**Comparing Quantities**

**Profit & Loss and Simple Interest****Important points**

Simple Interest

1. C.P. = Cost Price (Amount at which goods are purchased)

2. Sales Tax, VAT, cartage etc.are included in cost price.

3. S.P. = Selling Price (Amount at which goods are sold)

4. If SP > CP then it is Profit or Gain. Profit = SP - CP

5. If CP > SP then it is Loss. Loss = CP - SP

6. Profit of Loss per cent is always calculated on the cost price.

7. Similarly SP and CP can be computed as:

Simple Interest

8. Principal (P): Money borrowed or invested is called Principal or sum.

9. Interest: Additional money earned or paid on the invested money.

10. Rate (R%) = Interest on Rs 100 for a year.

11. Simple Interest (SI) = Interest calculated uniformly on the original principal throughout the loan or invested period.

12. Amount = Principal + Interest

13. If P = principal, R = rate% per annum, T = number of years,

SI = P × R × T/100

**NCERT EXERCISE 8.3**

**Q1: 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.

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

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

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

Answer:

(a) SP = ₹325

CP = ₹250

Profit = 325 - 250 = ₹

**75**

Profit% = (Profit/CP) × 100 = (75/250) × 100 =

**30%**

(b) SP = ₹13500

CP = ₹12000

SP > CP ⇒ Profit

Profit = 13500 - 12000 = ₹1500

Profit% = (Profit/CP) × 100 =(1500/12000) ×100 =

**₹**

**12.5%**

(c) SP = ₹3000

CP = ₹2500

SP > CP ⇒ Profit

Profit = SP - CP = 3000 - 2500 = ₹500

Profit% = (Profit/CP) × 100 = (500/2500) × 100 =

**20%**

(d) SP = ₹150

CP = ₹250

CP > SP ⇒ Loss

Loss = CP - SP = 250 - 150 = ₹100

Loss% = (Loss/CP) × 100 = (100/250) × 100 =

**40%**

**Q2: Convert each part of the ratio to percentage:**

**(a) 3:1 (b) 2:3:5 (c) 1:4 (d) 1:2:5**

Answer:

(a) 3:1

Total parts = 3 + 1 = 4

1st part in % = (3/4) × 100 = 75%

2nd part in % = (1/4) × 100 = 25%

(b) 2:3:5

Total Parts = 2 + 3 + 5 = 10

1st part = (2/10) × 100 = 20%

2nd part = (3/10) × 100 = 30%

3rd part = (5/10) × 100 = 50%

(c) 1:4

Total Parts = 1 + 4 = 5

1st part = (1/5) × 100 = 20%

2nd part = (4/5) × 100 = 80%

(d) 1:2:5

Total parts = 1 + 2 + 5 = 8

1st part = (1/8) × 100 =12.5%

2nd part = (2/8) × 100 = 25%

3rd part = (5/8) × 100 =62.5%

**Q3: The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.**

Answer: Initial population = 25000

Final population = 24500

Decrease in population = Final Population - Initial Population

= 24500 - 25000 = -500

(-ve value indicates decrease).

Percentage decrease = (decrease/initial value) × 100

= (-500/25000) × 100 =

**-2%**

**Q4: 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?**

Answer: Initial price of car = ₹ 350000

Increased price of car = ₹ 370000

Actual Increase = Increased Value - Initial Value = 370000 - 350000

= ₹ 20000

Percent Increase = (Actual Increase/Initial Value) × 100

= (20000/350000) × 100 =

**40/7%**

**Q5: I buy a T.V. for Rs 10,000 and sell it at a profit of 20%. How much money do I get for it?**

Answer: CP = ₹ 10000

Profit% = 20

SP = ?

SP = (100 + Profit%) × CP / 100 = (100+20) × 10000/100

= 120 × 100 =

**₹ 12000**

**Q6: Juhi sells a washing machine for Rs 13,500. She loses 20% in the bargain. What was the price at which she bought it?**

Answer: SP of washing machine = ₹ 13500

loss = 20%

CP = ?

CP =[100/(100 - loss%)] × SP = [100/(100 - 20)] × 13500

= (100/80) × 13500 = ₹16875

**Q7: (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 3g, what is the weight of the chalk stick?**

Answer:

(i) Calcium:Carbon:Oxygen = 10:3:12

Total Parts = 10 + 3 + 12 = 25

Calcium % in Chalk = (10/25) × 100 = 40%

Carbon % in Chalk = (3/25) × 100 = 12%

Oxygen % in Chalk = (12/25) × 100 = 48%

(ii) Weight of Carbon = 3g

Let the weight of chalk = n grams

Carbon % in Chalk = 12%

⇒ 12% of n = 3g

12/100 × n = 3

n = 3 × 100/12 =

**25g**

**Q8: Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sell it for?**

Answer: CP of the book = ₹ 275

Loss% = 15%

SP = [(100-loss%)/100] × CP = [(100 - 15)/100] × 275

= 85 × 275/100 = 23375/100 = ₹

**233.75**

**Q9: Find the amount to be paid at the end of 3 years in each case:**

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

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

Answer:

(a) Principal = ₹1200, R = 12% T = 3years

SI = P×R×T/100 = 1200 × 12 × 3 /100 = ₹432

Amount = Principal + SI = 1200 + 432 = ₹

**1632**

(b) Principal = ₹7500, R = 5% and T = 3 years

SI = P×R×T/100 =7500 × 5 × 3/ 100 = ₹1125

Amount = Principal + SI =7500 + 1125 = ₹8625

**Q10: What rate gives Rs 280 as interest on a sum of Rs 56,000 in 2 years?**

Answer: Principal = ₹56000

SI = ₹280

Time (T) = 2 years

Rate (R) = ?

R = (SI × 100)/ (P × T)

= (280 × 100) / (56000 × 2) = 28/(56 x 2) = 1/4 =

**0.25%**

**Q11: If Meena gives an interest of Rs 45 for one year at 9% rate p.a.. What is the sum she has borrowed?**

Answer: SI = ₹ 45

Rate = 9% p.a.

Time = 1 year

Principal = ?

P = (SI × 100)/ (R × T) = (45 × 100)/ (9 × 1) = ₹

**500**

__Misc. Questions__

**Q12: By selling California apples at Rs 384 a kilogram, a shopkeeper incurs a loss of 4%. At what price mush he sell so that he get a profit of 10%**

Answer: SP =₹

__384 and loss = 4%__

CP = [100/(100 - loss%)] × SP

∴ CP = [100/(100 - 4)] × 384

CP = 100/96 × 384 = 100 × 4 = ₹ 400

CP = ₹ 400

Gain = 10%

SP = [(100+gain%)/100] × CP = [(100+10)/100] × 400

= 110/100 × 400 = ₹

**440**

(In progress...)

this is helpful but give harder sums

ReplyDeleteIts good like to see more

ReplyDeleteit is a little hard if they would have given the sums it would be better i liked it

ReplyDelete:-p

easy to understand. specially for a school student

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteThis comment has been removed by the author.

ReplyDeletenice and correct answers

ReplyDeletevery helpful and understanding

ReplyDeletebut would be better if given harder sums