Maths-
General
Easy
Question
Ayush is choosing between two health clubs. Health club 1: Membership R s 22 and
Monthly fee R s 24.50. Health club 2: Membership R s 47.00 , monthly fee R s 18.25
. After how many months will the total cost for each health club be the same ?
Hint:
The concept used in the question is of the quadratic equation.
The correct answer is: Hence, after 4 months the total cost of each health club will be equal.
Answer:
- Step by step explanation:
○ Given:
Health club 1: Membership R s 22 and Monthly fee R s 24.50
Health club 2: Membership R s 47.00 , monthly fee R s 18.25.
○ Step 1:
○ Let the number of months after which the total cost is equal be x.
So, for Health club 1:
After x months, total cost will be =
Membership Rs 22 + fee for x months
22 + 24.50x
Health club 2:
After x months, total cost will be =
Membership R s 47 + fee for x months
47 + 18.25x
○ Step 2:
○ Equalize both costs to get number of months
22 + 24.50x = 47 + 18.25x
24.50x - 18.25x= 47 - 22
6.25x = 25
x = 
x = 4
- Final Answer:
Hence, after 4 months the total cost of each health club will be equal.
Health club 2:
After x months, total cost will be =
○ Step 2:
○ Equalize both costs to get number of months
Related Questions to study
Maths-
Find the gradient and y- Intercept of the line 
Hint:
Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.
Step by step solution:
The given equation of the line is
x + 2y = 14
We need to convert this equation in the slope-intercept form of the line, which is
y = mx + c
Rewriting the given equation, we have
2y = 14 - x
Dividing by 2, we get

Simplifying, we get

Comparing the above equation with , we get

Thus, we get
Gradient =
y-intercept = 7
Note:
We can find the slope and y-intercept directly from the general form of the equation too; slope =
and y-intercept =
, where the general form of equation of a line is ax + by + c = 0. Using this method, be careful to check that the equation is in general form before applying the formula.
Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.
Step by step solution:
The given equation of the line is
x + 2y = 14
We need to convert this equation in the slope-intercept form of the line, which is
y = mx + c
Rewriting the given equation, we have
2y = 14 - x
Dividing by 2, we get
Simplifying, we get
Comparing the above equation with , we get
Thus, we get
Gradient =
y-intercept = 7
Note:
We can find the slope and y-intercept directly from the general form of the equation too; slope =
Find the gradient and y- Intercept of the line 
Maths-General
Hint:
Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.
Step by step solution:
The given equation of the line is
x + 2y = 14
We need to convert this equation in the slope-intercept form of the line, which is
y = mx + c
Rewriting the given equation, we have
2y = 14 - x
Dividing by 2, we get

Simplifying, we get

Comparing the above equation with , we get

Thus, we get
Gradient =
y-intercept = 7
Note:
We can find the slope and y-intercept directly from the general form of the equation too; slope =
and y-intercept =
, where the general form of equation of a line is ax + by + c = 0. Using this method, be careful to check that the equation is in general form before applying the formula.
Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.
Step by step solution:
The given equation of the line is
x + 2y = 14
We need to convert this equation in the slope-intercept form of the line, which is
y = mx + c
Rewriting the given equation, we have
2y = 14 - x
Dividing by 2, we get
Simplifying, we get
Comparing the above equation with , we get
Thus, we get
Gradient =
y-intercept = 7
Note:
We can find the slope and y-intercept directly from the general form of the equation too; slope =
Maths-
We can express any constant in the variable form without changing its value as
Explanation:
We have given a constant k
We have to find how we can express any constant 𝑘 in the variable form without changing its value
We know that x0 = 1
So After multiplying it with any constant, it will not change its value.
So,kx0 is the answer
Hence, Option C is correct.
- We have given a constant
- We have to find how we can express any constant 𝑘 in the variable form without changing its value.
We have given a constant k
We have to find how we can express any constant 𝑘 in the variable form without changing its value
We know that x0 = 1
So After multiplying it with any constant, it will not change its value.
So,kx0 is the answer
Hence, Option C is correct.
We can express any constant in the variable form without changing its value as
Maths-General
Explanation:
We have given a constant k
We have to find how we can express any constant 𝑘 in the variable form without changing its value
We know that x0 = 1
So After multiplying it with any constant, it will not change its value.
So,kx0 is the answer
Hence, Option C is correct.
- We have given a constant
- We have to find how we can express any constant 𝑘 in the variable form without changing its value.
We have given a constant k
We have to find how we can express any constant 𝑘 in the variable form without changing its value
We know that x0 = 1
So After multiplying it with any constant, it will not change its value.
So,kx0 is the answer
Hence, Option C is correct.
Maths-
x0 = ?
x0 = ?
Maths-General
Maths-
State and prove the Perpendicular Bisector Theorem.
Answer:

Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)

So, according to SAS rule

Hence,
CA = CB
So, any point on perpendicular bisector is at equal distance from end points of line segment.
Hence proved.
- To prove:
- Perpendicular Bisector Theorem:
- Proof:
- Statement:
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step 1:
Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
So, according to SAS rule
Hence,
CA = CB
So, any point on perpendicular bisector is at equal distance from end points of line segment.
Hence proved.
State and prove the Perpendicular Bisector Theorem.
Maths-General
Answer:

Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)

So, according to SAS rule

Hence,
CA = CB
So, any point on perpendicular bisector is at equal distance from end points of line segment.
Hence proved.
- To prove:
- Perpendicular Bisector Theorem:
- Proof:
- Statement:
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step 1:
Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
So, according to SAS rule
Hence,
CA = CB
So, any point on perpendicular bisector is at equal distance from end points of line segment.
Hence proved.
Maths-
What is a monomial? Explain with an example.
Explanation:
A monomial is an expression with only one term.
The examples are: 3x, 6y
- We have to define monomial by giving an example.
A monomial is an expression with only one term.
The examples are: 3x, 6y
What is a monomial? Explain with an example.
Maths-General
Explanation:
A monomial is an expression with only one term.
The examples are: 3x, 6y
- We have to define monomial by giving an example.
A monomial is an expression with only one term.
The examples are: 3x, 6y
Maths-
Find the equation of a line that passes through
and 
Hint:
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

Step by step solution:
Let the given points be denoted by
(a, b) = (-3, 1)
(c, d) = (2, -14)
The equation of a line passing through two points (a, b) and (c, d) is

Using the above points, we have

Simplifying the above equation, we have


Cross multiplying, we get
5(y + 14) = -15(x - 2)
Expanding the factors, we have
5y + 70 = -15x + 30
Taking all the terms in the left hand side, we have
15x + 5y + 70 - 30 = 0
Finally, the equation of the line is
15x + 5y + 40 = 0
Dividing the equation throughout by 5, we get
3x + y + 8 = 0
This is the required equation.
Note:
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is
Step by step solution:
Let the given points be denoted by
(a, b) = (-3, 1)
(c, d) = (2, -14)
The equation of a line passing through two points (a, b) and (c, d) is
Using the above points, we have
Simplifying the above equation, we have
Cross multiplying, we get
5(y + 14) = -15(x - 2)
Expanding the factors, we have
5y + 70 = -15x + 30
Taking all the terms in the left hand side, we have
15x + 5y + 70 - 30 = 0
Finally, the equation of the line is
15x + 5y + 40 = 0
Dividing the equation throughout by 5, we get
3x + y + 8 = 0
This is the required equation.
Note:
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.
Find the equation of a line that passes through
and 
Maths-General
Hint:
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

Step by step solution:
Let the given points be denoted by
(a, b) = (-3, 1)
(c, d) = (2, -14)
The equation of a line passing through two points (a, b) and (c, d) is

Using the above points, we have

Simplifying the above equation, we have


Cross multiplying, we get
5(y + 14) = -15(x - 2)
Expanding the factors, we have
5y + 70 = -15x + 30
Taking all the terms in the left hand side, we have
15x + 5y + 70 - 30 = 0
Finally, the equation of the line is
15x + 5y + 40 = 0
Dividing the equation throughout by 5, we get
3x + y + 8 = 0
This is the required equation.
Note:
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is
Step by step solution:
Let the given points be denoted by
(a, b) = (-3, 1)
(c, d) = (2, -14)
The equation of a line passing through two points (a, b) and (c, d) is
Using the above points, we have
Simplifying the above equation, we have
Cross multiplying, we get
5(y + 14) = -15(x - 2)
Expanding the factors, we have
5y + 70 = -15x + 30
Taking all the terms in the left hand side, we have
15x + 5y + 70 - 30 = 0
Finally, the equation of the line is
15x + 5y + 40 = 0
Dividing the equation throughout by 5, we get
3x + y + 8 = 0
This is the required equation.
Note:
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.
Maths-
The degree of 25x2y23 is
Explanation:
We have given an expression
We know that degree is highest power of variable present in polynomial.
So,
The degree of is
Degree of is
Degree of is
So, Total degree is
Hence, Option B is correct.
- We have been given a polynomial expression in the question for which we have to find its degree.
We have given an expression
We know that degree is highest power of variable present in polynomial.
So,
The degree of is
Degree of is
Degree of is
So, Total degree is
Hence, Option B is correct.
The degree of 25x2y23 is
Maths-General
Explanation:
We have given an expression
We know that degree is highest power of variable present in polynomial.
So,
The degree of is
Degree of is
Degree of is
So, Total degree is
Hence, Option B is correct.
- We have been given a polynomial expression in the question for which we have to find its degree.
We have given an expression
We know that degree is highest power of variable present in polynomial.
So,
The degree of is
Degree of is
Degree of is
So, Total degree is
Hence, Option B is correct.
Maths-
Ritu earns R s 680 in commission and is paid R s 10.25 per hour. Karina earns R s
410 in commissions and is paid R s 12.50 per hour. What will you find if you solve
for x in the equation 10.25x + 680 = 12.5x + 480
Answer:
10.25x + 680 = 12.5x + 480
○ Step 1:
○ Solve the equation.
○ Group the like terms
10.25x + 680 = 12.5x + 480
680 - 480 = 12.5x - 10.25x
200 = 2.25x
○ Step 2:
○ Divide both side with 2.25

88.88 = x
- Hint:
- Step by step explanation:
10.25x + 680 = 12.5x + 480
○ Step 1:
○ Solve the equation.
○ Group the like terms
10.25x + 680 = 12.5x + 480
○ Step 2:
○ Divide both side with 2.25
- Final Answer:
Ritu earns R s 680 in commission and is paid R s 10.25 per hour. Karina earns R s
410 in commissions and is paid R s 12.50 per hour. What will you find if you solve
for x in the equation 10.25x + 680 = 12.5x + 480
Maths-General
Answer:
10.25x + 680 = 12.5x + 480
○ Step 1:
○ Solve the equation.
○ Group the like terms
10.25x + 680 = 12.5x + 480
680 - 480 = 12.5x - 10.25x
200 = 2.25x
○ Step 2:
○ Divide both side with 2.25

88.88 = x
- Hint:
- Step by step explanation:
10.25x + 680 = 12.5x + 480
○ Step 1:
○ Solve the equation.
○ Group the like terms
10.25x + 680 = 12.5x + 480
○ Step 2:
○ Divide both side with 2.25
- Final Answer:
Maths-
If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then the triangle is ____ isosceles.
Answer:

Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
ADC = BDC = 90o
So, according to SAS rule
ACD BCD
Hence,
CA = CB
So, triangle is always isosceles.
Hence proved.
- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step by step explanation:
- Step 1:
Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
ADC = BDC = 90o
So, according to SAS rule
ACD BCD
Hence,
CA = CB
So, triangle is always isosceles.
Hence proved.
- Final Answer:
If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then the triangle is ____ isosceles.
Maths-General
Answer:

Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
ADC = BDC = 90o
So, according to SAS rule
ACD BCD
Hence,
CA = CB
So, triangle is always isosceles.
Hence proved.
- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step by step explanation:
- Step 1:
Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
ADC = BDC = 90o
So, according to SAS rule
ACD BCD
Hence,
CA = CB
So, triangle is always isosceles.
Hence proved.
- Final Answer:
Maths-
Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B. On which of the following segments must P be located?
Answer:
So,
As P is equidistant from A and B it lies on perpendicular bisector on AB.
- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step by step explanation:
- Given:
- Step 1:
- In △ ABC,
So,
As P is equidistant from A and B it lies on perpendicular bisector on AB.
- Final Answer:
Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B. On which of the following segments must P be located?
Maths-General
Answer:
So,
As P is equidistant from A and B it lies on perpendicular bisector on AB.
- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step by step explanation:
- Given:
- Step 1:
- In △ ABC,
So,
As P is equidistant from A and B it lies on perpendicular bisector on AB.
- Final Answer:
Maths-
A constant has a degree__________.
Explanation:
The degree of a constant is Zero.
By definition of degree we know that it is highest power of variable present in polynomial.
But for constant there is no polynomial.
So, The degree of constant is Zero
Hence, Option B is correct.
- We have been given a statement in the question for which we have to fill the blank by choosing the appropriate answer from the given four options.
The degree of a constant is Zero.
By definition of degree we know that it is highest power of variable present in polynomial.
But for constant there is no polynomial.
So, The degree of constant is Zero
Hence, Option B is correct.
A constant has a degree__________.
Maths-General
Explanation:
The degree of a constant is Zero.
By definition of degree we know that it is highest power of variable present in polynomial.
But for constant there is no polynomial.
So, The degree of constant is Zero
Hence, Option B is correct.
- We have been given a statement in the question for which we have to fill the blank by choosing the appropriate answer from the given four options.
The degree of a constant is Zero.
By definition of degree we know that it is highest power of variable present in polynomial.
But for constant there is no polynomial.
So, The degree of constant is Zero
Hence, Option B is correct.
Maths-
Graph the equation 
Hint:
To plot the graph of an equation, first we make a table of points satisfying that equation. Then we draw an x-axis and y-axis on the graph. After that we scale both the axis according the values we get in the table. Lastly, we plot the points from the table on the graph and join them to get the required curve.
Step by step solution:
The given equation is
y = 3x − 5
First we make a table of points satisfying the equation.
Putting x = 0 in the above equation, we will get, y = -5
Similarly, putting x = 1, in the above equation, we get y = −2
Continuing this way, we have
For x = -1, we get y = -8
For x = 2, we get y = 1
For x = -2, we get y = -11
Making a table of all these points, we have

Now we plot these points on the graph.

After plotting the points, we join them with a line to get the graph of the equation.
Note:
We can find the tabular values for any points of x and then plot them on the graph. But we usually choose values for which calculating y is easier. This makes plotting the graph simpler. We can also find the values by putting different values of y in the equation to get different values for x. Either way, we need points satisfying the equation to plot its graph
To plot the graph of an equation, first we make a table of points satisfying that equation. Then we draw an x-axis and y-axis on the graph. After that we scale both the axis according the values we get in the table. Lastly, we plot the points from the table on the graph and join them to get the required curve.
Step by step solution:
The given equation is
y = 3x − 5
First we make a table of points satisfying the equation.
Putting x = 0 in the above equation, we will get, y = -5
Similarly, putting x = 1, in the above equation, we get y = −2
Continuing this way, we have
For x = -1, we get y = -8
For x = 2, we get y = 1
For x = -2, we get y = -11
Making a table of all these points, we have
Now we plot these points on the graph.
After plotting the points, we join them with a line to get the graph of the equation.
Note:
We can find the tabular values for any points of x and then plot them on the graph. But we usually choose values for which calculating y is easier. This makes plotting the graph simpler. We can also find the values by putting different values of y in the equation to get different values for x. Either way, we need points satisfying the equation to plot its graph
Graph the equation 
Maths-General
Hint:
To plot the graph of an equation, first we make a table of points satisfying that equation. Then we draw an x-axis and y-axis on the graph. After that we scale both the axis according the values we get in the table. Lastly, we plot the points from the table on the graph and join them to get the required curve.
Step by step solution:
The given equation is
y = 3x − 5
First we make a table of points satisfying the equation.
Putting x = 0 in the above equation, we will get, y = -5
Similarly, putting x = 1, in the above equation, we get y = −2
Continuing this way, we have
For x = -1, we get y = -8
For x = 2, we get y = 1
For x = -2, we get y = -11
Making a table of all these points, we have

Now we plot these points on the graph.

After plotting the points, we join them with a line to get the graph of the equation.
Note:
We can find the tabular values for any points of x and then plot them on the graph. But we usually choose values for which calculating y is easier. This makes plotting the graph simpler. We can also find the values by putting different values of y in the equation to get different values for x. Either way, we need points satisfying the equation to plot its graph
To plot the graph of an equation, first we make a table of points satisfying that equation. Then we draw an x-axis and y-axis on the graph. After that we scale both the axis according the values we get in the table. Lastly, we plot the points from the table on the graph and join them to get the required curve.
Step by step solution:
The given equation is
y = 3x − 5
First we make a table of points satisfying the equation.
Putting x = 0 in the above equation, we will get, y = -5
Similarly, putting x = 1, in the above equation, we get y = −2
Continuing this way, we have
For x = -1, we get y = -8
For x = 2, we get y = 1
For x = -2, we get y = -11
Making a table of all these points, we have
Now we plot these points on the graph.
After plotting the points, we join them with a line to get the graph of the equation.
Note:
We can find the tabular values for any points of x and then plot them on the graph. But we usually choose values for which calculating y is easier. This makes plotting the graph simpler. We can also find the values by putting different values of y in the equation to get different values for x. Either way, we need points satisfying the equation to plot its graph
Maths-
A= 5x-3y +2z, B= 4x-2y+3z, C= 6x-4y -4z, Find A-B+C
Answer:
A= 5x-3y +2z,
B= 4x-2y+3z,
C= 6x-4y -4z,
Find A-B+C
○ Step 1:
○ Put values of A, B and C in (A-B+C)
We get,
A-B+C
(5x-3y +2z) - (4x-2y+3z) + (6x-4y -4z)
○ Step 2:
○ Group like terms
(5x-4x +6x) - (3y-2y+4y) + (2z-3z -4z)
7x - 5y - 5z
- Hint:
- Step by step explanation:
A= 5x-3y +2z,
B= 4x-2y+3z,
C= 6x-4y -4z,
Find A-B+C
○ Step 1:
○ Put values of A, B and C in (A-B+C)
We get,
A-B+C
○ Step 2:
○ Group like terms
- Final Answer:
A= 5x-3y +2z, B= 4x-2y+3z, C= 6x-4y -4z, Find A-B+C
Maths-General
Answer:
A= 5x-3y +2z,
B= 4x-2y+3z,
C= 6x-4y -4z,
Find A-B+C
○ Step 1:
○ Put values of A, B and C in (A-B+C)
We get,
A-B+C
(5x-3y +2z) - (4x-2y+3z) + (6x-4y -4z)
○ Step 2:
○ Group like terms
(5x-4x +6x) - (3y-2y+4y) + (2z-3z -4z)
7x - 5y - 5z
- Hint:
- Step by step explanation:
A= 5x-3y +2z,
B= 4x-2y+3z,
C= 6x-4y -4z,
Find A-B+C
○ Step 1:
○ Put values of A, B and C in (A-B+C)
We get,
A-B+C
○ Step 2:
○ Group like terms
- Final Answer:
Maths-
For an obtuse triangle, the circumcentre lies
Answer:
For obtuse triangle, the circumcentre lies outside the triangle.
For obtuse triangle, the circumcentre lies outside the triangle.
For an obtuse triangle, the circumcentre lies
Maths-General
Answer:
For obtuse triangle, the circumcentre lies outside the triangle.
For obtuse triangle, the circumcentre lies outside the triangle.
Maths-
For a right triangle, the circumcentre lies
Answer:
For a right triangle, the circumcentre lies on the triangle.
a. on the triangle.
For a right triangle, the circumcentre lies on the triangle.
a. on the triangle.
For a right triangle, the circumcentre lies
Maths-General
Answer:
For a right triangle, the circumcentre lies on the triangle.
a. on the triangle.
For a right triangle, the circumcentre lies on the triangle.
a. on the triangle.