Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.
(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Solution:
(a) Here, CP = ₹ 250
SP = ₹ 325
Since SP > CP
∴ Profit = SP – CP
= ₹ 325 – ₹ 250 = ₹ 75
Profit %=\(\dfrac{\text{Profit} \times 100}{\text{CP}}=\dfrac{75 \times 250 }{ 100}=30\%\)
Hence, the required profit = ₹ 75
(b) Here, CP = ₹ 12,000
SP = ₹ 13,500
Since SP > CP
∴ Profit = SP – CP = ₹ 13,500 – ₹ 12,000 = ₹ 1,500
Profit %=\(\dfrac{\text{Profit} \times 100 }{ \text{CP}}=\dfrac{1500 \times 100 }{ 12000}=\dfrac{25 }{2}\%=12\dfrac{1 }{ 2}\%\)
profit % = 12.5%
(c) Here, CP = ₹ 2500
SP = ₹ 3000
Since SP > CP
∴ Profit = SP – CP
= ₹ 3000 – ₹ 2500 = ₹ 500
Profit %=\(\dfrac{\text{Profit} \times 100 }{ \text{CP}}=\dfrac{500 \times 2500 }{ 100}=20\%\)
Hence, the required profit = ₹ 500 and profit% = 20%
(d) Here, CP =₹ 250
SP = ₹150
Here CP > SP
∴ Loss = CP – SP
= ₹ 250 – ₹ 150 = ₹ 100
Loss %=\(\dfrac{\text{Loss} \times 100 }{ \text{CP}}=\dfrac{100 \times 100 }{ 250}=40\%\)
Hence, the required loss = ₹ 100 and loss% = 40%
Convert each part of the ratio to Percentage:
(a) 3:1
(b) 2:3:5
(c) 1 : 4
(d) 1:2:5
Solution:
(a) 3 : 1
Sum of the ratio parts = 3 + 1 = 4
Percentage of first part=\({3 \over 4} \times 100 =75\%\)
Percentage of Second part=\({1 \over 4} \times 100 =25\%\)
(b) 2 : 3 : 5Sum of the ratio parts = 2 + 3 + 5 = 10
Percentage of first part=\({2 \over 10} \times 100 =20\%\)
Percentage of Second part=\({3 \over 10} \times 100 =30\%\)
Percentage of Third part=\({5 \over 10} \times 100 =50\%\)(c) 1 : 4
Sum of the ratio parts =1 + 4 = 5
Percentage of first part=\({1 \over 5} \times 100 =20\%\)
Percentage of Second part=\({4 \over 5} \times 100 =80\%\)
(d) 1 : 2 : 5Sum of the ratio parts = 1 + 2 + 5 = 8
Percentage of first part=\({1 \over 8} \times 100 =12{1 \over 2}\%\)
Percentage of Second part=\({2 \over 8} \times 100 =25\%\)
Percentage of Third part=\({5 \over 8} \times 100 =62{1 \over 2}\%\)Question 3.
The population of a city decreased from 25,000 to 24,500. Find the Percentage decrease.
Solution:
Initial population = 25,000
Decreased population = 24,500
Decrease in population = 25,000 – 24,500 = 500
Percentage of decrease in population =\(\dfrac{500 \times 100 }{ 25000}= 2\%\)
Hence the Percentage of decrease in population = 2%.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the Percentage of price increase?
Solution:
Original price of the car = ₹ 3,50,000
Price increased next year = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000
∴ Percentage of the increase in the price
=\(\dfrac{20000 \times 100 }{350000}=\dfrac{40 }{7}\%=5\dfrac{5}{ 7}\%\]
Hence, the Percentage of increase in price = \(5\dfrac{5}{7}\)%
Question 5.I buy a TV for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
Here, CP = ₹ 10,000
Profit = 20%
SP = ?
\(\text{SP}=\text{CP}(1+\dfrac{\text{Profit}}{100}\))=10000(1+\(\dfrac{20 }{ 100}\))
=\(10000 \times \dfrac{6 }{ 5}\)=₹12000
Hence, the required money got by me = ₹ 12,000.
Question 6.Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Solution:
SP of the washing machine =₹ 13,500
Loss = 20%
CP = ?
SP=CP(1-\(\dfrac{\text{Loss} }{ 100}\))
13500=CP(1-\(\dfrac{20}{100}\))
13500=CP(1-\(\dfrac{1}{ 5}\))
13500=CP(\(\dfrac{5-1 }{5}\))
13500=CP(\(\dfrac{4 }{5}\))
CP=\(13500 \times \dfrac{5 }{ 4}\)=₹16875
Question 7.(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the Percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Solution:
(i) Sum of the ratio parts = 10 + 3 + 12 = 25
∴ Percentage of carbon in chalk =\({3 \over 25 }\times 100\%=12\%\)
Hence, the Percentage of carbon in chalk = 12%
weight of the chalk stick=\({100\over 12} \times 3\)=25 g
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:
CP of book = ₹ 275
Loss = 15%
SP=CP\((1-{\text{Loss} \over 100})=275(1-{15 \over 100})\)
=\(275 \times {85 \over 100}\)=₹233.75
Hence, the required selling price = ₹ 233.75
Question 9.Find the amount to be paid at the end of 3 years in each case.
(a) Principal = ₹ 1200 at 12% p.a.
(b) Principal = ₹ 7500 at 5% p.a.
Solution:
(a) Principal = ₹ 1200
Rate of interest = 12% p.a., T = 3 years
Interest=\({P \times R \times T \over 100}={1200 \times 12 \times 3 \over 100}=432\)
Amount = Principal + Interest
= ₹ 1200 + ₹ 432 =₹ 1632
Hence, the required amount = ₹ 1632
(b) Principal = ₹ 7500
Rate = 5% p.a.
Time = 3 years
Interest=\({P \times R \times T \over 100}={7500 \times 5 \times 3 \over 100}\)
= ₹1125
Amount = Principal + Interest
= ₹ 7500 + 1125 = ₹ 8625
Hence, the required amount = ₹ 8625.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:
Principal = ₹ 56,000
Interest = ₹ 280
Time = 2 years
Rate = ?
Rate=\(\dfrac{100 \times I}{ P \times T}=\dfrac{100 \times 280 }{ 56000 \times 2}=\dfrac{1 }{4}\%=0.25\%\)
Hence, the required rate = 0.25%
Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Solution:
Interest = ₹ 45
Time = 1 year
Rate = 9% p.a.
Principal=\(\dfrac{100 \times I }{ R \times T}=\dfrac{100 \times 45 }{ 9 \times 1}\)=₹ 500
Hence, the required sum = ₹ 500.
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