Exercise 8.3 Page: 171

**1. 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 Rs 250 and sold for Rs 325.**

**Solution:-**

From the question, it is given that

Cost price of gardening shears = Rs 250

Selling price of gardening shears = Rs 325

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= Rs (325 – 250)

= Rs 75

Profit % = {(Profit/CP) × 100}

= {(75/250) × 100}

= {7500/250}

= 750/25

= 30%

**(b) A refrigerator bought for Rs 12,000 and sold at Rs 13,500.**

**Solution:-**

From the question, it is given that

Cost price of refrigerator = Rs 12000

Selling price of refrigerator = Rs 13500

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= Rs (13500 – 12000)

= Rs 1500

Profit % = {(Profit/CP) × 100}

= {(1500/12000) × 100}

= {150000/12000}

= 150/12

= 12.5%

**(c) A cupboard bought for Rs 2,500 and sold at Rs 3,000.**

**Solution:-**

From the question, it is given that

Cost price of cupboard = Rs 2500

Selling price of cupboard = Rs 3000

Since (SP) > (CP), so there is a profit

Profit = (SP) – (CP)

= Rs (3000 – 2500)

= Rs 500

Profit % = {(Profit/CP) × 100}

= {(500/2500) × 100}

= {50000/2500}

= 500/25

= 20%

**(d) A skirt bought for Rs 250 and sold at Rs 150.**

**Solution:-**

Since (SP) < (CP), so there is a loss

Loss = (CP) – (SP)

= Rs (250 – 150)

= Rs 100

Loss % = {(Loss/CP) × 100}

= {(100/250) × 100}

= {10000/250}

= 40%

**2. Convert each part of the ratio to percentage:**

**(a) 3 : 1**

**Solution:-**

We have to find total parts by adding the given ratio = 3 + 1 = 4

1^{st} part = ¾ = (¾) × 100 %

= 3 × 25%

= 75%

2^{nd }part = ¼ = (¼) × 100%

= 1 × 25

= 25%

**(b) 2: 3: 5**

**Solution:-**

We have to find total parts by adding the given ratio = 2 + 3 + 5 = 10

1^{st} part = 2/10 = (2/10) × 100 %

= 2 × 10%

= 20%

2^{nd }part = 3/10 = (3/10) × 100%

= 3 × 10

= 30%

3^{rd }part = 5/10 = (5/10) × 100%

= 5 × 10

= 50%

**(c) 1:4**

**Solution:-**

We have to find total parts by adding the given ratio = 1 + 4 = 5

1^{st} part = (1/5) = (1/5) × 100 %

= 1 × 20%

= 20%

2^{nd }part = (4/5) = (4/5) × 100%

= 4 × 20

= 80%

**(d) 1: 2: 5**

**Solution:-**

We have to find total parts by adding the given ratio = 1 + 2 + 5 = 8

1^{st} part = 1/8 = (1/8) × 100 %

= (100/8) %

= 12.5%

2^{nd }part = 2/8 = (2/8) × 100%

= (200/8)

= 25%

3^{rd }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:-**

From the question, it is given that

Initial population of the city = 25000

Final population of the city = 24500

Population decrease = Initial population – Final population

= 25000 – 24500

= 500

Then,

Percentage decrease in population = (population decrease/Initial population) × 100

= (500/25000) × 100

= (50000/25000)

= 50/25

= 2%

**4. Arun bought a car for Rs 3,50,000. The next year, the price went upto Rs 3,70,000. What was the Percentage of price increase?**

**Solution:-**

From the question, it is given that

Arun bought a car for = Rs 350000

The price of the car in the next year, went up to = Rs 370000

Then increase in price of car = Rs 370000 – Rs 350000

= Rs 20000

The percentage of price increase = (Rs 20000/ Rs 350000) × 100

= (2/35) × 100

= 200/35

= 40/7

By using the formula, we have:

CP = Rs {(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:-**

From the question it is given that,

The ratio of calcium, carbon and oxygen in chalk = 10: 3: 12

So, total part = 10 + 3 + 12 = 25

In that 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:-**

From the question it is given that,

Weight of carbon in the chalk = 3g

Let us assume the weight of the stick be x

Then,

12% of x = 3

(12/100) × (x) = 3

X = 3 × (100/12)

X = 1 × (100/4)

X = 25g

∴The weight of the stick is 25g.

**8. Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sell it for?**

**Solution:-**

From the question, it is given that

Cost price of book = Rs 275

Percentage of loss = 15%

Now, we have to find the selling price book,

By using the formula, we have:

SP = {((100 – loss %) /100) × CP)}

= {((100 – 15) /100) × 275)}

= {(85 /100) × 275}

= 23375/100

= Rs 233.75

**9. Find the amount to be paid at the end of 3 years in each case:**

**(a) Principal = Rs 1,200 at 12% p.a.**

**Solution:-**

Given: – Principal (P) = Rs 1200, Rate (R) = 12% p.a. and Time (T) = 3years.

If interest is calculated uniformly on the original principal throughout the loan period, it is called Simple interest (SI).

SI = (P × R × T)/100

= (1200 × 12 × 3)/ 100

= (12 × 12 × 3)/ 1

= Rs432

Amount = (principal + SI)

= (1200 + 432)

= Rs 1632

**(b) Principal = Rs 7,500 at 5% p.a.**

**Solution:-**

Given: – Principal (P) = Rs 7500, Rate (R) = 5% p.a. and Time (T) = 3years.

If interest is calculated uniformly on the original principal throughout the loan period, it is called Simple interest (SI).

SI = (P × R × T)/100

= (7500 × 5 × 3)/ 100

= (75 × 5 × 3)/ 1

= Rs 1125

Amount = (principal + SI)

= (7500 + 1125)

= Rs 8625

**10. What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years?**

**Solution:-**

Given: – P = Rs 56000, SI = Rs 280, t = 2 years.

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 Rs 45 for one year at 9% rate p.a. What is the sum she has borrowed?**

**Solution:-**

From the question it is given that, SI = Rs 45, R = 9%, T = 1 year, P =?

SI = (P × R × T)/100

45 = (P × 9 × 1)/ 100

P = (45 ×100)/ 9

= 5 × 100

= Rs 500

Hence, she borrowed Rs 500.