CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Solve the following equations:
(a) 2y + \(\frac{5}{2}=\frac{37}{2}\)
(b) 5t + 28 = 10
(c) \(\frac{\mathrm{a}}{5}\) + 3 = 2
(d) \(\frac{\mathrm{q}}{4}\) + 7 = 5
(e) \(\frac{5}{2}\) x = -10
(f) \(\frac{5}{2}\) x = \(\frac{25}{4}\)
(g) 7m + \(\frac{19}{2}\) = 13
(h) 6z + 10 = -2
(i) \(\frac{31}{2}=\frac{2}{3}\)
(j) \(\frac{2 \mathrm{~b}}{3}\) – 5 = 3
Answer:
(a) we have
2y + \(\frac{5}{2}=\frac{37}{2}\)
Transposing \(\frac{5}{2}\) from L.H.S to R.H.S
2y = \(\frac{37}{2}-\frac{5}{2}\)
2y = \(\frac{37-5}{2}\)
2y = \(\frac{32}{2}\)
2y = 16.

(b) we have
5t + 28 = 10
Transposing 28 from L.H.S to R.H.S
5t = 10 – 28
5t = – 18
Dividing both sides by 5, we get
\(\frac{5 \mathrm{t}}{5}=\frac{-18}{5}\)
t = \(\frac{-18}{5}\)
The solution is t = \(\frac{-18}{5}\) or -3 \(\frac{3}{5}\)
Dividing both sides by 2, we get
\(\frac{2 y}{2}=\frac{16}{2}\)
y = 8
y = 8 is the required solution.

(c) we have a
\(\frac{\mathrm{a}}{5}\) + 3 = 2
Transposing 3 from L.H.S to R.H.S
\(\frac{\mathrm{a}}{5}\) = 2 – 3
\(\frac{\mathrm{a}}{5}\) = -1
Multiplying both sides by 5, we get
\(\frac{\mathrm{a} \times 5}{5}\) = -1 x 5
a = – 5
The solution is a = – 5.

(d) We have
\(\frac{\mathrm{q}}{4}\) + 7 = 5
Transposing 7 to R.H.S
\(\frac{\mathrm{q}}{4}\) = 5 – 7
\(\frac{\mathrm{q}}{4}\) = -2
Multiplying both sides by 4, we get
\(\frac{\mathrm{q}}{4}\) × 4 = – 2 × (4)
q = -8
The solution is q = – 8.

(e) We have
\(\frac{5}{2}\) x = – 10
Multiplying both sides by 2, we get
\(\frac{5}{2}\) x x 2 = -10 × 2
5x = – 20
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{-20}{5}\)
The solution is x = – 4.

(f) We have
\(\frac{5}{2}\) x = \(\frac{25}{4}\)
Multiplying both sides by 2, we get
\(\frac{5}{2}\) x × 2 = \(\frac{25}{4}\) × 2
5x = \(\frac{25}{2}\)
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{25}{2 \times 5}\)
x = \(\frac{5}{2}\)
The solution is x= \(\frac{5}{2}\) or x = 2 \(\frac{1}{2}\).

(g) We have
7m + \(\frac{19}{2}\) = 13
Transposing \(\frac{19}{2}\) to R.H.S
7m = 13 – \(\frac{19}{2}\)
7m = \(\frac{26-19}{2}\)
7m = \(\frac{7}{2}\)

(h) We have
6z + 10 = – 2
Transposing 10 to R.H.S
6z = – 2 – 10
6z = – 12
Dividing both sides by 6, we get
\(\frac{6 z}{6}=\frac{-12}{6}\)
z = – 2
The solution is z = – 2.
Dividing both sides by 7, we get
\(\frac{7 \mathrm{~m}}{7}=\frac{7}{2 \times 7}\)
m = \(\frac{1}{2}\)
The solution is m = \(\frac{1}{2}\) .

(i) We have
\(\frac{31}{2}=\frac{2}{3}\)
Multiplying both sides by 2, we get
\(\frac{31}{2}\) x 2 = \(\frac{2}{3}\) x 2
31 = \(\frac{4}{3}\)
Dividing both sides by 3, we get
\(\frac{31}{3}=\frac{4}{3 \times 3}\)
1 = \(\frac{4}{9}\)
The solution is 1 = \(\frac{4}{9}\)

(j) We have
\(\frac{2 \mathrm{~b}}{3}\) – 5 = 3
Transposing – 5 to R.H.S
\(\frac{2 \mathrm{~b}}{3}\) = 3 + 5
\(\frac{2 \mathrm{~b}}{3}\) = 8
Multiplying both sides by 3, we get
\(\frac{2 \mathrm{~b}}{3}\) × 3 = 8 × 3
2b = 24
Dividing both sides by 2, we get
\(\frac{2 b}{2}=\frac{24}{2}\)
b = 12
The solution is b = 12.

Question 2.
Solve the following equations:
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = – 21
(d) – 4(2 + x) = 8
(e) 4(2 – x) = 8
Answer:
(a) 2(x + 4) = 12
Dividing both sides by 2, we get
\(\frac{2(x+4)}{2}=\frac{12}{2}\) ⇒ x + 4 = 6
Transposing 4 to R.H.S
x = 6 – 4 = 2

(b) 3(n – 5) = 21
Dividing both sides by 3, we get
\(\frac{3(n-5)}{3}=\frac{21}{3}\)
⇒ n – 5 = 7
Transposing – 5 to R.H.S
n = 7 + 5 = 12

(c) 3(n – 5) = -21
Dividing both sides by 3, we get
\(\frac{3(n-5)}{3}=\frac{-21}{3}\)
⇒ n – 5 = -7
Transposing – 5 to R.H.S
n = -7 + 5 = -2

(d) – 4(2 + x) =8
Dividing both sides by -4, we get
\(\frac{-4(2+x)}{-4}=\frac{8}{-4}\)
⇒ 2 + x = – 2
Transposing 2 to R.H.S
x = -2 -2 = -4

(e) 4(2 – x) = 8
Dividing both sides by 4, we get 4(2-x) _ 8
\(\frac{4(2-\mathrm{x})}{4}=\frac{8}{4}\)
⇒ 2 – x = 2
Transposing 2 to R.H.S
-x = 2 – 2 = 0
Multiplying both sides by (- 1), we get
-x × (-1) = 0 × (- 1)
x = 0

Question 3.
Solve the following 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)
Answer:
(a) 4 = 5 (p – 2)
Interchanging the sides, we get 5 ( p – 2) =4
Dividing both sides by 5, we get
\(\frac{5(p-2)}{5}=\frac{4}{5}\)
⇒ p – 2 = \(\frac{4}{5}\)
Transposing -2 to R.H.S
p = \(\frac{4}{5}\) + 2
= \(\frac{4+10}{5}=\frac{14}{5}\) = 2\(\frac{4}{5}\)

(b) – 4 = 5(p – 2)
Interchanging the sides, we get }
5 ( p – 2) = -4
Dividing both sides by 5, we get
\(\frac{5(p-2)}{5}=\frac{-4}{5}\)
p – 2 = \(\frac{-4}{5}\)
Transposing – 2 to R.H.S
p = \(\frac{-4}{5}\) + 2 = \(\frac{-4+10}{5}\)
= \(\frac{6}{5}\) = 1\(\frac{1}{5}\)

(c) 16 = 4 + 3(t + 2)
Interchanging the sides, we get 4 + 3 (t + 2) = 16
Transposing 4 to R.H.S
3(t + 2) – 16 – 4 = 12
Dividing both sides by 3, we get
\(\frac{3(t+2)}{3}=\frac{12}{3}\)
⇒ t + 2 = 4
Transposing 2 to R.H.S
t = 4 – 2 = 2

(d) 4 + 5 (p – 1) = 34
Transposing 4 to R.H.S
5 (p – 1) = 34 – 4 = 3
Dividing both sides by 5, we get
\(\frac{5(\mathrm{p}-1)}{5}=\frac{30}{5}\)
p – 1 = 6
Transposing -1 to R.H.S
p = 6 + 1 = 7

(e) 0 = 16 + 4(m – 6)
Interchanging the sides we get
16 + 4(m – 6) = 0
Transposing 16 to R.H.S
4(m – 6) = 0 – 16 = – 16
Dividing both sides by 4 we get
\(\frac{4(\mathrm{~m}-6)}{4}=\frac{-16}{4}\)
⇒ m – 6 = – 4
Transposing – 6 to R.H.S
m = -4 + 6 = 2

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2
Answer:
(a) Starting with x = 2
(1) Multiplying both sides by 3, we get
3x = 6
Subtracting 7 from both sides, we get
3x – 7 = 6 – 7
3x – 7 = – 1

(2) x = 2
Adding 5 on both sides, we get
x + 5 = 2 + 5
x + 5 = 7
Multiplying both sides by 4, we get
4(x + 5) = 7 × 4
4 (x + 5) = 28

(b) Starting with x = – 2
(1) x = – 2
Adding 11 to both sides, we get
x + 11 = – 2 + 11
x + 11 = 9

(2) x = – 2
Multiplying both sides by 5, we get
5x = – 10
Adding 3 on both sides, we get
5x + 3 = – 10 + 3
5x + 3 = – 7

(3) x = 2
Dividing both sides by 5, we get
\(\frac{x}{5}=\frac{2}{5}\)
Subtracting 2 from both sides, we get
\(\frac{\mathrm{x}}{5}\) – 2 = \(\frac{2}{5}\) -2
\(\frac{\mathrm{x}}{5}\) – 2 = \(\frac{2-10}{5}\)
\(\frac{\mathrm{x}}{5}\) – 2 = \(\frac{-8}{5}\)

(3) x = – 2
Dividing both sides by 7, we get
\(\frac{x}{7}=\frac{-2}{7}\)
Add 3 on both sides, we get
\(\frac{\mathrm{X}}{7}\) + 3 = \(\frac{-2}{7}\) + 3
= \(\frac{-2+21}{7}\)
\(\frac{\mathrm{X}}{7}\) + 3 = \(\frac{19}{7}\)

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2

Question 1.
Find the area of each of the following parallelograms:


Answer:
Here, Base(b) = 7cm
Height (h) = 4 cm
∴ Area of the parallelogram
= b × h sq. units
= 7 × 4 cm²
= 28 cm²

(b) Here Base (b) = 5 cm
Height (h) = 3 cm
∴ Area of the parallelogram
= b x h sq. units
= 5 x 3 cm² =15 cm²

(c) Here Base (b) = 2.5 cm
Height (h) = 3.5 cm
∴ Area of the parallelogram
= b x h sq. units
= 2.5 x 3.5 cm²
= 8.75 cm²

(d) Here Base (b) = 5 cm
Height (h) = 4.8 cm
∴ Area of the parallelogram
= b x h sq. units
= 5 x 4.8 cm²
= 24 cm²

(e) Here Base (b) = 2 cm
Height (h) =4.4 cm
∴ Area of the parallelogram
= b x h sq. units
= 2 x 4.4 cm²
= 8.8 cm²

Question 2.
Find the area of each of the following triangles:

Answer:
(a) Here, Base (b) = 4cm
Height (h) = 3 cm
Area of the triangle
= \(\frac { 1 }{ 2 }\) x b x h sq.m
= \(\frac { 1 }{ 2 }\) x 4 x 3 cm²
= 6 cm²

(b) Here base (b) = 5 cm
Height (h) = 3.2 cm
Area of the triangle
= \(\frac { 1 }{ 2 }\) x b x h sq. units
= \(\frac { 1 }{ 2 }\) x 5 x 3.2 cm²
= 8 cm²

(c) Here base (b) = 3 cm
Height = 4 cm
Area of the triangle = \(\frac { 1 }{ 2 }\) x b x h sq. units
= \(\frac { 1 }{ 2 }\) x 3 x 4 cm²
= 6 cm²

(d) Here base (b) = 3 cm
Height (h) = 2m
Area of the triangle
= \(\frac { 1 }{ 2 }\) x b x h sq units
= \(\frac { 1 }{ 2 }\) x 3 x 2 cm² = 3 cm²

Question 3.
Find the missing values:

Answer:
(a) Here, base of the parallelogram
(b) = 20 cm
Let the height ‘h’
Area of the Parallelogram = 246 cm²
b x h = 246
20 x h = 246
h = \(\frac{246}{20}\)
\(\frac{123}{10}\) = 12.3 cm
∴ The missing value height = 12.3 cm

(b) Here, height (h) = 15 cm
Let the base of the parallelogram be ‘b’
Area of a parallelogram = 154.5 cm²
b x h = 154.5 cm²
b = \(\frac{154.5}{15}\) = 12.3
= \(\frac{1545}{15 \times 10}\) = 12.3cm
= \(\frac{103}{10}\)10.3
∴ The missing value base = 10.3 cm.

(c) Here, height (h) = 8.4 cm
Let the base of the parallelogram be ‘b’
Area of the parallelogram = 48.72 cm²
b × h = 48.72
b × 8.4 =48.72
b = \(\frac{48.72}{8.4}=\frac{48.72}{8.4}\)
= 5.8 cm
Thus, the missing value base = 5.8 cm.

(d) Here, base (b) = 15.6 cm
Let the height of the parallelogram
be ‘h’
Area of the parallelogram = 16.38 cm²
b × h = 16.38
15.6 × h = 16.38
h = \(\frac{16.38}{15.6}\)
= \(\frac{163.8}{156}\)
= 1.05 cm
Thus, the missing value (height)
= 1.05 cm.

Question 4.
Find the missing values:

Answer:
(i) Let the height of the triangle be ‘h’
Here base (b) = 15 cm
Area of the triangle = 87 cm²
\(\frac { 1 }{ 2 }\) × b × h =87 2
\(\frac { 1 }{ 2 }\) × 15 × h =87
h = \(\frac{87 \times 2}{15}\)
= \(\frac{29 \times 2}{5}\)
= \(\frac{58}{5}\)
= 11.6 cm
∴ The missing value height = 11.6 cm

(ii) Here Height = 31.4 mm
Let the base be “b”
Area of a triangle = 1256 mm²
\(\frac { 1 }{ 2 }\) × b × h = 1256 2
\(\frac { 1 }{ 2 }\) × b × 31.4 = 1256
b = \(\frac{1256 \times 2}{31.4}\)
b = \(\frac{1256 \times 2 \times 10}{314}\)
= 4 × 2 × 10 mm
= 80 mm.
∴ The missing value base = 80 mm.

(iii) Let the height of the triangle be ‘h’
base (b) = 22 cm.
Area of the triangle = 170.5 cm²
\(\frac { 1 }{ 2 }\) × b × h = 170.5
\(\frac { 1 }{ 2 }\) × 22 × h = 170.5
11 × h = 170.5
h = \(\frac{170.5}{11}\)
= 15.5 cm
∴ The missing value height = 15.5 cm.

Question 5.
PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS.
If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm

Answer:
Here, Base (SR) = 12 cm
Corresponding height (QM) = 7.6 cm
(a) Area of the parallelogram
= b × h sq units
= 12 × 7.6 cm²
= 91.2 cm²

(b) Base of the parallelogram (PS)
= 8 cm
Area of the parallelogram = 91.2 cm²
b × h = 91.2
8 × h = 91.2
h = \(\frac{91.2}{8}\) = 11.4 cm
The height QN =11.4 cm.

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Answer:
Area of the parallelogram ABCD
= Base x× height
= AD × BM
Area of a parallelogram
= 1470cm²
∴ AD × BM = 1470cm²
49 × BM = 1470
BM = \(\frac{1470}{49}\)
BM = \(\frac{210}{7}\) = 30 cm
Area of the parallelogram ABCD = 1470cm²
AB × DL = 1470
35 × DL = 1470
DL = \(\frac{1470}{35}\)
= \(\frac{210}{7}\) = 42 cm
∴ Length of BM = 30 cm
Length of DL = 42 cm]

Question 7.
ΔABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC =13 cm and AC = 12 cm, find the area of ΔABC. Also, find the length of AD.
Answer:
Area of AABC = \(\frac { 1 }{ 2 }\) × b × h sq. units 2 H
(base = 5 cm and height =12 cm)

= \(\frac { 1 }{ 2 }\) × 5 × 12 cm² = 5 × 6 cm² = 30 cm²
Area of the AABC = \(\frac { 1 }{ 2 }\) x base x height
30 = \(\frac { 1 }{ 2 }\) × 13 × AD
AD = \(\frac{30 \times 2}{13}\) cm = \(\frac{60}{13}\) cm
Hence, length of AD = \(\frac{60}{13}\) cm = 4.6 cm.

Question 8.
∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?
Answer:
Here base BC = 9 cm
Corresponding height AD = 6 cm
∴ Area of ∆ABC
= \(\frac { 1 }{ 2 }\) × base × height
= \(\frac { 1 }{ 2 }\) × BC × AD
= \(\frac { 1 }{ 2 }\) × 9 × 6 cm² = 27 cm²

Let the height from C to AB be ‘h’
= \(\frac { 1 }{ 2 }\) × AB × h = 27
= \(\frac { 1 }{ 2 }\) × 7.5 × h = 27
h = \(\frac{27 \times 3}{7.5}\)
= \(\frac{27 \times 2 \times 10}{75}\)
= 7.2 cm
∴ The height CE = 7.2 cm.

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1

Question 1.
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land, if 1 m2 of the land costs ₹ 10,000.
Answer:
(i) Length of the rectangular piece
(1) = 500 m
Breadth of the rectangular piece
(b) = 300 m
Area = 1 × b sq. unit
= 500 × 300 m²
= 150000m²

(ii) Cost of the land = ₹ 150000 × 10, 000
= ₹ 1,50,00,00,000

Question 2.
Find the area of a square park whose perimeter is 320 m.
Answer:
Perimeter of the square park = 320 m
4 × side = 320
side = \(\frac{320}{4}\) = 80 m
Area of the square park = side × side = 80 × 80 m²
= 6400 m²

Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.
Answer:
Length of the rectangular plot = 22 m
Area of the rectangular plot = 440 m²
1 × b = 440
22 × b = 440
b = \(\frac{440}{22}\) = 20m
Breadth of the rectangle = 20 m
Perimeter of the rectangular plot = 2 (l + b)
= 2(22 + 20) m
= 2 × 42 = 84 m
∴ Perimeter of the rectangular plot is 84 m.

Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also, find the area.
Ans: Length of the rectangular sheet (1) = 35
Perimeter of the rectangular sheet = 100 cm
2(1 + b) = 100
2(35+ b) = 100
70 + 2b = 100
2b = 100 – 70
2b = 30 , 30
b = \(\frac { 30 }{ 2 }\) = 15 cm
Area of the rectangular sheet
= 1 × b sq. units
= 35 × 15 cm²
= 525 cm²
Thus,
(i) Breadth of the rectangular sheet = 15 cm
(ii) Area of the rectangular sheet = 525 cm²

Question 5.
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Answer:
Side of the square park = 60 m
Area of the square park = 60 × 60 m²
= 3600 m²
Length of the rectangular park = 90 m
Area of the rectangular Park
= Area of the square park = 3600 m²
1 × b = 3600
90 × b = 3600
b = \(\frac { 3600 }{ 90 }\)
= 40 m
∴ Breadth of the rectangular park = 40 m.

Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also, find which shape encloses more area?

Answer:
Length of the wire
(l) = 40 cm
Breadth of the wire
(b) = 22 cm

Area of the rectangle = 1 × b
= 40 × 22 cm²
= 880 cm²

Perimeter of the rectangular wire = 2(1 + b)
= 2(40 + 22)cm
= 124 cm.

Perimeter of the square = Perimeter of the rectangle
4 × side = 124
side = \(\frac{124}{4}\) = 31 cm
Side of a square =31 cm
Area of the square = 31 × 31 cm²
= 961 cm²
961 cm² > 880 cm²
Area of the square is more than the area of rectangle.

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also, find the area of the rectangle.
Answer:
Breadth of the rectangle (b) = 30 cm
Perimeter of a rectangle =130 cm
2 (l + b) = 130 cm
2(l + 30) = 130
2l + 60 = 130
2l = 130 – 60
2l = 70
l = \(\frac{70}{2}\) = 35cm
Length of the rectangle = 35 cm
Area of the rectangle = l × b sq units = 35 × 30 cm²
= 1050 cm²
∴ Area of the rectangle

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m².

Answer:
Length of the wall (l) = 4.5 m
Breadth of the wall (b) = 3.6 m
∴ Area of the wall = l × b sq. units
= 4.5 × 3.6 m²
= \(\frac{45}{10} \times \frac{36}{10}\) m²
= \(\frac{1620}{100}\) m²
= 16.2 m²
Area of the door = 2 m × 1 m
(l = 2 m, b = 1 m) = 2 m²
Area to be white washed =
Area of the wall – Area of the door
= 16.2 m² – 2 m²
= 14.2 m²
Rate of white washing per m² = ₹ 20
∴ Cost of while washing = ₹ 20 × 14.2
= ₹ 284

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions

NCERT In-text Question Page No. 230

Question 1.
Describe how the following expressions are obtained:
(i) 7xy + 5
(ii) x²
(ii) 4x² – 5x
Answer:
(i) 7xy + 5
To obtain this expression, we first multiply two variables x and y, i.e. x × y = xy. Then we multiply their product by 7 to get 7xy. Next, we add 5 to 7xy to obtain 7xy + 5.

(ii) x:y
First, x is multiplied by itself i.e., x × x = x² Then multiply x² by y to get the required expression, i.e. x² × y = x²y

(iii) 4x² – 5x
Here, the variable x is multiplied by itself,
i. e. x × x = x²
Then x² is multiplied by, i.e. x² × 4 = 4x²
Next, we multiply the variable x by 5, i.e. x × 5 = 5x
Now, we subtract 5x from 4x² to get 4x² – 5x

NCERT In-text Question Page No. 231

Question 1.
What are the terms in the followig expressions? Show how the terms are formed. Draw a tree diagram for each expression:
(i) 8y + 3x²
(ii) 7mn – 4
(iii) 2x²y
Answer:
(i) 8y + 3x²
Terms of 8y + 3x² are: 8y and 3x² the term 8y is formed by multiplying the variable y by 8.
The term 3x² is formed by first multiplying the variable x with itself to get x × x = x² and then multiplying x² by 3.
Tree diagram:

Terms of 7mn – 4 are 7 mn and -4. The term 7 mn formed by first multiplying variables m and n to get m × n = mn
Then multiplying mn by constant 7 to get 7 × mn = 7 mn. The term -4 is a constant.
Tree diagram:

(iii) 2x²y
This expression has only one term, i.e. 2x²y. To form this term first we multiply the variable x by itself to get x × x = x². Then the variable y is multiplied by x² to get
y × x² = x²y
Next, the product x²y is multiplied by 2 to ‘ get
2 × x²y = 2x²y
Tree diagram:

Question 11.
Write three expressions each having 4
terms.
Answer:
(i) 2x³ – 4x² + 9xy + 8
(ii) 6x³ + 9y² – 3xy² – 12
(iii) 9x² – 3x + 12xy – 1

NCERT In-text Question Page No. 231

Question 1.
Identify the coefficients of the terms of following expressions:
Answer:
(i) 4x – 3y
(ii) a + b + 5
(iii) 2y + 5
(iv) 2xy
Answer:
(i) 4x – 3y
The coefficient of x in 4x is 4.
The coefficient of y in -3y is (-3)

(ii) a + b + 5
The coefficient of a is 1.
The coefficient of b is 1.
The coefficient of 5 is 1.

(iii) 2y + 5
The coefficient of y in 2y is 2.
The coefficient of 5 is 1.

(iii) 2xy
The coefficient of xy in 2xy is 2.
The coefficient of y in 2xy is 2x.
The coefficient of x in 2xy is 2y.

NCERT In-text Question Page No. 233

Question 1.
Group the like terms together from the following:
12x, 12, -25x, -25, -25y, 1, x, 12y, y
Answer:
We have:
(i) 12x, -25x and x are like terms.
(ii) -25y, 12y and y are like terms.
(iii) 12, -25 and 1 are like terms.

NCERT In-text Question Page No. 233

Question 1.
Classify the following expressions as a monomial, a binomial or a trianomial: a, a + b, ab + a + b, ab + a + b -5, xy, xy + 5, 5x²– x + 2,4pq – 3q + 5p, 7,4m – 7n + 10, 4mn + 7.
Answer:
(i) a is containing 1 term. It is a monomial.
(ii) a + b is containing 2 terms. It is a binomial.
(iii) ab + a + b is containing 3 terms, it is a trinomial.
(iv) ab + a + b – 5 is containing 4 terms. It is a polynomial.
(v) xy is containing only 1 term. It is a monomial.
(vi) xy + 5 is containing 2 terms. It is a binomial.
(vii) 5x² – x + 2 is containing 3 terms. It is a trinomial.
(viii) 4pq – 3q + 5p is containing 3 terms. It is a trinomial.
(ix) 7 is containing only 1 term. It is a monomial.
(x) 4 m – 7n + 10 is containing 3 terms. It is a trinomial.
(xi) 4mn + 7 is containing 2 terms. It is a binomial.

NCERT In-text Question Page No. 236

Question 1.
Think of at least two situations in which you need to form two algebraic expressions and add or subtract them.
Answer:
Situation-I
Rohan’s per month earning is twice the per month earning of Bibha. Kavita’s per month earnings is RS. 400 more than the sum of Rohan’s and Vibha’s per month earnings. What is the montly earning of Kavita?
Situation-II
Mahesh has twice the number of toys Kanta has: Lata has 5 toys more than twice the number of toys Mahesh and Kanta together have. What is the number of toys that Kanta has?

NCERT In-text Question Page No. 238

Question 1.
Add and subtract:
(i) m – n, m + n
(ii) mn + 5-2, mn + 3
Answer:
(i) Adding (m – n) and (m + n), we have:
(m-n) + (m + n) = m- n + m + n
Collecting like terms, we have:
(m + m) + (-n + n)
= (1 + 1)m + (-1 + 1)n
= 2m + (0)n
= 2m
Now subtracting (m – n) from (m + n), we have:
(m + n)-(m-n) = m + n- m + n
= (m – m) + (n + n)
= (1 – 1)m + (1 + 1)n
= 0m + 2n
= 2n

(ii) Adding mn + 5-2 and mn + 3, we have:
(mn + 5 – 2) + (mn + 3)
= mn + 5 – 2 + mn + 3
= (mn + mn) + (5 – 2 + 3)
= (1 + 1)mn + (6)
= 2 mn + 6
Now, subtracting (mn + 3) from (mn + 5 – 2), we have:
(mn + 5 – 2) – (mn + 3)
= mn + 5 – 2 – mn – 3
= (mn – mn) + (5 – 2 – 3)
= (0)mn + 0 = 0 + 0 = 0

NCERT In-text Question Page No. 245

Question 1.
Make similar pattern with basic figure as shown

(The number of segments required to make the figure is given to the right. Also, the expression for the number of segments required to make n shapes is also given.)
Answer:
Discover more such patterns as follow.

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 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 and 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 6, H, 0.
Answer:

Expression for number of line segments to form ‘n’ figures of the kind 5 is 5n + 1
For 5 figures, like segments = 5×5 + 1 = 25+1 = 26
For. 10 figures, the line segments = 5×10+1 = 50+ = 51
For 100 figures, line segments = 5×100+1 = 500 + 1 = 501

Expression for number of line segments to form ‘n figures of the kind 4 is 3n + 1
For 5 figures, line segment = 3 x 5 + 1 = 15 + 1 = 16
For 10 figures line segments = 3 x 10+1 = 30+ 1 = 31
For 100 figures line segments = 3 x 100 +1 = 300 + 1 = 301

Expression of number of line segments to form ‘n figures of the kind B is 5n + 2
For 5 figures line segments = 5×5 + 2 = 25+ 2 = 27
For 10 figures line segments = 5x 10 + 2 = 50+ 2 = 52
For 100 figures, line segments = 5x 100 + 2 = 500 + 2 = 502

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

Answer:
(i) For the expression 2n – 1
100th term = 2 x 100 – 1
= 200 – 1 = 199

(ii) For the expression 3n + 2
5th term = 3 (5) + 2 = 15 + 2 = 17
10th term = 3 (10) + 2 = 30 + 2 = 32
100th term = 3 (100) + 2 = 300 + 2 = 302

(iii) For the expression 4n + 1
5th term = 4 x 5 +1 = 20+ 1 = 21
10th term = 4 x 10 +1= 40+1 = 41
100th term = 4 x 100 + 1 = 400 + 1 = 401

(iv) For the expression 7n + 20
5th term = 7 x 5 + 20 = 35 + 20 = 55
10th term = 7 x 10 + 20 = 70 + 20 = 90
100th term = 7 x 100 + 20 = 700 + 20 = 720

(v) For the expression n² + 1
5th term = 52 + 1 = 25 + 1 = 26
10th term = 102 + 1 = 100 + 1 = 101

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Give first 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
(d) x + 6 = 2
(e) y – 4 = -7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Answer:
(a) x – 1 = 0
Adding 1 on both sides, we get.
x – 1 + 1 = 0 + 1
x = 1
The solution is x = 1.

(b) x + 1 = 0
Subtracting 1 from both sides, we get.
x + 1 – 1 = 0 – 1
x = -1.
The solution is x = – 1

(c) x – 1 = 5
Adding 1 to both sides, we get.
x – 1 + 1 = 5 + 1
x = 6
The solution is x = 6.

(d) x + 6 = 2
Subtracting 6 from both sides, we get
x + 6 – 6 = 2 – 6
x = – 4
The solution is x = – 4.

(e) y – 4 = – 7
Adding 4 on both sides, we get
y – 4 + 4 = -7 + 4
y = – 3
The solution is y = – 3.

(f) y – 4 = 4
Adding + 4 on both sides, we get
y – 4 + 4 = 4 + 4
y = 8
The solution is y = 8.

(g) y + 4 = 4
Subtracting 4 from both sides, we get
y + 4 – 4 = 4 – 4
y = 0
The solution is y = 0.

(h) y + 4 = – 4
Adding – 4 on both sides, we get
y + 4 – 4 = -4 -4
y = – 8
The solution is y = – 8.

Question 2.
Give first step you will use to separate the variable and then solve the equation:
(a) 31 = 42
(b) \(\frac{\mathrm{b}}{2}\) = 6
(c) \(\frac{\mathrm{p}}{7}\) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac{z}{3}=\frac{5}{4}\)
(g) \(\frac{a}{5}=\frac{7}{15}\)
(h) 20t = -10
Answer:
(a) 31 = 42
Dividing both sides by 3, we get
\(\frac{31}{3}=\frac{42}{3}\)
1 = 14
The solution is 1 = 14.

(b) \(\frac{\mathrm{b}}{2}\) = 6
Multiplying both sides by 2, we get. b
\(\frac{\mathrm{b}}{2}\) × 2 = 6 × 2
b = 12
The solution is b = 12.

(c) \(\frac{\mathrm{p}}{7}\) = 4
Multiplying both sides by 7, we get
\(\frac{\mathrm{p}}{7}\) × 7 = 4 × 7
p = 28
The solution is p = 28.

(d) 4x = 25
Dividing both sides by 4, we get
\(\frac{4 x}{4}=\frac{25}{4}\)
x = \(\frac{25}{4}\)
The solution is x = \(\frac{25}{4}\) or x = 6 \(\frac{1}{4}\) .

(e) 8y = 36
Dividing both sides by 8, we get
\(\frac{8 y}{8}=\frac{36}{8}\)
y = \(\frac{9}{2}\)
The solution is y = \(\frac{9}{2}\) or y = 4 \(\frac{1}{2}\)

(f) (f) \(\frac{z}{3}=\frac{5}{4}\)
Multiplying both sides by 3, we get
\(\frac{z}{3}\) × 3 = \(\frac{5}{4}\) × 3
z = \(\frac{25}{4}\)
The solution is z = \(\frac{15}{4}\) or y = 3 \(\frac{3}{4}\)

(g) \(\frac{a}{5}=\frac{7}{15}\)
Multiplying both sides by 5, we get
\(\frac{a}{5}\) × 5 = \(\frac{7}{15}\) × 5
a = \(\frac{7}{3}\)
The solution is a = \(\frac{7}{3}\) or a = 2 \(\frac{1}{3}\).

(h) 20t = – 10
Dividing both sides by 20, we get
\(\frac{20 \mathrm{t}}{20}=\frac{-10}{20}\)
t = \(\frac{-1}{2}\)
The solution is t = \(\frac{-1}{2}\).

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) \(\frac{20 \mathrm{p}}{3}\) = 40
(d) \(\frac{3 \mathrm{p}}{10}\) = 6
Answer:
(a) 3n – 2 = 46
Adding 2 on both sides, we get
3n – 2 + 2 =46 + 2
3n = 48
Dividing both sides by 3, we get
\(\frac{3 n}{3}=\frac{48}{3}\)
n = 16
The solution is n = 16.

(b) 5m + 7 = 17
Adding – 7 on both sides, we get
5m + 7 – 7 = 17 – 7
5m = 10
Dividing both sides by 5, we get
\(\frac{5 m}{5}=\frac{10}{5}\)
m = 2
The solution is m = 2

(c) \(\frac{20 \mathrm{p}}{3}\) = 40
Multiplying both sides by 3, we get.
\(\frac{20 \mathrm{p}}{3}\) × 3 = 40 × 3
20 p = 120
Dividing both sides by 20 we get.
\(\frac{20 p}{20}=\frac{120}{20}\)
p = 6
The solution is p = 6.

(d) \(\frac{3 \mathrm{p}}{10}\) = 6
Multiplying both sides by 10, we get 3p
\(\frac{3 \mathrm{p}}{10}\) × 10 = 6 × 10
3p = 60
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{60}{3}\)
p = 20
The solution is p = 20.

Question 4.
Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) \(\frac{\mathrm{p}}{4}\)
(d) \(\frac{-p}{3}\) = 5
(e) \(\frac{3 \mathrm{p}}{4}\) = 6
(f) 3s = – 9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Answer:
(a) 10p = 100
Dividing both sides by 10, we get
\(\frac{10 \mathrm{p}}{10}=\frac{100}{10}\)
p = 10 10p = 90
The solution is p = 10.

(b) 10p + 10 = 100
10p + 10 – 10= 100 – 10
Dividing both sides by 10, we get
\(\frac{10 p}{10}=\frac{90}{10}\)
p = 9
The solution is p = 9.

(c) \(\frac{\mathrm{p}}{4}\) = 5
Multiplying both sides by 4, we get
\(\frac{\mathrm{p}}{4}\) × 4 = 5 × 4
p = 20
The solution is p = 20.

(d) \(\frac{-p}{3}\) = 5
Multiplying both sides by 3, we get
-p = 15
\(\frac{-p}{3}\) × 3 = 5 × 3
Multiplying both sides by (- 1)
– p × (- 1)= 15 × (- 1)
p = – 15
The solution is p = -15.

(e) \(\frac{3 \mathrm{p}}{4}\) = 6
Multiplying both sides by 4, we get
\(\frac{3 \mathrm{p}}{4}\) × 4 = 6 × 4
3p = 24
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{24}{3}\)
p = 8
The solution is p = 8.

(f) 3s = – 9
Dividing both sides by 3, we get
\(\frac{3 s}{3}=\frac{-9}{3}\)
s = – 3
The solution is s = – 3.

(g) 3s + 12 = 0
Subtracting 12 from both sides, we get
3s + 12- 12 = 0 – 12
3s = – 12
Dividing both sides by 3, we get
\(\frac{3 s}{3}=\frac{-12}{3}\)
s = – 4
The solution is s = – 4.

(h) 3s =0
Dividing both sides by 3, we get
\(\frac{3 s}{3}=\frac{0}{3}\)
s = 0
The solution is s = 0.

(i) 2q = 6
Dividing both sides by 2, we get
\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)
q = 3
The solution is q = 3.

(j) 2q – 6 = 0
Adding 6 on both sides, we get
2q – 6 + 6= 0 + 6
2q = 6
Dividing both sides by 2, we get
\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)
q = 3
The solution is q = 3.

(k) 2q + 6 =0
Subtracting 6 from both sides, we get
2q + 6 – 6 = 0 – 6
2q = – 6
Dividing both sides by 2, we get
\(\frac{2 q}{2}=\frac{-6}{2}\)
q = -3
The solution is q = – 3.

(l) 2q + 6 =12
Subtracting 6 from both sides, we get
2q + 6 – 6 = 12 – 6
2q = 6
Dividing both sides by 2, we get
\(\frac{2 \mathrm{q}}{2}=\frac{6}{2}\)
q = 3
The solution is q = 3.