Author Archives: mekhalaschool

Introduction
There are several changes we come across daily e.g dissolving sugar in water, or flattening a metal rod by beating it. These involve changes in the form of the substance. Changes can be classified as:
(i) Physical
(ii) Chemical

Physical properties
Physical properties include size, shape, colour and state (solid/liquid/gas) of a substance.

Physical change

• Any change to the physical properties of a substance is called a physical change.
• Physical changes are usually reversible as no new substance is formed. It is the same substance but with changed physical properties.

Chemical change

• A change in which one or more new substances are formed is called as a chemical change.
• Usually a chemical change involves a chemical reaction, which forms new products.
• Example : Rusting of Iron, or burning wood.

Metallic Oxides

Formation of metal oxides are examples of chemical changes. They are formed by the reaction of oxygen in air.
– Burning of Magnesium ribbon:
– 2Mg + O₂ → 2MgO
– The product formed is the oxide of magnesium, which is in the form of ash. It does not look anything like the magnesium ribbon used for burning.

Reaction of metallic oxides with water

• Reaction of metal oxides with water form metal hydroxides.
• Example dissolving Magnesium oxide in water, by stirring the ash very well with water.
• MgO + H₂O → Mg(OH)₂
• The product formed is basic in nature and turns red litmus paper → blue

Reaction between baking soda and vinegar

When a pinch of baking soda is added to vinegar we hear a hissing sound and observe the formation of bubbles.

– Vinegar (Acetic Acid) + Baking Soda (Sodium bicarbonate)  → CO₂ (Carbon dioxide) + Other products
The carbon dioxide produced during the reaction of Vinegar and baking soda, when passed through lime water gives calcium carbonate, as follows:
– CO₂ + Ca(OH)₂ (lime water) → Calcium Carbonate (CaCO₃) + H₂O
– The calcium carbonate turns lime water milky.

Observations that indicate a chemical change

• Heat or light is absorbed or given out during a chemical reaction.
• Production of sound
• Production of gases or precipitates
• Production of smell
• A colour change may occur

Rusting

• When substances made of Iron are exposed to oxygen and moisture in the atmosphere, it forms a red layer, which is called rust.
• The formation of rust can be represented by the following reaction:
• 4Fe + 3O₂ → 2Fe₂O₃. The chemical formula for rust is Fe₂O₃.nH₂O. More the moisture in the air, quicker the formation of rust

Galvanization
– The process of depositing zinc on the surface of Iron to prevent rusting is called as galvanisation.
– Example: Iron water pipes are galvanised. Ships are made out of iron which is galvanised. Due to the presence of salts in seawater, the process of rusting is hastened. Hence ships need to replace their iron body every year.

Crystallisation
– The process of separation of salts from their solution is called as crystallisation. It is a purification technique that purifies seawater or separates crystals from impure samples. It is a physical change.

Matter:- Matter is any substance that has mass and takes up some space by
its volume. Everything present around us living or non living are matter.

Characteristics of Particles of Matter
(1)They are very small in size.
(2)They have space between them.
(3)They are in constant random motion.
(4)They always attract each other.

States of matter:- There are three states of matter namelySolids, Liquids, and Gases.

Solids
Solid is one of the three fundamental states of matter. The atoms in
a solid are closely packed together leaving negligible space between
the constituent atoms. Solids have a definite shape, size, volume and
density.
Examples of solids:- metals, minerals such as quartz and mica, etc.

Liquids
Liquid is another state of a matter. A liquid can be defined as a nearly
incompressible fluid that means this state of the matter cannot be
compressed under very high pressure. Liquids don’t have a definite
shape i.e. it can take the shape of the container in which it is kept
but the volume and density remain the same.
Examples of liquids:- water, vegetable oil, mercury, etc.

Gases
Gas is also one of the states of matter. A gas can be defined as a
highly compressible fluid that means this state of matter can be
compressed to a high extent under pressure. Gases neither have a
definite shape nor have a definite volume. Since they don’t have
definite volume their density is also not definite.
Examples of gases are:- oxygen, nitrogen, helium, etc.

Change of states of matter
Matters can change from one state to another state by means of
various processes such as melting, freezing, evaporation,
condensation and sublimation.

Melting:– It is the process by which a substance changes from solid
to liquid phase. It requires the action of heat or pressure onto the
solid matter to change it to liquid.
Example:- melting of ice to form water.

Freezing:- It is the process by which a liquid matter changes to solid. It
requires cooling the liquid substance below its freezing point.
Example:- forming of ice from water.

Condensation:– It is the process by which a gaseous state changes to a
liquid state due to cooling is known as condensation.
Example:- forming of water droplets on the lid of the vessel while cooking
food.

Evaporation:- It is the process by which a liquid changes to gaseous state
below its boiling point is known as evaporation.
Example:- when an open container filled with water is kept under the sun it
dries out after sometime.

Sublimation:– It is the process by which a solid substance directly changes
to a gaseous state without going through the liquid phase is known as
sublimation.
Example:- heating of camphor(Kapoor).

Effect of temperature on the states of matter

Variables and Expressions

Variable is a quantity that can take any value, its value is not fixed. It is a symbol for a number whose value is unknown yet.

Expressions are formed by performing operations like addition, subtraction, multiplication and division on the variables.

Example: 6x – 3 is an expression in variable x.

Algebraic Equation

An equation is a condition on a variable such that two expressions in the variable should have equal value.

Example: 8x−8=16 is an equation.

The value of the variable in an equation for which the equation is satisfied is called the solution of the equation.​

​​​​​​Example: The solution for the equation 2x−3=5 is x=4.

More about Equations

Mathematical Operations on Expressions

• Addition of variables: (3x+4z)+(5y+6)
• Subtraction of variables: (4x−7y)−(6y+5)
• Multiplication of variables: (5xy+6)×7x
• Division of variables: (8xz+5z)/(5x-6y)

Solving an Equation

Solving an equation involves performing the same operations on the expressions on either side of the “=” sign so that the value of the variable is found without disturbing the balance.
Example : Solve 2x+4=10
Consider 2x+4=10
⇒2x+4−4=10−4  [Subtracting 4 from both LHS and RHS] ⇒2x=6
⇒2x/2=6/2 [Dividing both LHS and RHS by 2] ⇒x=3

Methods of Solving an Equation

Method 1: performing the same operations on the expressions on either side of the “=” sign so that the value of the variable is found without disturbing the balance.
Opertions involve Adding, subtracting, multipling or dividing on both sides.

Example: x+2=6
Subtract 2 from LHS and RHS
⇒ LHS: x+2−2=x
⇒ RHS: 6−2=4
But LHS = RHS
⇒ x = 4

Method 2: Transposing
It involves moving the terms to one side of the equation to find out the value of the variable.
​​​​​​​When terms move from one side to another they change their sign.
Example: x+2=6
Transpose (+2) from LHS to RHS
⇒x=6−2
⇒x=4

Applying Equations

Forming Equation from Solution

Given a solution, many equations can be constructed.

Example:  Given solution: x = 3
Multiply both sides by 4,
⇒ 4x=4×3
Add -5 to both sides,
⇒ 4x−5=12−5
⇒ 4x−5=7
Similarly, more equations can be constructed.

Applications (Word problem)

Example: Ram’s father is 3 times as old as his son Ram. After 15 years, he will be twice the age of his son. Form an equation and solve it.
Solution: Let Ram’s age be x.
⇒ His father’s age is 3x.
After 15 years:
3x+15=2(x+15)
On solving,
3x+15=2x+30
3x−2x=30−15
x=15
∴ Ram’s age is 15 and his dad’s age is 45.

Introduction

It is a combination of constant and variables kept together through operations like addition,
subtraction etc.
Variable
It refers to the unknown quantities that can change or vary. It is generally an English
alphabet.
For example: In the expression 4x+5, x is the variable.

Constant
Constant is a quantity which has a fixed value.
For example: in the expression 4x+5, 5 is the constant.

Coefficient
It is a constant value used with the variable.
For example: in the expression 4x+5, 4 is coefficient.

Terms
The different parts of the algebraic expression are separated by the fundamental
operations.
For example: in the expression 4x+5, ‘4x’ and ‘5’ the terms

Like terms
Terms which contain same variables are called like terms
For example: 6pq and 2pq are like terms

Unlike terms
Terms having different variables are called unlike terms.
For example: 4x and -2y are unlike terms.

Types of Algebraic expression:

Monomial Expression
An algebraic expression containing only one term is known as a monomial.
For example
2×4, 2xy, etc.

Binomial Expression
A binomial expression has two unlike terms
For examples
2xy + 4, xyz + x2, etc.

Polynomial Expression
In general, an expression with more than one terms with non-negative integral exponents of
a variable is known as a polynomial.
Examples
ab + bc + ca, etc.

Addition and Subtraction of Algebraic Expression

Addition and Subtraction of like terms
Like terms can be added or subtracted by simply adding or subtracting the coefficients of
the terms.
For example: 5x + 3x = 8x , 5x – 3x = 2x

Addition and Subtraction of unlike terms
We write the unlike terms with the sign of addition and subtraction.
For example: adding 6x, 17y, 5
Result will be 6x+17y+5
Subtracting 7x from 3y
Result will be 3y-7x

Finding the value of an expression
Finding the numerical value of given expression.
Value of variable is given here.
Example: Find the value of 7x + y, given x=2, y=6
7x + y = 7(2) + 6 = 14 + 6 = 20

What is Simple Interest?

Simple Interest (S.I) is the method of calculating the interest amount for some principal amount of money. Have you ever borrowed money from your siblings when your pocket money is exhausted? Or lent him maybe? What happens when you borrow money? You use that money for the purpose you had borrowed it in the first place. After that, you return the money whenever you get the next month’s pocket money from your parents. This is how borrowing and lending work at home.

But in the real world, money is not free to borrow. You often have to borrow money from banks in the form of a loan. During payback, apart from the loan amount, you pay some more money that depends on the loan amount as well as the time for which you borrow. This is called simple interest. This term finds extensive usage in banking.

Simple Interest Formula

The formula for simple interest helps you find the interest amount if the principal amount, rate of interest and time periods are given.

Simple interest formula is given as:

Where SI = simple interest

P = principal

R = interest rate (in percentage)

T = time duration (in years)

In order to calculate the total amount, the following formula is used:

Amount (A) = Principal (P) + Interest (I)

Where,

Amount (A) is the total money paid back at the end of the time period for which it was borrowed.

The total amount formula in case of simple interest can also be written as:

A = P(1 + RT)

Here,

A = Total amount after the given time period

P = Principal amount or the initial loan amount

R = Rate of interest (per annum)

T = Time (in years)

Click here to get the simple interest calculator for quick computations.

Simple Interest Formula For Months

The formula to calculate the simple interest on a yearly basis has been given above. Now, let us see the formula to calculate the interest for months. Suppose P be the principal amount, R be the rate of interest per annum and n be the time (in months), then the formula can be written as:

Simple Interest for n months = (P × n × R)/ (12 ×100)

The list of formulas of simple interest for when the time period is given in years, months and days are tabulated below:

Time	Simple interest Formula	Explanation
Years	PTR/100	T = Number of years
Months	(P × n × R)/ (12 ×100)	n = Number of months
Days	(P × d × R)/ (365 ×100)	d = Number of days (non-leap year)

Difference Between Simple Interest and Compound Interest

There is another type of interest called compound interest. The major difference between simple and compound interest is that simple interest is based on the principal amount of a deposit or a loan whereas compound interest is based on the principal amount and interest that accumulates in every period of time. Let’s see one simple example to understand the concept of simple interest.

Simple Interest Problems

Let us see some simple interest examples using the simple interest formula in maths.

Example 1:

Rishav takes a loan of Rs 10000 from a bank for a period of 1 year. The rate of interest is 10% per annum. Find the interest and the amount he has to pay at the end of a year.

Solution:

Here, the loan sum = P = Rs 10000

Rate of interest per year = R = 10%

Time for which it is borrowed = T = 1 year

Thus, simple interest for a year,  SI = (P × R ×T) / 100 = (10000 × 10 ×1) / 100 = Rs 1000

Amount that Rishav has to pay to the bank at the end of the year = Principal + Interest = 10000 + 1000 = Rs 11,000

Example 2:

Namita borrowed Rs 50,000 for 3 years at the rate of 3.5% per annum. Find the interest accumulated at the end of 3 years.

Solution:

P = Rs 50,000

R = 3.5%

T = 3 years

 SI = (P × R ×T) / 100 = (50,000× 3.5 ×3) / 100 = Rs 5250 

Example 3:

Mohit pays Rs 9000 as an amount on the sum of Rs 7000 that he had borrowed for 2 years. Find the rate of interest.

Solution:

A = Rs 9000

P = Rs 7000

SI = A – P = 9000 – 7000 = Rs 2000

T = 2 years

R = ?

SI = (P × R ×T) / 100 

R = (SI  × 100) /(P× T)

R = (2000  × 100 /7000 × 2)  =14.29 % 

Thus,  R  = 14.29%

Practice Questions

1. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9% per annum in 5 years. What is the sum? 
2. A sum of Rs. 725 is lent at the beginning of a year at a specific rate of interest. After eight months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both loans. What was the actual rate of interest?
3. Simple interest on a certain sum is 16/25 of the sum. Find the rate of interest and time if both are numerically equal.

Profit and Loss Basic Concepts

Let us learn profit and loss concepts in maths. It is well explained in terms of cost price and selling price.

Profit(P)

The amount gained by selling a product with more than its cost price.

Loss(L)

The amount the seller incurs after selling the product less than its cost price, is mentioned as a loss.

Cost Price (CP)

The amount paid for a product or commodity to purchase it is called a cost price. Also, denoted as CP. This cost price is further classified into two different categories:

• Fixed Cost: The fixed cost is constant, it doesn’t vary under any circumstances
• Variable Cost: It could vary depending as per the number of units

Selling Price (SP)

The amount for which the product is sold is called Selling Price. It is usually denoted as SP. Also, sometimes called a sale price.

Marked Price Formula (MP)

This is basically labelled by shopkeepers to offer a discount to the customers in such a way that,

Discount = Marked Price – Selling PriceAnd Discount Percentage = (Discount/Marked price) x 100

Profit and Loss Formulas

Now let us find profit formula and loss formula.

• The profit or gain is equal to the selling price minus cost price.
• Loss is equal to cost price minus selling price.

Profit or Gain = Selling price – Cost PriceLoss = Cost Price – Selling Price

The formula for the profit and loss percentage is:

Profit percentage = (Profit /Cost Price) x 100Loss percentage = (Loss / Cost price) x 100

Profit and Loss Examples

• If a shopkeeper brings a cloth for Rs.100 and sells it for Rs.120, then he has made a profit of Rs.20/-.
• If a salesperson has bought a textile material for Rs.300 and he has to sell it for Rs.250/-, then he has gone through a loss of Rs.50/-.
• Suppose, Ram brings a football for Rs. 500/- and he sells it to his friend for Rs. 600/-, then Ram has made a profit of Rs.100 with the gain percentage of 20%.

These are some common examples of the profit and loss concept in real life, which we observe regularly.

Profit and Loss Tricks

You have learned until now how to calculate profit as well as loss and also the percentage of them. Now let us learn some tricks or formulas to solve maths problems based on gain and loss, starting from the general formulas.

1. Profit, P = SP – CP; SP>CP
2. Loss, L = CP – SP; CP>SP
3. P% = (P/CP) x 100
4. L% = (L/CP) x 100
5. SP = {(100 + P%)/100} x CP
6. SP = {(100 – L%)/100} x CP
7. CP = {100/(100 + P%)} x SP
8. CP = {100/(100 – L%)} x SP
9. Discount = MP – SP
10. SP = MP -Discount
11. For false weight, profit percentage will be P% = (True weight – false weight/ false weight) x 100.
12. When there are two successful profits say m% and n%, then the net percentage profit equals to (m+n+mn)/100
13. When the profit is m% and loss is n%, then the net % profit or loss will be: (m-n-mn)/100
14. If a product is sold at m% profit and then again sold at n% profit then the actual cost price of the product will be: CP = [100 x 100 x P/(100+m)(100+n)]. In case of loss, CP = [100 x 100 x P/(100-m)(100-n)]
15. If P% and L% are equal then, P = L and %loss = P²/100

Let us explain the above-given formulas with examples.

Solved Problems

Q. 1: Suppose a shopkeeper has bought 1 kg of apples for 100 rs. And sold it for Rs. 120 per kg. How much is the profit gained by him?

Solution:

Cost Price for apples is 100 rs.

Selling Price for apples is 120 rs.

Then profit gained by shopkeeper is ; P = SP – CP

P = 120 – 100 = Rs. 20/-

Q.2: For the above example calculate the percentage of the profit gained by the shopkeeper.

Solution:

We know, Profit percentage = (Profit /Cost Price) x 100

Therefore, Profit percentage = (20/100) x 100 = 20%.

Q.3: A man buys a fan for Rs. 1000 and sells it at a loss of 15%. What is the selling price of the fan?

Solution: Cost Price of the fan is Rs.1000

Loss percentage is 15%

As we know, Loss percentage = (Loss/Cost Price) x 100

15 = (Loss/1000) x 100

Therefore, Loss = 150 rs.

As we know,

Loss = Cost Price – Selling Price

So, Selling Price = Cost Price – Loss

= 1000 – 150

Selling Price = R.850/-

Q.4: If a pen cost Rs.50 after 10% discount, then what is the actual price or marked price of the pen?

Solution: MP x (100 – 10) /100 = 50

MP x (90/100) = 50

MP = (50 x 100)/90

MP = Rs. 55.55/-

Points to remember:

• For-profit, the selling price should be more than the cost price
• For loss, cost price should be more than the selling price.
• The percentage value for profit and loss is calculated in terms of cost price

 

Percentages are an important part of our everyday lives. We use percentages in different areas. Such as % discount in shopping malls, % rainfall of the year, bank interest rates, % housing loan, % of the girl passed in 10th class, % rate of petrol and diesel. Percentages are a useful way of comparing fractions with different denominators. Percentages give information which is often easier to understand than fractions.

Example: of the total number of students in the class are male, is quite difficult to interpret, while the statement 45% of the total number of students in the class are male is easier to understand

Meaning of Percentage:

Percent: The word percent is derived from the Latin word ‘percentum’ which means ‘per hundred’.

Percentage: Percentage are numerators of fractions with denominator 100. The percentage is denoted by the symbol ‘%’ i.e., 1 % means 1 out of 100. It can be written as 1%= 1/100=0.01

Example: When we say Rakesh gives 20% of his income as income tax, this means that he gives ₹ 20 out of every hundred rupees of his income.

To understand this let us consider the following example.

Aniket is fond of collecting souvenirs. He collected 100 souvenirs from different countries. From the table can you tell the percentage of souvenirs of any country?

Percentages when the total is not a hundred
In such cases, we need to convert the fractions to an equivalent fraction with denominator 100.
Any fraction can be expressed into a percentage by multiplying it by 100.

We multiply the fraction by 100/100. This does not change the value of the fraction.

Example: Write   as a percent. 

=×= (× 100) % = ( )%=12%​

Example: Write  as a percent.

 =  x  = (× 100) % = ( )%=5%

Example: There are 50 students in class VII, and 35 go to the picnic. Calculate the percentage of the number of students who go to the picnic?
35 students of the 50 go to the picnic.

Percentage of the number of students in class VII =  ×
= (× 100) % = = 70%

So, 70% of the students go to the picnic

Converting fractional numbers to Percentage

To convert a fractional number into percentages, multiply the fraction by 100 and put the % sign.

Example: Convert the fractional number   into percent.

 =  ×  = ( × 100)%= ( )%= 6%

Example: Convert the fractional number  into percent.

 =  ×  = ( × 100)%= ( )%=12.5%

Converting Decimals to Percentage

To convert a decimal number into a percentage, multiply the decimal number by 100 and put the % sign.

Express each of the following decimal number into percent.

(i) 0.55 (ii) 0.07 (iii) 2.2
(i) 0.55 = 0.55 × 100% = × 100% = 55 %

(ii) 0.07 = 0.07 × 100% = × 100% = 7%

(iii) 2.2 = 2.2 × 100% = × 100% = 220%

Converting Percentages to fraction or decimals

To convert a percentage into a fraction or a decimal, divide it by 100%.

Express 140 % percentage into a fraction.
140 % =  =  = 

Express 140 % percentage into a decimal.

140 % =  = 1.40

Express 3 % percentage into a decimal.

3 % = = 0. 03

Express 20 % percentage into a fraction.

20 % =  = 

Percentage – Concept of Whole

The percentage is also used as the concept of the whole. Suppose if we add parts together it always gives a whole (100%). Similarly, if the whole is divided into parts it combines and gives you a whole or 100%.

Example: As we know the air is a mixture of gases and that gases are present in the atmosphere in different percentages. When we add the percentage of different gases in the atmosphere we get 100%

So, if we are given one part, we can always find out the other part. Suppose Rashmi has 1 whole pizza, it consists of 8 parts out of these Rashmi gives 4 parts to her brother, we can always find out the remaining parts of the pizza.

Example: In a school, 30% students are girls, find the percentage of boys. 

Total number of students in the school = 100%

Percentage of girls in a school = 30%

The percentage of boys in a school = Total number of students in the school (100%) – Percentage of girls in a school

The percentage of boys in a school = 100% – 30% = 70%

What is Unitary Method?

The unitary method is a method in which you find the value of a unit and then the value of a required number of units. What can units and values be?

Suppose you go to the market to purchase 6 apples. The shopkeeper tells you that he is selling 10 apples for Rs 100. In this case, the apples are the units, and the cost of the apples is the value. While solving a problem using the unitary method, it is important to recognize the units and values.

For simplification, always write the things to be calculated on the right-hand side and things known on the left-hand side. In the above problem, we know the amount of the number of apples and the value of the apples is unknown. It should be noted that the concept of ratio and proportion is used for problems related to this method.

Example of Unitary Method

Consider another example; a car runs 150 km on 15 litres of fuel, how many kilometres will it run on 10 litres of fuel?

In the above question, try and identify units (known) and values (unknown).

Kilometre = Unknown (Right Hand Side)

No of litres of fuel = Known (Left Hand Side)

Now we will try and solve this problem.

15 litres = 150 km

1 litre = 150/15 = 10 km

10 litres = 10 x 10 = 100 km

The car will run 100 kilometres on 10 litres of fuel.

Unitary Method in Ratio and Proportion

If we need to find the ratio of one quantity with respect to another quantity, then we need to use the unitary method. Let us understand with the help of examples.

Example: Income of Amir is Rs 12000 per month, and that of Amit is Rs 191520 per annum. If the monthly expenditure of each of them is Rs 9960 per month, find the ratio of their savings.

Solution: Savings of Amir per month = Rs (12000 – 9960) = Rs 2040

In 12 month Amit earn = Rs.191520

Income of Amit per month =  Rs 191520/12 = Rs. 15960

Savings of Amit per month = Rs (15960 – 9960) =  Rs 6000

Therefore, the ratio of savings of Amir and Amit  = 2040:6000 = 17:50

Types of Unitary Method

In the unitary method, the value of a unit quantity is calculated first to calculate the value of other units. It has two types of variations.

1. Direct Variation
2. Inverse Variation

Direct Variation

In direct variation, increase or decrease in one quantity will cause an increase or decrease in another quantity. For instance, an increase in the number of goods will cost more price. 

Also, the amount of work done by a single man will be less than the amount of work done by a group of men. Hence, if we increase the number of men, the work will increase.

Indirect Variation

It is the inverse of direct variation. If we increase a quantity, then the value of another quantity gets decrease. For example, if we increase the speed, then we can cover the distance in less time. So, with an increase in speed, the travelling time will decrease.

Applications of Unitary Method

The unitary method finds its practical application everywhere ranging from problems of speed, distance, time to the problems related to calculating the cost of materials.

• The method is used for evaluating the price of a good.
• It is used to find the time taken by a vehicle or a person to cover some distance in an hour.
• It is used in business to determine profit and loss.

Unitary Method Speed Distance Time

Let us take unitary method problems for speed distance time and for time and work.

Illustration: A car travelling at a speed of 140 kmph covers 420 km. How much time will it take to cover 280 km?

Solution: First, we need to find the time required to cover 420 km.

Speed = Distance/Time

140 = 420/T

T = 3 hours

Applying the unitary method,

420 km = 3 hours

1 km = 3/420 hour

280 km = (3/420) x 280 = 2 hours

Unitary Method For Time and Work

Example: “A” finishes his work in 15 days while “B” takes 10 days. How many days will the same work be done if they work together?

Solution:

If A takes 15 days to finish his work then,

A’s 1 day of work = 1/15

Similarly, B’s 1 day of work = 1/10

Now, total work is done by A and B in 1 day = 1/15 + 1/10

Taking LCM(15, 10), we have,

1 day’s work of A and B = (2+3)/30

1 day’s work of  (A + B) = ⅙

Thus, A and B can finish the work in 6 days if they work together.

Unitary Method Questions

1. 12 workers finish a job in 20 hours. How many workers will be required to finish the same work in 15 hours?
2. If the annual rent of a flat is Rs. 3600, calculate the rent of 7 months.
3. If 56 books weigh 8 Kg, calculate the weight of 152 books.
4. If 5 cars can carry 325 people, find out the total number of people which 8 cars can carry.
5. Rakesh completes 5/8 of a job in 20 days. How many more days will he take to finish the job at his current rate?

Ratio–

Ratio means the comparison of two terms by dividing one by the other.Continued

Ratio–If the ratio of two numbers is a:b and the ratio of two other quantities is b : c then a: b:c is said to be continued ratio.

For example –If a:b = 5 : 9 and b : c = 9:14, then a : b : c = 5 : 9 : 14 is a continue ratio.

Compound Ratio-For two or more ratios, if we take antecedent as product of antecedents of the ratios and consequent as product of consequents of the ratios, then the ratio thus formed is called mixed or compound ratio. The compound ratio of m : n and p :q is mp : nq.When two or more ratios are multiplies term wise, the ratio thus obtained is called compound ratio.

For example:The compounded ratio of the two ratios a : b and c : d is the ratio ac : bd, For ratios m : n and p :q; the compound ratio is (m × p) : (n × q).

Proportion–

Four quantities a, b, c and d are said to be in proportion if a : b = c : d. It can also be written as a:b :: c : d.

For example –8,16 and 19, 38 are in proportionAs 8 : 16 = 1:219 : 38 = 1:2

Volume:
It is measure or calculated of the space occupied by something.
SI unit: m3 and cm3
1 m3 =100,0000 cm3
Volume of any object is depends on its dimensions.
In case of cube having 1 cm side it will be written as,
= 1 cm x 1 cm x 1 cm = 1 cm3

Capacity and Volume:
The space available to hold something is termed as capacity.
Capacity is the amount a container can hold. Eg. A jug or mug
Volume is the measured of the space taken up by something .
Metric units : litres (L) , centiliters (cL) , milliliters (mL)
There are 100 centiliters in 1 litre
There are 1000 milliliter in 1 litre

Let’s study the correlation between different volume units:
1m3=106
cm3 = 103
L
1 L= 103
cm3
1 mL= 1cc ( cubic centimeter)

Measuring the volume of irregular solids:
Stone is tied to string and it is partially immersed in water containing jar
:. Volume of stone = volume of water displaced
Measuring the volume of liquids:
A liquid takes the shape of the container. Jars and cylinders having fixed volume are used.

Area :
It is the amount of space that exist as shape. It is always measure in square units , such as
square inches, square feet etc.

Measurement of Area :
It is depends on the basic shape of object. In case of rectangle, Area can be calculated as
product of length (l) times width (w).
Area of irregular shapes depends on the, whole area is divided into smaller shapes such that
we can calculated by formula and at end adding up all area of shapes.
Density:
The density of a substance is its mass per unit volume. The symbol most often used for
density is (ρ) termed as rho.

Speed :
Average speed is measured by comparing the distance travelled and the time taken for
journey.

SI unit: meter/second, kilometer/hour

Distance-time graphs:

If an object is moving along a straight line, the distance covered by that object can be
represented by a distance-time graph.
In a distance-time graph, the gradient or sharp point increase change of the line is equal to
the speed of the object.
The greater the gradient or the steeper the line on graph, indicating the faster the object is
moving.

Q. A man covers 52 m distance in 10 s , 48 m in next 10 s and 25 m in next 5 s.
Calculate the average speed?

Solution:
Total distance covered by man = (52 + 48 + 25) = 125 m
Total time taken = (10 + 10 + 5 ) = 25 s

125
25
= 5 m/s
(The average speed of man in above example is found to be 5 m/s)