Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36660. How much trouser material did he buy?

Asked by Sakshi | 1 year ago |  181

1 Answer

Solution :-

Let 2x m of trouser material and 3x m of shirt material be bought by him. Per metre selling price of trouser material = Rs(90 + \(\frac{90 \times 12}{100}\)) = Rs 100.80

Per metre selling price of shirt material = Rs(50 + \(\frac{50 \times 10}{100}\)) = Rs 55

Given that, total amount of selling = Rs 36660

100.80 × (2x) + 55 × (3x) = 36660

201.60x + 165x = 36660

366.60x= 36660

Dividing both sides by 366.60, we obtain

x = 100

Trouser material = 2x m = (2 × 100) m = 200 m

Answered by Aaryan | 1 year ago

Related Questions

Solve the inequations and graph their solutions on a number line  – 1 < (\( \dfrac{x}{2}\)) + 1 ≤ 3, x ε l

Class 8 Maths Linear Equations in One Variable View Answer

Solve (\( \dfrac{x}{3}\)) + (\( \dfrac{1}{4}\)) < (\( \dfrac{x}{6}\)) + (\( \dfrac{1}{2}\)), x ε W. Also represent its solution on the number line.

Class 8 Maths Linear Equations in One Variable View Answer

If the replacement set is {-3, -2, -1, 0, 1, 2, 3}, solve the inequation \(\dfrac {(3x – 1) }{2} < 2\). Represent its solution on the number line.

Class 8 Maths Linear Equations in One Variable View Answer

Solve the inequations (\( \dfrac{3}{2}\)) – (\( \dfrac{x}{2}\)) > – 1, x ε N

Class 8 Maths Linear Equations in One Variable View Answer