1. Tell what is the profit or loss in the following transactions. Also, find profit percent or loss per cent in each case.
(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
Solution:
Given,
Cost price (CP) of gardening shears = ₹ 250
Selling price (SP) of gardening shears = ₹ 325
Here, (SP) > (CP)
So, there is a profit
Now,
Profit = (SP) – (CP)
= ₹ (325 – 250)
= ₹ 75
Profit % = {(Profit/CP) × 100}
= {(75/250) × 100}
= {7500/250}
= 750/25
= 30%
(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
Solution:
Given,
Cost price of refrigerator = ₹ 12000
Selling price of refrigerator = ₹ 13500
Here, (SP) > (CP), so there is a profit
Then,
Profit = (SP) – (CP)
= ₹ (13500 – 12000)
= ₹ 1500
Now,
Profit % = {(Profit/CP) × 100}
= {(1500/12000) × 100}
= {150000/12000}
= 150/12
= 12.5%
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
Solution:
Given,
Cost price of cupboard = ₹ 2500
Selling price of cupboard = ₹ 3000
Here, (SP) > (CP), so there is a profit
Profit = (SP) – (CP)
= ₹ (3000 – 2500)
= ₹ 500
Now,
Profit % = {(Profit/CP) × 100}
= {(500/2500) × 100}
= {50000/2500}
= 500/25
= 20%
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Solution:
CP of skirt = ₹ 250
SP of skirt = ₹ 150
Here, (SP) < (CP), so there is a loss
Loss = (CP) – (SP)
= ₹ (250 – 150)
= ₹ 100
Now,
Loss % = {(Loss/CP) × 100}
= {(100/250) × 100}
= {10000/250}
= 40%
2. Convert each part of the ratio to percentage:
(a) 3 : 1
Solution:
First, we add the given ratio to find total parts,
3 + 1 = 4
Now,
1st part = 3/4 = (3/4) × 100 %
= 3 × 25%
= 75%
2nd part = 1/4 = (1/4) × 100%
= 1 × 25
= 25%
(b) 2: 3: 5
Solution:
First, we add the given ratio to find total parts
2 + 3 + 5 = 10
Now,
1st part = 2/10 = (2/10) × 100 %
= 2 × 10%
= 20%
2nd part = 3/10 = (3/10) × 100%
= 3 × 10
= 30%
3rd part = 5/10 = (5/10) × 100%
= 5 × 10
= 50%
(c) 1:4
Solution:
First, we add the given ratio to find total parts
1 + 4 = 5
Now,
1st part = (1/5) = (1/5) × 100 %
= 1 × 20%
= 20%
2nd part = (4/5) = (4/5) × 100%
= 4 × 20
= 80%
(d) 1: 2: 5
Solution:
First, we add the given ratio to find total parts
1 + 2 + 5 = 8
Now,
1st part = 1/8 = (1/8) × 100 %
= (100/8) %
= 12.5%
2nd part = 2/8 = (2/8) × 100%
= (200/8)
= 25%
3rd part = 5/8 = (5/8) × 100%
= (500/8)
= 62.5%
3. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Given,
The initial population of the city = 25000
The final population of the city = 24500
So,
Population decrease = Initial population – Final population
= 25000 – 24500
= 500
Now,
Percentage decrease in population = (population decrease/Initial population) × 100
= (500/25000) × 100
= (50000/25000)
= 50/25
= 2%
4. 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:
Given,
Arun bought a car for = ₹ 350000
The next year, the price of the car went up to = ₹ 370000
Then,
Increase in price of car = ₹ 370000 – ₹ 350000
= ₹ 20000
Now,
The percentage of price increase = (₹ 20000/ ₹ 350000) × 100
= (2/35) × 100
= 200/35
= 40/7
= 5.714 %
5. I buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
Given,
Cost price (CP) of the T.V. = ₹ 10000
Percentage of profit = 20%
Then,
Profit = (20/100) × 10000
= ₹ 2000
Now,
Selling price (SP) of the T.V. = cost price + profit
= 10000 + 2000
= ₹ 12000
Hence, I will get it for ₹ 12000.
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:
Given,
Selling price of washing machine = ₹ 13500
Percentage of loss = 20%
Now,
we have to find the cost price washing machine
We know that,
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 20)) × 13500}
= {(100/ 80) × 13500}
= {1350000/80}
= {135000/8}
= ₹ 16875
7. (i) Chalk contains calcium, carbon, and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.
Solution:
Given,
The ratio of calcium, carbon, and oxygen in chalk = 10: 3: 12
So, we have to add them to find the total part
10 + 3 + 12 = 25
In the total part amount of carbon = 3/25
Then,
Percentage of carbon = (3/25) × 100
= 3 × 4
= 12 %
(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?
Solution:
Given,
Weight of carbon in the chalk = 3g
Now,
Let the weight of the stick be x
According to the question,
12% of x = 3
(12/100) × (x) = 3
X = 3 × (100/12)
X = 1 × (100/4)
X = 25g
Hence, the weight of the stick is 25g.
8. Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:
Given,
Cost price of book = ₹ 275
Percentage of loss = 15%
Now, we have to find the selling price book,
We know that,
SP = {((100 – loss %) /100) × CP)}
= {((100 – 15) /100) × 275)}
= {(85 /100) × 275}
= 23375/100
= ₹ 233.75
9. Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ₹ 1,200 at 12% p.a.
Solution:
Given,
Principal (P) = ₹ 1200
Rate (R) = 12% p.a.
Time (T) = 3years.
Then,
SI = (P × R × T)/100
= (1200 × 12 × 3)/ 100
= (12 × 12 × 3)/ 1
= ₹432
Now,
Amount = (principal + SI)
= (1200 + 432)
= ₹ 1632
(b) Principal = ₹ 7,500 at 5% p.a.
Solution:
Given,
Principal (P) = ₹ 7500
Rate (R) = 5% p.a.
Time (T) = 3years.
Then,
SI = (P × R × T)/100
= (7500 × 5 × 3)/ 100
= (75 × 5 × 3)/ 1
= ₹ 1125
Now,
Amount = (principal + SI)
= (7500 + 1125)
= ₹ 8625
10. What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:
Given,
Principal (P) = ₹ 56000
Simple Interest (SI) = ₹ 280
Time (T) = 2 years.
From the formula, we know that,
R = (100 × SI) / (P × T)
= (100 × 280)/ (56000 × 2)
= (1 × 28) / (56 × 2)
= (1 × 14) / (56 × 1)
= (1 × 1) / (4 × 1)
= (1/ 4)
= 0.25%
11. If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Solution:
Given,
Simple Interest (SI) = ₹ 45
Rate (R) = 9%
(Time)T = 1 year
Principal (P) =?
From the formula, we know that,
SI = (P × R × T)/100
45 = (P × 9 × 1)/ 100
P = (45 ×100)/ 9
= 5 × 100
= ₹ 500
Hence, she borrowed ₹ 500.
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