Course Content
Chapter 01 – Sets
A set is a list of objects in no particular order; they could be numbers, letters, or even words. A Venn diagram is a way of representing sets visually.
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Chapter 02 – Rational Numbers
In mathematics, a rational number is a number that can be expressed as the quotient or fraction p/q of two integers, a numerator p, and a non-zero denominator q. In this chapter, we will learn to represent rational numbers on a number line and perform arithmetic operations.
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Chapter 03 – Decimals
Decimals are a set of numbers lying between integers on a number line. They are just another way to represent fractions in mathematics. In this chapter, we will learn about the conversion of decimals to rational numbers, the kinds of decimals, and absolute values.
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Chapter 04 – Exponents
The exponent of a number says how many times to use that number in a multiplication. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily. In this chapter, we are going to discuss the six important laws of exponents.
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Chapter 05 – Square Root of Positive Numbers
Square root, in mathematics, is a factor of a number that, when multiplied by itself, gives the original number. In this chapter, we will learn about what makes perfect squares and will find the roots of positive numbers by considering real-life scenarios.
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Chapter 06 – Direct and Inverse Variation
Variation means change. With direct variation, numbers change proportionately in the same direction, while with inverse variation, they change in opposite directions. In this chapter, we will earn how to solve direct and inverse variation problems, explore their definitions, and work examples to understand the equations and techniques for solving them. Also, we learn to find the continued ratio for two or more ratios.
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Chapter 07 – Financial Arithmetic
Financial mathematics describes the application of mathematics and mathematical modeling to solve financial problems. In this chapter, we will learn about the concept of taxation, profit/markups, zakat & ushr, and how they relate to our daily life.
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Chapter 08 – Algebraic Expressions
Algebraic expressions are the idea of expressing numbers using letters or alphabets without specifying their actual values. The algebraic equations which are valid for all values of variables in them are called algebraic identities. In this chapter, we will learn to perform operations on polynomials and to factorize an algebraic equation by using identities.
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Chapter 09 – Linear Equations
Linear equations are equations having variables with power 1. ax+b = 0 is an example with one variable where x is the variable, and a and b are real numbers. In this chapter, we will learn the definition, type of solutions, and how to solve these equations with one variable and two variables using different methods along with examples.
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Chapter 10 – Fundamentals of Geometry
Geometry is the study of different types of shapes, figures, and sizes in Maths or real life. In geometry, we learn about different angles, transformations, and similarities in the figures. It is important to know and understand some basic concepts. We will learn about working in different numbers of dimensions, and about some of the most fundamental concepts in geometry, including points, lines, and planes.
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Chapter 11 – Practical Geometry
The practical Geometry chapter will teach you about lines and to construct two-dimensional given different kinds of measurements. A quadrilateral is a closed two-dimensional shape that has four sides and four angles. Any four-sided closed shape such as square, rectangle, rhombus, parallelogram, trapezium, etc. And a closed two-dimensional shape that has 3 sides and 3 angles is known as a triangle.
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Chapter 12 – Circumference, Area and Volume
This topic comes under analytical geometry and the formulas for the volume and the surface area of the sphere were first discovered by Archimedes. In this chapter, we will learn about the area and volume of two-dimensional and three-dimensional shapes.
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Chapter 13 – Information Handling
Frequency distribution, in statistics, is a graph or data set organized to show the frequency of occurrence of each possible outcome of a repeatable event observed many times. And, a pie chart is a way of representing data in a circular graph. Pie slices of the chart show the relative size of the data. In this chapter, we will learn to construct the frequency distribution table, some new pie chart vocabulary, and learn to construct the pie chart to represent the data.
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Mathematics – VII
About Lesson

Important Formulas

  • Speed = Distance / Time
  • Time = Distance / speed
  • Distance = speed x time

 

If the ratio of the speed of A and B is a:b, then the ratio of the time taken by them to cover the same distance is 1/a : 1/b or b:a

 

Suppose a man covers a distance at x km/h and an equal distance at y km/h. then the average speed during the whole journey is (2xy/x+y) km/h

 

An object that moves at a constant rate is said to be in uniform motion.

The formula d = rt gives the relationship between distance d, rater, and time t.

Uniform motion problems may involve objects going in the same direction, opposite directions, or round trips.

 

Examples:

1. Walking at the rate of 4 km/h a man covers a certain distance in 2 hr 45 min. Running at a speed of 16.5 km/h the man will cover the same distance.

Solution:

Distance = Speed x time

4x(11/4) = 11km

New Speed = 16.5 km/h

Therefore time = D/S = 11/16.5 = 40 min

 

2. A train covers a distance in 50 min if it runs at a speed of 48 km/h on average. The speed at which the train must run to reduce the time of journey to 40 min will be.

Solution:

Time = 50/60 hr = 5/6hr

Speed = 48 m/h

distance = S x T = 48 x 5/6 = 40 km

time = 40/60hr = 2/3hr

New speed = 40* 3/2 km/h = 60 km/h

 

3. Two boys starting from the same place walk at a rate of 5 km/h and 5.5 km/h respectively. What time will they take to be 8.5 km apart, if they walk in the same direction?

Solution:

The relative speed of

the boys = 5.5 km/h – 5km/h = 0.5 km/h

Distance between them is 8.5 km

Time= 8.5km / 0.5 km/h = 17 hrs

 

4. In covering a distance of 30 km, Saad takes 2 hours more than Rohan. If Saad doubles his speed, then he would take 1 hour less than Rohan. Saad speed is?

Solution:

Let Saad’s speed be X km/hr.

Then, 30/x – 30/2x = 3

6x = 30

x = 5 km/hr.

 

5. A train 140 m long running at 60 kmph. In how much time will it pass a platform 260 m long?

Solution:

Distance travelled = 140 + 260 m = 400 m,

Speed = 60 x 5/8 = 50/3 m

Time = 400 x 3/50 = 24 Seconds

 

Time and Work

Work is the quantity of energy transferred from one system to another but for question based on this topic. OR Problems on work are based on the application of the concept of the ratio of time and speed.

Above mentioned definition of work throws light on three important points.

 

  1. Work = 1 ( as it is always measured as a whole) = Distance
  2. Rate at which work is done = speed
  3. Number of days required to do the work = Time

 

Formulas:

  • Work from Days:  If A can do a piece of work in n days, then A’s 1 day’s work = 1/n
  • Days from Work:  If A’s 1 day’s work = 1/n then A can finish the work in n days.

 

Examples:

1. If 5 women or 8 girls can do work in 84 days. In how many days can 10 women and 5 girls can do the same work?

Solution:

Given that 5 women is equal to 8 girls to complete a work.

So, 10 women = 16 girls.

Therefore 10 women + 5 girls = 16 girls + 5 girls = 21 girls.

8 girls can do a work in 84 days then 21 girls can do a work in (8 x 84/21) = 32 days.

Therefore 10 women and 5 girls can a work in 32 days

 

2. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long it take both A & B, working together but independently, to do the same job?

Solution:

A’s one hour work = 1/8. B’s one hour work = 1/10.

(A+B)’s one hour work

= 1/8 + 1/10 = 9/40. Both A & B can finish the work in 40/9  days.

 

3. A can finish a work in 18 days and B can do the same work in half the time taken by A. Then, working together, what part of the same work they can finish in a day?

Solution:

Given that B alone can complete the same work in days = half the time taken by A  

half the time taken by A = 9 days

A’s one day work = 1/18

B’s one day work = 1/9

(A+B)’s one day work = 1/18 + 1/9 = 3/18 = 1/6

 

4. A is twice as good a workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?

Solution:

If A takes x days to do a work then B takes 2x days to do the same work

1/x + 1/2x = 1/18

= 3/2x = 1/18

x = 27 days.

Hence, A alone can finish the work in 27 days.

 

5. A can do a certain work in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?

Solution:

Ratio of time taken by A & B = 160 : 100 = 8 : 5

Suppose B alone takes x days to do the job.

Then, 8 : 5 :: 12 : x

= 8x = 5 x 12

= x = 15/2 days.

Exercise Files
Speed, Time, and Distance Worksheet.pdf
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Speed, Time, and Distance Worksheet1.pdf
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Speed, Time, and Distance Worksheet2.pdf
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Speed, Time, and Distance Worksheet4.pdf
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