Hasan buys two kinds of clothes for school uniforms, shirt material that costs him rs 90 per metre and trouser material that costs him rs 50 per metre.For every 3 metres of shirt material he buys 2 metres of trouser material.He sells the materials at 12% and 10% profit respectively.His total sale is ₹36,600.How much trouser material does he buy?
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Now to find the maximum capacity of the container we will have to find the highest factor of 850 and 680. So, from (1) and (2) we can see that the highest common factor is 2×5×17=170. Therefore, 170 litres is the maximum capacity of a container which can measure the petrol of either tanker in an exact number of times.
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Let material bought for trousers and shirts be 2x meters and 3x meters respectively.
Cost price of 2x metres of trouser material at Rs. 90 per metre =Rs(90×2x)=Rs180x and
Cost price of 3x metres of shirt material at Rs. 50 per metre =Rs(50×3x)=Rs150x
Given, profit in case of shirt material = 12%
∴ Its S.P. =Rs(150*112/100)=Rs 168x
Also, profit in case of trouser material = 10%
∴ Its S.P. =Rs(180*110/100)=Rs 198x
∴ Total selling price =Rs.(168x+198x)
=Rs.366x
But the total selling price is given as Rs. 36600.
Therefore, 366x=36600
x=36660/3666=100
Thus, the trouser material purchased
=2×100m
=200m
Cost price of 2x metres of trouser material at Rs. 90 per metre =Rs(90×2x)=Rs180x and
Cost price of 3x metres of shirt material at Rs. 50 per metre =Rs(50×3x)=Rs150x
Given, profit in case of shirt material = 12%
∴ Its S.P. =Rs(150*112/100)=Rs 168x
Also, profit in case of trouser material = 10%
∴ Its S.P. =Rs(180*110/100)=Rs 198x
∴ Total selling price =Rs.(168x+198x)
=Rs.366x
But the total selling price is given as Rs. 36600.
Therefore, 366x=36600
x=36660/3666=100
Thus, the trouser material purchased
=2×100m
=200m
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