Question

Hasan buys two kinds of cloth material for school uniform shirt material that cost him rupees 50 per metre and trouser material that cost him rupees 90 per metre for every 3 metre of the shirt material he buys 2 metre of the trouser material he sells the material at 12% and 10% profit respectively is total sale is 36600 rupees how much trouser material did he buy​

Answer 1

Answer:

assume that Hasan bought x meters of trouser material

so the shirt material Hasan bought is 3/2x = 1.5x

Cost of the trouser material is 90x and cost of shirt material is 50*1.5 x = 75x

He sells shirt material at a profit of 12%. So sales price is (12+100)*75x/100 = 84x

He sells trouser material for a profit of 10% so sales price is (10+100)*90x/100 = 99x

Now total sale is 36,600

99x + 84x = 36,600

183x = 36600

x = 200

He bought 200 m of trouser material

Answer 2

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: \: \: cast \: \: of \: \: {1m} \: \: shirt \: \: material \: \: = \: \: Rs \: {50}$
• $\: \: \: \: \: \: \: cast \: \: of \: \: {1m} \: \: trouser \: \: material \: \: = \: \: Rs \: {90}$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: find \: \: the \: \: length \: \: and \: \: width \: \: of \: \: the \: \: rectangle \: ?$

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: Let \: \: the \: \: length \: \: of \: \: shirt \: \: matrial \: \: be \: \: {2x}.$
• $\: \: \: \: \: \: \: Let \: \: the \: \: length \: \: of \: \: trouser \: \: matrial \: \: be \: \: {3x}.$

$\: \: \: \: \: \: \: \therefore \: S.P \: \: of \: \: {1m} \: \: of \: \: trouser \: \: matrial \: \: \leadsto \: \: {[{90} \: \times \: {2x} \: + \: \frac{12}{100} \: \times \: {90} \: \times \: {2x}]}$

$\: \: \: \: \: \: \: \leadsto \: \: {[{180x} \: + \: \frac{108x}{5}]}$

$\: \: \: \: \: \: \: \leadsto \: \: {[\frac{{900x} \: + \: {108x}}{5}]}$

$\: \: \: \: \: \: \: \leadsto \: \: {[\frac{1008x}{5}]}$

$\: \: \: \: \: \: \: \therefore \: S.P \: \: of \: \: {1m} \: \: of \: \: trouser \: \: matrial \: \: \leadsto \: \: {[{50} \: \times \: {3x} \: + \: \frac{10}{100} \: \times \: {50} \: \times \: {3x}]}$

$\: \: \: \: \: \: \: \leadsto \: \: {[{50} \: \times \: {3x} \: + \: \frac{10}{2} \: \times \: {3x}]}$

$\: \: \: \: \: \: \: \leadsto \: \: {[{150x} \: + \: {5} \: \times \: {3x}]}$

$\: \: \: \: \: \: \: \leadsto \: \: {[{150x} \: + \: {15x}]}$

$\: \: \: \: \: \: \: \leadsto \: \: {[{165x}]}$

• $\: \: \: \: \: \: \: Total \: \: sale \: \: = \: \: Rs \: {36600}$

$\: \: \: \: \: \: \: \therefore S.P \: \: of \: \: trouser \: \: matrial \: \: + \: \: S.P \: \: of \: \: shirt \: \: matrial \: \: = \: \: Total \: \: sale.$

$\: \: \: \: \: \: \: \leadsto \: \: \frac{1008x}{5} \: + \: {165x} \: \: = \: \: {36600}$

$\: \: \: \: \: \: \: \leadsto \: \: \frac{{1008x} \: + \: {825x}}{5} \: \: = \: \: {36600}$

$\: \: \: \: \: \: \: \leadsto \: \: \frac{1833x}{5} \: \: = \: \: {36600}$

$\: \: \: \: \: \: \: \leadsto \: \: {1833x} \: \: = \: \: {36600} \: \times \: {5}$

$\: \: \: \: \: \: \: \leadsto \: \: {1833x} \: \: = \: \: {183,025}$

$\: \: \: \: \: \: \: \leadsto \: \: {x} \: \: = \: \: \frac{183,025}{1833}$

$\: \: \: \: \: \: \: \leadsto \: \: {x} \: \: = \: \: {99.84}$

$\: \: \: \: \: \: \: \therefore The \: \: value \: \: of \: \: x \: \: is \: \: assumption \: \: {100}$

• $\: \: \: \: \: \: \: shirt \: \: material \: \: \leadsto \: \: {2x}$

$\: \: \: \: \: \: \: \leadsto \: \: {2} \: \times \: {100}$

$\: \: \: \: \: \: \: \leadsto \: \: {200}$

• $\: \: \: \: \: \: \: trouser \: \: material \: \: \leadsto \: \: {3x}$

$\: \: \: \: \: \: \: \leadsto \: \: {3} \: \times \: {100}$

$\: \: \: \: \: \: \: \leadsto \: \: {300}$

• $Hence, \\ \: \: \: \: \: \: \: shirt \: \: material \: \: \leadsto \: \: {200} \\ \: \: \: \: \: \: \: trouser \: \: material \: \: \leadsto \: \: {300.}$

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