=8.8 kg

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He selled wheat at 33% is 8.8 kg

Answer 2

Answer:

8.8 kg.

Explanation:

Given:

= x + ( 1 - 23 / 100 ) + ( 22 - x ) ( 1 + 3 / 100 ) = 22 ( 1 + 27 / 100 ).

= x ( 123 / 100 ) - x ( 133 / 100 ) + ( 22 × 133 ) / 100 = ( 22 × 127 / 100 ).

Then

→ 10x / 100 + 2926 / 100 = 2794 / 100

→ 10x / 100 = 2794 / 100 - 2926 / 100

→ 10x / 100 = - 132 / 100

→ x = 132 / 10 = 13.2 kg

Now 22 - x = 33 (%)

33% is the profit percentage.

→ 22 - x = 33%

→ 22 - 13.2

→ 8.8 kg

Therefore, 8.8 kg is the weat sold at 33% profit.