Class 8 Exam > Class 8 Videos > Examples: Word problems (Part 1)
Examples: Word problems (Part 1) Video Lecture - Class 8
FAQs on Examples: Word problems (Part 1) Video Lecture - Class 8
| 1. What are some examples of word problems for Class 8 mathematics? |  |
Ans. Some examples of word problems for Class 8 mathematics include:
1. A fruit vendor sells 25 apples for $50. How much does each apple cost?
2. A rectangle has a length of 12 cm and a width of 8 cm. What is its area?
3. A train travels at a speed of 60 km/h. How far will it travel in 3 hours?
4. If a box of chocolates costs $15 and contains 20 chocolates, what is the cost of each chocolate?
5. The sum of two numbers is 48. If one number is 20, what is the other number?
| 2. How can word problems help improve mathematical skills in Class 8? | |
Ans. Word problems help improve mathematical skills in Class 8 by:
1. Developing problem-solving abilities: Word problems require students to analyze the given information, identify the mathematical operations needed, and solve the problem step by step.
2. Enhancing critical thinking: Word problems require students to think critically and apply mathematical concepts to real-life situations, promoting a deeper understanding of the subject.
3. Strengthening mathematical reasoning: Word problems challenge students to use logical reasoning to arrive at the correct solution, helping them develop their mathematical reasoning skills.
4. Connecting math to the real world: Word problems provide practical applications of mathematical concepts, making math more relatable and showing its relevance in everyday life.
5. Building confidence: Successfully solving word problems boosts students' confidence in their mathematical abilities and encourages them to tackle more complex problems.
| 3. How can students approach solving word problems effectively? | |
Ans. Students can approach solving word problems effectively by following these steps:
1. Read the problem carefully: Understand the given information and identify what is being asked in the problem.
2. Identify the mathematical operations: Determine which mathematical operations (addition, subtraction, multiplication, division, etc.) are required to solve the problem.
3. Create an equation or model: Translate the problem into a mathematical equation or model that represents the given information and the unknown.
4. Solve the equation or model: Use the appropriate mathematical operations to solve the equation or model and find the solution.
5. Check the solution: Verify if the solution satisfies the conditions stated in the problem and makes sense in the given context.
6. Communicate the solution: Clearly state the final answer, including any necessary units or explanations, to effectively communicate the solution.
| 4. What are some strategies to improve word problem-solving skills in Class 8? | |
Ans. Some strategies to improve word problem-solving skills in Class 8 include:
1. Practice regularly: Solve a variety of word problems regularly to become familiar with different problem structures and improve problem-solving abilities.
2. Break down the problem: Divide complex word problems into smaller, manageable parts to better understand the given information and identify the required operations.
3. Use visual aids: Draw diagrams, charts, or models to visualize the problem and help make sense of the given information.
4. Work with peers or a tutor: Collaborate with classmates or seek guidance from a tutor to discuss different approaches and gain different perspectives on problem-solving techniques.
5. Reflect on mistakes: Analyze mistakes made in solving word problems and learn from them to avoid similar errors in the future.
6. Apply real-life scenarios: Relate word problems to real-life situations to understand the practical applications of mathematical concepts and enhance problem-solving skills.
| 5. Where can Class 8 students find additional resources for practicing word problems? | |
Ans. Class 8 students can find additional resources for practicing word problems in the following places:
1. Textbooks: Class 8 mathematics textbooks often contain a variety of word problems at different difficulty levels for practice.
2. Online educational platforms: Numerous websites and educational platforms offer interactive math lessons, quizzes, and practice exercises specifically designed for Class 8 students.
3. Math workbooks: Workbooks specifically focused on word problems provide additional practice opportunities with step-by-step solutions and explanations.
4. Mobile applications: There are various math-related mobile applications that provide word problem-solving exercises and practice games for Class 8 students.
5. Tutoring centers: Local tutoring centers or online tutoring services offer personalized guidance and practice sessions for improving word problem-solving skills.
Text Transcript from Video
hello friends this video on linear
equations in one variable part nine is
brought to you by exam feel calm no more
to your prom exam so now that we have
learned how to solve linear equations
the next question that you might be
thinking is why do we need to solve such
equations okay fine
so we understood that in our real life
we come across situations where we can
make use of these equations but why do
we need to solve them so here we will
look at some of the applications of
linear equations that what are those
kind of questions which keep happening
around us those are the problems around
us and we need to solve them and to
solve them we need to solve linear
equations so let us look at one example
of such problems let's say that you had
some money which you never bothered to
calculate but you just had some money
maybe your father gave you some money
some time back your grandma gave you
some money some time back so you
collected all of them but you never
bothered to calculate them so you had
some amount of money and you spend and
you spend half of the money on buying
books and then you see that you are
finally left with rupees 30 so how much
money did you have so now that it was
just a matter of chance that you never
bother to calculate it before but now
that you have already spent the half of
it so you have exactly spent the half of
it in buying books so you are left with
rupees 30 so how would you actually
calculate that how much money did you
have so here I will show you how we can
make use of linear equation so let us
assume because since we don't know how
much money you had so let us assume that
you had rupees X now X is a variable so
it can be anywhere right now depending
on the equation there it is it will take
a specific value now as per this
question you had X money and how much
money did you spend on buying books half
of X that means X by two and how much
money are you left with that it would be
nothing but X minus X
- because thanks bye to you have spent
so let's see you have ten rupees so you
have spent five in peace that means ten
minus five is what you will be left with
similarly here you add X money you spend
X by two so you should be left with X
minus X by two and in the question it is
given that this value the money that is
left that is equal to 30 so basically
what is this this is nothing but a
linear equation in one variable and we
already learned how to solve a linear
equation in one variable so let us try
to solve this equation only then we will
be able to calculate the value of x so
when you try to solve it this is X by 2
is equal to 30 or X is equal to 2 into
30 which is equal to 60 so that means
you had rupees 60 so that's the solution
so now for solving these kind of
problems and these problems happen in
our life so to find these problems we
make use of linear equations so first we
form the equation and then we try to
solve it so let us look at another
example let's say that there is a
problem statement where you know that
there are two brothers hurry and happy
and their ages are in the ratio of 5 is
to 7 so hurry is the 1 so since the
ratio is 5 is to 7 that means hurry is
younger than handle so let's say this is
hurry and this is having now we tell you
now it's that the state problem
statement is given like a puzzle that 4
years from now the ratio of their ages
would be 3 is to 4 so you have to find
their present ages now you see the prep
the ages they their ratio is only given
so you have to calculate their exact
ages
so how how will you do that so here also
let us assume that they are present ages
because the present ages are in the
ratio 5 is to 7 so let us
say that the present ages are 5 X + 7 X
so we assume that Hadees present age is
5 X and hurry patents present H is 7 X
now why did we choose 5 X + 7 X because
in this case also the ratio would be 5 X
by 7 X X and X good cancer so the ratio
would be 5 is to 7 so we have just taken
this variable to calculate their exact
ages like so we have I've shown to this
now as per our assumption what would be
their age after 4 years so after 4 years
what happens to everybody's age your age
increases by 4 so hurry's age would be 5
X + 4 similarly had his age would be 7 X
+ 4 so this little fellow will become 5
X + 4 similarly this fellow would become
7 X + 4 so therefore as per our
assumption the ratio of their ages after
4 years would be how much so their ratio
would be 5 X + 4 divided by 7 X + 4 so
as per what we are calculating using X
so this should be the ratio but as per
the question after four years the ratio
should be 3 is to 4 so according to
question this value which we have
calculated as u mu X is actually equal
to 3 by 4 so that is what our question
says right so now if we are able to
solve this equation and find out the
value of x then very easily we will be
able to calculate their present ages so
let's do that
so how will you solve this equation this
is again in the form of numerator by
denominator equal to 2 not numerator by
denominator okay so we will cross
multiply so cross multiplication would
give 4 into 5 X + 4 is equal to 3 into 7
X + 4 so this would be 20 X + 16 is
equal to 21 X + 12 therefore 20 X - 21 X
is equal to 12 - 16 or - a
is equal to minus four therefore we can
say X is equal to four so now that we
have calculated the value of x we can
calculate their present ages so herding
his present age would be 5 into X that
is 5 into 4 which is 20 years and what
would be had his present age his age
would be 7 into X that the 7 into 4
which is 28 years therefore we can say
that having is 28 years old and 30 is 20
years old so using linear equations we
had solved the beans type of puzzles as
well thank you please visit exam fee or
come for free quality education you can
learn with a simple four-step learning
process wherein you can watch video
lessons you can ask your questions you
can refer notes and you can take a free
online test we have content for class 6
to 12 on physics chemistry mathematics
and biology along with practical videos
so please subscribe to our channel for
daily updates thank you
equations in one variable part nine is
brought to you by exam feel calm no more
to your prom exam so now that we have
learned how to solve linear equations
the next question that you might be
thinking is why do we need to solve such
equations okay fine
so we understood that in our real life
we come across situations where we can
make use of these equations but why do
we need to solve them so here we will
look at some of the applications of
linear equations that what are those
kind of questions which keep happening
around us those are the problems around
us and we need to solve them and to
solve them we need to solve linear
equations so let us look at one example
of such problems let's say that you had
some money which you never bothered to
calculate but you just had some money
maybe your father gave you some money
some time back your grandma gave you
some money some time back so you
collected all of them but you never
bothered to calculate them so you had
some amount of money and you spend and
you spend half of the money on buying
books and then you see that you are
finally left with rupees 30 so how much
money did you have so now that it was
just a matter of chance that you never
bother to calculate it before but now
that you have already spent the half of
it so you have exactly spent the half of
it in buying books so you are left with
rupees 30 so how would you actually
calculate that how much money did you
have so here I will show you how we can
make use of linear equation so let us
assume because since we don't know how
much money you had so let us assume that
you had rupees X now X is a variable so
it can be anywhere right now depending
on the equation there it is it will take
a specific value now as per this
question you had X money and how much
money did you spend on buying books half
of X that means X by two and how much
money are you left with that it would be
nothing but X minus X
- because thanks bye to you have spent
so let's see you have ten rupees so you
have spent five in peace that means ten
minus five is what you will be left with
similarly here you add X money you spend
X by two so you should be left with X
minus X by two and in the question it is
given that this value the money that is
left that is equal to 30 so basically
what is this this is nothing but a
linear equation in one variable and we
already learned how to solve a linear
equation in one variable so let us try
to solve this equation only then we will
be able to calculate the value of x so
when you try to solve it this is X by 2
is equal to 30 or X is equal to 2 into
30 which is equal to 60 so that means
you had rupees 60 so that's the solution
so now for solving these kind of
problems and these problems happen in
our life so to find these problems we
make use of linear equations so first we
form the equation and then we try to
solve it so let us look at another
example let's say that there is a
problem statement where you know that
there are two brothers hurry and happy
and their ages are in the ratio of 5 is
to 7 so hurry is the 1 so since the
ratio is 5 is to 7 that means hurry is
younger than handle so let's say this is
hurry and this is having now we tell you
now it's that the state problem
statement is given like a puzzle that 4
years from now the ratio of their ages
would be 3 is to 4 so you have to find
their present ages now you see the prep
the ages they their ratio is only given
so you have to calculate their exact
ages
so how how will you do that so here also
let us assume that they are present ages
because the present ages are in the
ratio 5 is to 7 so let us
say that the present ages are 5 X + 7 X
so we assume that Hadees present age is
5 X and hurry patents present H is 7 X
now why did we choose 5 X + 7 X because
in this case also the ratio would be 5 X
by 7 X X and X good cancer so the ratio
would be 5 is to 7 so we have just taken
this variable to calculate their exact
ages like so we have I've shown to this
now as per our assumption what would be
their age after 4 years so after 4 years
what happens to everybody's age your age
increases by 4 so hurry's age would be 5
X + 4 similarly had his age would be 7 X
+ 4 so this little fellow will become 5
X + 4 similarly this fellow would become
7 X + 4 so therefore as per our
assumption the ratio of their ages after
4 years would be how much so their ratio
would be 5 X + 4 divided by 7 X + 4 so
as per what we are calculating using X
so this should be the ratio but as per
the question after four years the ratio
should be 3 is to 4 so according to
question this value which we have
calculated as u mu X is actually equal
to 3 by 4 so that is what our question
says right so now if we are able to
solve this equation and find out the
value of x then very easily we will be
able to calculate their present ages so
let's do that
so how will you solve this equation this
is again in the form of numerator by
denominator equal to 2 not numerator by
denominator okay so we will cross
multiply so cross multiplication would
give 4 into 5 X + 4 is equal to 3 into 7
X + 4 so this would be 20 X + 16 is
equal to 21 X + 12 therefore 20 X - 21 X
is equal to 12 - 16 or - a
is equal to minus four therefore we can
say X is equal to four so now that we
have calculated the value of x we can
calculate their present ages so herding
his present age would be 5 into X that
is 5 into 4 which is 20 years and what
would be had his present age his age
would be 7 into X that the 7 into 4
which is 28 years therefore we can say
that having is 28 years old and 30 is 20
years old so using linear equations we
had solved the beans type of puzzles as
well thank you please visit exam fee or
come for free quality education you can
learn with a simple four-step learning
process wherein you can watch video
lessons you can ask your questions you
can refer notes and you can take a free
online test we have content for class 6
to 12 on physics chemistry mathematics
and biology along with practical videos
so please subscribe to our channel for
daily updates thank you
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Nov 15, 2024 Last updated |
Examples: Word problems (Part 1) for Class 82024 is part of Class 8 preparation. The notes and questions for Examples: Word problems (Part 1)
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