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Maharashtra State BoardSSC (English Medium) 6th Standard

Using the figures given below, frame problems based on profit percent or loss percent.​ ​Cost price ₹ 8600, transport charges ₹ 250, porterage ₹ 150, selling price ₹ 10000

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Question

Using the figures given below, frame problems based on profit percent or loss percent.

​Cost price ₹ 8600, transport charges ₹ 250, porterage ₹ 150, selling price ₹ 10000

Sum
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Solution

Problem: Vinay bought a machine from a factory for Rs 8600. He spent Rs 250 on transport and Rs 150 on porterage. If he sold it to a customer for Rs 10000, what was his profit percent or loss percent?

Cost price of machine = ₹ 8600

Money spent on transport = ₹ 250

Money spent on porterage = ₹ 150

Total expenses = cost of machine + transport charges + porterage charges

= 8600 + 250 + 150

= ₹ 9000

That is, total cost price = ₹ 9000

Selling price of the machine = ₹ 10000

Since selling price is more than total cost price, so there is a profit.

Profit = Selling price − total cost price

= 10000 − 9000

= ₹ 1000

Let x % be the Vinay's profit in this transaction. Then,

`"x"/100 = "Profit"/"total cost price" = 1000/9000`

`=> "x"/100 xx 100 = 1000/9000 xx 100`

`=> "x" = 100/9 = 11 1/9`

∴ Vinay made a profit of `11 1/9%` in this transaction.

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Chapter 13: Profit – Loss - Practice Set 34 [Page 72]

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Balbharati Mathematics [English] Standard 6 Maharashtra State Board
Chapter 13 Profit – Loss
Practice Set 34 | Q 5 | Page 72
Balbharati Mathematics Integrated [English] Standard 6 Maharashtra State Board
Chapter 13 Profit-Loss
Practice Set 34 | Q 5. | Page 67

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