# Comparing Quantities Class 8 Notes: Chapter 8

Comparing Quantities Class 8 Notes given here has been carefully put together by experts to help students understand all the concepts given in chapter 8 clearly and at the same time allow them to practice sums effectively. The notes are further designed to help students complete timely revisions and score better marks in the exams.

## Introduction to Fraction and Ratios

### Fractions and Ratios

A fraction represents a part of a whole which consists of numerators and denominators and it is the division of two same quantities.
Eg: $\frac{3}{5}$

Ratio is the comparison of one value to the other or the comparison of two different quantities.
Eg:3:5

## Finding the Increase or Decrease in Percent

### Finding Increase or Decrease Percentage in Situations

Finding new number, when there is increase in percentage.
New number = original number + (increase in percentage × number)
Ex : The Cost of a mobile phone  is Rs 15,000. Find the new price if there is a increaseof 5%
New price = original price + 5% of original price
New price = 15,000+$\frac{5}{100}$ ×15,000
New price = 15,000+750 = 15,750
Here Rs 750 is increase in the price.

The new number can be found out using,
New number = original number × percentage increase
Ex : New price = 15,000×105÷100=15,000×1.05=15,750

Finding new number, when there is decrease in percentage.
New number = original number – (decrease in percentage × number)
Also, New number = original number × percentage decrease
Ex : The Cost of a mobile phone  is Rs 15,000. Find the new price if there is a decrease of 5%
New price = 15,000×95÷100=15,000×0.95=14,250

## Finding Discounts

### Finding SP without Finding Discount Percentage

A reduction (decrease) on the marked price is known as discount.
If the discount is given in numbers then it is calculated by
Discount = Marked price – Sale price

If the discount is given in percentage then it is calculated by
Discount = Discount % of Marked price

### Finding Discounts

If the discount is given in numbers.
Example : Marked price of a shirt is Rs 535. Its selling price is Rs 495. Find the discount.
Solution : Discount = Marked price – Sale price
Discount = Rs 535 – Rs 495 = Rs 40

If the discount is given in percentage.
Example : A toy priced Rs 500 is available at a discount of 5%. Find the discount.
Solution : Discount = Discount % of Marked price
Discount = 5% of 500=$\frac{5}{100}$ × 500
Discount = Rs 25

### Estimation of Amounts (In Percentages)

Estimating amounts when there is a discount or hike on the marked price.

Example : Anil bought a  pair of shoes priced Rs 650, at a discount of 10%. Find the billing amount.
Solution : Billing amount = Marked price – discount
Billing amount = Rs 650−$\frac{10}{100}$ ×650
Billing amount = Rs 650Rs 65=Rs 585

Example : Shilpa bought a new mobile​​​​​​ for Rs 15,000. She has to pay 2% as delivery charges.
Find the billing amount.
Solution : Billing amount = Marked price + Hike
Billing amount = Rs 15,000+$\frac{2}{100}$ × 15000
Billing amount = Rs 15,000+Rs 300=Rs 15,300

## Prices Related to Buying and Selling

### Prices / Charges Related to Buying and Selling

Profit = Selling price – Cost price

Profit % =$\frac{Profit}{Cost\;price}$×100

Loss = Cost price – Selling price

Loss % =$\frac{Loss}{Cost\;price}$×100

### Finding Prices / Charges Related to Buying and Selling

Example : A shopkeeper sold a T.V priced Rs 12,000 at Rs 13,500. Find his profit percentage.
Profit = Selling price – Cost price
Profit = Rs 13,500Rs 12,000=Rs 1,500
Profit % =$\frac{Profit}{Cost\;price}$×100
Profit % = $\frac{1500}{12000}$×100=12.5%

Example : Amit sold his laptop, priced Rs 20,000 at Rs 18,000. Find his loss percentage.
Loss = Cost price – Selling price
Loss = Rs 20,000Rs 18,000=Rs 2000
Loss % =$\frac{Loss}{Cost\;price}$×100
Loss% = $\frac{2000}{20,000}$×100=10%

## Sales Tax and Value Added Tax

### Sales Tax / VAT

Sales tax or value added tax(VAT) is the tax that should be paid to the government on sale of an item
and it is added to the bill amount.
Normally, VAT is included in the price of items like groceries.

### Finding Sales Tax / VAT

Sales tax or VAT =  Tax % of  Selling price

Billing Amount = Selling price + VAT

Example : Megha bought a wrist watch for Rs 1,200 and VAT is charged at 8%. Calculate the VAT and billing amount.
Solution : VAT = Tax % of selling price
VAT = 8% of 1,200=$\frac{8}{100}$×1200=Rs 96
Billing amount = S.P + VAT = Rs 1,200 + Rs 96 = Rs 1296.

## Simple and Compound Interest

### SI

Simple interest is the extra money charged on a loan where the principal amount will be fixed for a  particular time period.
Interest is the extra money that a bank gives for saving or depositing money with them.
Similarly, when anybody borrow money, they pay interest.

Simple interest =$\frac{P.T.R}{100}$, where

P is the principal amount
T is the number of years.
R is the interest rate

### Calculating CI

Compound interest is the interest, calculated on the principal and the interest for the previous period.
The principal amount  increases with every time period, as the interest payable is added to the principal.

Eg : Find CI on Rs 10,000 for 2 years at an interest rate of 5%.

Ans : Interest for the 1st year
For 1st year, P = 10,000, T = 1 year, R = 5%
I1=$\frac{P.T.R}{100}$ = $\frac{10000.1.5}{100}$ =Rs 500
A=P+I1=10,000+500=10,500

Interest for the 2nd year
For 2nd year, P = 10,500, T = 1 year, R = 5%
I2=$\frac{P.T.R}{100}$=$\frac{10500.1.5}{100}$ =Rs 525

C.I=I1+I2=Rs 500+Rs 525=Rs 1025

## Deducing a Formula for Compound Interest

### Formula for CI

Calculation of compound interest can be generalized.

let P1 be the sum on which the interest is compounded annually at the rate of R
Then the interest  for the 1st year,
I1=$\frac{P_{1}. 1.R} {100}$ =$\frac{P_{1}.R} {100}$
A1=P1+I1=P1+$\frac{P_{1}.R}{100}$
A1=P1(1+$\frac{R}{100}$)=P2

For 2nd year,
P2=P1(1+$\frac{R}{100}$),T=1 year and R=R%
I2=$\frac{P_{2}.1.R} {100}$=$\frac{P_{2}.R}{100}$
I2=P1(1+$\frac{R}{100}$)×$\frac{R}{100}$
I2=$\frac{P_{1} R}{100}$(1+$\frac{R}{100}$)
A2=P2+I2
A2=P1(1+$\frac{R}{100}$)+$\frac{P_{1}R}{100}$(1+$\frac{R}{100}$)
A2=P1(1+$\frac{R}{100}$)(1+$\frac{R}{100}$)  [taking P1(1+$\frac{R}{100}$) as common ] A2=P1(1+$\frac{R}{100}$)2

Continuing this way, the amount at the end of n years will be,
An=P(1+$\frac{R}{100}$)n

i.e., A=P(1+$\frac{R}{100}$)n

Where, P is the principal amount, R is the rate of interest and n is the number of years.
We get the formula for the amount to be paid at the end of n years.
Compound Interest can be calculated using the formula,

CI=AP

## Rate Compounded Annually and Half Yearly

### Rate Compounded Annually or Half-Yearly

If interest is compounded annually,
time span, n = 1 year, here the principal amount varies yearly.
Principal amount (A=P+I1)  for first year will serve as the principal for the second year.

If interest is compounded half – yearly,
time span, n  = 6 months, here the principal amount varies half – yearly.
Principal amount (A=P+I1) for first 6 months will be the principal for the next 6 months.

### Finding CI When Rate Compounded Annually or Semi – Annually

When compound interest is compounded annually,

A=P(1+$\frac{R}{100}$)n

C.I=AP

Where, P is the principal amount, R is the rate of interest and n is the number of years.

When compound interest is compounded half yearly,
the interest rate will be half of the annual interest rate and the time period will be doubled.

A=P(1+$\frac{R}{200}$)2n

C.I=AP

Where, P is the principal amount, R is the rate of interest and n is the number of years.

## Application of Compound Interest

### Application of Formula of CI

Application of compound interest are :

1.To calculate the growth rate of population (increase or decrease).
2. To calculate change in the  price of an item (increase or decrease).

Example : If the population of a town increases 2% annually and the present population is 3,26,40,000, find its population after 2 years.

Solution. P = 3,26,40,000   n = 2 years, R = 2%

Therefore,    Population after 2 years

A=P(1+$\frac{R}{100}$)n

A=32640000(1+$\frac{2}{100}$)2

A=32640000×($\frac{51}{50}$)2

A=32640000×$\frac{51}{50}$ × $\frac{51}{50}$
A=13056×51×51

A=33958656

The population after 2 years is 3,39,58,656

Example : A motorcycle is bought at Rs 1,60,000. Its value depreciates at the rate of 10% per annum. Find its value after 2 years.

Solution. P = 1,60,000  n = 2 years, R = 10%

A=P(1−$\frac{R}{100}$)n

A=160000×(1−$\frac{10}{100}$)2

A=160000×$\frac{9}{10}$ × $\frac{9}{10}$
A=129600

The value of the motorcycle after 2 years is Rs 1,29,600.