Solution:
Given, CP = Rs. 250 and SP = Rs. 260
We have,
Profit = SP-CP = Rs. 260 - Rs. 250 = Rs. 10
We also have,
Profit per cent \(\ = \frac{profit}{CP} \) ×100%
\(\ = \frac{10}{250} \) ×100% = 4 %
2. Buying a pen at Rs. 60, at what price should it be sold to get a profit of 15% ?
Solution:
Given, CP = Rs. 60
SP = ?
Also given, percentage of profit, p = 15 %
\(\ ⇨ \frac{SP-CC}{CP}×100 = 15 \)
\(\ ⇨ \frac{SP-60}{60}×100 = 15 \)
\(\ ⇨ (SP-60)×100 = 15×60 \)
\(\ ⇨ SP - 60 = \frac{900}{100} \)
\(\ ⇨ SP = 9+60 \)
\(\ ⇨ SP = 69 \)
⸫ The required SP should be Rs. 69.
Aternative Mathod
Given, CP = Rs. 60
percentage of profit, p = 15 %
We have,
\(\ SP = \left(\frac{100+p}{100}\right)×CP \)
\(\ ⇨ SP = \left(\frac{100+15}{100}\right)× 60 \)
\(\ ⇨ SP = \left(\frac{115}{100}\right)× 60 \)
\(\ ⇨ SP = 1.15× 60 \)
\(\ ⇨ SP = 69 \)
⸫ The required SP should be Rs. 69.
3. Ramen sold a mobile for Rs. 13500 at a loss of 20%. Find the CP of the mobile.
Solution:
Given, SP = Rs. 13500
loss percentage, l = 20%
We have,
\(\ CP = \left(\frac{100}{100-20}\right)×13500 \)
\(\ ⇨ CP = \left(\frac{100}{80}\right)×13500 \)
\(\ ⇨ CP = \left(\frac{100}{8}\right)×1350 \)
\(\ ⇨ CP = 25×675 \)
\(\ ⇨ CP = 16875 \) (Rs.)
4. If the SP of 10 pens is equal to the CP of 8 pens, calculate the loss or gain percentage.
Solution:
Let the SP of the one pen be \(\ x \)
⸫ the SP of the 10 pens be \(\ 10x \) ...(1)
A/Q, CP of 8 pens \(\ = 10x \) [using (1)]
⸫ CP of one pen \(\ = \frac{10x}{8} \)
⸪ \(\ \frac{10x}{8} > x \) , i.e., CP>SP
⸫ There will be a loss.
Loss \(\ = \frac{10x}{8}-x = \frac{10x-8x}{8} = \frac{2x}{8}= \frac{x}{4} \)
⸫ Percentage of loss \(\ = \frac{Loss~ in~ one~ pen}{CP~ of~ one~ pen} \)×100%
\(\ = \frac{\frac{x}{4}}{\frac{10x}{8}} \)×100%
\(\ = \frac{x}{4}×\frac{8}{10x} \)×100%
\(\ = 20 \)%
5. A cycle bought for Rs. 5000 is again sold at a profit of 12%. Find the SP of the cycle.
Solution:
Given, CP = Rs. 5000
percentage of profit, p = 12 %
We have,
\(\ SP = \left(\frac{100+p}{100}\right)×CP \)
\(\ ⇨ SP = \left(\frac{100+12}{100}\right)× 5000 \)
\(\ ⇨ SP = \left(\frac{112}{100}\right)× 5000 \)
\(\ ⇨ SP = 112× 50 \)
\(\ ⇨ SP = 5600 \) (Rs.)
6. Kamal bought a water filter for Rs. 4500 and sold it at Rs. 4230. Calculate the loss per cent.
Solution:
Given, CP = Rs. 4500 and SP = Rs. 4230
We have,
Loss per cent \(\ = \frac{CP-SP}{CP} \) ×100%
\(\ = \frac{4500-4230}{4500} \) ×100%
\(\ = \frac{270}{45} \)%
\(\ = 6 \)%
7. A shopkeeper sold a watch for Rs. 785 at a loss of 5%. What is the CP of the watch?
Solution:
Given, SP = Rs. 785
loss percentage, l = 5%
We have,
\(\ CP = \left(\frac{100}{100-5}\right)×785 \)
\(\ ⇨ CP = \left(\frac{100}{95}\right)×785 \)
\(\ ⇨ CP = \left(\frac{20}{19}\right)×785 \)
\(\ ⇨ CP = 1.052×785 \)
\(\ ⇨ CP = 826.32 \) (Rs.)
8. If the selling price of 10 items is equal to the cost price of 11 items of the same types, find the profit or loss percentage.
Solution:
Let the SP of the one item be \(\ x \)
⸫ the SP of the 10 items be \(\ 10x \) ...(1)
A/Q, CP of 11 items \(\ = 10x \) [using (1)]
⸫ CP of one item \(\ = \frac{10x}{11} \)
⸪ \(\ x > \frac{10x}{11} \) , i.e., SP>CP
⸫ There will be a profit.
Profit \(\ = x - \frac{10x}{11}= \frac{11x-10x}{11} = \frac{x}{11} \)
⸫ Percentage of loss \(\ = \frac{\frac{x}{11}}{\frac{10x}{11}} \)×100%
\(\ = \frac{x}{11}×\frac{11}{10x} \)×100%
\(\ = 10 \)%
9. A man bought two cars for Rs. 99000 each. He sold one of them at a profit of 10% and one at a loss of 10%. Calculate the profit or loss percentage in the whole transaction.
Solution:
As the percentage of profit and loss for both the cars are same which is 10%.
So, there will be a loss by \(\ \frac{10^2}{100} \)% \(\ = \frac{100}{100} \)% = 1 %
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