A dealer sold 6 sewing machines for Rs. 63,000 with a profit of 5%. For how much should be sell 8 machines if he intends to earn 15% profit?
Rs. 92,000
The question asks us to determine the selling price needed for 8 sewing machines to achieve a 15% profit, given the selling price and profit percentage for 6 machines.
To solve this, we first need to find the original cost price (CP) of the machines. The cost price is the base value upon which profit or loss is calculated. Once we know the cost price per machine, we can calculate the total cost price for 8 machines and then determine the selling price required for a 15% profit.
We know the selling price (SP) of 6 sewing machines is Rs. 63,000 and the profit earned is 5%.
The relationship between SP, CP, and Profit Percentage is:
\( SP = CP \times (1 + \frac{Profit\%}{100}) \)
Plugging in the given values:
\( 63000 = CP_{6 \text{ machines}} \times (1 + \frac{5}{100}) \)
\( 63000 = CP_{6 \text{ machines}} \times (1 + 0.05) \)
\( 63000 = CP_{6 \text{ machines}} \times 1.05 \)
Now, we can find the CP of 6 machines:
\( CP_{6 \text{ machines}} = \frac{63000}{1.05} \)
\( CP_{6 \text{ machines}} = \frac{63000 \times 100}{105} \)
\( CP_{6 \text{ machines}} = \frac{6300000}{105} \)
Performing the division:
\( CP_{6 \text{ machines}} = 60000 \)
So, the cost price of 6 sewing machines is Rs. 60,000.
If 6 machines cost Rs. 60,000, the cost price of one machine is:
\( CP_{1 \text{ machine}} = \frac{CP_{6 \text{ machines}}}{6} \)
\( CP_{1 \text{ machine}} = \frac{60000}{6} \)
\( CP_{1 \text{ machine}} = 10000 \)
The cost price of one sewing machine is Rs. 10,000.
To find the cost price of 8 machines, we multiply the cost price of one machine by 8:
\( CP_{8 \text{ machines}} = CP_{1 \text{ machine}} \times 8 \)
\( CP_{8 \text{ machines}} = 10000 \times 8 \)
\( CP_{8 \text{ machines}} = 80000 \)
The cost price of 8 sewing machines is Rs. 80,000.
Now we want to find the selling price (SP) for these 8 machines to earn a 15% profit on their cost price (Rs. 80,000).
Using the formula \( SP = CP \times (1 + \frac{Profit\%}{100}) \):
\( SP_{8 \text{ machines}} = CP_{8 \text{ machines}} \times (1 + \frac{15}{100}) \)
\( SP_{8 \text{ machines}} = 80000 \times (1 + 0.15) \)
\( SP_{8 \text{ machines}} = 80000 \times 1.15 \)
Performing the multiplication:
\( SP_{8 \text{ machines}} = 80000 \times 1.15 \)
\( SP_{8 \text{ machines}} = 92000 \)
The dealer should sell 8 sewing machines for Rs. 92,000 to earn a 15% profit.
| Item | Value |
|---|---|
| SP of 6 machines | Rs. 63,000 |
| Profit % on 6 machines | 5% |
| CP of 6 machines | Rs. 60,000 |
| CP of 1 machine | Rs. 10,000 |
| CP of 8 machines | Rs. 80,000 |
| Desired Profit % on 8 machines | 15% |
| Required SP for 8 machines | Rs. 92,000 |
To earn a 15% profit, the dealer should sell 8 sewing machines for Rs. 92,000.
| Term | Definition | Formula Example |
|---|---|---|
| Cost Price (CP) | The original price at which an item is bought. | - |
| Selling Price (SP) | The price at which an item is sold. | - |
| Profit | When SP > CP. Calculated as SP - CP. | \( \text{Profit} = \text{SP} - \text{CP} \) |
| Loss | When SP < CP. Calculated as CP - SP. | \( \text{Loss} = \text{CP} - \text{SP} \) |
| Profit % | Profit expressed as a percentage of the CP. | \( \text{Profit}\% = (\frac{\text{Profit}}{\text{CP}}) \times 100 \) |
| Loss % | Loss expressed as a percentage of the CP. | \( \text{Loss}\% = (\frac{\text{Loss}}{\text{CP}}) \times 100 \) |
| SP with Profit | Calculating SP when CP and Profit% are known. | \( \text{SP} = \text{CP} \times (1 + \frac{\text{Profit}\%}{100}) \) |
| SP with Loss | Calculating SP when CP and Loss% are known. | \( \text{SP} = \text{CP} \times (1 - \frac{\text{Loss}\%}{100}) \) |
| CP from SP & Profit % | Calculating CP when SP and Profit% are known. | \( \text{CP} = \frac{\text{SP}}{1 + \frac{\text{Profit}\%}{100}} \) |
This problem effectively uses the unitary method concept. By first finding the cost price of a single unit (one sewing machine), we can then easily scale that cost price to any number of units (8 machines in this case).
This approach is useful in many percentage-based problems involving multiple items. It breaks down a complex problem into simpler, per-unit calculations before scaling back up to the desired total quantity.
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