CBSE Class 7

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind

Solution:

Let the number of digits formed be n, Then, the number of segments required to form n digits is given by the algebraic expression 5n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 1 = 25 + 1 = 26
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 1 = 50 + 1 = 51
3. the number of segments required to form 100 digits of this kind = 5 × 100 + 1 = 500 + 1 = 501.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 3n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 3 × 5 + 1 = 15 + 1 = 16
2. the number of segments required to form 10 digits of this kind = 3 × 10 + 1 = 30 + 1 = 31
3. The number of segments required to form 100 digits of this kind = 3 × 100 + 1 = 300 + 301.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 5n + 2.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 2 = 25 + 2 = 27
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 2 = 50 + 2 = 52
3. the number of segments required to form loo digits of this kind = 5 × 100 + 2 = 500 + 2 = 502.

Question 2.
Use the given algebraic expression to complete the table of number patterns.

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of :

1. m – 2
2. 3m – 5
3. 9 – 5m
4. 3m² – 2m – 7
5. \(\frac { 5m }{ 2 } \) – 4

Solution:

Question 2.
If p = – 2, find the value of :

1. 4p + 7
2. – 3p² + 4p + 7
3. – 2p³ – 3p² + 4p + 7.

Solution:

Question 3.
Find the value of the following expressions, when x = – 1 :

1. 2x – 7
2. – x + 2
3. x² + 2x + 1
4. 2x² – x – 2.

Solution:

Question 4.
If a = 2, b = – 2, find the value of :

1. a² + b²
2. a² + ab + b²
3. a² – b².

Solution:

Question 5.
When a = 0, b = – 1, find the value of the given expressions :

1. 2a + 2b
2. 2a² + b² + 1
3. 2a²b + 2ab² + ab
4. a² + ab + 2.

Solution:

Question 6.
Simplify the expressions and find the value ifx is equal to 2.

1. x + 7 + 4 (x – 5)
2. 3 (x + 2) + 5x – 7
3. 6x + 5 (x – 2)
4. 4 (2x – 1) + 3x + 11.

Solution:

Question 7.
Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.

1. 3x – 5 – x + 9
2. 2 – 8x + 4x + 4
3. 3a + 5 – 8a + 1
4. 10 – 3b – 4 – 5b
5. 2a – 2b – 4 – 5 + a.

Solution:

Question 8.
(i) If z = 10, find the value of z³ – 3(z – 10).
(ii) If p = -10, find the value of p² – 2p – 100.
Solution:

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3 2(a² + ab) + 3 – ab.
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms:

1. 21b – 32 + 7b – 20b
2. – z² + 13z² – 5z + 7z³ – 15z
3. p – (p – q) – q – (q – p)
4. 3a – 2b – ab – (a – b + ab) + 3ab + b – a
5. 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
6. (3y² + 5y – 4) – (8y – y² – 4).

Solution:

Question 2.
Add:

Solution:

Question 3.
Subtract:

1. -5y² from y²
2. 6xy from – 12xy
3. (a – b) from (a + b)
4. a (b – 5) from b (5 – a)
5. -m² + 5mn from 4m² – 3mn + 8
6. -x² + 10x – 5 from 5x – 10
7. 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
8. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq.

Solution:

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 76 + 16?
Solution:

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:

Question 6.
(a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2, 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 11.4 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 11.4.

• Perimeter and Area Class 7 Ex 11.1
• Perimeter and Area Class 7 Ex 11.2
• Perimeter and Area Class 7 Ex 11.3
• Perimeter and Area Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4

Question 1.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also, find the area of the garden in a hectare.
Solution:

Question 2.
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Solution:
PQ = 125 m + 3m + 3m = 131 m
QR = 65m + 3m + 3m = 71 m

Question 3.
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:

Question 4.
A verandah of width 2.25 m is constructed all along the outside of a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah
(ii) the cost of cementing the floor of the verandah at the rate of ₹200 per m².
Solution:

Question 5.
A path 1 rn wide is built along the border and inside a square garden of side 30 m. Find.
(i) the area of the path.
(ii) the cost of planting grass in the remaining portion of the garden at the rate of f 40 per m².
Solution:

Question 6.
Two crossroads, each of width 10 m, cut at right angles through the centre of a rectangular park of the length 700 m and the breadth 300 m parallel to its sides. Find the area of the roads. Also, find the area of the park excluding crossroads. Give the answer in hectares.
Solution:

Question 7.
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of ₹ 110 per m².
Solution:
(i) Area covered by the roads = Area of the rectangle PQRS + Area of the rectangle EFGH – Area of the square KLMN
= (90 × 3 + 60 × 3 – 3 × 3) m²
= (270 + 180 – 9) m²
= 441 m²

(ii) Cost of constructing the roads at the rate of ₹ 110 per m² ₹ 441 × 110 = ₹ 48510.

Question 8.
Pragya wrapped a cord around the circular pipe of the radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around the square box of the side 4 cm (also shown. Did she hate any cord left? = ( π = 3.14)

Solution:
Radius of the circular pipe (r) = 4 cm
∴ Circumference of the circular pipe = 2πr = 2 × 3.14 × 4 cm = 25.12 cm
Perimeter of the square box of the side 4 cm = 4 × side = 4 × 4 cm = 16 cm
∵ 25.12 cm > 16 cm
∴ She had extra cord left and the length of the cord left = (25.12 – 16) cm = 9.12 cm.

Question 9.
The following figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land.
(ii) the area of the flower bed.
(iii) the area of the lawn excluding the area of the flower bed.
(iv) the circumference of the flower bed.

Solution:

Question 10.
In the following figures, find the area of the shaded portions.

Solution:

Question 11.
Find the area of the quadrilateral ABCD.
Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC.
Solution:
Area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ADC

We hope the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4, 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

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.