Category: Class 7

Students can go through AP Board 7th Class Maths Notes Chapter 7 Data Handling to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 7 Data Handling

→ Data: Information which is in the form of numbers or words and helps in taking decisions or drawing conclusions is called data. Tables and graphs are the ways in which data is presented.
The numerical entries in the data are called ‘observations’.

→ The average or Arithmetic Mean or Mean
A.M = \(\frac{\text { Sum of all observations }}{\text { Number of observations }}\)

→ (i.e.,) A.M. is equal to sum of all the observations of a data set divided by the number of observations. It lies between the lowest and highest values of the data.

→ Mode: An observation of data that occurs most frequently is called the mode of the data. A data may have one or more modes and sometimes none.

→ Median: Median is simply the middle observation, when all observations are arranged in ascending or descending order. In case of even number of observations median is the average of middle observations.

→ Mean, mode, median are representative values for a data set.

→ When all values of data set are increased or decreased by a certain number, the mean also increases or decreases by the same number.

→ Data can be presented in bar graphs / double bar graphs or pie chart.

→ Bar graph: Bar graph are made up of bars of uniform width which can be drawn horizontally or vertically with equal spacing between them. The length of each bar tells us the frequency of the particular item. We take convenient scale for the length of bar graph.

→ Double bar graph: It presents two observations side by side.

→ Pie chart: A circle is divided into sectors to represent the given data.
Angle subtended by the sector at the centre of the circle is directly proportional to each observation.

Students can go through AP Board 7th Class Maths Notes Chapter 6 Ratio – Applications to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 6 Ratio – Applications

→ Ratio: A ratio is an ordered comparison of quantities of the same units.
We use the symbol ‘:’ to represent a ratio. The ratio of two quantities ‘a’ and ‘b’ is a : b and we read it as “a is to b”. The two quantities ‘a’ and ‘b’ are called the terms of the ratio. The first quantity ‘a’ is called first term or antecedent and the second quantity ‘b’ is called consequent.

→ Proportion: If two ratios are equal, then the four terms of the ratios are said to be in proportion. We use the symbol : : (is as)
If two ratios a : b and c : d are equal, we write a : b :: c : d or a : b = c : d
Here ‘a’, ‘d’ are called extremes and b, c are called means.

→ Unitary Method: The method in which we first find the value of one unit and then the value of the required number of units is known as unitary method.
Eg: If the cost of 5 pens is Rs. 85; then the cost of 12 pens is ……… ?
Solution. Cost of 5 pens = Rs. 85
Cost of 1 pen = \(\frac{85}{5}\) = Rs. 17
∴ Cost of 12 pens = 12 × 17 = Rs. 204

→ Direct proportion: If in two quantities, when one quantity increases, the other also increase or vice-versa then the two quantities are said to be in direct proportion.
Eg: The number of books and their cost are in direct proportion.
As the number of books increases, the cost also increases.

→ Ratios also appear in the form of percentages.

→ The word percent means “per every hundred” or for a hundred. The symbol % is used to denote percentage.

→ To convert a quantity into its equivalent percentage

• express it as a fraction.
• multiply it with 100.
• assign % symbol.

Eg: A man purchased an article for Rs. 80 and sells it for Rs. 100. Find his gain percent.
Solution. Cost price = Rs. 80
Selling price = Rs. 100
gain = Rs. 20
gain as a fraction = \(\frac{20}{80}\)
gain as percent = \(\frac{20}{80}\) × 100 = 25%

→ When C.P > S.P there incurs loss.

→ When C.P < S.P there is gain.

→ When C.P = S.P neither loss nor gain.

→ Loss = C.P – S.P gain = S.P – C.P

→ Discount is always expressed as some percentage of marked price.

→ In general when P is principle; R% is rate of interest per annum and I is the interest, then
I = R% of P
I = R% of P for T years

Students can go through AP Board 7th Class Maths Notes Chapter 5 Triangle and Its Properties to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 5 Triangle and Its Properties

→ Triangles can be classified according to properties of their sides and angles. Based on sides, triangles are of three types.

→ Equilateral triangle: A triangle in which all the three sides are equal is called an equilateral triangle. In △ABC
AB = BC = CA, also ∠A = ∠B = ∠C In an equilateral triangle each angle is equal to 60°.

→ Isosceles triangle: A triangle in which two sides are equal is called an isosceles triangle.
In △PQR
PQ = PR also ∠Q = ∠R

The non-equal side in an isosceles triangle may be taken as base of the triangle.

→ Scalene triangle: A triangle in which no two sides are equal is called a scalene triangle.

In △BAT
BA ≠ AT ≠ BT also ∠B ≠ ∠A ≠ ∠T.

→ Based on angles, triangles can be classified into three types.

→ Acute angled triangle: A triangle in which all the three angles are acute is called an acute-angled triangle.

In △TAP,
∠T, ∠A, ∠P are acute angles.

→ Obtuse angled triangle: A triangle in which one angle is obtuse is called an obtuse angled triangle.

In △FAN,
∠A is obtuse angle.
A triangle cannot have more than one obtuse angle,

→ Right angled triangle: A triangle in which one angle is a right angle is called a right angled triangle.

In △COT
ZO is right angle (i.e) 90°.
A triangle cannot have more than one right angle,

→ Right angled isosceles triangle: A triangle in which one angle is right angle and two sides are equal is called a right angled isosceles triangle.

In △POT,
PO = OT and ∠O = 90° also
∠P = ∠T = 45°

→ Family of triangles – Flow chart

→ Relation between sides of a triangle
In any triangle the sum of the lengths of any two sides is greater than the length of the third side.

In △TIN,
TI + IN > TN; TN + NI > TI; TI + TN > IN
Also the difference between lengths of any two sides of the triangle is less than the length of the third side.
In △TIN, TI > TN – NI; IN > TI – TN; TN > IN – TI

→ Altitutes of a triangle
The length of a line segment drawn from a vertex to its opposite side and is perpendicular to it is called an altitude or height of the triangle. An altitude can be drawn from each vertex.

Altitude of a triangle may be in its interior or exterior.

→ Medians of a triangle
A line segment joining a vertex and the mid-point of its opposite side is called a median.

A triangle has three medians.
The medians of a triangle are concurrent.
The point of concurrence of medians of a triangle is called the centroid of the triangle.
In △ABC, D, E and F are mid-points of the sides AB, BC and AC.
AE, CD and BF are mid-points.
G is the centroid.

→ Angle – sum property of a triangle: The sum of interior angles of a triangle is equal to 180° or two right angles.

In △BET,
∠B + ∠E + ∠T = 180°

→ Exterior angle of a triangle

→ When one side of a triangle is produced, the angle thus formed is called an exterior angle.

In △COL; the side OL is produced to D.
∠CLD is an exterior angle.
The exterior angle of a triangle is equal to the sum of the interior opposite angles.
∠COL + ∠OCL = ∠CLD

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions Exercise 4

Question 1.
Add the following algebraic expressions using both horizontal and vertical methods. Did
you get the same answer with both methods.
(i) x² – 2xy + 3y²; 5y² + 3xy – 6x²
(ii) 4a² + 5b² + 6ab 3ab ; 6a² – 2b² ; 4b² – 5ab
(iii) 2x + 9y – 7z ; 3y + z + 3x ;2x – 4y – z
(iv) 2x² – 6x + 3 ; – 3x² – x – 4 ; 1 + 2x – 3x²
Solution:
(i) x² – 2xy + 3y²; 5y² + 3xy – 6x²
Addition by Horizontal method
(x² – 2xy + 3y²) + ( – 6x² + 3xy + 5y²)
= (x² – 6x²) + ( – 2xy + 3xy) + (3y² + 5y²)
= – 5x² + xy + 8y² .

ii) 4a² + 5b² + 6ab;3ab; 6a² – 2b²; 4a² – 5ab
Horizontal Method
(4a² + 5b² + 6ab) + 3ab + (6a² – 2b²) (4a² – 5ab)
= (4a² + 6a² + 4a²) + (5b² – 2b²) + (6ab + 3ab – 5ab)
= 14a² + 3b² + 4ab

iii) 2x² + 9y – 7z ; 3y + z + 3x; 2x4y – z
Horizontal Method
= (2x + – 7z) + (3y + z + 3x) + (2x – 4y – z)
= (2x + 3x + 2x) + (9y + 3y_4y) + ( – 7z + z – z)
= 7x + 8y – 7z

iv) 2x² – 6x – 3; – 3x² – x – 4; 1 + 2x – 3x²
Horizontal Method
(2x² – 6x + 3) + ( – 3x² – x – 4) + (1 + 2x – 3x²)
(2x² – 3x² – 3x²) + (- 6x – x + 2x) + (3 – 4 + 1)
= – 4x² – 5x + 0
= – 4x² – 5x

In all the above sums
(i), (ii), (iii), (iv) we got same answer in both horizontal and vertical methods.

Question 2.
Simpli1’: 2x² + 5x – 1 + 8x + x² + 7 – 6x + 3 – 3x²
Solution:
Given algebraic expression is
2x² + 5x – 1 + 8x + x² + 7 – 6x + 3 – 3x²
= (2x² + x² – 3x²) + (5x + 8x – 6x) . ( – 1 + 7 + 3)
= 0 + 7x + 9 = 7x + 9

Question 3.
Find the perimeter of the following rectangle?

Solution:
The perimeter of a rectangle = 2 (length + breadth)
= 2 (6x + y + 3x – 2y)
= 2 (9x – y)
∴ P = (18x – 2y) units

Question 4.
Find the perimeter of a triangle whose sides are 2a + 3b, b – a, 4a – 2b.

Solution:
Perimeter of the triangle ABC = \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CA}}\)
= (b – a) + (3a + 2b) + (4a – 2b)
=( – a + 3a + 4a) +(b + 2b – 2b)
= (6a + b) units

Question 5.
Subtract the second expression from the first expression
(i) 2a + b, a – b
(ii) x + 2y + z , – x – y – 3z
(iii) 3a² – 8ab – 2b², 3a² – 4ab+6b²
(iv) 4pq – 6p² – 2q², 9p²
(v) 7 – 2x – 3x², 2x² – 5 x – 3
(vi) 5x² – 3xy – 7y² , 3x² – xy – 2y²
(vii) 6m³ + 4m² + 7m – 3 , 3m³ + 4
Solution:
(i) 2a + b, a – b = 2a + b – a + b
= (2a – a) + (b + b)
= a + 2b

ii) (x + 2y + z) – ( – x – y – 3z) = x + + z + x + y + 3z
=(x + x) + (2y + y) +(z + 3z)
= 2x + 3y + 4z

iii) (3a² – 8ab – 2b²) – (3a² – 4ab + 6b²) = 3a² – 8ab – 2b² – 3a² +4ab – 6b²
= (3a² – 3a²) + ( – 8ab + 4ab) + (2b² – 6b²)
= 0 – 4ab – 8b²
= – 4ab – 8b²

iv) (4pq – 6p² – 2q²) – (9p²) = 4pq – 6p² – 2q² – 9p²
= 4pq – 15p² – 2q²

v) (7 – 2x – 3x²) – (2x² – 5x – 3)=7 – 2x – 3x² – 2x² + 5x + 3
= ( – 3x² – 2x²) + ( – 2x + 5x) + (7 + 3)
= – 5x² + 3x + 10

vi) (5x² – 3xy – 7y²) – (3x² – xy – 2y²) = 5x²– 3xy – 7y² – 3x² + xy + 2y²
= (5x² – 3x²) + ( – 3xy + xy) + ( – 7y² + 2y²)
= 2x² – 2xy – 5y²

viii) (6m³ + 4m² + 7m – 3) – (3m³ + 4)= 6m³ + 4m² + 7m – 3 – 3m3 – 4
= (6m³ – 3m³) + 4m² + 7m + (- 3 – 4)
= 3m³ + 4m² + 7m – 7

Question 6.
Subtract the sum of x² – 5xy + 2y² and y² – 2xy – 3x² from the sum of 6x² – 8xy – y² and 2xy – 2y² – x².
Solution:
The sum of x² – 5xy + 2y² and y² – 2xy – 3x² = (x² – 5xy + 2y²) (y – 2xy – 3x²)
= (x² – 3x²) + ( – 5xy – 2xy) + (2y² + y²)
= – 2x² – 7xy + 3y² ………………….(1)

The sum of 6x² – 8xy – y² and
2xy – 2y² – x² = (6x² – 8xy – y²) + (2xy – 2y² – x²)
= (6x² – x²) + ( – 8xy + 2xy) + (-y² – 2y²)
= 5x² – 6xy – 3y² ………………. (2)

From (1) and (2)
(2) – (1)= (5x² – 6xy – 3y²) – ( – 2x² – 7xy + 3y²)
= 5x² – 6xy – 3y² + 2x² + 7xy – 3y²
= (5x² + 2x²) + ( – 6xy + 7xy) + ( – 3y² – 3y²)
= 7x² + xy – 6y²

Question 7.
What should be added to 1 + 2x – 3x² to get x² – x – 1?
Solution:
Let the added algebraic expression may be ‘A say
(1 + 2x – 3x²) + A = x² – x – 1
=A =(x² – x – 1) – (1 + 2x – 3x²)
= x² – x – 1 – 1 – 2x + 3x²
= (x² + 3x²) + ( – x – 2x) +( – 1 – 1)
∴ A = 4x² – 3x – 2
∴ The required added expression (A) = 4x² – 3x – 2

Question 8.
What should be taken away from 3x² – 4y² + 5xy +20 to get – x² – y² + 6xy + 20.
Solution:
Let the subtracted algebraic expression may be B’ say
3x² – 4y² + 5xy – B = – x² – y² + 6xy + 20
B = (3x² – 4y² + 5xy) – (- x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + x² +y² – 6xy – 20
= (3x² + x²) ( – 4y² + y²) + (5xy – 6xy) – 20
∴ B = 4x² – 3y² – xy – 20
∴ The required subtracted expression is B = 4x² – 3y² – xy – 20

Question 9.
The sum of 3 expressions is 8 + 13a + 7a². Two of them are 2a² + 3a + 2 and 3a2 – 4a + 1. Find the third expression.
Solution:
Given that f he sum of 3 expressions is 8 + 13a + 7a² ……………..(1)
Two of them are 2a² + 3a + 2 and 3a² – 4a + 1.
∴ The sum of above two expressions (2a² + 3a + 2) + (3a² – 4a + 1)
= 2a² + 3a + 2 + 3a² – 4a + 1.
= (2a² + 3a²) + (3a – 4a) + (2 + 1)
= 5a² – a + 3
∴ The required 3rd expression (1) – (2)
(1) – (2) =(7a² + 13a + 8) – (5a² – a + 3) = 7a² + 13a + 8 – – 5a² – a – 3
=(7a² – 5a²) +(13a + a)+ (8 – 3)
= 2a² + 14a + 5

Question 10.
If A = 4x² + y² – 6xy;
B = 3y²+ 12x² + 8xy;
C = 6x² + 8y² + 6xy
Find (i) A + B + C (ii) (A – B) – C
Solution:
(i) im 7
(ii) (A – B) – C
(A – B) = (4x² + y² – 6xy) – (3y² + 12x² + 8xy)
= (4x² – 12x²) + (y² – 3y²) + ( – 6xy – 8xy)
A – B = – 8x² – 2y² – 14xy
∴ (A – B) – C = ( – 8x² – 2y² – 14xy) – (6x² + 8y² + 6xy)
= ( – 8x² – 6x²) + -2y² – 8y²) + ( – 14xy – 6xy)
= – 14x² – 10y² – 20xy
∴ (A – B) – C = – (14x² + 10y² + 20xy)

iii) 2A + B
2A = 2(4x² + y² – 6xy) = 8x² + 2y² – 12xy
∴ 2A + B = (8x² + 2y² – 12xy) + (3y² + 12x² + 8xy)
= (8x² + 12x²) + (2y² + 3y²) + ( – 12xy + 8xy)
∴ 2A + B = 20x² + 5y² – 4xy

iv) A – 3B
38 = 3(3y² + 12x² + 8xy) = 9y² + 36x² + 24xy
∴ A – 3B = (4x² + y² – 6xy) – (9y² + 36x² + 24xy)
= (4x² – 36x²) (y² – 9y²) ( 6xy – 24 xy)
= 32x² – 8y² – 30xy
∴ A – 3B = – 32x² – 8y² – 30xy (or)
= – [32x² + 8y + 30xy]

Students can go through AP Board 7th Class Maths Notes Chapter 4 Lines and Angles to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 4 Lines and Angles

→ Complementary angles: If the sum of two angles is 90°, the angles are called complementary angles.
Eg: 55°, 35° are complementary angles.

→ Linear pair of angles: When a ray stands on a line, the pair of angles thus formed are called linear pair of angles and their sum is 180°. In the figure ∠x and ∠y are called linear pair of angles.

→ Adjacent angles: Two angles with a common vertex and a common arm are called adjacent angles. Here the non-common arms lie on either sides of the common arm.
Eg: ∠AOB and ∠BOC are adjacent angles.

→ Supplementary angles : Two angles are said to be supplementary if their sum is 180°. Eg : (100°, 80°), (110°, 70°), (60°, 120°)

→ Vertically opposite angles: Two angles are said to be vertically opposite angles if they are formed by two intersecting lines and are not adjacent.
Eg: ∠AOC and ∠BOD are vertically opposite angles.

→ A line which intersects two or more lines at distinct points is called a transversal. In the figures ‘t’ is a transversal.

In the figure (i) t is not a transversal as it doesn’t intersect other two lines at two distinct points.

→ When a transversal intersects a pair of lines, 8 angles are formed.
Here l, m are two lines and ‘t’ is a transversal.

The angles are ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8.
∠3, ∠4, ∠5, ∠6 are called interior angles and ∠1, ∠2, ∠7 and ∠8 are called exterior angles.

→ When a transversal intersects a pair of lines the following pairs of angles are called corresponding angles.

i) One angle is interior and the other is exterior.
ii) Not adjacent angles.
iii) Two angles are on the same side of the transversal.

→ The following pairs are called alternate interior angles
i) both are interior
ii) not-adjacent
iii) either sides of the transversal

→ The following pairs are called alternate exterior angles
i) both are exterior
ii) not-adjacent
iii) on the either sides of the transversal

→ The following pairs are called interior angles on the same side of the transversal.

→ When a pair of parallel lines intersected by a transversal the pairs of corresponding angles are equal.
∠1 = ∠5
∠2 = ∠6
∠3 = ∠7
∠4 = ∠8

The pairs of alternate interior angles are equal.
∠3 = ∠5
∠4 = ∠6
The pairs of alternate exterior angles are equal.
∠1 = ∠7
∠2 = ∠8
The interior angles on the same side of the transversal are supplementary.
∠3 + ∠6 = 180°
∠4 + ∠5 = 180°
The exterior angles on the same side of the transversal are supplementary.
∠1 + ∠8 = ∠2 + ∠7 = 180°

→ Conversely when a transversal intersects a pair of lines in such way
i) making pairs of corresponding angles equal
(or)
ii) making alternate interior angles equal.
(or)
iii) making alternate exterior angles equal.
(or)
iv) making angles on the same side of the transversal interior/exterior supplementary then the lines are parallel.

Students can go through AP Board 7th Class Maths Notes Chapter 3 Simple Equations to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 3 Simple Equations

→ Simple equations help in solving various problems in daily life.
Eg: After 5 years Ramesh’s age is 15 years. What is his present age?
Solution. Let Ramesh’s present age be x years
After 5 years Ramesh’s age = x + 5
By problem, x + 5 = 15
x = 15 – 5 = 10 years
∴ Ramesh’s present age = 10 years

→ To balance an equation
a) Same number can be added on both sides.
b) Same number can be subtracted from both sides.
c) Multiply both sides with same number.
d) Divide both sides with same number.
So that the equality remains unaltered.

→ An equation remains same if the L.H.S and R.H.S are interchanged.

→ To solve a simple equation we transform term from one side to another.
While transforming term from one side to another
‘+’ quantity becomes ‘-‘ quantity
‘-‘ quantity becomes ‘+’ quantity
‘×’ quantity becomes ‘÷’ quantity
‘÷’ quantity becomes ‘×’ quantity
(i.e.) when the terms are transposed they get opposite signs and the term which multiplies one side, divides the other side.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions Exercise 3

Question 1.
Find the length of the line segment PR in the following figure in terms of ’a’.

Solution:
The length of PR = \(\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}\)
= 3a + 2a
= 5a units

Question 2.
(i) Find the perimeter of the following triangle.

(ii) Find the perimeter of the following rectangle.

Solution:
i) The perimeter of the triangle = \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CA}}\)
= 2x + 6x + 5x
= 13x units
ii) The perimeter of the rectangle ABCD = \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CD}}+\overline{\mathrm{DA}}\)
or
= 2 (l + b)
= 2 (3x + 2x)
= 2 × 5x
= 10x units

Question 3.
Subtract the second terni from first term.
(i) 8x, 5x
(ii) 5p, 11p
(iii) 13m², 2m²
Solution:
i) 8x – 5x = 3x
ii) 5p – 11p = -6p
iii) 13m² – 2m² = 11m²

Question 4.
Find the value of following monomials, if x =1.
(i) -x
(ii) 4x
(iii) -2x²
Solution:
1) If x = 1 ⇒ – x = – (1)= – 1
ii) If x = 1 ⇒ 4x = 4 x 1 = 4
iii) If x= 1 ⇒ – 2x² = – 2(1)² = – 2 × 1= – 2

Question 5.
Simplify and find the value of 4x + x – 2x² + x – 1 when x = -1.
Solution:
Given expression = 4x + x – 2x² + x – 1
= 4x + x + x – 2x² – 1
= – 2x² + 6x – 1
If x = – 1 then, the value of – 2x² + 6x – 1 = – 2 (-1)2 + 6( – 1) – 1
= – 2(1) – 6 – 1
= – 2 – 6 – 1
= – 9

Question 6.
Write the expression 5x² – 4 – 3x² + 6x + 8 + 5x – 13 in its simplified form. Find its value when x = -2
Solution:
The given algebraic expression = 5x² – 4 – 3x² + 6x + 8 + 5x – 13.
= (5x² – 3x²) (6x + 5x) + (-4 – 13)
= 2x² + 11x – 17
When x = – 2. then the value of 2x²+ 11x – 17 = 2(- 2)² + 11(-2) – 17
= 2 × 4 – 22 – 17
= 8 – 39
= – 31

Question 7.
If x = 1; y = 2 find the values of the following expressions
(i) 4x – 3y + 5
(ii) x² + y²
(iii) xy + 3y – 9
Solution:
i) 4x – 3y+5
If x = 1, y = 2 then the value of 4x – 3y + 5 = 4(1) – 3(2) + 5
= 4 – 6 + 5
= 9 – 6 = 3

ii) x² + y²
If x = 1, y = 2, the value of x² + y²= (1)² + (2)²
=(1 × 1) + (2 × 2)
= 5

iii) xy + 3y – 9
If x = 1; y = 2,the value of xy + 3y – 9 = 1 × 2 + 3(2) – 9
= 2 + 6 – 9
= 8 – 9

Question 8.
Area of a rectangle is given by A = l × b . If l = 9cm, b = 6cm, find its area?
Solution:
Formula for area of a rectangle (A) = l × b
If l = 9cm, b = 6cm
then the area of the rectangle = l × b
= 9 × 6
= 54 cm²

Question 9.
Simple interest is given by I = \(\frac{P T R}{100}\) . If P = ₹ 900, T =2 years; and R =5%, find the simple interest.
Solution:
Given that the formula for simple interest (I) = \(\frac{P T R}{100}\)
P = Rs.900, T = 2years, R = 5%
∴ The required simple interest (I) = \(\frac{\mathrm{PTR}}{100}=\frac{900 \times 2 \times 5}{100}\) = 9 x 2 x 5 = Rs. 90

Question 10.
The relationship between speed (s), distance (d) and time (t) is given by S = – . Find the
value of s, if d = 135 meters and t = 10 seconds.
Solution:
Given that s = \(\frac{\mathrm{d}}{\mathrm{t}}\) and d = 135 meters, t = 10 seconds
∴ The value of s = \(\frac{\mathrm{d}}{\mathrm{t}}=\frac{135}{10}\)
∴ speed (s) = 13.5 m/sec.

Students can go through AP Board 7th Class Maths Notes Chapter 2 Fractions, Decimals and Rational numbers to understand and remember the concepts easily.

AP State Board Syllabus 7th Class Maths Notes Chapter 2 Fractions, Decimals and Rational numbers

→ A proper fraction is a fraction that represents a part of a whole i.e., a fraction in which numerator is less than the denominator is called a proper fraction.
Eg: \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{5}{6}\), \(\frac{8}{13}\),….. etc

→ An improper fraction is a fraction that represents a whole or more than a whole i.e., a fraction in which the numerator is more than or equal to the denominator is called an improper fraction.
Eg: \(\frac{5}{3}\), \(\frac{4}{3}\), \(\frac{8}{7}\), \(\frac{11}{5}\), ….. etc

→ Fractions can be represented pictorially.
Eg:

→ Like fractions can be compared by their numerators.

→ Unlike fractions can be compared by converting them into like fractions.

→ An equivalent fraction of a given fraction can be obtained by multiplying its numerator and denominator by same number.
Eg: Equivalent fraction for \(\frac{3}{5}\) is

→ To multiply a fraction with a whole number; we take the product of the numerator and the whole number as the new numerator, keeping the denominator the same.
Eg:

→ Product of the fractions = \(\frac{\text { Product of Numerators }}{\text { Product of Denominators }}\)
Eg: \(\frac{5}{3}\) × \(\frac{4}{7}\) = \(\frac{20}{21}\)

→ In mathematical computation ‘of’ means multiplication.
Eg: \(\frac{1}{3}\) of 24 = \(\frac{1}{3}\) × 24 = 8

→ The product of two proper fractions is less than each of the fraction in multiplication.
Eg:

→ The product of a proper and improper fraction is less than the improper fraction and greater than the proper fraction.
Eg: \(\frac{3}{4}\) × \(\frac{7}{5}\) = \(\frac{21}{20}\)
Here \(\frac{3}{4}\) < \(\frac{21}{20}\) and \(\frac{7}{5}\) > \(\frac{21}{20}\)

→ The product of two improper fractions is greater than each of the fractions.
Eg: \(\frac{7}{5}\) × \(\frac{3}{2}\) = \(\frac{21}{20}\)
Here \(\frac{7}{5}\) < \(\frac{21}{10}\) and \(\frac{3}{2}\) < \(\frac{21}{20}\)

→ To divide a whole number with a fraction; multiply the whole number by the reciprocal of the given fraction and vice versa.
Eg:

→ To divide one fraction by another, multiply the first fraction with the reciprocal of 2nd fraction.
Eg: \(\frac{3}{5}\) ÷ \(\frac{5}{8}\) = \(\frac{3}{5}\) × \(\frac{8}{5}\)

→ To multiply a decimal by 10,100,1000,, we move the decimal point in the number to the right side as many places as there are zeros in the numbers 10, 100, 1000 ……
Eg:
1.125 × 10 = 11.25
1.125 × 100 = 112.5
1.125 × 1000 = 1125
1.255 × 10,000 = 12,550

→ To multiply two decimal numbers.
i) multiply them as whole numbers.
ii) count the total number of digits in decimal places and add them.
iii) place the decimal point in the product by counting the sum of digits from its right most place.
Eg: 6.25 × 3.14
i) 625 × 314 = 196250
ii) sum of the number of digits in decimal places = 2 + 2 = 4
iii) 19.6250

→ To divide a decimal number by numbers like 10,100,1000, …… etc. we shift the decimal point in the decimal number to the left by as many places as there are zeros in 10, 100,1000 etc.
Eg: 435.873 ÷ 10 = 43.5873
4551.3 ÷ 100 = 45.513
8374.2 ÷ 1000 = 8.3742
24.82 ÷ 1000 = 0.02482

→ To divide a decimal number by a whole number
i) divide them as whole numbers
ii) place the decimal point in the quotient as in the decimal number.
Eg: 86.5 ÷ 5
i) 865 ÷ 5 = 173
ii) 17.3
To divide a decimal number by another,
i) shift the decimal to the right by equal number of places in both to convert the denominator to a whole number.
ii) divide them as in above
Eg: 6.25 ÷ 2.5

→ The numbers in the form \(\frac{p}{q}\) where p, q are integers and q ≠ 0 are called rational numbers.

→ The set of rational numbers is represented by Q.

→ Q includes all integers, positive fractional numbers and negative fractional numbers.

→ All rational numbers can be represented on a number line.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions Exercise 2

Question 1.
Identify and write the like terms in each of the following groups.
(i) a², b², -2a², c², 4a
(ii) 3a, 4x, – yz, 2z
(iii) -2xy², x²y, 5y²x, x²z
(iv) 7p, 8pq, -5pq, -2p, 3p
Solution:
i) Group of like terms : [a², -2a²]
ii) Group of like terms: {-yz, 2zy}
iii) Group of like terms: {-2xy², 5y²x}
iv) Group of like terms: {7p, -2p, 3p} , : {8pq,-5pq}

Question 2.
State whether the expression is a numerical expression or an algebraic expression.
(i) x + 1
(ii) 3m²
(iii) -30 + 16
(iv) 4p² – 5q²
(v) 96
(vi) x² – 5yz
(vii) 215x²yz
(viii) 95 ÷ 5 x 2
(ix) 2 + m + n
(x) 310 + 15 + 62
(xi) 11 a² + 6b² – 5
Solution:
Algebraic expression
(i) x + 1
(ii) 3m²
(iv) 4p² – 5q²
(vi) x² – 5yz
(vii) 215x²yz
(ix) 2 + m + n
(x) 310 + 15 + 62
(xi) 11 a² + 6b² – 5

Numerical expression
(iii) -30 + 16
(v) 96
(viii) 95 ÷ 5 x 2
(x) 310 + 15 + 62

Question 3.
State whether the algebraic expression given below is monomial, binomial, trinomial or multinornial.
(i) y²
(ii) 4y — 7z
(iii) 1 + x + x²
(iv) 7mn
(v) a² + b²
(vi) 100 xyz
(vii) ax + 9
(viii) p² – 3pq + r
(ix) 3y² – x²y² + 4x
(x) 7x² – 2xy + 9y² – 11
Solution:

Algebraic Expression	Its name
(i) y2	Monomial
(ii) 4y – 7z	Binomial
(iii) 1 + x + x2	Trinomial
(iv) 7mn	Monomial
(v) a2 + b2	Binomial
(vi) 100 xyz	Monomial
(vii) ax + 9	Binomial
(viii) p2 – 3pq + r	Trinom ial
(ix) 3y2 – x2y2 + 4x	Trinomial

Question 4.
What is the degree of each of the monomials.
(i) 7y
(ii) -xy²
(iii) xy²z²
(iv) -11y²z²
(v) 3mn
(vi) -5pq²
Solution:

Monomial	Degree
(i) 7y	1
(ii) -xy2	1 + 2 = 3
(iii) xy2z2	1 + 2 + 2 = 5
(iv) -11y2z2	2 + 2 = 4
(v) 3mn	1 + 1 = 2
(vi) -5pq2	1 + 2 = 3

Question 5.
Find the degree of each algebraic expression.
(i) 3x – 15
(ii) xy + yz
(iii) 2y²z + 9yz – 7z – 11x²y²
(iv) 2y²z + 10yz
(v) pq + p²q – p²q²
(vi) ax² + bx + c
Solution:

Algebraic Expression	Degree
(i) 3x – 15	1
(ii) xy + yz	2
(iii) 2y2z + 9yz – 7z – 11x2y2	4
(iv) 2y2z + 10yz	3
(v) pq + p2q – p2q2	4
(vi) ax2 + bx + c	2

Question 6.
Write any two Algebraic expressions with the same degree.
Solution:
Algebraic expression, with the same degree are
i) ax² + bx + c .
ii) 4x² – 5x – 1

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions Exercise 1

Question 1.
Find the rule which gives the number of matchsticks required to make the following patterns
(i) A pattern of letter ‘H’
(ii) A pattern of letter ‘V’
Solution:
i) A pattern of letter ‘H’
The letter ‘H’ is formed with 3 matchsticks.
∴ Its pattern should be 3, 6, 9, 12 (or)
3 × 1, 3 × 2, 3 × 3, 3 × 4 ……………….. 3 x n
∴ The required pattern is 3 × n = 3n

ii) A pattern of letter ‘V’
The letter ‘V’ is formed with 2 matchsticks.
∴ Its pattern should be 2, 4. 6. 8 (or)
2 × 1, 2 × 2, 2 × 3 …………………2 × n
∴ The required pattern is 2 × n = 2n

Question 2.
Given below is a pattern made from coloured tiles and white tiles.

(i) Draw the next two figures in the pattern above.
(ii) Fill the table given below and express the pattern in the form of an algebraic expression.

(iii) Fill the table given below and express the pattern in the form of an algebraic expression.

Solution:



∴ The required algebraic expression is “4”

∴ The required algebraic expression is “4+1”
If n = 1 ⇒ 4n + 1 = 4(1) + 1 = 5
n = 2 ⇒ 4 × 2 + 1 = 8 + 1 = 9
n = 3 ⇒ 4 × 3 + 1 = 12 + 1 =13 ……………….. soon.

Question 3.
Write the following statements using variables, constants and arithmetic operations.
(i) 6 more than p
(ii) ‘x’ is reduced by 4
(iii) 8 subtracted from y
(iv) q multiplied by ‘-5’
(v) y divided by 4
(vi) One-fourth of the product of ’p’ and ‘q’
(vii) 5 added to the three times of ’z’
(viii) x multiplied by 5 and added to ‘10’
(ix) 5 subtracted from two times of ‘y’
(x) y multiplied by 10 and added to 13
Solution:

Sentence	Algebraic expression
1. 6 more than p	p + 6
2. x is reduced by 4	x – 4
3. 8 subtracted from y	y – 8
4. q multiplied by ‘-5’	-5q
5. y divided by 4	\(\frac{y}{4}\)
6. One-fourth of the product of ‘p’ and ‘q’	\(\frac{\mathrm{pq}}{4}\)
7. 5 added to the three times of z	3z + 5
8. ‘x’ multiplied by 5 and added to 10	5x + 10
9. 5 subtracted from two times of ‘y’	2y –  5
10. y multiplied by 10 and added to 13	10y + 13

Question 4.
Write the following expressions in Statements.
(i) x + 3
(ii) y – 7
(iii) 10l
(iv) \(\frac{x}{5}\)
(v) 3m + 11
(vi) 2y – 5
Solution:
Expression Statements

Expression	Statements
i) x + 3	x is added to 3
ii) y – 7	7 is subtracted from y.
iii) 10l	l is multiplied by 10.
iv) \(\frac{\mathrm{x}}{5}\)	x is divided by 5
v) 3m + 11	m is multiplied by 3 and added to 11
vi) 2y – 5	y is multiplied by 2 and 5 ¡s subtracted from the product

Question 5.
Some situations are given below. State the number in situations is a variable or constant?
Example: Our age – its value keeps on changing so it is an example of a variable quantity.
(i) The number of days in the month of January
(ii) The temperature of a day
(iii) Length of your classroom
(iv) Height of the growing plant
Solution:
i) No. of days ¡n the month of January are fixed in every year. So, “Number of days” is a constant.
ii) The temperature of a day changes every minute. So (temperature) it ¡s a variable.
iii) The length of the classroom is fixed. So it is a constant.
iv) The height of a growing plant changes in every month. So it is a variable.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 9 Construction of Triangles Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Construction of Triangles Exercise 4

Question 1.
Construct a right-angled ΔABC such that ∠B = 90°, AB = 8 cm and AC = 10 cm.
Solution:
∠B = 90°, AB = 8cm, AC = 10 cm.

Step -1: Draw a rough sketch of the right angled triangle and label it with the given measurements.
Step -2: Draw a line segment AB of length 8 cm.
Step -3: Draw a ray \(\overrightarrow{\mathrm{BX}}\) perpendicular to AB to B.
Step -4 : Draw an arc from A with radius 10 cm to intersect \(\overrightarrow{\mathrm{BX}}\) at C.
Step -5: Join A, C to get the required ΔABC.

Question 2.
Construct a ΔPQR, right-angled at R, hypotenuse is 5 cm and one of its adjacent sides is 4cm.
Solution:
∠R=90D,QR=4cm,PQ=5cm.

Step -1: Draw a rough sketch of the right angled triangle and label it with the given measurements.
Step -2: Draw a line segment QR of length 4cm.
Step -3: Draw a ray \(\overrightarrow{\mathrm{RY}}\) perpendicular to QR at R.
Step -4: Draw an arc from Q with radius 5 cm to intersect \(\overrightarrow{\mathrm{RY}}\) at P.
Step -5: Join P, Q to get the required ΔPQR.

Question 3.
Construct an isosceles right-angled ΔXYZ in which ∠Y = 90° and the two sides are 5 cm each.
Solution:
∠Y = 90°,XY = YZ = 5cm.

Step- 1: Draw a rough sketch of a right angled triangle and label it with given measurements.
Step -2: Draw a line segment XY of length 5cm.
Step -3: Draw a ray \(\overrightarrow{\mathrm{YK}}\) perpendicular to XY at Y.
Step -4: Draw an arc from Y with radius 5 cm to intersect \(\overrightarrow{\mathrm{YK}}\) at Z.
Step -5: Join Z, X to get the required ΔXYZ.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 9 Construction of Triangles Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Construction of Triangles Exercise 3

Question 1.
Construct Δ NET with measurement NE = 6.4 cm, ∠N = 50° and ∠E = 100°.
Solution:
NE = 6.4 cm, ∠N = 50° and ∠E = 100°

Step -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a Line segment NE of length 6.4 cm.
Step -3: Draw a ray \(\overrightarrow{\mathrm{NX}}\) maktng an angle 50° at N.
Step -4: Draw a ray \(\overrightarrow{\mathrm{EY}}\) making an angle 100° at E. Extend the ray \(\overrightarrow{\mathrm{NX}}\) if necessary to intersect ray \(\overrightarrow{\mathrm{EY}}\)
Step – 5: Mark the intersecting point of the two rays as T.
We have the required ΔNET.

Question 2.
Construct ΔPQR such that QR =6 cm, ∠Q = ∠R = 60°. Measure the other two sides of the triangle and name the triangle.
Solution:


QR = 6cm ∠Q = ∠R = 60°.
Step -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a line segment QR of length 6 cm.
Step -3: Draw a ray \(\overrightarrow{\mathrm{QX}}\) making an angle 60° at Q
Step -4: Draw a ray \(\overrightarrow{\mathrm{RY}}\) making an angle 60° at R.
Step -5: Mark the intersecting point of the two rays as P. PQ = 6cm and PR = 6 cm
∴ The triangle is an equilateral triangle.

Question 3.
Construct ΔRUN in which RN = 5cm, ∠R = ∠N = 45°. Measure the other angle and other sides. Name the triangle.
Solution:
RN = 5cm, ∠R = ∠N = 45°.


Step -1: Draw a rough sketch of a triangle and label it with the given measurements.
Step -2: Draw a line segment RN of length 5cm.
Step -3: Draw a ray \(\overrightarrow{\mathrm{RX}}\) making an angle 45° at R.
Step -4: Draw a ray \(\overrightarrow{\mathrm{NY}}\) making an angle 45° at N.
Step -5: Mark the intersecting point of the two rays as U. ΔRUN is the required triangle
∠U = 90°, RU = UN = 3.5 cm.
∴ The triangle is an isosceles right angled triangle.