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NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1.

• Integers Class 7 Ex 1.2
• Integers Class 7 Ex 1.3
• Integers Class 7 Ex 1.4
• Integers Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 1
Chapter Name	Integers
Exercise	Ex 1.1
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1

Question 1.
Following number line shows the temperature in degree Celsius (°C) at different places on a particular day:

(a) Observe this number line and write the temperature of the places marked on it.
(b) What is the temperature difference between the hottest and the coldest places among the above?
(c) What is the temperature difference between Lahul-Spiti and Srinagar?
(d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?
Solution:
(a)

(b) The hottest place is Bangalore (22 °C) and the coldest place is Lahulspiti (-8°C). The temperature difference between the hottest and the coldest places
= 22 °C – (-8 °C)
= 22 °C + 8 °C
= 30 °C

(c) Temperature difference between Lahul-Spiti and Srinagar
= Temperature of Srinagar
– Temperature of Lahul-Spiti
= – 2°C – (- 8°C)
= – 2°C + 8°C
= 6°C

(d) Yes, we can say that the temperature of Srinagar and Shimla took together is less than the temperature at Shimla as -2 + 5 = 3 and 3 < 5.
This temperature is not less than the temperature at Srinagar.

Question 2.
In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, -5, -10, 15 and 10, what was his total at the end?
Solution:
Total at the end
= 25 + (- 5) + (- 10) + 15 + 10
= (25 + 15 + 10) + {(- 5) + (- 10))
= 50 + (- 15) = 35

Question 3.
At Srinagar, the temperature was – 5°C on Monday and then it dropped by 2°C on Tues¬day. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?
Solution:
At Srinagar, the temperature was on Monday = -5 °C
Since the temperature was dropped by 2 °C on Tuesday, therefore, the temperature was on Tuesday = (-5-2) °C = -7°C
Also, on Wednesday the temperature rose by 4 °C.
∴ Temperature on Wednesday= (-7 + 4) °C = -3 °C

Question 4.
A plane is flying at the height of 5000 m above sea level. At a particular point, it is e×actly above a submarine floating 1200 m below sea level. What is the vertical distance between them?

Solution:
The vertical distance between the plane and submarine.
= 5000 m – (- 1200 m)
= 5000 m + 1200 m
= 6200 m.

Question 5.
Mohan deposits ₹ 2,000 in his bank account and withdraws ₹1,642 from it, the next day. If the withdrawal of the amount from the account is represented by a negative integer, then how will you represent the amount deposited?
Find the balance in Mohan’s account after the withdrawal.
Solution:
The amount deposited will be represented by a positive integer.
Balance in Mohan’s account after withdrawal
= (+ ₹ 2000) + (- ₹ 1642)
= ₹ (2000-1642)
= ₹ 358

Question 6.
Rita goes 20 km towards the east from point A to point B. From B, she moves 30 km towards the west along the same road. If the distance towards the east is represented by a positive integer then, how will you represent the distance travelled towards the west? By which integer will you represent her final position from A?

Solution:
The distance towards the west = -30 km
Her final position from A
= + 20 km + (- 30) km
= – (30 – 20) km
= – 10 km.

Question 7.
In a magic square each row, column, and diagonal have the same sum. Check which of the following is a magic square.

Solution:

I. Row, Sum = 5 + (- 1) + (- 4)
= 5 + (- 5) = 0
II. Row, Sum = (- 5) + (- 2) + 7
= (- 7) + 7 = 0
III. Row, Sum = 0 + 3 + (- 3)
= 0 + 0 = 0
I. Column, Sum = 5 + (- 5) + 0
= 0 + 0 = 0
II. Column, Sum = (- 1) + (- 2) + 3
= (- 3) + 3 = 0
III. Column, Sum = (- 4) + 7 + (- 3)
= 7 + (- 4) + (- 3)
= 7 + (- 7) = 0
One Diagonal, Sum = 5 + (- 2) + (- 3) = 5 + (- 5) = 0
Other Diagonal, Sum = 0 + (- 2) + (- 4) = 0 + (- 6) = -6 ≠ 0
Therefore, the given square is not a magic square.

I. Row, Sum = 1 + (- 10) + 0 = – 9
II. Row, Sum = (- 4) + (- 3) + (- 2) = – 9
III. Row, Sum = (- 6) + 4 + (- 7)
= 4 + (- 6) + (- 7)
= 4 + (- 13)
= – (13-4) = -9
I. Column, Sum = 1 + (- 4) + (- 6)
= 1 + (- 10)
= -(10 – 1) = – 9
II. Column, Sum = (- 10) + (- 3) + 4
= (-13)+ 4
= – (13-4) = -9
III. Column, Sum = 0 + (- 2) + (- 7)
= 0 + (- 9) = – 9
One Diagonal, Sum = 1 + (- 3) + (- 7)
= 1 + (- 10)
= -(10-1) = -9
Other Diagonal, Sum = (- 6) + (- 3) + 0
= (- 9) + 0 = – 9
Since each row, column, and diagonal ‘ have the same sum, therefore, the given square is a magic square.

Question 8.
Verify a – (- b) = a + b for the following values of a and b.
(i) a = 21, b = 18
(ii) a = 118, b = 125
(iii) a = 75, b = 84
(iv) a = 28, b = 11.
Solution:
(i) a = 21, b = 18
L.H.S. = a – (- b) = 21 – (- 18) = 21 + 18 = 39 …..(1)
R.H.S. = a + b = 21 + 18 = 39 …..(2)
From (1) and (2), we get a -(-b) = a + b

(ii) a = 118, b = 125
L.H.S. = a – (- b) = 118 – (- 125) = 118 + 125 = 243 …(1)
R.H.S. = a + b = 118 + 125 = 243 …(2)
From (1) and (2), we get a -(- b) = a + b

(iii) a = 75, b = 84
L.H.S. = a – (- b) = 75 – (- 84) = 75 + 84 = 159 …(1)
R.H.S. = a + b = 75 + 84 = 159 …(2)
From (1) and (2), we get a – (- b) = a + b

(iv) a = 28, b = 11
L.H.S. = a – (- b) = 28 -(-11) = 28 + 11 = 39 …(1)
R.H.S. = a + b = 28 + 11 = 39 …(2)
From (1) and (2), we get a – (- b) = a + b.

Question 9.
Use the sign of >, < or = in the bo× to make the statements true.
(a) (- 8) + (-4) …… (- 8) – (- 4)
(b) (-3) +7 – (19) …… 15-8 +(-9)
(c) 23 – 41 + 11 …… 23-41- 11
(d) 39 + (-24) – (15) …… 36+ (-52) – (- 36)
(e) – 231 + 79 + 51 …… -399 + 159 + 81.
Solution:
(a) L.H.S. = (- 8) + (- 4)
= – (8 + 4) = – 12
R.H.S. = (- 8) – (- 4)
= – 8 + 4 = – (8 – 4) = – 4
∴ (-8) + (-4) < (-8)-(-4)

(b) L.H.S. = (-3) + 7 – (19)
= + 4 – (19)
= + 4 – 19
= -15
R.H.S. = 15 – 8 + (- 9)
= 7 + (-9) = 7- 9 = -2
∴ (- 3) + 7 – (19) < 15 – 8 + (- 9)

(c) L.H.S. = 23 – 41 + 11
= 23 + 11- 41
= 34 – 41 = – (41 – 34)
= – 7
R.H.S. = 23 – 41 – 11
= 23 – (41 + 11)
= 23 – 52
= – (52 – 23)
= – 29
∴ 23 – 41 + > 23 – 41 – 11

(d) L.H.S. = 39 + (- 24) – (15)
= 39 – 24 – (15)
= 15 – (15) = 0
R.H.S. = 36 + (- 52) – (- 36)
= – (52 – 36) – (- 36)
= – 16 – (- 36)
= – 16 + 36 = 20
∴ 39 + (- 24) – (15) < 36 + (- 52) – (- 36)

(e) L.H.S. = – 231 + 79 + 51
= – 231 + 130
= – (231 – 130) = – 101
R.H.S. = – 399 + 159 + 81
= – 399 + 240
= – (399 – 240)
= – 159
∴ – 231 + 79 + 51 > – 399 + 159 + 81.

Question 10.
A water tank has stepped inside it. A monkey is sitting on the topmost step (i.e., the first step The water level is at the ninth step.

(i) He jumps 3 steps down and 2 steps up. In how many jumps will he reach the water level?
(ii) After drinking water he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?
(iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following;
(a) -3 + 2 – …………. = -8
(b) 4 – 2 + = 8.
In (a) the sum (-8) represents going down by eight steps. So, what will the sum 8 in (b) represent?
Solution:
(i) While going down the monkey jumps 3 steps down and then jumps back 2 steps up. To reach the water level he is to jump as under:
-3+ 2 -3 + 2 – 3 +2 – 3 + 2 – 3 + 2 – 3 = -8 Hence, he takes 11 jumps to reach the water level.
(ii) After drinking water, he jumps back as under to reach the top step as under : 4 – 2+ 4 – 2+ 4 = 8
Hence, he takes 5 jumps to reach back the top.
(iii) (a) – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = -8
(b) 4 – 2 + 4 – 2 + 4 = 8
In (b) the sum 8 represents going up 8 steps.

We hope the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1.

• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.1
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1

Question 1.
Draw a line, say, AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
Solution:
Steps of Construction

1. Draw a line AB.
2. Take a point C outside it.
3. Take any point D on AB.
4. Join C to D.
   
5. with D as centre and a convenient radius, draw an arc cutting AB at F and CD at E.
6. Now with C as centre and the same radius as in step 5, draw an arc GH cutting CD at I.
7. Place the pointed tip of the compasses at F and adjust the opening so that the pencil tip is at E.
8. With the same opening as in step 7 and with I as centre, draw an arc cutting the arc GH at J.
9. Now join CJ to draw a line ‘KL’. Then KL is the required line.

Question 2.
Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
Solution:
Steps of Construction

1. Draw a line l.
2. Take any point A on line l.
   
3. Construct an angle of 90° at point A of line l and draw a line AL perpendicular to line l.
4. Mark a point X on AL such that AX = 4 cm.
5. At X construct an angle of 90° and draw a line XC perpendicular to line AL.
6. Then line XC (line m) is the required line through X such that m || l.

Question 3.
Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
Solution:
Steps of Construction

1. Draw a line l and take a point P not on it.
2. Take any point Q on l.
3. Join Q to P.
   
4. Draw a line m parallel to line l, as shown in the figure. Then line m || line l.
5. Join P to any point Q on l.
6. Choose any point R on m.
7. Join R to Q.
8. Through R, draw a line n parallel to the line PQ.
9. Let the line n meet the line l at S.
10. Then, the shape enclosed by the two sets of parallel lines is a parallelogram.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3.

• Perimeter and Area Class 7 Ex 11.1
• Perimeter and Area Class 7 Ex 11.2
• Perimeter and Area Class 7 Ex 11.4
• Perimeter and Area Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 11
Chapter Name	Perimeter and Area
Exercise	Ex 11.3
Number of Questions Solved	17
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

Question 1.
Find the circumference of the circles with the following radus (Take π = \(\frac { 22 }{ 7 } \))
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:

Question 2.
Find the area of the following circles, given that: ( Take π = \(\frac { 22 }{ 7 } \) )
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm.
Solution:

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:
Circumference of the circular sheet = 154 m
Let the radius of the circular sheet be r cm
Then, its circumference = 2nr m According to the question,
Circumference = 2πr = 154

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase if he makes 2 rounds of offense. Also find the cost of the rope, if it costs ₹ 4 per meter. ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:
The diameter of the circular garden (r) = 21 m
Radius of the circular garden (r) = \(\frac { 21 }{ 2 } \) m
∴ Circumference of the circular garden = 2πr
= 2 × \(\frac { 22 }{ 7 } \) × \(\frac { 21 }{ 2 } \) m = 66m
⇒ Length of the rope needed to make 1 round of fence = 66 m
⇒ Length of the rope needed to make 2 rounds of fence
= 66 × 2 m = 132 m
Cost of rope per meter = ₹ 4
∴ Cost of the rope = ₹ 132 × 4 = ₹ 528.

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Solution:

Here, Outer radius, r = 4 cm
Inner radius, r = 3 cm
Area of the remaining sheet = Outer area – Inner area
= π (R² – r²) = 3.14 (4² – 3²) cm²
= 3.14 (16 – 9) cm²
= 3.14 × 7 cm² = 21.98 cm²

Question 6.
Saima wants to put lace on the edge of a circular table cover of a diameter of 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)
Solution:
Diameter of the table cover = 1.5 m
⇒ Radius of the table cover (r) = \(\frac { 1.5 }{ 2 } \) m
⇒ Circumference of the table cover = 2πr
= 2 × 3.14 × \(\frac { 1.5 }{ 2 } \) m = 4.71 m
⇒ Length of the lace required = 4.71 m
∵ Cost of lace per meter = ₹ 15
∴ Cost of the lace = ₹ 4.71 × 15 = ₹ 70.65

Question 7.
Find the perimeter of the following figure, which is a semicircle including its diameter.
Solution:

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m². (Take π = 3.14)
Solution:
Diameter of the table-top = 1.6 m
⇒ Radius of the table-top (r) = \(\frac { 1.6 }{ 2 } \) m = 0.8 m
∴ Area of the table-top = πr²
= 3.14 × (0.8)2 m²
= 3.14 × 0.64 m²
= 2.0096 m²
∵ Rate of polishing = ₹ 15 per m²
∴ Cost of polishing the table-top = ₹ 2.0096 × 15
= ₹ 30.144
= ₹ 30.14 (approx.).

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its side? Which figure encloses more area, the circle or the square? ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the following figure). Find the area of the remaining sheet. ( Take π = \(\frac { 22 }{ 7 } \) )

Solution:

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? (Take π = 3.14)
Solution:

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Solution:

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Solution:

Question 14.
A circular flower garden has an area of about 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Taken π = 3.14)
Solution:
The circular area of the sprinkler = πr²
= 3.14 × 12 × 12
= 3.14 × 144 = 452.16 m²
Area of the circular flower garden = 314 m²
Since the area of the circular flower garden is smaller than by sprinkler
Therefore, the sprinkler will water the entire garden.

Question 15.
Find the circumference of the inner and the outer circles as shown in the following figure? (Take π = 3.14)

Solution:
Radius of inner circle = 19 – 10 = 9 m
∴ Circumference of the inner circle = 2 πr = 2 × 3.14 × 9 m = 56.52 cm
The radius of the outer circle = 19 m
∴ Circumference of the outer circle = 2πr = 2 × 3.14 × 19 m = 119.32 m.

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? ( Take π = \(\frac { 22 }{ 7 } \) )
Solution:

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Solution:
We know that the minute hand describes one complete revolution in one hour.
∴ Distance covered by its tip = Circumference of the circle of radius 15 cm
= (2 × 3.14 × 15) cm
= 94.2 cm

We hope the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3, drop a comment below and we will get back to you at the earliest.