NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2.

• Exponents and Powers Class 7 Ex 13.1
• Exponents and Powers Class 7 Ex 13.3
• Exponents and Powers Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

Question 1.
Using laws of exponents, simplify and write the answer in exponential form :

Solution:

Question 2.
Simplify and express each of the following in exponential form:

Solution:

Question 3.
Say true or false and justify your answer:

Solution:

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768.
Solution:

Question 5.
Simplify

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2, 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 12.4 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.