NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 4
Chapter Name Simple Equations
Exercise Ex 4.3
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

Question 1.
Solve the following equations:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 1
Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 2
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 3
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 4
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 5
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 6
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 7
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 8
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 9
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 10

Question 2.
Solve the following equations:

(a) 2 (x + 4) = 12
(6) 3 (n – 5) = 21
(c) 3 (n – 5) = -21
(d) -4 (2 + x) = 8
(e) 4 (2 – x) = 8

Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 11
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 12
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 13
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 14

Question 3.
Solve the /bilowing equations:

(a) 4 = 5 (p – 2)
(b) -4 = 5 (p – 2)
(c) 16 = 4 + 3 (t + 2)
(d) 4 + 5 (p – 1) = 34
(e) 0 = 16 + 4 (m – 6)

Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 15
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 16
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 17
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 18
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 19
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 20

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2.
Solution:
(a) 1. Start with x = 2
Multiply both sides by 3, 3x = 6
Subtract 2 from both sides, 3x – 2 = 4 …(1)

2. Start with x = 2
Multiply both sides by 4, 4x = 8
Add 5 to both sides, 4x + 5 = 13 …(2)

3. Start with x = 2 Multiply both sides by 5 5x = 10
Subtract 1 from both sides, 5x – 1 = 9 …(3)

(b) First equation:
Start with x = -2
Multiply both sides by 2, 2x = -4
Subtract 3 from both sides, 2x – 3 = -7

Second equation:
Start with x = – 2
Multiply both sides by – 5, – 5x = 10
Add 10 to both sides, 10 – 5x = 20

Third equation:
Start with x = -2
Divide both sides by 2, \(\frac { x }{ 2 } \) = -1
Add 3 to both sides, \(\frac { x }{ 2 } \) + 3 = 2

We hope the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3, 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 4.2

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 4
Chapter Name Simple Equations
Exercise Ex 4.2
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

Question 1.
Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Solution:
(a) The given equation is x – 1 = 0
Add 1 to both sides,
x – 1 + 1 = 0 + 1 ⇒ x = 1
It is the required solution.
Check. Put the solution x = 1 back into the equation.
L.H.S. = x – 1 = 1 = 1 – 0 = R.H.S.
The solution is thus checked for its correctness.

(b) The given equation is x + 1 = 0
Subtract 1 from both sides, x + 1 – 1 = 0 – 1 ⇒ x = – 1
It is the required solution.
Check. Put the solution x = – 1 back into the equation.
L.H.S. = x + 1 = (-1)+1
= 0 = R.H.S.
The solution is thus checked for its correctness.

(c) The given equation is
x – 1 = 5
Add 1 to both sides,
x + 1 – 1 = 5 + 1 ⇒ x = 6
It is the required solution
Check. Put the solution x = 6 back into the equation.
L.H.S. = x – 1 = 6 – 1 = 5 = R.H.S.
The solution is thus checked for its correctness.

(d) The given equation is x + 6 = 2
Subtract 6 from both sides, x + 6 – 6 = 2 – 6 ⇒ x = – 4
It is the required solution.
Check. Put the solution x = – 4 back into the equation.
L.H.S. = x + 6 = – 4 + 6 = 2 = R.H.S.
The solution is thus checked for its correctness.

(e) The given equation is y – 4 = – 7
Add 4 to both sides, y – 4 + 4 = – 7 + 4 ⇒ y = – 3
It is the required solution.
Check. Put the solution
L.H.S. = y – 4 = – 3 – 4 = – 7 = R.H.S.
The solution is thus checked for its correctness.

(f) The given equation is y – 4 = 4
Add 4 to both sides,
y – 4 + 4 = 4 + 4 ⇒ y = 8
It is the required solution.
Check. Put the solution y = 8 back into the equation.
L.H.S. = y – 4 = 8 – 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(g) The given equation is y + 4 = 4
Subtract 4 from both sides, y + 4 – 4 = 4 – 4 ⇒ y = 0
It is the required solution.
Check. Put the solution y = 0 back into the equation.
L.H.S. =y + 4 = 0 + 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(h) The given equation is y + 4 = – 4
Subtract 4 from both sides, y + 4 – 4 = – 4 – 4 ⇒ y = -8
It is the required solution.
Check. Put the solution y = – 8 back into the equation.
L.H.S. = y + 4 = -8 + 4 = – 4 = R.H.S.
The solution is thus checked for its correctness.

Question 2.
Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
(b) \(\frac { b }{ 2 } \) = 6
(c) \(\frac { p }{ 7 } \) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac { z }{ 3 } \) = \(\frac { 5 }{ 4 } \)
(g) \(\frac { a }{ 5 } \) = \(\frac { 7 }{ 15 } \)
(h) 20t = – 10

Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 1
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 2
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 3
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 4
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 5
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 6

Question 3.
Give the steps you will use to separate the variable and then solve the equation :
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) \(\frac { 20p }{ 3 } \) = 40
(d) \(\frac { 3p }{ 10 } \) = 6
Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 7NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 7
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 8
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 9
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 10

Question 4.
Solve the following equations:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 11
Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 12
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 13
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 14
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 15
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 16
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 17
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 18

We hope the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 6
Chapter Name The Triangle and its Properties
Exercise Ex 6.4
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4

Question 1.
Is it possible to have a triangle with the following sides?

  1. 2 cm, 3 cm, 5 cm
  2. 3 cm, 6 cm, 7 cm
  3. 6 cm, 3 cm, 2 cm.

Solution:

  1. Since, 2 + 3 > 5
    So the given side lengths cannot form a triangle.
  2. We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3
    i. e., the sum of any two sides is greater than the third side.
    So, these side lengths form a triangle.
  3. We have, 6 + 3 > 2, 3 + 2 \(\ngtr \) 6
    So, the given side lengths cannot form a triangle.

Question 2.
Take any point O in the interior of a triangle PQR. Is

  1. OP + OQ > PQ ?
  2. OQ + OR > QR?
  3. OR + OP > RP ?

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 1

Solution:

  1. Yes ! OP + OQ > PQ …(1)
    Sum of the lengths of any two sides of a triangle is greater than the length of the third side
  2. Yes! OQ + OR > QR …(2)
    Sum of the lengths of any two sides of a triangle is greater than the length of the third side
  3. Yes! OR + OP > RP …(3)
    Sum of the lengths of any two sides of a triangle is greater than the length of the third side

Question 3.
AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM?
(Consider the sides of triangles ∆ ABM and ∆ AMC.)
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 2
Solution:
Using triangle inequality property in triangles ABM and AMC, we have
AB + BM > AM …(1) and, AC + MC > AM …(2)
Adding (1) and (2) on both sides, we get
AB + (BM + MC) + AC > AM + AM
⇒ AB + BC + AC > 2AM

Question 4.
ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD ?
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 3
Solution:
In ∆ ABC, AB + BC > AC …(1)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ ACD, CD + DA > AC …(2)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (1) and (2),
AB + BC + CD + DA > 2AC …(3)
In ∆ ABD, AB + DA > BD …(4)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ BCD, BC + CD > BD …(5)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (4) and (5),
AB + BC + CD + DA > 2BD …(6)
Adding (3) and (6),
2 [AB + BC + CD + DA] > 2 (AC + BD)
⇒ AB + BC + CD + DA > AC + BD.

Question 5.
ABCD is a quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 4
Solution:
In ∆ OAB, OA + OB > AB ….(1)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
In ∆ OBC, OB + OC > BC ….(2)
Sum of the lengths of any two sides of a triangle la greater than the length of the third side.
In ∆ OCA,OC + OA > CA ….(3)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
In ∆ OAD, OA + OD > AD ….(4)
Sum of the lengths of any two sides of a triangle is greater than the length of the third side
Adding (1), (2), (3) and (4),
2(OA + OB + OC + OD) > AB + BC + CD + DA
⇒ AB + BC + CD + DA < 2 (OA + OB + OC + OD)
⇒ AB + BC + CD + DA < 2(OA + OC + OB + OD)
⇒ AB + BC + CD + DA < 2 (AC + BD).

Question 6.
The lengths of the two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Solution:
Let x cm be the length of the third side.
∴ Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
∴ We should have
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4 5
∴ The length of the third side should be any length between 3 cm and 27 cm.

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and it’s Properties Ex 6.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 6
Chapter Name The Triangle and its Properties
Exercise Ex 6.3
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3

Question 1.
Find the value of the unknown x in the following diagrams:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 1
Solution:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 2
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 3
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 4
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 5

Question 2.
Find the values of the unknowns x and y in the following diagrams:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 6
Solution:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 7
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 8
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 9
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 10

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NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 6
Chapter Name The Triangle and its Properties
Exercise Ex 6.2
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2

Question 1.
Find the value of the unknown exterior angle x in the following diagrams:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 1
Solution:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 2
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 3

Question 2.
Find the value of the unknown interior angle x in the following figures:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 4
Solution:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 5
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2 6

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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 3
Chapter Name Data Handling
Exercise Ex 3.3
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3

Question 1.
Use the bar graph to answer the following questions.
(a) Which is the most popular pet?
(b) How many children have dog as a pet?
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 1
Solution:
(a) The most popular pet is ‘Cat’.
(b) 8 children have a dog as a pet.

Question 2.
Read the bar graph which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 2

  1. About how many books were sold in 1989?1990?1992?
  2. ln which year were about 475 books sold? About 225 books sold?
  3. In which years were fewer than 250 books sold?
  4. Can you explain how you would estimate the number of books sold in 1989?

Solution:
Clearly, from the given graph, we have

  1. Number of books sold in the year
    1989: 170 (approx.)
    1990: 475 (approx.)
    1992: 225 (approx.)
  2. In the year 1990, about 475 books were sold. In the year 1992, about 225 books were sold.
  3. Fewer than 250 books were sold in the years 1989 and 1992.
  4. It can be estimated using the height of the bar such that the height of 1 cm = 100 books.

Question 3.
The number of children in six different classes is given below. Represent the data on a bar graph.
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 3
(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the students of the class eighth.
Solution:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 4
(a) Start the scale at 0. The greatest value in the data is 135, so end the scale at a value greater than 135, such as 140. Use equal divisions along the axes, such as increments of 10.
We know that all the bars would lie between 0 and 140.
We choose the scale such that the length between 0 and 140 is neither too long nor too small.
Here, we take 1 unit for 10 children.
(b) (i) Fifth class has the maximum number of children. Tenth class has the minimum number of children.
(ii) Ratio of students of class sixth to eighth = 120 : 100 = \(\frac { 12 }{ 100 } \) = \(\frac { 6 }{ 5 } \) = 6 : 5

Question 4.
The performance of a student in 1st Term and, 2nd Term is given. Draw a double bar graph choosing the appropriate scale and answer the following :
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 5

  1. In which subject, has the child improved his performance the most?
  2. In which subject is the improvement the least?
  3. Has the performance gone down in any subject?

Solution:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 6

  1. The child improved his performance the most in the subject of Maths.
  2. The improvement is the least in the subject of S. Science.
  3. Yes! The performance has gone down in the subject of Hindi.

Question 5.
Consider this data collected from a survey of a colony.
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 7

  1. Draw a double graph choosing an appropriate scale.
    What do you infer from the bar graph?
  2. Which sport is most popular?
  3. Which is more preferred, watching or participating in sports?

Solution:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 8

  1. It is inferred that more people prefer cricket and fewer athletics.
  2. The most popular sport is cricket.
  3. Watching is more preferred than participating.

Question 6.
Take the data giving the minimum and the maximum temperature of various cities given at the beginning of this chapter. Plot a double bar graph using the data and answer the following:

  1. Which city has the largest difference in the minimum and maximum temperature on the given date?
  2. Which is the hottest city and which is the coldest city?
  3. Name two cities where the maximum temperature of one was less than the minimum temperature of the other.
  4. Name the city which has the least difference between its minimum and the maximum temperature.

Solution:
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 9
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 10

  1. The city Jammu has the largest difference in the minimum and maximum temperature on the given date.
  2. Jammu is the hottest city and Bangalore is the coldest city.
  3. The name of the two cities where the maximum temperature of one was less than the minimum temperature of the other is Bangalore and Jaipur or Bangalore and Ahmedabad.
  4. Mumbai has the least difference between its minimum and maximum temperature.

We hope the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3, 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 3.4

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 3
Chapter Name Data Handling
Exercise Ex 3.4
Number of Questions Solved 3
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4

Question 1.
Tell whether the following is certain to happen, impossible, can happen but not certain.

  1. You are older today than yesterday.
  2. A tossed coin will land heads up.
  3. A die when tossed shall land up with 8 on top.
  4. The next traffic light seen will be green.
  5. Tomorrow will be a cloudy day.

Solution:

  1. Certain to happen
  2. Can happen but not certain
  3. Impossible
  4. Can happen but not certain
  5. Can happen but not certain

Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.

  1. What is the probability of drawing a marble with the number 2?
  2. What is the probability of drawing a marble with the number 5?

Solution:
Out of 6 marbles, one can be drawn in 6 ways. So, the total number of events = 6

1. The marble with the number 2 can be obtained only in one way.
∴ Required probability = \(\frac { 1 }{ 6 } \)

2. The marble with the number 5 can be obtained only in one way.
∴ Required probability = \(\frac { 1 }{ 6 } \)

Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Solution:
On tossing a coin, the possible outcomes are head (H) or tail (T).
Required probability = \(\frac { 1 }{ 2 } \)

We hope the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4, 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 2.7

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 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 2.7.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 2
Chapter Name Fractions and Decimals
Exercise Ex 2.7
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

Question 1.
Find:

  1. 0.4 ÷ 2
  2. 0.35 ÷ 5
  3. 2.48 ÷ 4
  4. 65.4 ÷ 6
  5. 651.2 ÷ 4
  6. 114.49 ÷ 7
  7. 3.96 ÷ 4
  8. 0.80 ÷ 5

Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 1
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 2

Question 2.
Find:

  1. 4.8 ÷ 10
  2. 52.5 ÷ 10
  3. 0.7 ÷ 10
  4. 33.1 ÷ 10
  5. 272.23 ÷ 1o
  6. 0.56 ÷ 10
  7. 3.97 ÷ 10

Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 3

Question 3.
Find:

  1. 2.7 ÷ 100
  2. 0.3 ÷ 100
  3. 0.78 ÷ 100
  4. 432.6 ÷ 100
  5. 23.6 ÷ 100
  6. 98.53 ÷ 100

Solution:

  1. 2.7 ÷ 100 = 0.027
  2. 0.3 ÷ 100 = 0.003
  3. 0.78 ÷ 100 = 0.0078
  4. 432.6 ÷ 100 = 4.326
  5. 23.6 ÷ 100 = 0.236
  6. 98.53 ÷ 100 = 0.9853.

Question 4.
Find:

  1. 7.9 ÷ 1000
  2. 26.3 ÷ 2000
  3. 38.53 ÷ 1000
  4. 128.9 ÷ 2000
  5. 0.5 ÷ 2000

Solution:

  1. 7.9 ÷ 1000 = 0.0079
  2. 26.3 ÷ 1000 = 0.0263
  3. 38.53 ÷ 1000 = 0.03853
  4. 128.9 ÷ 1000 = 0.1289
  5. 0.5 ÷ 1000 = 0.0005

Question 5.
Find:

  1. 7 ÷ 3.5
  2. 36 ÷ 0.2
  3. 3.25 ÷ 0.5
  4. 30.94 ÷ 0.7
  5. 0.5 ÷ 0.25
  6. 7.75 ÷ 0.25
  7. 76.5 ÷ 0.15
  8. 37.8 ÷ 1.4
  9. 2.73 ÷ 1.3

Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 4
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 5

Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?
Solution:
∵ Distance covered in 2.4 litres of petrol = 43.2 km
∴ Distance travelled in 1 litre of petrol = 43.2 ÷ 2.4
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 6

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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 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 2.6.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 2
Chapter Name Fractions and Decimals
Exercise Ex 2.6
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

Question 1.
Find:

  1. 0.2 × 6
  2. 8 × 4.6
  3. 2.71 × 5
  4. 20.1 × 4
  5. 0.05 × 7
  6. 211.02 × 4
  7. 2 × 0.86

Solution:

  1. 0.2 × 6 = 1.2
  2. 8 × 4.6 = : 36.8
  3. 2.71 × 5 = 13.55
  4. 20.1 × 4 = 80.4
  5. 0.05 × 7 = 0.35
  6. 211.02 × 4 = 844.08
  7. 2 × 0.86 = 1.72

Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
∴ Area of the rectangle
= Length × Breadth
= 5.7 × 3 = 17.1 cm2

Question 3.
Find:

  1. 1.3 × 10
  2. 36.8 × 10
  3. 153.7 × 10
  4. 168.07 × 10
  5. 31.1 × 100
  6. 156.1 × 100
  7. 3.62 × 100
  8. 43.07 × 100
  9. 0.5 × 10
  10. 0.08 × 10
  11. 0.9 × 100
  12. 0.03 × 1000

Solution:

  1. 1.3 × 10 = 13.0
  2. 36.8 × 10 = 368.0
  3. 153.7 × 10 = 1537.0
  4. 168.07 × 10 = 1680.70
  5. 31.1 × 100 = = 3110.0
  6. 156.1 × 100 = 15610.0
  7. 3.62 × 100 = = 362.00
  8. 43.07 × 100 = 4307.00
  9. 0.5 × 10 = 5 .0
  10. 0.08 × 10 = 0.80
  11. 0.9 × 100 = 90.0
  12. 0.03 × 1000 = 30.0.

Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
Distance covered in one liter of petrol = 55.3 km
Distance covered in 10 litres of petrol = (55.3 × 10) km = 553 km

Question 5.
Find:

  1. 2.5 × 0.3
  2. 0.1 × 51.7
  3. 0.2 × 316.8
  4. 1.3 × 3.1
  5. 0.5 × 0.05
  6. 11.2 × 0.15
  7. 1.07 × 0.02
  8. 10.05 × 1.05
  9. 101.01 × 0.01
  10. 100.01 × 1.1

Solution:

  1. 2.5 × 0.3 = 0.75
  2. 0.1 × 51.7 = 5 .17
  3. 0.2 × 316.8 = 63.36
  4. 1.3 × 3.1 = 4.03
  5. 0.5 × 0.05 = 0 .025
  6. 11.2 × 0.15 = 1.680
  7. 1.07 × 0.02 = 0.0214
  8. 10.05 × 1.05 = 10.5525
  9. 101.01 × 0.01 = 1.0101
  10. 100.01 × 1.1 = 110.011

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6, 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 2.5

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 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 2.5.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 2
Chapter Name Fractions and Decimals
Exercise Ex 2.5
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Which is greater?

  1. 0.5 or 0.05
  2. 0.7 or 0.5
  3. 7 or 0.7
  4. 1.37 or 1.49
  5. 2.03 or 2.30
  6. 0.8 or 0.88

Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 1

Question 2.
Express as rupees using decimals:

  1. 7 paise
  2. 7 rupees 7 paise
  3. 77 rupees 77 paise
  4. 50 paise
  5. 235 paise.

Solution:

  1. 7 paise = ₹ 0.07
  2. 7 rupees 7 paise = ₹ 7.07
  3. 77 rupees 77 paise = ₹ 77.77
  4. 50 paise – ₹ 0.50
  5. 235 paise = ₹ 2.35

Question 3.

  1. Express 5 cm in metre and kilometre
  2. Express 35 mm in cm, m and km?

Solution:

  1. 5 cm = 0.05 m = 0.00005 km
  2. 35 mm = 3.5 cm = 0.035 m = 0.000035 km

Question 4.
Express in kg:

  1. 200 g
  2. 3470 g
  3. 4 kg 8 g

Solution:

  1. 200 g = 0.200 kg = 0.2 kg
  2. 3470 g = 3.470 kg
  3. 4 kg 8 g = 4.008 kg.

Question 5.
Write the following decimal numbers in the expanded form:

  1. 20.03
  2. 2.03
  3. 200.03
  4. 2.034

Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 2

Question 6.
Write the place value of 2 in the following decimal numbers:

  1. 2.56
  2. 21.37
  3. 10.25
  4. 9.42
  5. 63.352

Solution:
(i) Place value of 2 in the decimal number 2.56 = 2 × 1 = 2
(ii) Place value of 2 in decimal number 21.37 = 2 × 10 = 20
(iii) Place value of 2 in the decimal number
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 3

Question 7.
Dinesh went from place A to place t B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. f Who travelled more and by how much?
Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 4
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 5
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 6

Question 8.
Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Solution:
For Shyama
Apples bought = 5 kg 300 g = 5.300 kg
Mangoes bought = 3 kg 250 g = 3.250 kg
∴ Fruits bought = Apples bought
+ Mangoes bought = 5.300 kg + 3.250 kg
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 7
For Sarala
Oranges bought = 4 kg 800 g = 4.800 kg
Bananas bought = 4 kg 150 g = 4.150 kg
∴ Fruits bought = Oranges bought
+ Bananas bought
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 8

Question 9.
How much less is 28 km than 42.6 km?
Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 9
So, 28 km is less than 42.6 km by 14.6 km.

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5, 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 2.4

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 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 2.4.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 2
Chapter Name Fractions and Decimals
Exercise Ex 2.4
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4

Question 1.
Find:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 1
Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 2
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 3
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 4

Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 5
Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 6
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 7

Question 3.
Find:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 8

Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 9

Question 4.
Find:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 10
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 11
Solution:
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 12
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 13

We hope the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4, drop a comment below and we will get back to you at the earliest.