NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1.

• Lines and Angles Class  7 Ex 5.2
• Lines and Angles Class  7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 5
Chapter Name	Lines and Angles
Exercise	Ex 5.1, Ex 5.2.
Number of Questions Solved	14
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1

Question 1.
Find the complement of each of the following angles:

Solution:
Since, the sum of the measures of an angle and its complement is 90°, therefore,

1. The complement of an angle of measure 20° is the angle of (90° – 20°), f.e., 70°.
2. The complement of an angle of measure 63° is the angle of (90° – 63°), i.e., 27°.
3. The complement of an angle of measure 57° is the angle of (90° – 57°), i.e., 33°.

Question 2.
Find the supplement of each of the following angles:

Solution:

1. Supplement of the angle 105° = 180° – 105° = 75°
2. Supplement of the angle 87° = 180° – 87° = 93°
3. Supplement of the angle 154° = 180° – 154° = 26°

Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary.

1. 65°, 115°
2. 63°, 27°
3. 112°, 68°
4. 130°, 50°
5. 45°,45°
6. 80°, 10°.

Solution:

1. Since, 65°+ 115° = 180°
   So, this pair of angles are supplementary.
2. Since, 63°+ 27° = 90°
   So, this pair of angles are complementary.
3. Since, 112° + 68° = 180°
   So, this pair of angles are supplementary.
4. Since, 130°+50° = 180°
   So, this pair of angles are supplementary.
5. Since, 45°+ 45° = 90°
   So, this pair of angles are complementary.
6. Since, 80°+ 10° = 90°
   So, this pair of angles are complementary.

Question 4.
Find the angle which is equal to its complement.
Solution:
Let the measure of the angle be x°. Then, the measure of its complement is given to be x°.
Since, the sum of the measures of an angle and its complement is 90°, therefore,
x° + x° = 90°
⇒ 2x° = 90°
⇒ x° = 45°
Thus, the required angle is 45°.

Question 5.
Find the angle which is equal to its supplement.
Solution:
Let the measure of the angle be x°. Then,
a measure of its supplement = x°
Since, the sum of the measures of an angle and its supplement is 180°, therefore,
x° + x° = 180°
⇒ 2x° =180°
⇒ x° = 90°
Hence, the required angle is 90°.

Question 6.
In the given figure, ∠ 1 and ∠ 2 are supplementary angles.

If ∠1 is decreased, what changes should take place in ∠ 2 so that both the angles still remain supplementary?
Solution:
∠ 2 will increase as much as ∠ 1 decreases.

Question 7.
Can two angles be supplementary if both of them are:

1. acute?
2. obtuse?
3. right?

Solution:

1. No! two acute angles cannot be a supplement.
2. No! Two obtuse angles cannot be supplementary.
3. Yes! Two right angles are always supplementary.

Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°.
Solution:
Since the sum of the measure of ah angle and its complement is 90°.
∴ The complement of an angle of measures 45° + x°,
where x > 0 is the angle of [90° – (45° + x°)] = 90° – 45° – x°= 45° – x°.
Clearly, 45° + x° > 45° – x°
Hence, the complement of an angle > 45° is less than 45°.

Question 9.
In the adjoining figure:

1. Is ∠1 adjacent to ∠2 ?
2. Is ∠ AOC adjacent to ∠ AOE?
3. Do ∠ COE and ∠ EOD form a linear pair?
4. Are ∠ BOD and ∠ DOA supplementary?
5. Is ∠ 1 vertically opposite to ∠ 4?
6. What is the vertically opposite angle of ∠ 5?

Solution:

1. Yes ! ∠ 1 is adjacent to ∠ 2.
2. No ! ∠ AOC is not adjacent to ∠ AOE.
3. Yes! ∠ COE and ∠ EOD form a linear pair.
4. Yes ! ∠ BOD and ∠ DOA are supplementary.
5. Yes ! ∠ 1 is vertically opposite to ∠ 4.
6. The vertically opposite angle of ∠ 5 is ∠ 2 + ∠ 3, i.e., ∠ COB.

Question 10.
Indicate which pairs of angles are:

1. Vertically opposite angles.
2. Linear pairs.

Solution:

1. The pair of vertically opposite angles are ∠1, ∠4; ∠5, ∠2 + ∠3.
2. The pair of linear angles are ∠1, ∠5; ∠4, ∠5.

Question 11.
In the following figure, is ∠ 1 adjacent to ∠ 2? Give reasons.

Solution:
∠1 is not adjacent to ∠2 because they have no common vertex.

Question 12.
Find the values of the angles x, y, and z in each of the following:

Solution:

Question 13.
Fill in the blanks:

1. If two angles are complementary, then the sum of their measures is
2. If two angles are supplementary, then the sum of their measures is
3. Two angles forming a linear pair are
4. If two adjacent angles are supplementary, they form a
5. If two lines intersect at a point, then the vertically opposite angles are always
6. If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are

Solution:

1. 90°
2. 180°
3. supplementary
4. linear pair
5. equal
6. obtuse angles

Question 14.
In the adjoining figure, name the following pairs of angles.

1. Obtuse vertically opposite angles
2. Adjacent complementary angles
3. Equal supplementary angles
4. Unequal supplementary angles
5. Adjacent angles that do not form a linear pair.

Solution:

1. Obtuse vertically opposite angles are ∠AOD and ∠BOC.
2. Adjacent complementary angles are ∠BOA and ∠AOE.
3. Equal supplementary angles are ∠BOE and ∠EOD.
4. Unequal supplementary angles are ∠BOA and ∠AOD, ∠BOC and ∠COD, ∠EOA, and ∠EOC.
5. Adjacent angles that do not form a linear pair are ∠AOB and ∠AOE, ∠AOE and ∠EOD; ∠EOD and ∠COD.

We hope the NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1.

• Simple Equations Class 7 Ex 4.2
• Simple Equations Class 7 Ex 4.3
• Simple Equations Class 7 Ex 4.4
• Simple Equations Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 4
Chapter Name	Simple Equations
Exercise	Ex 4.1
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

Question 1.
Complete the last column of the table.

Solution:

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not.

(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = – 2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0).

Solution:
(a) n + 5 = 19 (n = 1)
L.H.S. = n + 5 = 1 + 5 | when n = 1 = 5
R.H.S. = 19
∵ L.H.S. ≠ R.H.S.
∴ n = 1 is not a solution to the given equation n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)
L.H.S. = 7n + 5 = 7(- 2) + 5 | when n = – 2 = – 14 + 5 = – 9
R.H.S. = 19
∵ L.H.S. ≠ R.H.S.
∴ n = – 2 is not a solution to the given equation 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)
L.H.S. = In + 5 = 7(2) + 5 | when n = 2 = 14 + 5 = 19 = R.H.S.
∴ n = 2 is a solution to the given equation 7n + 5 = 19.

(d) 4p – 3 = 13 (p = 1)
L.H.S. = 4p – 3 = 4(1) – 3 | when p = 1 = 4 – 3 = 1
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = 1 is not a solution to the given equation 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)
L.H.S. = 4p – 3 = 4(- 4) – 3 , | when p = – 4 = – 16 – 3 = – 19
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = – 4 is not a solution to the given equation
4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)
L.H.S. = 4 (p) – 3 = 4(0) – 3 | when p = 0 = 0 – 3 = – 3
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = 0 is not a solution to the given equation 4p – 3 = 13.

Question 3.
Solve the following equations by trial and error method.

1. 5p + 2 = 17
2. 3m – 14 = 4.

Solution:
(i) 5p + 2 = 17

So, p = 3 is the solution of the given equation 5p + 2 = 17.

(ii) 3m – 14 = 4

So, m = 6 is the solution of the given equation 3m – 14 = 4.

Question 4.
Write equations for the following statements.

1. The sum of numbers x and 4 is 9.
2. 2 subtracted from y is 8.
3. Ten times a is 70.
4. The number b divided by 5 gives 6.
5. Three-fourth oft is 15.
6. Seven times m plus 7 gets you 77.
7. One-fourth of a number x minus 4 gives 4.
8. If you take away 6 from 6 times y, you get 60.
9. If you add 3 to one-third of z, you get 30.

Solution:

1. x + 4 = 9
2. y – 2 = 8
3. 10 a = 70
4. b ÷ 5 = 6
5. \(\frac { 3 }{ 4 } \) × t = 15
6. 7m + 7 = 77
7. \(\frac { 1 }{ 4 } \) × x – 4 = 4
8. 6y – 6 = 60
9. \(\frac { 1 }{ 3 } \) × z + 3 = 30

Question 5.
Write the following equations in statement forms:

1. p + 4 = 15
2. m – 7 = 3
3. 2m = 7
4. \(\frac { m }{ 5 } \) = 3
5. \(\frac { 3m }{ 5 } \) = 6
6. 3p + 4 = 25
7. 4p – 2 = 18
8. \(\frac { p }{ 2 } \) + 2 = 8.

Solution:

1. The sum of p and 4 is 15.
2. 7 subtracted from m is 3.
3. Twice a number m is 7.
4. One-fifth of a number m is 3.
5. Three-fifth of a number m is 6.
6. Three times a number p, when added to 4, gives 25.
7. 2 subtracted from four times a number p is 18.
8. Add 2 to half of a number p to get 8.

Question 6.
Set up an equation in the following cases:

1. Irfan says that he has 7 marbles more than five times the marbles Permit has. Irfan has 37 marbles. (Take m to be the number of Permit’s marbles.)
2. Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
3. The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
4. In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of triangle is 180 degrees).

Solution:
(i) Let the number of marbles with Parmit be m.
Then, 7 added to 5 times mis 5m + 7
It is given that 7 marbles more than five times the marble is 37. Thus, the equation obtained is 5m + 7 = 37.

(ii) Let Laxmi’s age be y years. Then, 4 added to 3 times y is 3y + 4
It is given that the father is 4 years older than 3 times Laxmi’s age. His age is 49years.
Then, we have the following equation : 3y + 4 = 49

(iii) Let the lowest marks be l. Then, twice the lowest marks plus 7 is 2l +7
It is given that, the highest marks 87 obtained by a student is twice the lowest marks plus 7.
So, we have the following equation : 2l + 7 = 87

(iv) Let the base angle be b. Then, the vertex angle = 2b.
Since, sum of the angles of a triangle is 180°
∴ b + b + 2b = 180°
⇒ 4b = 180°
which is the required equation.

We hope the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1.

• Data Handling Class 7 Ex 3.2
• Data Handling Class 7 Ex 3.3
• Data Handling Class 7 Ex 3.4
• Data Handling Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 3
Chapter Name	Data Handling
Exercise	Ex 3.1
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

Question 1.
Find the range of heights of any ten students of your class.
Solution:
Let the heights (in cm) of 10 students in the class be 150, 152, 151, 148, 149, 149, 150, 152, 153, 146.
Arranging the heights in ascending order, we have 146, 148, 149, 149, 150, 150, 151, 152, 152, 153.
Range of height of students = 153 – 146 = 7

Question 2.
Organize the following marks in a class assessment, in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Solution:

(i) Highest number is 9.
(ii) Lowest number is 1.
(iii) Range of the data = Highest observation – Lowest observation
= 9 – 1
= 8

Question 3.
Find the mean of the first five whole numbers.
Solution:
The first 5 whole numbers are 0, 1, 2, 3, and 4.
Their arithmetic mean

Question 4.
A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.
Solution:

Question 5.
Following table shows the points of each player scored in four games:

Now answer the following questions:

1. Find the mean to determine A’s average number of points scored per game.
2. To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
3. B played in all four games. How would you find the mean?
4. Who is the best performer?

Solution:

So, A’s average number of points scored per game is 12.5.
(ii) To find the mean number of points per game for C, we shall divide the total points by 3 because the number of games under consideration is 4 but ‘C’ did not play game 3.

Question 6.
The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the:

1. Highest and the lowest marks obtained by the students.
2. Range of the marks obtained.
3. Mean marks obtained by the group.

Solution:

1. Highest marks obtained by the students = 95
   Lowest marks obtained by the students = 39
2. Range of the marks obtained = Highest marks – Lowest marks = 95 – 39 = 56
3. Mean marks obtained by the group
   

Question 7.
The enrolment of a school during six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820.
Find the mean enrolment of the school for this period.
Solution:
Mean enrolment of the school for this period.

Question 8.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Solution:
(i) Range of the rainfall = Highest rainfall – Lowest rainfall = 20.5 mm – 0.0 mm = 20.5 mm
(ii) Mean rainfall for the week

(iii) The rainfall was less than the mean rainfall on 5 days.

Question 9.
The heights of 10 girls were measured in cm and the results are as follows:
135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
(i) What is the height of the tallest girl?
(i) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height?

Solution:
(i) Height of the tallest girl = 151 cm
(ii) Height of the shortest girl = 128 cm
(iii) Range of the data

(v) 5 girls have heights more than the mean height.

We hope the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1.

• Fractions and Decimals Class 7 Ex 2.2
• Fractions and Decimals Class 7 Ex 2.3
• Fractions and Decimals Class 7 Ex 2.4
• Fractions and Decimals Class 7 Ex 2.5
• Fractions and Decimals Class 7 Ex 2.6
• Fractions and Decimals Class 7 Ex 2.7
• Fractions and Decimals Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 2
Chapter Name	Fractions and Decimals
Exercise	Ex 2.1
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 1.
Solve :

Solution:

Question 2.
Arrange the following in descending order

Solution:
(i) Converting the given fractions into like fractions, we have


(ii) Converting the given fractions into like fractions, we have

Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

Solution:
Yes ! this is a magic square.

Question 4.
A rectangular sheet of paper is 12 \(\frac { 1 }{ 2 } \) cm long 10 \(\frac { 2 }{ 3 } \) cm wide. find its perimeter
Solution:
Perimeter of the rectangular sheet of paper

Question 5.
Find the perimeters of (i) A ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Solution:
(i) Perimeter of ∆ ABE = AB + BE + EA

(ii) Perimeter of the rectangle BCDE = BC + CD + DE + EB

So, the perimeter of A ABE is greater than the perimeter of the rectangle BCDE.

Question 6.
Salil wants to put a picture in a frame. The picture is 7 \(\frac { 3 }{ 5 } \) cm wide. To fit in the frame the picture cannot be more than 7 \(\frac { 3 }{ 10 } \) cm wide. How much should the picture be trimmed?
Solution:
The picture should be trimmed by

Question 7.
Ritu are \(\frac { 3 }{ 5 } \) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
Solution:
Part of the apple ate by Somu

Question 8.
Michael finished colouring a picture in \(\frac { 7 }{ 12 } \) hour. Vaibhav finished colouring the same picture in \(\frac { 3 }{ 4 } \) hour. Who worked longer? By what fraction was it longer?
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1, drop a comment below and we will get back to you at the earliest.