Class 7

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 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 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

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 3 Data Handling MCQ Questions

Multiple Choice Questions :

Question 1.
Mean of first five natural numbers is :
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 2.
The mode of the data :
3, 5, 1, 2, 2, 3, 2, 0 is :
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
(b) 2

Question 3.
The probability of possible event is :
(a) 0
(b) 1
(c) 2
(d) -1
Answer:
(b) 1

Question 4.
The probability of impossible event is :
(a) 1
(b) -1
(c) 0
(d) None of these
Answer:
(c) 0

Question 5.
The probability of selecting letter S from the word ‘STUDENT’ is :
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {1}{4}\)
(d) \(\frac {1}{7}\)
Answer:
(d) \(\frac {1}{7}\)

Fill in the blanks :

Question 1.
The mean of first five prime numbers is ……………..
Answer:
5.6

Question 2.
The median of the data: 3, 1, 5, 6, 3, 4, 5 is ……………..
Answer:
5

Question 3.
Mode of the data : 1, 0, 1, 2, 3, 1, 2, is ……………..
Answer:
1

Question 4.
The probability of getting a head or tail is ……………..
Answer:
\(\frac {1}{2}\)

Question 5.
When a die is thrown the probability of getting a number 5 is ……………..
Answer:
\(\frac {1}{6}\)

Write True or False :

Question 1.
Mean of the first five whole numbers is 2. (True/False)
Answer:
True

Question 2.
Mode of data : 1, 1, 2,4, 3, 2, 1 is 2. (True/False)
Answer:
False

Question 3.
Median of data : 1, 2, 3, 4, 5 is 3. (True/False)
Answer:
True

Question 4.
An outcome is the result of an experiment. (True/False)
Answer:
True

Question 5.
Events that have many probabilities can have probability between 0 and 1. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4

1. State whether the following is certain to happen, impossible to happen, may happen.

Question (i).
Two hundred people sit in a Maruti car.
Answer:
Impossible to happen.

Question (ii).
You are older than yesterday.
Answer:
Certain to happen.

Question (iii).
A tossed coin will land heads up.
Answer:
Can happen but not certain.

Question (iv).
A die when rolled shall land up 8 on top.
Answer:
Impossible to happen.

Question (v).
Tommorrow will be a cloudy day.
Answer:
Can happen but not certain.

Question (vi).
India will win the next test series.
Answer:
Can happen but not certain.

Question (vii).
The next traffic light seen will be green.
Answer:
Can happen but not certain.

2. There are 6 marbles in a box with numbers 1 to 6 marked on them.
(i) What is the probability of drawing a marble with number 5 ?
(ii) What is the probability of drawing a marble with number 2 ?
Solution:
Sample space : {1, 2, 3, 4, 5, 6}
There are 6 equally likely possible outcomes.
Outcomes : 1, 2, 3, 4, 5 or 6.
(i) Probability of drawing a marble with number 5 = \(\frac {1}{6}\)
(ii) Probability of drawing a marble with number 2 = \(\frac {1}{6}\)

3. There are two teams A and B. A coin is flipped to decide which team starts the game. What is the probability that team A will start ?
Solution:
When a coin is tossed once. The number of possible outcomes is head or tail
Sample space : [H, T]
The probability of getting head or tail is equal and is \(\frac {1}{2}\) for each.
∴ The probability that team A will start = \(\frac {1}{2}\)

4. A bag contains 3 red and 7 green balls. One ball is drawn at random from the bag. Find the probability of getting (i) a red ball (ii) a green ball.
Solution:
Total number of balls in the bag = 3 + 7 = 10
(i) The event is getting a red ball.
Probability (getting a red ball) = \(\frac {3}{10}\)

(ii) The event is getting a green ball.
Probability (getting a green ball) = \(\frac {7}{10}\)

5. Multiple Choice Questions :

Question (i).
The probability of an impossible event is : ………………
(a) -1
(b) 0
(c) \(\frac {1}{2}\)
(d) 1.
Answer:
(b) 0

Question (ii).
The probability of selecting letter G from the world ‘GIRL’ is :
(a) 1
(b) \(\frac {1}{2}\)
(c) \(\frac {1}{4}\)
(d) \(\frac {1}{3}\)
Answer:
(c) \(\frac {1}{4}\)

Question (iii).
When a die is thrown, the probability of getting a number 4 is :
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{3}\)
(c) \(\frac {4}{6}\)
(d) \(\frac {1}{6}\)
Answer:
(d) \(\frac {1}{6}\)

Question (iv).
A bag contains 5 white balls and 10 black balls. The probability of drawing a white ball from the bag is :
(a) \(\frac {5}{10}\)
(b) \(\frac {5}{15}\)
(c) \(\frac {10}{15}\)
(d) 1
Answer:
(b) \(\frac {5}{15}\)

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3

1. Following data gives total marks (out of 600) obtained by six students of a particular class. Represent the data on a bar graph.

Solution:
(i) To choose an appropriate scale we make equal division taking increments of 100.
Thus, 1 unit represent 100 marks.

(ii) Now represent the data on the bar graph.

2. The following bar graph shows the number of books sold by a bookstore during five consecutive years. Read the bar graph and answer the following questions :

Question (i).
About how many books were sold in 2008, 2009 and 2011 years ?
Solution:
140; 360; 180,

Question (ii).
In which year about 475 books were sold ? And in which year about 225 books were sold ?
Solution:
2012; 2010.

3. Two hundred students of 6th and 7th class were asked to name their favourite colour so as to decide upon what should be the colour of their school building. The results are shown in the table :

Represent the data on a graph.
Answer the following questions with the help of bar graph :

Question (i).
Which is the most preferred colour ?
Answer:
Choose a suitable scale as follows :
Start the scale at 0. The greatest value in data is 55, so end the scale at a value greater than 55, such as 60.

Question (ii).
Which is the least preferred colour ?
Answer:
Use equal divisions along the vertical axis, such as increments = 10. All the bars would between 0 and 60. We choose the scale such that the length between 0 and 60 is neither too long nor too small. Here we take 1 unit for 10 students.

Question (iii).
How many colours are there in all ? What are they ?
Answer:
We then draw and label the graph as shown :

From the graph we conclude that :

• Blue is the most preferred colour Because the bar representing Blue is the tallest).
• Green is the least preferred colour (Because the bar representing Green is the shortest).
• There are five colours. They are Red, Green, Blue, Yellow and Orange.

4. Consider the following data collected from a survey of a colony :

Draw a double bar graph choosing an appropriate scale.
What do you infer from the bar graph ?
(i) Which sports is the most popular ?
(ii) Which is more preferred, watching or participating in sports ?
Solution:
Take different sports along X = axis and number of persons watching and participating favourite sports Y-axis.
Scale. Take 1 unit height along Y-axis = 200 persons. The double bar graphs representing the given data is shown below.

5. The following table shows the time (in hours) spent by a student of class VII in a day.

Draw a bar graph to represent the above data. What do you infer from the above table ?
Solution:
Choose a suitable scale as follows :
(i) Start the scale at 0. The greatest value in the data is 8, so end the scale at a value greater than 8, such as 9. Take 1 unit of height of bars.
(ii) Use equal division along vertical axis, such as increments is 10. All the bars would be between 0 and 9.
(iii) We then draw and label the graph as below :

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 3 Data Handling Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2

1. Find the median of the following data :
3, 1, 5, 6, 3, 4, 5
Solution:
We arrange the data in ascending order.
1, 3, 3, 4, 5, 5, 6
Here, 3 and 5 both occur twice.
Therefore, both the numbers 3 and 5 are modes of the given data.

2. Find the mode of the following numbers :
2, 2, 2, 3, 4, 5, 5, 5, 6, 6, 8
Solution:
The given data is :
2, 2, 2, 3, 4, 5, 5, 5, 6, 6, 8
Median is the middle observation
∴ 5 is the median

3. The scores in mathematics test (out of 25) of 15 students are as follows :
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20.
Find the mean, mode and median of this data ? Are they same ?
Solution:
Mean =

= \(\frac {263}{15}\)
= 17.53
Now we arrange the data in ascending order
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Here 20 occurs more frequently
∴ Mode = 20
Median is the middle observation
∴ 20 is the median.
Yes, both Mode and Median are same.

4. The weight (in kg) of 15 students of class are :
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47

Question (i).
Find the mode and median of this data.
Solution:
We arrange the data in ascending order
32, 25, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
Mode. Here 38 and 43 occurs more frequently i.e. 3 times
∴ Mode = 38 and 43
Median is the middle observation
∴ 40 is median.

Question (ii).
Is there more than one mode ?
Solution:
Yes, there are two mode i.e. 38 and 43

5. Find the mode and median of the following data :
13, 16, 12, 14, 19, 12, 14, 13, 14
Solution:
We arrange the data in ascending order
12, 12, 13, 13, 14, 14, 14, 16, 19
Here 14 occurs more frequently
∴ Mode = 14
Median is middle observation
∴ 14 is median.

6. Find the mode of the following data :
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14.
Solution:
We arrange the data in ascending order
12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 18, 19.
Here 15 occurs more frequently
∴ Mode =15

7. Multiple Choice Questions :

Question (i).
The mode of the data :
3, 5, 1, 2, 0, 2, 3, 5, 0, 2, 1, 6 is :
(a) 6
(b) 3
(c) 2
(d) 1.
Answer:
(c) 2

Question (ii).
A cricketer scored 38, 79, 25, 52, 0, 8, 100 runs in seven innings, the range of the runs scored is :
(a) 100
(b) 92
(c) 52
(d) 38.
Answer:
(a) 100

Question (iii).
Which of the following is not a central tendency of a data ?
(a) Mean
(b) Median
(c) Mode
(d) Range.
Answer:
(a) Mean

Question (iv).
The mean of 3, 1, 5, 7 and 9 is :
(a) 6
(b) 4
(c) 5
(d) 0.
Answer:
(c) 5