Class 7

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.2

1. State, if a triangle is possible with the following angles.

Question (a).
35°, 70°, 65°
Answer:
No Reason :
Sum of three angles
= 35° + 70° + 65° = 170°
But, we know that sum of angles of a triangle is always 180°
∴ A triangle cannot have angles 35°, 70° and 90°.

Question (b).
70°, 50°, 60°
Answer:
Yes
Reason :
Sum of three angles
= 70° + 50° + 60°
= 180°
By angle sum property
∴ A triangle can have angles 70°, 50° and 60°.

Question (c).
90°, 80°, 20°
Answer:
No
Reason :
Sum of three angles
= 90° + 80° + 20°
=190°
But, we know that sum of angles of a triangle is always 180°
(Angle sum properly)
∴ A triangle cannot have angles 90°, 80° and 20°.

Question (d).
60°, 60°, 60°
Answer:
Yes
Reason :
Sum of three angles
= 60° + 60° + 60°
= 180°
by angle sum property.
∴ A triangle can have angles 60°, 60° and 60°.

Question (e).
90°, 90°, 90°
Answer:
No Reason :
Sum of three angles
= 90° + 90° + 90°
= 270°
But, we know that sum of angles of a triangle is always 180° (Angle sum properly)
∴ A triangle cannot have angles 90°, 90° and 90°.

2. Find the value of x in the following figures :

Question (i).

Answer:
By angle sum property of a triangle
x + 53° + 60° = 180°
x + 113° = 180°
x = 180° – 113°
x = 67°

Question (ii).

Answer:
By angle sum property of a triangle
90° + x + 42° = 180°
132° + x = 180°
x = 180° – 132°
x = 48°

Question (iii).

Answer:
By angle sum property of a triangle
x + x + 70° = 180°
2x + 70° = 180°
2x = 180° – 70°
2x = 110°
x = \(\frac{110^{\circ}}{2}\)
x = 55°

Question (iv).

Answer:
By angle sum property of a triangle
x + 3x + 2x = 180°
6x = 180°
x = \(\frac{180^{\circ}}{6}\)
x = 30°

Question (v).

Answer:
By angle sum property of a triangle
x + x + x = 180°
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°

Question (vi).

Answer:
By angle sum property of a triangle
x – 5° + 60° + x + 5° = 180°
2x + 60° = 180°
2x = 180° – 60°
2x = 120°
x = \(\frac{120^{\circ}}{2}\)
x = 60°

3. Find the values of x and y in the following figures :

Question (i).

Answer:
Since in ΔABC, BC is produced to D
∴ 60° + x = 110°
(By exterior angle property)
x = 110°- 60°
x = 50° ………. (1)
Now, in ΔABC
60° + x + y = 180°
(By angle sum property of triangle)
60° + 50° + y = 180° [(By using (1)]
110° + y = 180°
y = 180° – 110°
y = 70°
Hence, x = 50°,
y = 70°

Question (ii).

Answer:
In ΔPQR,
∠P + ∠Q + ∠R = 180°
60° + 40° + x = 180°
(By angle sum property of triangle)
100° + x = 180°
x = 180° – 100°
x = 80°
Now,in ΔPQR, QR is produced
∴ y = 60° + 40°
(By exterior angle property)
y = 100°
Hence, x = 80°,
y = 100°

Question (iii).

Answer:
∠ACB = ∠ECD
∴ x = 80°….(1)
(vertically opposite angles)
∠ACD + ∠ECD = 180° (Linear pair)
∴ ∠ACD + 80° = 180° [by using (1)]
∠ACD = 180°- 80°
∠ACD = 100° ….(2)
In ΔABC, BC is produced to D
∴ x + y = ∠ACD
(By exterior angle property)
80° + y = 100°
(By using (1) and (2))
y = 100° – 80°
y = 20°
Hence, x = 80° and y = 20°

Question (iv).

Answer:
By angle sum property of a triangle
∠L + ∠M + ∠N = 180°
y + 90° + y = 180°
2 y + 90° = 180°
2y = 180° – 90°
2y = 90°
y = \(\frac{90^{\circ}}{2}\)
y = 45°

Question (v).

Answer:
∠ABC = ∠HBI
(Vertically opposite angles)
∴ y = x …(1)
∠BAC = ∠GAF
(Vertically opposite angles)
∴ ∠BAC = x ….(2)
∠ACB = ∠EFD
(Vertically opp. angles)
∠ACB = x …(3)
Now, in ΔABC
∠BAC + ∠ABC + ∠ACB = 180°
(By angle sum property of triangle)
x + x + x = 180°
[by using (1), (2) and (3)]
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°
y = x
= 60° (by using (1) and (4))
Hence, x = 60°, y = 60°

Question (vi).

Answer:
In ΔPQR, QR is produced to S,
∴ 2x – 5° = 50° + x + 5°
(By exterior angle property.)
2x – 5°= 55° + x
2x – x = 55° + 5°
x = 60° ….(i)
Now, by angle sum property of a ΔPQR
50° + x + 5° + y = 180°
55° + 60° + y = 180°
115°+ y = 180°
y = 180° – 115°
y = 65°
Hence, x = 60° and y = 65°

4. The angles of a triangle are in the ratio 5:6:7. Find the measure of each of the angles.
Solution:
Let the measure of the given angles be
(5x)°, (6x)°, (7x)°
By angle sum property of a triangle
(5x)° + (6x)° + (7x)° = 180°
(18x)° = 180°
x = \(\frac{180^{\circ}}{18}\)
x = 10
Required angles
= (5 × 10)°, (6 × 10)°, (7 × 10)°
= 50°, 60°, 70°

5. One angle of a triangle is 60°. The other two angles are in the ratio 4 : 8. Find the angles.
Solution:
One angle of triangle = 60°
Let the other two angles be (4x)° and (8x)°
By angle sum property of a triangle
60° + (4x)° + (8x)° = 180°
60° + (12x)° = 180°
(12x)° = 180° – 60°
(12x)° = 120°
x = \(\frac{120^{\circ}}{12}\)
x = 10
Required angles = (4x)°, (8x)°
(4 × 10)°, (8 × 10)°
= 40°, 80°

6. In a triangle ABC, ZB = 50°, ∠C = 62°. Find ∠A.
Solution:
In a ΔABC, ∠B = 50°, ∠C = 62°
By angle sum property of a triangle
∠A + ∠B + ∠C = 180°
∠A + 50° + 62° = 180°
∠A + 112° = 180°
∠A = 180° – 112°
∠A = 68°

7. In a right angled triangle two acute angles are in the ratio 2 : 3. Find the angles.
Solution:
In a right angle triangle one angle = 90°
Let the other two angles be (2x)°, (3x)°
By angle sum property of a triangle.
90° + (2x)° + (3x)° = 180°
90° + (5x)° = 180°
(5x)° = 180° – 90°
(5x)° = 90°
x = \(\frac{90^{\circ}}{5}\)
x = 18
Required angles = (2x)°, (3x)°
= (2 × 18)°, (3 × 18)°
= 36°, 54°

8. Three angles of a triangle are (2x + 20)°, (x + 30)° and (2x – 10)°. Find the angles.
Solution:
Since, we know that the sum of angles of a triangle is always 180°
∴ (2x + 20)° + (x + 30)° + (2x – 10)° = 180°
(5x + 40)° = 180°
(5x)° = 180° – 40°
(5x)° = 140°
x = \(\frac{140^{\circ}}{5}\)
x = 28
Required angles
= (2x + 20)°, (2x + 30)° and (2x – 10)°
= (2 × 28 + 20)°, (28 + 30)° and (2 × 28 -10)°
= (56 + 20)°, (58)° and (56 – 10)°
= 76°, 58° and 46°

9. Multiple choice questions :

Question (i).
A triangle can have two …………….
(a) Acute angles
(b) Obtuse angles
(c) Right angles
(d) None of these.
Answer:
(a) Acute angles

Question (ii).
A triangle is possible with measure of angles
(a) 30°, 40°, 100°
(b) 60°, 60°, 70°
(c) 60°, 50°, 70°
(d) 90°, 89°, 92°
Answer:
(c) 60°, 50°, 70°

Question (iii).
One of the equal angles of an isosceles triangle is 45° then its third angle is
(a) 45°
(b) 60°
(c) 100°
(d) 90°
Answer:
(d) 90°

Question (iv).
The number of obtuse angles that a triangle can have
(a) 2
(b) 1
(c) 3
(d) 4.
Answer:
(b) 1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.1

1. In ΔABC, P is midpoint of BC, then
(i) BP = ……………..
(ii) AP is a …………….. of ΔABC
(iii) ∠ADC = ……………..
(iv) BD = BC (True/False)
(v) AD is an …………….. of ΔABC

Solution:
(i) PC
(ii) Median
(iii) 90°
(iv) False
(v) Altitude

2. (a) Draw AD, BE, CF three medians in a ΔABC.
(b) Draw an equilateral triangle and its medians. Also compare the lengths of the medians.
(c) Draw an isosceles triangle ABC in which AB = BC. Also draw its altitudes.
Solutions:
(a) We are given ΔABC D, E and F are mid points of the sides BC, CA and AB respectively. Join AD, BE and CF.
The AD, BE and CF are the required medians.

(b) Draw an equilateral triangle ABC, D,E and F are the mid points of sides BC, CA and AB respectively. On joining AD, BE and CF, we get the required medians AD, BE and CF. Measure the lengths of AD, BE and CF we observe that the three medians AD, BE and CF are equal in length.
∴ AD = BE = CF.

(c) Draw an isosceles ΔABC in which AB = BC Altitude can be drawn as below :
AD is the altitude from A to D.


3. Find the value of the unknown exterior angles.

Question (i).

Answer:
In the given triangle,
By exterior angle property of a triangle
Exterior angle = sum of interior opposed angles
x = 100° + 40°
∴ x = 140°

Question (ii).

Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 20° + 30°
∴ x = 50°

Question (iii).

Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 60° + 60°
∴ x = 120°

Question (iv).

Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 90° + 30°
∴ x = 120°

4. Find the value of x in the following figures.

Question (i).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
40° + x = 120°
x = 120° – 40°
x = 80°.

Question (ii).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 90° = 135°
x = 135° – 90°
x = 45°

Question (iii).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 80° = 130°
x = 130° – 80°
x = 50°

Question (iv).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 25° = 155°
x = 155° – 25°
x = 130°

5. Find the value of y in following figures.

Question (i).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
y + y = 140°
2y = 140°
y = \(\frac{140^{\circ}}{2}\)
y = 70°

Question (ii).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
y + 90° = 160°
y = 160° – 90°
y = 70°

Question (iii).

Answer:
By exterior angle property of a triangle
exterior angle = Sum of interior opp. angles
5y = y + 80°
5y – y = 80°
4y = 80°
y = \(\frac{80^{\circ}}{4}\)
y = 20°.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 5 Lines and Angles MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 5 Lines and Angles MCQ Questions

Multiple Choice Questions :

Question 1.
If a line intersects three lines, tell the number of intersecting point:
(a) 1
(b) 2
(c) 3
(d) A
Answer:
(c) 3

Question 2.
A line has :
(a) Two ends
(b) One end
(c) No end
(d) None of these
Answer:
(c) No end

Question 3.
The complementary angle of 45° is :
(a) 45°
(b) 135°
(c) 90°
(d) 180°
Answer:
(a) 45°

Question 4.
Supplementary angle of 100° is :
(a) 80°
(b) 100°
(c) 90°
(d) 180°
Answer:
(d) 180°

Fill in the blanks :

Question 1.
It two angles are complementary, then the sum of their measures is
Answer:
90°

Question 2.
If two angles are supplementary, then the sum of their measures is
Answer:
180°

Question 3.
The angle which is equal to its complement is
Answer:
45°

Question 4.
The angle which is equal to its supplement is
Answer:
90°

Question 5.
If two adjacent angles are supplementary they form a
Answer:
linear pair

Write True/False for the following :

Question 1.
Two acute angles can be complementary. (True/False)
Answer:
True

Question 2.
Two obtuse angles can be supplementary. (True/False)
Answer:
False

Question 3.
The complement of a right angle is also a right angle. (True/False)
Answer:
False

Question 4.
Adjacent angle can be complementary. (True/False)
Answer:
False

Question 5.
Complementry angles are always adjacent. (True/False)
Answer:
False

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 5 Lines and Angles Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.2

1. In the figure question identify the pair of angles as corresponding angles alternate interior angles, exterior alternate angles, adjacent angles, vertically opposite angles and co-interior angles, linear pair.
(i) ∠3 and ∠6
(ii) ∠3 and ∠7
(iii) ∠2 and ∠4
(iv) ∠2 and ∠7
(v) ∠1 and ∠8
(vi) ∠4 and ∠6
(vii) ∠1 and ∠5
(viii) ∠1 and ∠4
(ix) ∠5 and ∠7

Answer:
(i) Alternate interior angles.
(ii) Corresponding angles.
(iii) Adjacent angles.
(iv) Alternate exterior angles.
(v) Alternate exterior angles.
(vi) Co-interior angles.
(vii) Corresponding angles.
(viii) Vertically opposite angles.
(ix) Linear pair.

2. In the figure identify :

(i) The pairs of corresponding angle.
(ii) The pairs of alternate interior angles.
(iii) The pairs of interior angles on the same side of the transversal.
(iv) The pairs of vertically opposite angles.
Answer:
(i) ∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7, ∠4 and ∠8.
(ii) ∠1 and ∠7, ∠2 and ∠8.
(iii) ∠1 and ∠8, ∠2 and ∠7.
(iv) ∠1 and ∠3, ∠2 and ∠4, ∠5 and ∠7, ∠6 and ∠8.

3. In the given figures, the intersected lines are parallel to each other. Find the unknown angles.

Question (i).

Answer:
l || m and a is a transversal ∠b = 80°
[Alternate interior angles]
∠a = ∠b
[Vertically opposite angles]
∴ ∠a = 80° [∵ ∠b = 80°]
Also ∠c = 80°
[Vertically opposite ∠5]
Hence a = 80°, b = 80°, c = 80°

Question (ii).

Answer:
∠x° + 70°= 180° (Linear pair)
∠x° = 180° – 70°
∠x° = 110°
∠y° = 70°.
(Vertically opposite angles)
AB || CD and EF is a transversal
∴ ∠z° = 70°
[Alternate interior angles]
Hence x = 110°, y = 70° and z = 70°

Question (iii).

Answer:
110° + a = 180° (Linear pair)
∴ a = 180°- 110° = 70°
b = a
(Corresponding angles)
∴ b = 70°
d = b
(Vertically opposite angles)
∴ d = 70°
b + c = 180° (Linear pair)
70° + c = 180°
∴ c = 180° – 70° = 110°
Hence a = 70°, b = 70°, c = 110°, d = 70°

Question (iv).

Answer:
P + 75° = 180° (Linear pair)
∴ P = 180° – 75° = 105°
R = P
= 105°
(Vertically opposite angles)
Q =75°
(Vertically opposite angles)
AB || CD and EF is a transversal
S = R
(Alternate interior angles)
∴ S = 105°
T = Q
(Alternate interior angles) = 75°
Now U = T
= 75°
(Vertically opposite angles) V = S
(Vertically opposite angles) = 105°
Hence P = 105°, Q = 75°, R = 105°,
S = 105°, T = 75°, U = 75°, V = 105°

4. Find the value of x in the following figures if l || m

Question (i).

Answer:
l || m and n is a transversal
∴ 2x + 3x = 180°
[The pair of co-interior angles are supplementary]
or 5x = 180°
x = \(\frac{180^{\circ}}{5}\) = 36°

Question (ii).

Answer:
a = 5x
(Vertically opposite angles)
Since l || m and n is a transversal
∴ 4x + 5x= 180°
[The pair of co-interior angles are supplementary]

or 9x = 180°
∴ x = \(\frac{180^{\circ}}{9}\) = 20°

Question (iii).

Answer:
a = x
(Vertically opposite angles)
Now l || m and n is a transversal
a + 4x = 180°
[The pair of co-interior angles are supplementary]

∴ a + 4x = 180°
or x + 4x = 180°
or 5x = 180°
Or x = \(\frac{180^{\circ}}{5}\) = 36°

Question (iv).

Answer:
Since l || m and n is a transversal
[The pair of co-interior angles are supplementary]
∴ 5x + 4x = 180°
Or 9x = 180°
x = \(\frac{180^{\circ}}{9}\) = 20°

5. In the given figures arms of two angles are parallel find the following.

Question (a).
(i) ∠DGC
(ii) ∠DEF

Answer:
(i) AB || DE and BC is a transversal
∴ ∠DGC = ∠ABC
(Corresponding angles)
= 65° (∵ ∠ABC = 65°)

(ii) Since BC || EF and DE is the transversal.
∴ ∠DEF = ∠DGC
(Corresponding angles)
= 65° (∵ ∠DGC = 65°)

Question (b).
(i) ∠MNP
(ii) ∠RST

Answer:
(i) Since MN || RS and NP is a transversal
∴ ∠MNP = ∠RQP
(Corresponding angles)
= 70° (∵ ∠RQP = 70°)

(ii) Since NP || ST and RS is a transversal
∴ ∠RST = ∠RQP
(Corresponding angles)
= 70° (∵ ∠RQP = 70°)

6. In the following figure AB || CD and EF || GH, find the measure of ∠x and ∠y.

Solution:
Since AB || CD and EF is a transversal
∴ ∠y = 65°
(Corresponding angles)
Since EF || GH and AB is a transversal.
∴ ∠x = 65°
[alternate interior angles]
Therefore ∠x = 65° and ∠y = 65°

7. PQ ⊥ RS find the value of x in the following figure.

Solution:
Let O be the point of intersection of PQ and RS.
Now PQ and MN intersect each other at O
∴ ∠POM = ∠NOQ
(Vertically opposite angles)
= 3x° (∵ ∠WOQ = 3x°)
Now ∠POS = 90°
∴ ∠POM + ∠MOS = 90°
6x° + 3x° = 90°
9x° = 90°
x = \(\frac{90^{\circ}}{9}\) = 10°

8. In the given figure below, decide whether l is parallel to m.

Question (i).

Answer:
Here 123° + 47° = 170°
But the sum of the pair of co-interior angles is 180°
∴ l is not parallel to m.

Question (ii).

Answer:
Here 127° + 53° = 180°
∴ sum of the pair of co-interior angles is 180°.
Thus l parallel to m.

Question (iii).

Answer:
Since 80° + 80° = 160°
But the sum of the pairs of co-interior angles is 180°
Therefore l is not parallel to m.

Question (iv).

Answer:
115° and 65° are corresponding angles which are not equal.
Therefore l is not parallel to m.

9. Multiple Choice Questions :

Question (i).
A pair of complementary angles is
(a) 130°, 50°
(b) 35°, 55°
(c) 25°, 75°
(d) 27°, 53°
Answer:
(d) 27°, 53°

Question (ii).
A pair of supplementary angles is
(a) 55°, 115°
(b) 65°, 125°
(c) 47°, 133°
(d) 40°, 50°
Answer:
(b) 65°, 125°

Question (iii).
If one angle of a linear pair is acute, then the other angle is
(a) acute
(b) obtuse
(c) right
(d) straight.
Answer:
(b) obtuse

Question (iv).
In the adjoining figure, if l || m, then the value of x is

(a) 50°
(b) 60°
(c) 70°
(d) 45°
Answer:
(a) 50°

Question (v).
In the adjoining figure, if l || m, then

(a) 75°
(b) 95°
(c) 105°
(d) 115°
Answer:
(c) 105°

Question (vi).
In the adjoining figure, the value of x that will make the lines l and m parallel is

(a) 20
(b) 30
(c) 60
(d) 80
Answer:
(a) 20

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 5 Lines and Angles Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1

1. Name each of the following as acute, obtuse, right straight or a reflex angle.

Question (i).

Answer:
Right angle

Question (ii).

Answer:
Obtuse angle

Question (iii).

Answer:
Straight angle

Question (iv).

Answer:
Reflex angle

Question (v).

Answer:
Obtuse angle

Question (vi).

Answer:
Acute angle.

2. Write the complement of each of the following angles :

Question (i).
53°
Answer:
Complement of 53°
= (90° – 53°) = 37°.

Question (ii).
90°
Answer:
Complement of 90°
= (90° – 90°) = 0°.

Question (iii).
85°
Answer:
Complement of 85°
= (90° – 85°) = 5°.

Question (iv).
\(\frac {4}{9}\) of a right angle
Answer:
Complement of \(\frac {4}{9}\) of a right angle
i. e. 40° = (90° – 40°) = 50°

Question (v).
0°
Answer:
Complement of 0° = (90° – 0°)
= 90°.

3. Write the supplement of each of the following angle :

Question (i).
55°
Answer:
Supplement of 55°
= (180° – 55°) = 125°.

Question (ii).
105°
Answer:
Supplement of 105°
= (180° – 105°) = 75°.

Question (iii).
100°
Answer:
Supplement of 100°
= (180° – 100°) = 80°.

Question (iv).
\(\frac {2}{3}\) of a right angle
Answer:
\(\frac {2}{3}\) of a right angle
= \(\frac {2}{3}\) × 90° = 60°.
∴ Supplement of 60°
= (180° – 60°) = 120°.

Question (v).
\(\frac {1}{3}\) of 270°.
Answer:
Supplement of \(\frac {1}{3}\) of 270° i.e. 90°
= (180°- 90°) = 90°.

4. Identify the following pairs of angles as complementary or supplementary.

Question (i).
65° and 115°
Answer:
Since 65° + 115° = 180°.
∴ It is a pair of supplementary angles.

Question (ii).
112° and 68°
Answer:
Since 112° + 68° = 180°
∴ It is a pair of supplementary angles.

Question (iii).
63° and 27°
Answer:
Since 63° + 27° = 90°
∴ It is a pair of complementary angles.

Question (iv).
45° and 45°
Answer:
Since 45° + 45° = 90°
∴ It is a pair of complementary angles.

Question (v).
130° and 50°
Answer:
Since 130° + 50° = 180°.
∴ It is a pair of supplementary angles.

5. Two complementary angles are in the ratio of 4 : 5, find the angles.
Solution:
Ratio of angles = 4 : 5
Let two complementary angles are 4x and 5x
Their sum = 90°
∴ 4x + 5x = 90°
9x = 90°
x = \(\frac{90^{\circ}}{9}\) = 10°
∴ 1st angle = 4x = 4 × 10° = 40°
2nd angle = 5x = 4 × 10° = 50°

6. Two supplementary angles are in the ratio of 5 : 13, find the angles.
Solution:
Ratio of two supplementary angles = 5 : 13
Let 5x and 13x are two supplementary angles
Since their sum = 180°
∴ 5x + 13x = 180°
18x = 180°
x = \(\frac{180^{\circ}}{18}\) = 10°
∴ 1st angle = 5x = 5 × 10° = 50°.
2nd angle = 13x = 13 × 10° = 130°

7. Find the angle which is equal to its complement.
Solution:
Let the angle be = x
Therefore its complement = 90° – x
Since the angle is equal to its complement
∴ x = 90° – x
or x + x = 90°
or 2x = 90°
or x = \(\frac{90^{\circ}}{2}\)
or x = 45°
Therefore the required angle is 45°.

8. Find the angle which is equal to its supplement.
Solution:
Let the angle be x
Therefore its supplement = 180° – x
Since the angle is equal to its supplement
∴ x = 180° – x
or x + x = 180°
or 2x = 180°
or x = \(\frac{180^{\circ}}{2}\) = 90°
Therefore, the required angle is 90°.

9. In the given figure, AOB is straight line. Find the measure of ∠AOC.

Solution:
In the given figure AOB is straight line (see Fig.)
∴ ∠AOB = 180°
∴ ∠AOC + ∠BOC = 180°
or ∠AOC + 50° = 180°
[∵ ∠BOC = 50° (given)]
∴∠AOC = 180° – 50°
= 130°

10. In the given figure, MON is straight line find.
(i) ∠MOP
(ii) ∠NOP

Solution:
Since MON is straight line (see Fig.)
∴ ∠MON = 180°
∴ ∠MOP + ∠NOP = 180°
[∵ ∠MOP = x + 20°
∠NOP = x + 40°]
or 2x + 60° = 180°
or 2x = 180° – 60°
or 2x = 120°
or x = \(\frac{120^{\circ}}{2}\) = 60°.
(i) ∠MOP = x + 20°
= 60° + 20°
= 80°
(ii) ∠NOP = x + 40°
= 60° + 40°
= 100°

11. Find the value of x, y and z in each of following.

Question (i).

Solution:
In fig (i)
x = 100°
(Vertically opposite angles)
y = 80°
(Vertically opposite angles)

Question (ii).

Solution:
In fig (ii)
z = 60°
(Vertically opposite angles)
∠y + 60° = 180° (Linear pair)
or ∠y = 180° – 60°
or ∠y = 120°
x = y
(Vertically opposite angles)
= 120°

12. Find the value of x, y, z and p in each of following.

Question (i).

Solution:
In fig (i)
45° + x + 35°= 180° (Linear pair)
or x + 80° = 180°
or x = 180° – 80°
or x = 100°
y = 45° (Vertically opposite angles)
Also 45° + z = 180° (Linear pair)
z = 180° – 45°
= 135°.
Hence x = 100°,
y = 45°,
z = 135°

Question (ii).

Solution:
In fig (ii)
p + 65° + 55° = 180° (Linear pair)
p + 120°= 180°
∴ p = 180° -120°
i. e. p = 60°
x = 55°
(Vertically opposite angles)
y = 65°
(Vertically opposite angles)
z = p
= 60°
(Vertically opposite angles)
Hence x = 55°,
y = 65°,
z = 60°,
p = 60°

13. Multiple Choice Questions :

Question (i).
If two angles are complementary then the sum of their measure is …………..
(a) 180°
(b) 90°
(c) 360°
(d) None of these.
Answer:
(b) 90°

Question (ii).
Two angles are called ………….. if the sum of their measures is 180°.
(a) supplementary
(b) complementary
(c) right
(d) none of these.
Answer:
(a) supplementary

Question (iii).
If two adjacent angles are supplementary then, they form a …………..
(a) right angle
(b) vertically opposite angles
(c) linear pair
(d) corresponding angles.
Answer:
(c) linear pair

Question (iv).
If two lines intersect at a point, the vertically opposite angles are always …………..
(a) equal
(b) zero
(c) 90°
(d) none of these.
Answer:
(a) equal

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 4 Simple Equations MCQ Questions

Multiple Choice Questions :

Question 1.
Choose simple equation out of the following:
(a) 3x + 11
(b) 2x + 5 < 11
(c) x – 5 = 7x + 6
(d) \(\frac{5 x+6}{6}\)
Answer:
(c) x – 5 = 7x + 6

Question 2.
A quantity which takes a fixed numerical value is called :
(a) Constant
(b) Variable
(c) Equation
(d) Expression
Answer:
(a) Constant

Question 3.
In equation 5x = 25 the value of x is :
(a) 0
(b) 5
(c) -5
(d) 1
Answer:
(b) 5

Question 4.
In equation \(\frac{m}{3}\) = 2 the value of m is :
(a) 1
(b) 0
(c) 6
(d) -6
Answer:
(c) 6

Question 5.
In equation 7x + 5 = 19 the value of n is :
(a) 0
(b) -2
(c) 1
(d) 2
Answer:
(d) 2

Question 6.
In equation 4p – 3 = 13, the value of p is :
(a) 1
(b) 4
(c) 0
(d) -4
Answer:
(b) 4

Question 7.
The equation of the statement, the sum of number x and 4 is 9 is :
(a) x + 4 = 9
(b) x – 4 = 9
(c) x = 4 + 9
(d) x – 9 = 4.
Answer:
(a) x + 4 = 9

Question 8.
The equation of the statement, ‘seven times m plus 7 = gives 77’ is.
(a) 1m × 7 = 77
(b) 7m + 7 = 77
(c) 7m = 77 + 7
(d) m + 7 × 7 = 77
Answer:
(a) 1m × 7 = 77

Fill in the blanks :

Question 1.
A quantity which takes a fixed numerical value is called …………….
Answer:
Constant

Question 2.
The equation for the statement seven time a number is 42 is …………….
Answer:
7x = 42

Question 3.
If x + 4 = 15, then the value of x is …………….
Answer:
x = 11

Question 4.
If 2y – 6 = 4, then y is equal to …………….
Answer:
y = 5

Question 5.
If 8x – 4 = 28, then x is equal to …………….
Answer:
x = 4

Write True or False :

Question 1.
An equation of one variable is called linear equation. (True/False)
Answer:
True

Question 2.
If x – 3 = 1, then value of x is 2. (True/False)
Answer:
False

Question 3.
If 7m + 7 = 77, then value of m is 10. (True/False)
Answer:
True

Question 4.
If 3 subtracted from twice a number is 5, then the number is 4. (True/False)
Answer:
True

Question 5.
If one fourth of a number is 10 then the number is 40. (True/False)
Answer:
False

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Question 1.
If 7 is added to five times a number, the result is 57. Find the number.
Solution:
Let the required number = x
Five times the number = 5x
7 added to five times the number = 5x + 7
According to the problem
5x + 7 = 57
5x = 57 – 7
5x = 50
x = \(\frac {50}{5}\)
So, x = 10
Hence the required number is 10.

Question 2.
9 decreased from four times a number yields 43. Find the number.
Solution:
Let the required number = x
Four times the number = 4x
9 decreased from four times the number = 4x – 9
According to the problem
4x – 9 = 43
4x = 43 + 9
4x = 52
x = \(\frac {52}{4}\)
x = 13
Hence, the required number is 13.

Question 3.
If one-fifth of a number minus 4 gives 3, find the number.
Solution:
Let the required number = x
One fifth of the number = \(\frac {1}{5}\)x
One fifth of the number minus 4 = \(\frac {1}{5}\)x – 4
According to problem
\(\frac {1}{5}\)x – 4 = 3
\(\frac {1}{5}\)x = 3 + 4
\(\frac {1}{5}\)x = 7
x = 35
Hence the required number is 35.

Question 4.
In a class of 35 students, the number of girls is two-fifth the number of boys. Find the number of girls in the class.
Solution:
Let the number of boys = x
∴ number of girls = \(\frac {2}{5}\)x
Total number of students = 35
x + \(\frac {2}{5}\)x =35
\(\frac{5 x+2 x}{5}\) = 35
7x = 5 × 35
x = \(\frac{5 \times 35}{7}\)
x = 25
Therefore number of boys = 25
Number of girls = 35 – 25 = 10.

Question 5.
Sham’s father’s age is 5 years more than three times Sham’s age. Find Sham’s age, if his father is 44 years old.
Solution:
Let Sham’s age = x years
Then Sham’s father age = 3x + 5
But Sham’s fathers age = 44
According to question
3x + 5 = 44
3x = 44 – 5
3x = 39
Dividing both sides by 3
\(\frac{3 x}{3}=\frac{39}{3}\)
or x = 13
Hence Sham’s age is 13 years.

Question 6.
In an isosceles triangle the base angles are equal, the vertex angle is 40°. What are the base angles of the triangle ? (Remember, the sum of three angles of a triangle is 180°)
Solution:
Let each base angle of an isosceles triangle = x (in degrees)
Vertex angle = 40°
The sum of angles of a triangle = 180°
∴ x + x + 40° = 180°
2x = 180° – 40°
2x = 140°
Divide both sides by 2
\(\frac{2 x}{2}-\frac{140^{\circ}}{2}\)
Or x = 70°
Each equal angle is of 70°

Question 7.
Irfan says that he has 7 marbles more than five times the marbles Pannit has. Irfan has 37 marbles. How many marbles does Pannit have ?
Solution:
Let marbles Parmit has = x
Marbles Irfan has = 5x + 7
But Marbles Irfan has = 37
∴ 5x + 7 = 37
5x = 37 – 7
5x = 30
x = \(\frac {30}{5}\) = 6
Therefore Parmit has 6 marbles.

Question 8.
The length of a rectangle is 3 units more than its breadth and the perimeter is 22 units. Find the breadth and length of a rectangle.
Solution:
Let breadth of rectangle (l)
= x units
∵ length of rectangle (b) = (x + 3) units
∴ Perimeter of rectangle = 2(l + b)
= 2 (x + x + 3) units
= 2(2x + 3) units
According to the question
Perimeter = 22 units
2 (2x + 3) =22
\(\frac{2(2 x+3)}{2}=\frac{22}{2}\)
2x + 3 = 11
2x = 11 – 3
or 2x = 8
Dividing both sides by 2 we get
\(\frac{2 x}{2}=\frac{8}{2}\)
x = 4
∴ breadth = 4 units
Length = (4 + 3) units
= 7 units

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.3 Textbook Exercise Questions and Answers.

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

1. Solve each of the following equation.

Question (i).
6x + 10 = – 2
Answer:
Given equation is 6x + 10 = – 2
Transposing + 10 from L.H.S to R.H.S
we get
6x = -2 – 10
or 6x = -12
Dividing both sides by 6, we get
\(\frac{6 x}{6}=\frac{-12}{6}\)
or x = – 2, which is the required solution.

To check Put x = – 2 in the LHS of the equation 6x + 10 = – 2
L.H.S. = 6x + 10
= 6 × -2 + 10
= -12 + 10
= – 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
2y – 3 = 2
Answer:
Given equation is 2y – 3 = 2
Transposing – 3 from L.H.S. to R.H.S,
we get
2y = 2 + 3
or 2y = 5
Dividing both sides by 2, we get:
\(\frac{2 y}{2}=\frac{5}{2}\)
or y = \(\frac {5}{2}\), which is the required solution

To check. Put y = \(\frac {5}{2}\) in the L.H.S of the equation 2y – 3 = 2
L.H.S = 2y – 3 = 2 × \(\frac {5}{2}\) – 3
= 5 – 3 = 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iii).
\(\frac{a}{5}\) + 3 = 2
Answer:
Given equation is \(\frac{a}{5}\) + 3 = 2
Transposing + 3 from L.H.S to R.H.S., we get
\(\frac{a}{5}\) = 2 – 3
or \(\frac{a}{5}\) = -1
Multiplying both sides, by 5, we get
5 × \(\frac{a}{5}\) = 5 × – 1
or a = – 5, which is the required solution.

To Check: Put a = – 5 in the L.H.S of the equation
\(\frac{a}{5}\) + 3 = 2,
L.H.S. = \(\frac{a}{5}\) + 3
= \(\frac {-5}{5}\) + 3
= – 1 + 3
= 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
\(\frac{3 x}{2}=\frac{2}{3}\)
Answer:
Given equation is \(\frac{3 x}{2}=\frac{2}{3}\)
Multiplying both sides by 2, we get
2 × \(\frac{3 x}{2}\) = 2 × \(\frac {2}{3}\)
or 3x = \(\frac {4}{3}\)
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{4}{3} \times \frac{1}{3}\)
or x = \(\frac {4}{9}\), which is the required solution.

To Check. Put x = \(\frac {4}{9}\) in the L.H.S. of equation \(\frac{3 x}{2}=\frac{2}{3}\)
L.H.S. = \(\frac{3 x}{2}=\frac{3}{2} \times \frac{4}{9}\) = \(\frac {2}{3}\) = R.H.S.
∴L.H.S. = R.H.S.

Question (v).
\(\frac {5}{2}\)x = -5
Answer:
Given equation is \(\frac {5}{2}\) x = – 5
Multiplying both sides by 2, we get
2 × \(\frac {5}{2}\) x = 2 × – 5
or 5x = – 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{-10}{5}\)
or x = – 2, which is the required solution.

To Check. Put x = – 2 in L.H.S. of the equation \(\frac {5}{2}\)x = – 5
L.H.S. = \(\frac {5}{2}\)x = \(\frac {5}{2}\) × -2
= – 5 = R.H.S.
∴ L.H.S. = R.H.S.

Question (vi).
2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
Answer:
Given equation is 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
Subtract \(\frac {5}{2}\) from both sides, we get
2x + \(\frac {5}{2}\) – \(\frac {5}{2}\)
= \(\frac {37}{2}\) – \(\frac {5}{2}\)
or 2x = \(\frac{37-5}{2}\)
or 2x = \(\frac {32}{2}\)
or 2x = 16
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{16}{2}\)
or x = 8, which is the required solution.

To Check. Put x = 8 in the L.H.S. of the equation 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
L.H.S. = 2x + \(\frac {5}{2}\)
= 2 × 8 + \(\frac {5}{2}\)
= 16 + \(\frac {5}{2}\)
= \(\frac{32+5}{2}\)
= \(\frac {37}{2}\) = R.H.S.
∴ L.H.S. = R.H.S.

2. Solve the following equation

Question (i).
5 (x + 1) = 25
Answer:
Given equation is 5 (x + 1) = 25
Dividing both sides by 5 we get
\(\frac{5(x+1)}{5}=\frac{25}{5}\)
or x + 1 = 5
Transposing 1 from L.H.S. to R.H.S. we get
x = 5 – 1
or x = 4, which is the required solution.

To Check. Put x = 4 in the L.H.S. of the equation 5 (x + 1) = 25
L.H.S. = 5 (x + 1)
= 5 (4 + 1)
= 5 (5)
= 25 = R.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
2 (3x – 1) = 10
Answer:
Given equation is 2 (3x – 1) = 10
Dividing both sides by 2, we get
\(\frac{2(3 x-1)}{2}=\frac{10}{2}\)
or 3x – 1 = 5
Transposing – 1 from L.H.S. to R.H.S we get
3x = 5 + 1
3x = 6
Dividing both sides by 3, we get \(\frac{3 x}{3}=\frac{6}{3}\)
or x = 2, which is the required solution.

To Check. Put x = 2, in the L.H.S. of the equation 2 (3x – 1) = 10
L.H.S. = 2 (3x – 1) = 10
L.H.S = 2 (3x – 1) = 2 (3 × 2 – 1)
= 2 (6 – 1)
= 2 × 5
= 10 = R.H.S.
∴L.H.S. = R.H.S.

Question (iii).
4 (2 – x) = 8
Answer:
Given equation is 4 (2 – x) = 8
Dividing both sides by 4, we get
\(\frac{4(2-x)}{4}=\frac{8}{4}\)
or 2 – x= 2
Transposing 2 from L.H.S. to R.H.S. we get
-x = 2 – 2
or – x = 0
Multiplying both sides by – 1, we get
-x × – 1 = x – 1
or x = 0, which is the required solution.

To Check. Put x = 0 in the L.H.S. of the equation 4 (2 – x) = 8
L.H.S. = 4 (2 – x) = 4 (2 – 0)
= 4 × 2
= 8 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
– 4 (2 + x) = 8.
Answer:
Given equation is – 4 (2 + x) = 8
Dividing both sides by – 4, we get
\(\frac{-4(2+x)}{-4}=\frac{8}{-4}\)
Transposing 2 from L.H.S. to R.H.S. we get :
x = – 2 – 2
or x = – 4, which is the required solution

To Check. Put x = – 4 in the L.H.S. of equation – 4 (2 + x) = 8
L.H.S. = – 4 (2 + x) = – 4 [2 + (- 4)]
= – 4 (2 – 4)
= – 4 (- 2)
= 8 = R.H.S.
∴ L.H.S. = R.H.S.

3. Solve the following equations :

Question (i).
4 = 5 (x – 2)
Answer:
Given equation is 4 = 5 (x – 2)
or 4 = 5x – 10
Transposing 5x to L.H.S. and 4 to R.H.S.,
we get
– 5x = – 4 – 10
or – 5x = – 14
Dividing both sides by – 5, we get
\(\frac{-5 x}{-5}=\frac{-14}{-5}\)
or, x = \(\frac {14}{5}\), which is the required solution.

To Check. Put x = \(\frac {14}{5}\) in the R.H.S. of the equation 4 = 5 (x – 2)
R.H.S. = 5 (x – 2) = 5\(\left(\frac{14}{5}-2\right)\)
= 5\(\left(\frac{14-10}{5}\right)\)
= 5 \(\left(\frac{4}{5}\right)\)
= 4 = L.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
– 4 = 5 (x – 2)
Answer:
Given equation is – 4 = 5 (x – 2)
or – 4 = 5x – 10
Transposing -4 to R.H.S and 5x to L.H.S
we get
-5x = 4 – 10 or -5x = -6
Dividing both sides by – 5 we get
\(\frac{-5 x}{-5}=\frac{-6}{-5}\)
or x = \(\frac {6}{5}\), which is the required solution.

To Check. Put x = \(\frac {6}{5}\) in the R.H.S. of the equation – 4 = 5 (x – 2)
L.H.S. = 5 (x – 2)
= 5\(\left(\frac{6}{5}-2\right)\)
= 5\(\left(\frac{6-10}{5}\right)\)
= 5\(\left(\frac{-4}{5}\right)\)
= -4 = L.H.S.
L.H.S. = R.H.S.

Question (iii).
4 + 5 (p – 1) = 34
Answer:
Given equation is 4 + 5(p – 1) = 34
Transposing 4 to R.H.S. we get
5(p – 1) = 34 – 4
5(p – 1) = 30
Dividing both sides, by 5, we get
\(\frac{5(p-1)}{5}=\frac{30}{5}\)
p – 1=6
Transposing -1 to R.H.S. we get
p = 6 + 1
p = 7 which is the required solution.

To Check : Put p = 7 in L.H.S. of the equation 4 + 5 (p – 1) = 34
L.H.S. = 4 + 5 (p – 1)
= 4 + 5 (7 – 1)
= 4 + 5 (6)
= 4 + 30
= 34 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
6y – 1 = 2y + 1.
Answer:
Given equation is 6y – 1 = 2y + 1
Transposing – 1 to R.H.S. and 2y to L.H.S,
we get
6y – 2y = 1 + 1
or 4y = 2 or y = \(\frac {2}{4}\)
or y = \(\frac {1}{2}\), which is the required solution.

To Check Put y = \(\frac {1}{2}\) in both L.H.S. and R.H.S. of the equation
6y – 1 = 2y + 1
L.H.S. = 6y – 1 = 6 × \(\frac {1}{2}\) – 1 = 3 – 1 = 2
R.H.S. = 2y + 1 = 2 × \(\frac {1}{2}\) + 1 = 1 + 1 = 2.
∴ L.H.S. = R.H.S.

4.

Question (i).
Construct 3 equations starting with x = 2
Answer:
First Equation.
(i) Start with x = 2
Multiplying both sides by 10
10x = 20
Adding 2 to both sides
10x + 2 = 20 + 2
or 10x + 2 = 22
This has resulted in an equation.

Second Equation. Start with x = 2
Divide both sides by 5
∴ \(\frac{x}{5}=\frac{2}{5}\)
This has resulted in an equation.

Third Equation. Start with x = 2
Multiply both sides by 5, we get
5x = 5 × 2
or 5x = 10
Subtracting 3 from both sides, we get
5x – 4 = 10 – 3
or 5x – 3 = 7
This has resulted in an equation.

Question (ii).
Construct 3 equation starting with x = – 2
Answer:
First Equation. Start with x = – 2
Multiplying both sides with 3, we get
3x = – 6
This has resulted in an equation

Second Equation. Start with x = – 2
Multiplying both sides with 3, we get 3x = -6
Adding 7 to both sides, we get 3x + 7
= -6 + 7 or 3x + 7 = 1
This has resulted in an equation.

Third Equation. Start with x = – 2
Multiplying both side with 2 we get 3x = – 6
Adding 10 to both sides we get
3x+ 10 = -6 + 10
or 3x + 10 = 4
This has resulted in an equation.

Multiple Choice Questions :

5. If 7x + 4 = 39, then x is equal to :
(a) 6
(b) -4
(c) 5
(d) 8
Answer:
(c) 5

6. If 8m – 8 = 56 then m is equal to :
(a) -4
(b) -2
(c) -14
(d) 8
Answer:
(d) 8

7. Which of the following number satisfies the equation – 6 + x = -18 ?
(a) 10
(b) – 13
(c) – 12
(d) – 16.
Answer:
(a) 10

8. If \(\frac{x}{2}\) = 14, then the value of 2x + 6 is equal to :
(a) 62
(b) -64
(c) 16
(d) -62.
Answer:
(a) 62

9. If 3 subtracted from twice a number is 5, then the number is :
(a) -4
(b) -2
(c) 2
(d) 4
Answer:
(d) 4

10. If 5 added to thrice an integer is – 7, then the integer is :
(a) – 6
(b) – 5
(c) -4
(d) 4
Answer:
(c) -4

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

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

1. Write the first step that you will use to separate the variable and then solve the equation.

Question (i).
x + 1 = 0
Answer:
Given equation x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = -1
or x = – 1

Question (ii).
x – 1 = 5
Answer:
Given equation is x – 1 = 5
Adding 1 to both sides we get
x – 1 + 1 = 5 + 1
or x = 6
Thus x = 6 is the solution of the given equation

Question (iii).
x + 6 = 2
Answer:
Given equation is x + 6 = 2
Subtracting 6 from both sides, we get:
x + 6 – 6 = 2 – 6
or x = – 4
Thus, x = – 4 is the solution of the given equation.

Question (iv).
y + 4 = 4
Answer:
Given equation is y + 4 = 4
Subtracting 4 from both sides we get
y + 4 – 4 = 4 – 4
or y = 0
Thus, y = 0 is the solution of the given equation.

Question (v).
y – 3 = 3
Answer:
Given equation is y – 3 = 3
Adding 3 to both sides we get
y – 3 + 3 = 3 + 3
or y = 6
Thus, y = 6 is the solution of the given equation.

2. Write the first step that you will use to separate the variable and then sotye the equation :

Question (i).
3x = 15
Answer:
Given equation is 3x = 15
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{15}{3}\)
or x = 5

Question (ii).
\(\frac{P}{7}\) = 4
Answer:
Given equation is \(\frac{P}{7}\) = 4
Multiplying both sides by 7, we get
7 × \(\frac{P}{7}\) = 7 × 4
or p = 28
Thus, p = 28 is the solution of the given equation.

Question (iii).
8y = 36
Answer:
Given equation is 8y = 36
Dividing both sides by 8, we get
\(\frac{8 y}{8}=\frac{36}{8}\)
or y = \(\frac {9}{2}\)

Question (iv).
20x = – 10
Answer:
Given equation is
20x = – 10
Dividing both sides by 20
\(\frac{20 x}{20}=\frac{-10}{20}\)
or x = \(\frac {-1}{2}\)

3. Give the steps you will use to separate the variable and then solve the equation.

Question (i).
5x + 7 = 17
Answer:
Given equation is 5x + 7 = 17
Subtracting 7 from both sides, we get
5x + 7 – 7 = 17 – 7
or 5x = 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{10}{5}\)
or x = 2

Question (ii).
\(\frac{20 x}{3}\) = 40
Answer:
Given equation is \(\frac{20 x}{3}\) = 40
Multiplying both sides by 3, we get
3 × \(\frac{20 x}{3}\) = 3 × 40
or 20x = 3 × 40
Dividing both sides by 20, we get
\(\frac{20 x}{20}\) = \(\frac{3 \times 40}{20}\)
or x = 6

Question (iii).
3p – 2 = 46
Answer:
Given equation is 3p – 2 = 46
Adding 2 to both sides, we get
3p – 2 + 2 = 46 + 2
or 3 p = 48
Dividing both sides by 3, we get:
\(\frac{3 p}{3}=\frac{48}{3}\)
or p = 16

4. Solve the following equations :

Question (i).
10x + 10 = 100
Answer:
Given equation is 10x + 10 = 100
Subtracting 10 from both sides, we get
10x + 10 – 10 = 100 – 10
or 10x = 90
Dividing both sides by 10, we get
\(\frac{10 x}{10}=\frac{90}{10}\)
or x = 9
Thus x = 9 is the solution of the given equation.

Question (ii).
\(\frac{-p}{3}\) = 5
Answer:
Given equation is \(\frac{-p}{3}\) = 5
Multiplying both sides by – 3, we get
– 3 × \(\frac{-p}{3}\) = -3 × 5
or p = -15
Thus p = – 15 is the solution of the given equation.

Question (iii).
3x + 12 = 0
Answer:
Given equation is 3x + 12 = 0
Subtracting 12 from both sides, we get
3x + 12 – 12 = – 12
or 3x = – 12
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{-12}{3}\)
or x = -4
Thus x = – 4 is the solution of the given equation.

Question (iv).
2q – 6 = 0
Answer:
The given equation is 2q – 6 = 0
Adding 6 to both sides, we get
2q – 6 + 6 = 0 + 6
or 2q = 6
Dividing both sides by 2, we get
\(\frac{2 q}{2}=\frac{6}{2}\)
or q = 3
Thus, q = 3 is the solution of the given equation.

Question (v).
3p = 0
Answer:
The given equation is 3p = 0
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{0}{3}\)
or p = 0
Thus, p = 0 is the solution of the given equation.

Question (vi).
3s = -9
Answer:
The given equation is
3s = -9
Dividing both sides by 3, we get
\(\frac{3 s}{3}=-\frac{9}{3}\)
or s = – 3
Thus, s = – 3 is the solution of the given equation.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

1. Complete the following :

Solution:
(i) No
Reason : For x = 5
L.H.S. = x + 5 = 5 + 5 = 10
RHS = 0
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = 5

(ii) Yes
Reason : For x = -5
L.H.S. = x + 5
= -5 + 5 = 0
R.H.S. = 0
Since L.H.S. = R.H.S.
Therefore, given equation is satisfied for
x = – 5

(iii) NO
Reason : For x = 3
L.H.S. = x – 3
= 3 – 3 = 0
R.H.S. = 1
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = 3

(iv) No
Reason : x = – 3
L.H.S. = x – 3
= – 3 – 3 = -6
R.H.S. = 1
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = – 3

(v) Yes
Reason : For x = 5
L.H.S. = 2x
= 2 × 5 = 10
R.H.S. = 10
Since L.H.S. = R.H.S.
Therefore given equation is satisfied for
x = 5

(vi) No
Reason : For x = – 6
L.H.S. = \(\frac{x}{3}\)
= \(\frac {-6}{3}\)
= -2
R.H.S. = 2
Since L.H.S. ≠ R.H.S.
Therefore given equation is not satisfied for x = – 6

(vii) No,
Reason : For x = 0
L.H.S. = \(\frac{x}{2}\)
= \(\frac {0}{2}\) = 0
R.H.S. = 2
Since L.H.S. ≠ R.H.S.
Therefore given equation is not satisfied for x = 0

2. Check whether the value given in the brackets is a solution to the given equation or not.

Question (i).
x + 4 = 11 (x = 7)
Answer:
Yes
Check : For x = 1
L.H.S. = x + 4
= 7 + 4 = 11
R.H.S. = 11
Since L.H.S. = R.H.S.
Therefore x = 7 is the solution to the given equation.

Question (ii).
8x + 4 = 28 (x = 4)
Answer:
No
Check: Forx-4
L.H.S. = 8x + 4
= 8 × 4 + 4
= 32 + 4
= 36
R.H.S. = 28
Since L.H.S. ≠ R.H.S.
Therefore x = 4 is not solution to the given equation.

Question (iii).
3m – 3 = 0 (m = 1)
Answer:
Yes
Check : For m = 1
L.H.S. = 3m – 3
= 3 × 1 – 3 = 3 – 3 = 0
R.H.S.= 0
Since L.H.S. = R.H.S.
Therefore, m = 1 is the solution to the given equation.

Question (iv).
\(\frac{x}{5}\) – 4 = -1 (x = 15)
Answer:
Yes
Check : For x = 15
L.H.S. = \(\frac{x}{5}\) – 4
= \(\frac {15}{5}\) – 4 = 3 – 4 = – 1
R.H.S. = – 1
Since L.H.S. = R.H.S.
Therefore, x = 15 is the solution to the given equation.

Question (v).
4x – 3 = 13 (x = 0)
Answer:
No
Check : For x = 0
L.H.S. = 4x – 3
= 4 × 0 – 3
= 0 – 3
= -3
R.H.S. = 13
Since L.H.S. ≠ R.H.S.
Therefore x = 0 is not the solution to the given equation.

3. Solve the following equations by trial and error method

Question (i).
5x + 2 = 17
Answer:

Value of x	L.H.S	R.H.S
1	5 × 1 + 2 = 5 + 2 = 7	17
2	5 × 2 + 2 = 10 + 2 = 12	17
3	5 × 3 + 2 = 15 + 2 = 17	17

We observe that for x = 3, L.H.S. = R.H.S.
Hence x = 3 is the solution of the given equation.

Question (ii).
3p – 14 = 4
Answer:

Value of p	L.H.S.	R.H.S.
1	3 × 1 – 14 = 3 – 14 = -11	4
2	3 × 2 – 14 = 6 – 14 = -8	4
3	3 × 3- 14 = 9 – 14 = -5	4
4	3 × 4 – 14 = 12 – 14 = -2	4
5	3 × 5 – 14 = 15 – 14 = 1	4
6	3 × 6 – 14 = 18 – 14 = 4	4

We observe that for p = 6, L.H.S. = R.H.S.
Hence p = 6 is the solution of the given equation

4. Write equations for the following statements.

Question (i).
The sum of numbers x and 4 is 9
Answer:
x + 4 = 9

Question (ii).
3 subtracted from y gives 9
Answer:
y – 3 = 9

Question (iii).
Ten times x is 50
Answer:
10x = 50

Question (iv).
Nine times x plus 6 is 87
Answer:
9x + 6 = 87

Question (v).
One fifth of a number y minus 6 gives 3.
Answer:
\(\frac {1}{5}\)x – 6 = 3

5. Write the following equations in statement form :

Question (i).
x – 2 = 6
Answer:
2 substracted from x is 6

Question (ii).
3y – 2 = 10
Answer:
2 subtracted from 3 times a number y is 10.

Question (iii).
\(\frac{x}{6}\) = 6
Answer:
One sixth of a number x is 6

Question (iv).
7x – 15 = 34
Answer:
15 subtracted 7 times a number x is 34.

Question (v).
\(\frac{x}{2}\) + 2 = 8
Answer:
add 2 to half of a number x to get 8.

6. Write an equation for the following statements :

Question (i).
Raju’s father’s age is 4 years more than five times Raju’s age. Raju’s father is 54 years old.
Answer:
Let x years be Raju’s age
Five times Raju’s age is 5x years
His father’s age will be 4 years more than five times
Raju’s age = 5x + 4
But 4 years more han five times Raju’s age = Raju’s father’s age
Therefore 5x + 4 = 54

Question (ii).
A teacher tells that the highest marks obtained by a student in his class is twice the lowest marks plus 6. The highest score is 86. (Take the lowest score to be x).
Answer:
Let the lowest score to be x
Twice the lowest score plus 6 = 2x + 6
The highest score obtained by a student is twice the lowest score plus 6 = 2x + 6
But highest score = 86
Hence 2x + 6 = 86

Question (iii).
In an isosceles triangle, the vertex angle is twice either base angle (Let the base angle be x in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Answer:
Let the base angle be x (in degrees)
Therefore vertex angle is twice the base angle = 2x (in degrees)
Sum of three angles of a triangle = 180°
∴ x + x + 2x = 180° = 4x = 180°

Question (iv).
A shopkeeper sells mangoes in two types of boxes. One small and one large. The large box contains as many as 8 small boxes plus 4 loose mangoes. The number of mangoes in a large box is given to be 100.
Answer:
Let the mangoes in small box be x.
Large box contains mangoes
= 8 small box + 4
= 8x + 4
But mangoes in large box = 100
∴ 8x + 4 = 100