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Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 11 Ratio and Proportion MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 11 Ratio and Proportion MCQ Questions

Multiple Choice Questions.

Question 1.
The ratio of 24 seconds to 1 minute is :
(a) 2 : 5
(b) 24 : 1
(c) 5 : 2
(d) 1 : 24.
Answer:
(a) 2 : 5

Question 2.
The ratio of 2 m to 75 cm is :
(a) 2 : 75
(b) 75 : 2
(c) 8 : 3
(d) 3 : 8.
Answer:
(c) 8 : 3

Question 3.
The ratio of 1 year to 8 months is :
(a) 2 : 3
(b) 3 : 2
(c) 1 : 8
(d) 8 : 1.
Answer:
(b) 3 : 2

Question 4.
Divide ₹ 40 in 2 : 3.
(a) ₹ 20, ₹ 30
(b) ₹ 24, ₹ 16
(c) ₹ 30, ₹ 20
(d) ₹ 16, ₹ 24.
Answer:
(d) ₹ 16, ₹ 24.

Question 5.
Which of the following is equivalent ratio of 4 : 7.
(a) 28 : 42
(b) 28 : 49
(c) 20 : 49
(d) 20 : 42.
Answer:
(b) 28 : 49

Question 6.
Find a, if 8, a, 40, 65 are in proportion.
(a) 26
(b) 12
(c) 13
(d) 9.
Answer:
(c) 13

Question 7.
Find x if 12, 25, x, 75 are in proportion.
(a) 36
(b) 40
(c) 30
(d) 38.
Answer:
(a) 36

Question 8.
The cost of 12 pens is ₹ 108. Find the cost of 18 such pens.
(a) ₹ 152
(b) ₹ 216
(c) ₹ 162
(d) ₹ 144.
Answer:
(c) ₹ 162

Question 9.
Aslam earns ₹ 1680 in a week. In how many days, he will earn ₹ 2400?
(a) 10
(b) 8
(c) 12
(d) 9.
Answer:
(a) 10

Question 10.
A bus travels 90 km in 2\(\frac {1}{2}\) hours. How much distance it cover in 5 hours?
(a) 100 km
(b) 180 km
(c) 150 km
(d) 120 km.
Answer:
(b) 180 km

Question 11.
Which of the following complete the given figure?

(a) 35
(b) 45
(c) 15
(d) 61.
Answer:
(a) 35

Question 12.
Which of the following complete the given blank space?

(a) 24
(b) 26
(c) 18
(d) 20.
Answer:
(a) 24

Question 13.
Which of the following complete the given blank space?

(a) 7
(b) 3
(c) 2
(d) 4.
Answer:
(c) 2

Question 14.
Which of the following complete the given blank space?

(a) 5
(b) 6
(c) 4
(d) 2.
Answer:
(a) 5

Question 15.
Which of the following complete the given blank space?

(a) 20
(b) 25
(c) 35
(d) 45.
Answer:
(b) 25

Fill in the blanks:

Question (i)
The length of a room is 30 m and breadth is 20 m. The ratio of length to breadth is …………. .
Answer:
3 : 2

Question (ii)
Sheena has 25 marbles and her friend Shabnam has 30 marbles. The ratio of the marbles Sheena and Shabnam is ……………… .
Answer:
5 : 6

Question (iii)
Ratio of 50 m and 15 m is ………………. .
Answer:
10 : 3

Question (iv)
The comparison by division is known as ……………. .
Answer:
ratio

Question (v)
A ratio is a comparison of ……………. quantities.
Answer:
two

Write True/False:

Question (i)
The comparison by division is called ratio. (True/False)
Answer:
True

Question (ii)
The first and fourth term of a proportion called. Extreme Terms. (True/False)
Answer:
True

Question (iii)
The ratio 18 : 24 in the simplest form is 3 : 4. (True/False)
Answer:
True

Question (iv)
Ratio of 15 minutes to 40 minutes is 8 : 3. (True/False)
Answer:
False

Question (v)
20, 40, 25, 50 are in proportion. (True/False)
Answer:
True

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 11 Ratio and Proportion Ex 11.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 11 Ratio and Proportion Ex 11.3

1. The cost of 1 kg apples is ₹ 45. What is the cost of 7 kg apples?
Solution:
Cost of 1 kg apples = ₹ 45
Cost of 7 kg apples = ₹ 45 × 7
= ₹ 315

2. A car travels 224 km in 7 litres of petrol. How much distance will it cover in 1 litre?
Solution:
Distance covered in 7 litres = 224 km
Distance covered in 1 litres = \(\frac {224}{7}\)
= 32 km

3. A pipe can fill 10 water tanks in 12 hours. How much time will it take to fill 15 such water tanks?
Solution:
Time taken to fill 10 water tanks = 12 hours
Time taken to fill 1 water tank = \(\frac {12}{10}\) hours
Time taken to fill 15 water tanks = \(\frac {12}{10}\) × 15 hours
= 18 hours

4. The cost of 18 m cloth is ₹ 810. What is the cost of 25 m cloth?
Solution:
Cost of 18 m cloth = ₹ 810
Cost of 1 m cloth = ₹ \(\frac {810}{18}\)
Cost of 25 m cloth = ₹ \(\frac {810}{18}\) × 25
= ₹ 1125

5. The weight of 24 books is 6 kg. What is the weight of 36 such books?
Solution:
Weight of 24 books = 6 kg
Weight of 1 book = \(\frac {6}{24}\) kg
Weight of 36 books = \(\frac {6}{24}\) × 36 kg
= 9 kg

Aliter:

By cross product, we have
24 × x = 6 × 36
x = \(\frac{6 \times 36}{24}\)
⇒ x = 9
Hence, weight of 36 books is 9 kg

6. ‘A’ runs 28 km in 5 hours. How many kilometres does it run in 9 hours?
Solution:
A runs in 5 hours = 28 km
A runs in 1 hour = \(\frac {28}{5}\) km
A runs in 9 hours= \(\frac {28}{5}\) × 9 km
= \(\frac {252}{5}\) km
50.4 km

7. A 12 m high pole casts a shadow of 30 m. Find the height of the pole that casts a shadow of 45 m.
Solution:
If shadow cast is 30 m, then height of Pole = 12 m
If shadow cast is 1 m, then height of Pole = \(\frac {12}{30}\) m
If shadow cast is 45 m, then height of Pole
= \(\frac {12}{30}\) × 45 m
= 18 m

8. A man earns ₹ 11200 in 7 months.

Question (i)
How much will he earn in 18 months?
Solution:
A man earns in 7 months = ₹ 11200
A man earns in 1 month = ₹ \(\frac {11200}{7}\)
A man earns in 18 months = ₹ \(\frac {11200}{7}\) × 18
= ₹ 28800

Question (ii)
In how many months will he earn ₹ 40,000?
Solution:
A man earns ₹ 1600 = 1 month
A man earns ₹ 40000 = \(\frac {1}{1600}\) × 40000
= 25 months

9. If the cost of a dozen soaps is ₹ 153.60. What will be the cost of 16 such soaps?
Solution:
Cost of 12 soaps = ₹ 153.60
(1 dozen =12 pieces)
Cost of 1 soap = ₹ \(\frac {153.60}{12}\)
Cost of 16 soap = ₹ \(\frac {153.60}{12}\) × 16
= ₹ 204.80

10. Cost of 105 envelops is ₹ 35. How many envelops can be purchased for ₹ 10?
Solution:
Number of envelops purchased for ₹ 35 = 105
Number of envelops purchased for ₹ 1 = \(\frac {105}{35}\)
Number of envelops purchased for ₹ 10 = \(\frac {105}{35}\) × 10
= 30

11. A bus travels 90 km in 2\(\frac {1}{2}\) hours.

Question (i)
How much time is required to cover 54 km with the same speed?
Solution:
Time required to cover 90 km
= 2\(\frac {1}{2}\) hours
= \(\frac {5}{2}\) hours
Time required to cover 1 km
= \(\frac{5}{2} \times \frac{1}{90}\) hours
Time required to cover 54 km
= \frac{5}{2} \times \frac{1}{90} × 54 hours
= \(\frac {3}{2}\) hours
= 1\(\frac {1}{2}\) hours

Question (ii)
Find the distance covered in 4 hours with the same speed?
Solution:
Distance covered in \(\frac {5}{2}\) hours
= 90 km
Distance covered 1 hour
= 90 × \(\frac {2}{5}\) km
Distance covered 4 hours
= 4 × 90 × \(\frac {2}{5}\) km
= 144 km

12. Anshul made 57 runs in 6 overs. In how many overs he made 95 runs with same strike rate?
Solution:
Number of overs to make 57 runs = 6 overs
Number of overs to make 1 run = \(\frac {6}{57}\) over
Number of overs to make 95 runs = \(\frac {6}{57}\) × 95 overs
= 10 overs

13. Cost of 5 kg rice is ₹ 32.50.

Question (i)
What will be the cost of 14 kg such rice?
Solution:
Cost of 5 kg rice = ₹ 32.50
Cost of 1 kg rice = ₹ \(\frac {32.50}{5}\)
Cost of 14 kg rice = ₹ \(\frac {32.50}{5}\) × 14
= ₹ 91

Question (ii)
What quantity of rice can be purchased in ₹ 162.50?
Solution:
Qunatity of rice for ₹ 32.50 = 5 kg
Qunatity of rice for ₹ 1 = \(\frac {5}{32.50}\) kg
Qunatity of rice for ₹ 162.50 = \(\frac {5}{32.50}\) × 162.50
= 25 kg

14. If a cow grazes 21 sq. m of a field in 6 days. How much area will it graze in 27 days?
Solution:
Field grazed in 6 days = 21 sq. m
Field grazed in 1 day = \(\frac {21}{6}\) sq. m
Field grazed in 27 days = \(\frac {21}{6}\) × 27 sq. m
= 94.5 sq. m

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 11 Ratio and Proportion Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 11 Ratio and Proportion Ex 11.2

1. Determine if the following are in proportion:

Question (i)
20, 40, 25, 50
Solution:
Yes,

Product of Extremes = 20 × 50 = 1000
Product of Means = 40 × 25 = 1000
∴ Product of Extremes = Product of Means
Hence, 20, 40, 25, 50 are in proportion.

Question (ii)
35, 49, 55, 78
Solution:
No,

Product of Extremes = 35 × 78 = 2730
Product of Means = 49 × 55 = 2695
∴ Product of Extremes ≠ Product of Means
Hence, 35, 49, 55, 78 are not in proportion

Question (iii)
24, 30, 36, 45
Solution:
Yes,

Product of Extremes = 24 × 45 = 1080
Product of Means = 30 × 36 = 1080
Since Product of Extremes = Product of Means
Hence, 24, 30,36,45 are in proportion

Question (iv)
10, 22, 45, 99
Solution:
Yes,

Product of Extremes = 10 × 99 = 990
Product of Means = 22 × 45 = 990
∴ Product of Extremes = Product of Means
Hence, 10,22,45,99 are in proportion

Question (v)
32, 48, 70, 210.
Solution:
No,

Product of Extremes = 32 × 210 = 6720
Product of Means = 48 × 70 = 3360
Since Product of Extremes ≠ Product of Means
Hence, 32, 48, 70, 210 are not in proportion.

2. Do the following ratios forms a proportion:

Question (i)
5:9 and 20 : 36
Solution:
Yes,

Product of Extremes = 5 × 36 = 180
Product of Means = 9 × 20 = 180
Since Product of Extremes = Product of Means
Hence, 5 : 9 and 20 : 36 are in proportion

Question (ii)
24 : 36 and 32 : 48
Solution:
Yes,
First ratio = 24 : 36 (Dividing both terms by 12) = 2 : 3
Second ratio = 32 : 48 (Dividing both terms by 16) = 2 : 3
∴ Both ratios are equal.
Hence, 24 : 36 and 32 : 48 are in proportion

Question (iii)
32 : 40 and 36 : 42
Solution:
No,
First ratio = 32 : 40 (Dividing both terms by 8) = 4 : 5
Second ratio = 36 : 42 (Dividing both terms by 6) = 6 : 7
∴ Both ratios are not equal.
Hence, 32 : 40 and 36 : 42 are not in proportion

Question (iv)
27 : 18 and 3:2
Solution:
Yes,
First ratio = 27 : 18 (Dividing both terms by 9) = 3 : 2
Second ratio = 3 : 2
∴ Both ratios are equal.
Hence, 27 : 18 and 3 : 2 are in proportion

Question (v)
35 : 28 and 77 : 44.
Solution:
No,
First ratio = 35 : 28 (Dividing both terms by 7) = 5 : 4
Second ratio = 77 : 44 (Dividing both terms by 11) = 7 : 4
∴ Both ratios are not equal.
Hence, 35 : 28 and 77 : 44 are not in proportion.

3. State true or false of the following:

Question (i)
4 : 3 : : 36 : 37
Solution:
False,

Product of Extremes = 4 × 37 = 148
Product of Means = 3 × 36 = 108
∴ Product of Extremes ≠ Product of Means
Hence, it is false.

Question (ii)
16 : 4 : : 20 : 5
Solution:
True,

Product of Extremes = 16 × 5 = 80
Product of Means = 4 × 20 = 80
∴ Product of Extremes = Product of Means
Hence, it is true

Question (iii)
19 : 43 : : 8 : 21.
Solution:
False,

Product of Extremes = 19 × 21 = 399
Product of Means = 43 × 8 = 344
∴ Product of Extremes ≠ Product of Means
Hence, it is false.

4. Determine if the following ratios form a proportion:

Question (i)
40 cm : 1 m and ₹ 12 : ₹ 30
Solution:
Yes,
First ratio = 40 cm : 1 m
= 40 : 100
(Dividing both terms by 20) = 2:5
Second ratio = ₹ 12 : ₹ 30
= 12 : 30
(Dividing both terms by 6) = 2:5
∴ Both ratios are equal.
Hence, 40 cm : 1 m and ₹ 12 : ₹ 30 are in proportion.

Question (ii)
25 min : 1 hour and 40 km : 96 km
Solution:
Yes,
First ratio = 25 min : 1 hour
= 20 min : 60 min
= 25 : 60
(Dividing both terms by 5) = 5 : 12
Second ratio = 40 km : 96 km
= 40 : 96
(Dividing both terms by 8) = 5 : 12
∴ First ratio = Second ratio
Hence, 25 min : 1 hour and 40 km : 96 km are in proportion.

Question (iii)
₹ 4 : 35 paise and 8 kg : 9 kg.
Solution:
No,
First ratio = ₹ 4 : 35 paise
= 400 paise : 35 paise
= 400 : 35
(Dividing both terms by 5)
= 80 : 7
Second ratio = 8 kg : 9 kg
= 8 : 9
∴ First ratio ≠ Second ratio
Hence, ₹ 4 : 35 paise and 8 kg : 9 kg are not in proportion.

5. Find the value of ‘x’ in each case:

Question (i)
25 : x :: 15 : 6
Solution:
Since, given terms are in proportion
∴ Product of Extremes = Product of Means
⇒ 25 × 6 = x × 15
⇒ \(\frac{25 \times 6}{15}\) = x
⇒ x = 10

Question (ii)
28 : 49 :: x : 56
Solution:
28 : 49 : : x : 56
Since, given terms are in proportion
∴ Product of Extremes = Product of Means
⇒ 28 × 56 = 49 × x
⇒ \(\frac{28 \times 56}{49}\) = x
⇒ x = 32

Question (iii)
8 : 20 :: 10 : x.
Solution:
8 : 20 : : 10 : x
Since, given terms are in proportion
∴ Product of Extremes = Product of Means
⇒ 8 × x = 20 × 10
⇒ x = \(\frac{20 \times 10}{8}\)
⇒ x = 25

6. Check if the following terms are in continued proportion:

Question (i)
1, 4, 16
Solution:
For continued proportion, 1, 4, 16 can be written as 1, 4, 4, 16
∴ Product of Extremes = 1 × 16 = 16
Product of Means = 4 × 4 = 16
∴ Product of Extremes = Product of Means
Hence, 1, 4, 16 are in continued proportion

Question (ii)
3, 9, 27
Solution:
For continued proportion, 3, 9, 27 can be written as 3, 9, 9, 27
∴ Product of Extremes = 3 × 27 = 81
∴ Product of Means =9 × 9 = 81
∴ Product of Extremes = Product of Means
Hence, 3, 9, 27 are in continued proportion.

Question (iii)
5, 10, 20.
Solution:
For continued proportion, 5, 10, 20 can be written as 5, 10, 10, 20
∴ Product of Extremes = 5 × 20 = 100
∴ Product of Means = 10 × 10 = 100
∴ Product of Extremes = Product of Means
Hence, 5, 10, 20 are in continued proportion.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1

Solve the following equations.

Question 1.
x – 2 = 7
Solution:
x – 2 = 7
∴ x = 7 + 2 (Transposing – 2 to RHS)
∴ x = 9

Question 2.
y + 3 = 10
Solution:
y + 3 = 10
∴ y = 10 – 3 (Transposing 3 to RHS)
∴ y = 7

Question 3.
6 = z + 2
Solution:
6 = z + 2
∴ z + 2 = 6 (Interchanging both the sides)
∴ z = 6 – 2 (Transposing 2 to RHS)
∴ z = 4.

Question 4.
\(\frac {3}{7}\) + x = \(\frac {17}{7}\)
Solution:
\(\frac {3}{7}\) + x = \(\frac {17}{7}\)
∴ x = \(\frac{17}{7}-\frac{3}{7}\) (Transposing \(\frac {3}{7}\) to RHS)
∴ x = \(\frac{17-3}{7}\)
∴ x = \(\frac {14}{7}\)
∴ x = 2

Question 5.
6x = 12
Solution:
6x = 12
∴ \(\frac{6 x}{6}=\frac{12}{6}\) (Dividing both the sides by 6)
∴ x = 2

Question 6.
\(\frac{t}{5}\) = 10
Solution:
\(\frac{t}{5}\) = 10
∴ \(\frac{t}{5}\) × 5 = 10 × 5 (Multiplying both the sides by 5)
∴ t = 50

Question 7.
\(\frac{2 x}{3}\) = 15
Solution:
\(\frac{2 x}{3}\) = 15
∴ \(\frac{2 x}{3} \times \frac{3}{2}=18 \times \frac{3}{2}\) (Multiplying both the sides by \(\frac {3}{2}\))

Question 8.
1.6 = \(\frac{y}{1.5}\)
Solution:
1.6 = \(\frac{y}{1.5}\)
∴ 1.6 × 1.5 = \(\frac{y}{1.5}\) × 1.5 (Multiplying both the sides by 1.5)
∴ 2.4 = y (∵ 1.6 × 1.5 = 2.4)
∴ y = 2.4

Question 9.
7x – 9 = 16
Solution:
7x – 9 = 16
∴ 7x = 16 + 9 (Transposing – 9 to RHS)
∴ 7x = 25
∴ \(\frac{7 x}{7}=\frac{25}{7}\) (Dividing both the sides by 7)
∴ x = \(\frac {25}{7}\)

Question 10.
14y – 8 = 13
Solution:
14y – 8 = 13
∴ 14y = 13 + 8 (Transposing – 8 to RHS)
∴ 14y = 21
∴ \(\frac{14 y}{14}=\frac{21}{14}\) (Dividing both the sides by 14)
∴ y = \(\frac{7 \times 3}{7 \times 2}\)
∴ y = \(\frac {3}{2}\)

Question 11.
17 + 16p = 9
Solution:
17 + 16p = 9
∴ 6p = 9 – 17 (Transposing 17 to RHS)
∴ 6p = -8
∴ \(\frac{6 p}{6}=\frac{-8}{6}\) (Dividing both the sides by 6)
∴ p = \(\frac{-4 \times 2}{3 \times 2}\)
∴ p = –\(\frac {4}{3}\)

Question 12.
\(\frac{x}{3}+1=\frac{7}{15}\)
Solution:
\(\frac{x}{3}+1=\frac{7}{15}\)
∴ \(\frac{x}{3}=\frac{7}{15}-1\) (Transposing 1 to RHS)
∴ \(\frac{7-15}{15}\) (LCM = 15)
∴ \(\frac{x}{3}=\frac{-8}{15}\)
∴ \(\frac{x}{3} \times 3=\frac{-8}{15} \times 3\) (Multiplying both the sides by 3)
∴ x = –\(\frac {8}{5}\)

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.4

1. Draw a circle of the following radius:

Question (i)
3.5 cm
Solution:
Steps of construction.

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measure OA = 3.5 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (ii)
4 cm
Solution:
Steps of construction.

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measure OA = 4 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob.

Question (iii)
2.8 cm
Solution:
Steps of construction

1. Mark a point O on the page of your note book, where a circle is to be drawn.
2. Take compasses fixed with sharp pencil and measures OA = 2.8 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (iv)
4.7 cm
Solution:
Steps of construction

1. Mark a point O on the page of your note book.
2. Take compasses fixed with sharp pencil and measures OA = 4.7 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

Question (v)
5.2 cm.
Solution:
Steps of construction

1. Mark a point O on the page of your note book.
2. Take compasses fixed with sharp pencil and measures OA = 5.2 cm using a scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw a complete arc by holding the compasses from its knob, we get the required circle.

2. Draw a circle of diameter 6 cm.
Solution:
Steps of construction

1. Draw a line segment PQ = 6 cm.
2. Draw the perpendicular bisector of PQ intersecting PQ at O.
3. With O as centre and radius = OQ = 3 cm (= OP), draw a circle.
The circle thus drawn is the required circle.

3. With the same centre O, draw two concentric circles of radii 3.2 cm and 4.5 cm.
Solution:
Steps of Construction

1. Mark a point O on the page of your note book, where a circle is drawn.
2. Take compasses fixed with sharp pencil measuring OA = 4.5 cm using scale.
3. Without changing the opening of the compasses, keep the needle at point O and draw complete arc by holding the compasses from its knob.
After completing one round, we get circle I.
4. Again with the same centre O and new radius = 3.2 cm draw another circle II following the same step 3.

4. Draw a circle of radius 4.2 cm with centre at O. Mark three points A, B and C such that point A is on the circle, B is in the interior and C is in the exterior of the circle.
Solution:
Steps of Construction

1. Mark a point O on the page of your note book, where a circle is to be drawn
2. Take compasses fixed with sharp pencil and measure OA = 4.2 cm using scale (∴ A is on the circle).
3. Without changing the opening of the compasses, keep the needle at point O and draw complete arc by rotating the compasses from the knob. After completing one round, we get required circle.
4. Mark point B in the interior of the circle and point C in the exterior of the circle.

5. Draw a circle of radius 3 cm and draw any chord. Draw the perpendicular bisector of the chord. Does the perpendicular bisector passes through the centre?
Solution:
Steps of Construction.

1. Draw a circle with C as centre and radius 3 cm.
2. Draw AB the chord of the circle.
3. Draw PQ the perpendicular bisector of chord AB.
4. We see that the perpendicular bisector of chord AB passes through the centre C of the circle.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.3

1. Draw a line r and mark a point P on it. Construct a line perpendicular to r at point P.

Question (i)
Using a ruler and compasses.
Solution:
Using ruler and compasses

Steps of Construction.

1. Draw a line r and mark a point P on it.

2. Draw an arc from P to the line r of any suitable radius which intersects line r at A and B.

3. Draw arcs of any radius which is more than half of arc made in step (2) from A and B which intersect at Q.

4. Join PQ.

Thus PQ is perpendicular to AB or line l or PQ ⊥ A.
Here P is called foot of perpendicular.

Question (ii)
Using a ruler and a set square.
Solution:
Using a ruler and a set square

Steps of Construction

1. Draw a line r and a point P on it.
2. Place one of the edges of a ruler along the line l and hold if firmly.

3. Place the set square in such a way that one of its edges contaning the right angle coincides with the ruler.
4. Holding the ruler, slide the set square along the line l till the vertical side reaches the point P.

5. Firmly hold the set square in this position. Draw PQ along its vertical edge. Now PQ is the required perpendicular to l ie. PQ ⊥ r.

2. Draw a line p and mark a point z above it. Construct a line perpendicular to p, from the point z.

Question (i)
Using a ruler and compasses.
Solution:
1. Draw a line p and mark a point z not lying on it.
2. From point z draw an arc which intersects line p at two points P and Q.


3. Using any radius and taking P and Q as centre, draw two arcs that intersect at point say B. On the other side (a shown in figure).
4. Join AB to obtain altitude to the line p.

Thus xz is altitude to line p.
i.e. xz ⊥ p.

Question (ii)
Using a ruler and set square
Solution:
Steps of constructions:
1. Draw a line p and mark a point z which is not lying on it.
2. Place one of the edge of a ruler along the line p and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Holding the ruler firmly, slide the set square along the line p till its vertical side reaches the point z.

5. Firmly hold the set square in this position, Draw xz along its vertical edge. Now xz is the required altitude to p i.e. xz ⊥ p.

3. Draw a line AB and mark two points P and Q on either side of line AB, Construct two lines perpendicular to AB, from P and Q using a ruler and compasses.
Solution:
1. Draw a line AB and Mark two points P and Q on either side of AB.

2. From point P draw an arc which intersect line AB at two points C and D.
3. Using any radius and taking C and D as centre draw two arcs that intersects at point say E on the other side as shown in figures.

4. Join PE to obtain perpendicular to AB.
5. From point Q draw an arc which intersects AB at two points X and Y.

6. Using any radius and taking X and Y as centre draw two arcs that intersects at point say R on the other side of line AB as shown in figures.
7. Join QR to obtain perpendicular to AB.
Thus, PE ⊥ AB and QR ⊥ AB

4. Draw a line segment of 7 cm and draw perpendicular bisector of this line segment.
Solution:
Steps of Construction:
1. Draw a line segment AB = 7 cm.
2. With A as centre and radius more than half of AB, draw an arc on both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

5. Draw a line segment PQ = 6.8 cm and draw its perpendicular bisector XY which bisect PQ at M. Find the length of PM and QM. Is PM = QM ?
Solution:
Steps of Construction:

1. Draw a line segment PQ = 6.8 cm
2. With P as centre and radius more than half of PQ draw arcs on both sides of PQ.

3. Now with Q as centre and the same radius as in step 2 draw arcs intersecting the previous drawn arcs at A and B respectively.
4. Join AB intersecting PQ at M. Then M bisects the line segment.
5. Measure the length of PM and QM
PM = 3.4 cm and QM = 3.4 cm
∴ PM = QM.

6. Draw perpendicular bisector of line segment AB = 5.4 cm. Mark point X anywhere on perpendicular bisector Join X with A and B. Is AX = BX ?
Solution:
Steps of construction.
1. Draw a line segment AB = 5.4 cm.
2. With A as centre and radius more than half of AB, draw an arc in both sides of AB.

3. With B as centre and the same radius as in step 2, draw an arc intersecting the first arc at C and D.
4. Join CD intersecting AB at O.
Then CD is the perpendicular bisector of AB.
Mark any point X on the perpendicular bisector CD. Drawn. Then join AX and BX.
On examination, we find that AX = BX.

7. Draw perpendicular bisectors of line segment of the following lengths.

Question (i)
8.2 cm
Solution:
Steps of Construction.
1. Draw a line regment AB = 8.2 cm

2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw an arcs intersecting the previous arc at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (ii)
7.8 cm
Solution:
Steps of Construction.

1. Draw a line segment AB = 7.8 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

Question (iii)
6.5 cm.
Solution:
Steps of Construction.
1. Draw a line segment AB = 6.5 cm

2. With A as centre and radius more than half of AB draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2 draw arcs intersecting the previous arcs at C and D.
4. Join CD intersecting AB at O. Then CD is the perpendicular bisector of AB.

8. Draw a line segment of length 8 cm and divide it into four equal parts Using compasses. Measure each part.
Solution:
Steps of construction.

1. Draw a line segment AB of length 8 cm
2. With A as centre and radius more than half of AB, draw arcs on both sides of AB.
3. With B as centre and the same radius as in step 2, draw arcs intersecting the previous arcs at P and Q.
4. Join PQ intersecting AB at C then PQ is the perpendicular bisector of AB intersecting AB at C.
5. Similarly draw the perpendicular bisector of AC intersecting AC at D.
6. Draw the perpendicular bisector of CB intersecting CB at E.
By actual measurement, it can be verified that
AD = DC = CE = EB

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°.
Solution.
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
= \(\frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)}\)
= \(\frac{\sin 18^{\circ}}{\sin 18^{\circ}}\) = 1
[∵ cos (90° – θ) = sin θ]

(ii) \(\frac{\tan 26^{\circ}}{\cos 64^{\circ}}=\frac{\tan 26^{\circ}}{\cot \left(90^{\circ}-26^{\circ}\right)}\)
= \(\frac{\tan 26^{\circ}}{\tan 26^{\circ}}\) = 1
[∵ cot (90°- θ) = tan θ]

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
[∵ cos (90° – 0) = sin O]
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59°
=cosec 31° – sec (90° – 31°)
= cosec 31° – cosec 31°
[∵ sec (90° – θ) = cosec θ].

Question 2.
Show that:
(i) tan 4 tan 230 tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) L.H.S.
= tan 48° tan 23° tan 42° tan 67°
= tan 48° × tan 23° × tan (90° – 48°) × tan (90° – 23°)
= tan48° × tan 23° × cot48° × cot 23°
= tan 48C × tan 23° × \(\frac{1}{\tan 48^{\circ}}\) × \(\frac{1}{\tan 23^{\circ}}\) = 1
∴ L.H.S. = R.H.S.

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°
= cos 38° × cos (90 – 38°) – sin 38° × sin (90° – 38°)
= cos 38° × sin 38° – sin 38° × cos 38
= 0.
∴ L.H.S. = RH.S.

Question 3.
If tan 2A = cot (A – 18°) where 2A is an acute angle, find the value of A.
Solution:
Given: tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°)
[cot (90° – θ) = tan θ]
⇒ 90°- 2A = A – 18°
⇒ 3A = 108°
⇒A = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
Given that: tan A = cot B
⇒ tan A = tan(90° – B)
[∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B.
⇒ A + B = 90°..

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Given that: sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A — 20°)
[∵ cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20°
⇒ 5A = 110°
⇒ A = 22°.

Question 6.
If A, B and C interior angles of a triangle ABC, then show that: \(\sin \left(\frac{B+C}{2}\right)=\cos \left(\frac{A}{2}\right)\)
Solution:
Since, A, B and C are interior angles of a triangle
∴ A + B + C = 180°
[Sum of three angles of a triangle is 180°]
⇒ B + C = 180° – A
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}\)
⇒ \(\frac{\mathrm{B}+\mathrm{C}}{2}=\left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
Taking sin on both sides, we get
⇒ \(\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
[∵ sin (90° – θ) = cos θ].

Question 7.
Express sin 67° + cos 75° in terms of Trigonometric ratios of angles between 0° and 45°.
Solution:
Given that: sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
[∵ sin(90° – θ) = cos θ and cos (90° – θ) = sin θ].

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.6

1. Draw a line XY and point P not lying on XY. Draw a line parallel to XY passing through P with the help of ruler and compasses.
Solution:
Steps of Construction:
1. Draw a line XY and point P not lying on it.

2. Take any point Q, anywhere on line XY.
3. Join PQ.

4. Now take Q as centre, draw arc AB of any radius on XY. Similarly, draw an arc CD of same radius on line segment PQ from point P.

5. Measure arc AB with compasses.
6. Draw an arc equal to radius AB from point C witch intersect CD on E.

7. Join PE and produce it. So, the line l is the required line parallel to XY.

2. Draw a line p parallel to line m passing through a point A which is not lying on line m with the help of set squares.
Solution:
Steps of Construction:
1. Given a line m with point A not lying on it.
2. Place one of the edge of a ruler along the line m and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Hold the ruler firmly, slide the set square along the line m till its vertical side reaches the point A.

5. Firmly hold the set square in this position, take another set square and place it in such a way that one of its edges containing right angle concides with previous set square as shown.
6. Now draw a line p along side of second set square passing through A.

Thus p \(\text { ॥ } \) m passing through A.

3. Given a line AB and the point X is not lying on it. Draw a line parallel to AB passing through X.

Question (i)
By a ruler and compasses
Solution:
By a ruler and compasses:
Let us consider a line AB and point X not lying on it.

Steps of Construction:
1. Take any point, say Y anywhere on line AB.
2. Join XY.

3. Now take Y as centre, draw an arc PQ of any radius on AB. Similarly draw an arc RS of same radius on line segment XY from point X.

4. Measure arc PQ with compasses.
5. Draw an arc equal to radius PQ from point R which intersect RS on T.
6. Join XT and produce it.

So the line m is the required line parallel to AB.

Question (ii)
By set squares.
Solution:
By set squares.

Steps of Construction:
1. Given a line AB with point X not lying on it.
2. Place one of the edge of a ruler along the line AB and hold it firmly.

3. Place the set square in such a way that one of its edges containing the right angle coincides with the ruler.
4. Hold the ruler firmly, slide the set square along the line AB till its vertical side reaches the point X.

5. Firmly hold the set square in this position, take another set square and place it in such a way that one of its edges containing right angle concides with previous set square as shown.
6. Now draw a line l along side of second set square passing through X.

7. Thus l \(\text { ॥ } \) AB passing through X.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.5

1. Draw the following angles in both directions (Left and right) by protractor:

Question (i)
75°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on ray OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 75° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 75°.

If ray OA lies to the left of the centre (midpoint) of the baseline, start reading the angle on the outer scale from 0° and mark 75°. Join OB, then \(\angle AOB\) = 75°.

Question (ii)
110°
Solution:
Steps of Construction:
1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 110° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 110°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 110°. Join OB, then \(\angle AOB\) = 110°.

Question (iii)
62°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-62° base line along OA.

3. Mark a point B on the paper against the mark of 62° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 62°.

If the ray OA lies to the left of the centre (midpoint) of the baseline, start reading the angle on the outer scale from 0° and mark 62°. Join OB, then \(\angle AOB\) = 62°.

Question (iv)
165°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° baseline along OA.

3. Mark a point B on the paper against the mark of 165° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle
\(\angle AOB\) = 165°

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 165°. Join OB, then \(\angle AOB\) = 165°.

Question (v)
170°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on ray OA such that its centre lies on the initial point O and 0-480° base line along OA.

3. Mark a point B on the paper against the mark of 170° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 170°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 170°. Join OB, then \(\angle AOB\) = 170°.

Question (vi)
32°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 32° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 32°.
If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 32°. Join OB, then \(\angle AOB\) = 32°.

Question (vii)
128°
Solution:
Steps of Construction:

1. Draw a ray OA.

Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 128° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle
\(\angle AOB\) = 128°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 128°. Join OB, then \(\angle AOB\) = 128°.

Question (viii)
25°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 25° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 25°.

If the ray OA lies to the left of the centre (mid point) of the base line, start reading the angle on the outer scale from 0° and mark 25°. Join OB, then \(\angle AOB\) = 25°.

Question (ix)
80°
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 80° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 80°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 80°. Join OB, then \(\angle AOB\) = 80°.

Question (x)
135°.
Solution:
Steps of Construction:

1. Draw a ray OA.

2. Place the protractor on OA such that its centre lies on the initial point O and 0-180° base line along OA.

3. Mark a point B on the paper against the mark of 135° (inner scale) on the protractor.

4. Remove the protractor and join OB.
Thus, required angle \(\angle AOB\) = 135°.

If the ray OA lies to the left to the centre (mid point) of the bar line, start reading the angle on the outer scale from 0° and mark 135°. Join OB, then \(\angle AOB\) = 135°.

2. Bisect the following angles by compasses:

Question (i)
48°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 48°.
4. Join OB. Then \(\angle AOB\) = 48°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point
E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 48°.
Measure \(\angle AOB\) and \(\angle AOB\)
\(\angle AOB\) = \(\angle AOB\) = 24°.

Question (ii)
140°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 140°.
4. Join OB. Then \(\angle AOB\) = 140°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 140°.
On measurement \(\angle AOE\) = \(\angle BOE\) = 70°.

Question (iii)
75°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 75°.
4. Join OB. Then \(\angle AOB\) = 75°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 75°.
On measurement
\(\angle AOE\) = \(\angle BOE\) = 37.5°.

Question (iv)
64°
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 64°.
4. Join OB. Then \(\angle AOB\) = 64°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 64°.
On measurement
\(\angle AOE\) = \(\angle BOE\) = 32°.

Question (v)
124°.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 124°.
4. Join OB. Then \(\angle AOB\) = 124°.
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\) at C and D respectively.
6. With C as centre and radius more than half of CD. Draw an arc.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, then OE is the bisector of angle \(\angle AOB\) = 124°.
On measurement \(\angle AOE\) = \(\angle BOE\) = 62°.

3. Draw an angle of 80° and bisect it in to four equal parts by compasses.
Solution:
1. Draw a line OY of any length.
2. Place the centre of the protractor at O.
3. Starting with 0 mark a point X at 80°.
4. Join OX. Then \(\angle XOY\) = 80°.
5. With O as centre and using compass, draw an arc that cuts both rays of \(\angle XOY\). Name the point of intersection as X’ and Y’.
6. With Y’ as centre, draw an arc whose radius is more than half the length X’Y’.
7. With the same radius and with X’ as a centre, draw another arc which cut the first arc at point C.
8. With O as centre and using compass, draw an arc that cuts both rays of \(\angle COY\). Name the points of intersection as B and A.
9. With A as centre, draw an arc whose radius is more than half the length AB.
10. With the same radius and with B as centre, draw another arc which cuts the first arc at point S.
11. With O as centre and using compass, draw an arc that cuts both rays of \(\angle XOC\) . Name the points of intersection as D and E.
12. With E as centre, draw an arc whose radius is more than half the length DE.
13. With the same radius and with E as centre, draw another arc which bisects the first arc at T. Then OT is the bisector of \(\angle XOC\).
Thus \(\overline{\mathrm{OS}}, \overline{\mathrm{OC}} \text { and } \overline{\mathrm{OT}}\) divide, \(\angle AOB\) = 80° into four equal parts.

4. Draw a right angle and bisect it.
Solution:
1. Draw a ray OB.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point A at 90°.

4. Join OA. Then \(\angle AOB\) = 90°
5. With O as centre and using compasses, draw an arc that cuts both rays of \(\angle AOB\). Name the points of intersection as A’ and B’.
6. With B’ as centre, draw an arc whose radius is more than half of the length B’A’.
7. With the same radius and with A’ as a centre, draw another arc which cuts the first arc at point C. Join OC bisects \(\angle AOB\).

5. Draw the following angles by ruler and compasses:

Question (i)
30°
Solution:
To Construct angle of 30°

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.

3. With C as centre and same radius as before draw another arc cutting the previous arc at E.
4. Join OE and produce it to B. \(\angle AOB\) = 60°.
5. Bisect \(\angle AOB\).
Thus \(\angle AOM\) = \(\angle MOB\) = 30°.

Question (ii)
45°
Solution:
To Construct Angle of 45°:

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.

3. With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.
4. With P and Q as centres and any suitable radius (more than half of PQ) or even the same radius draw arc cutting each other at R.
5. Join OR and produce it to B. Then \(\angle AOB\) = 90°.
6. Bisect \(\angle AOB\).
7. OD is the bisector of \(\angle AOB\).
\(\angle BOD\) = \(\angle DOA\) = 45°.

Question (iii)
135°
Solution:
To Construct Angle of 135°.

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at point C.
3. With C as centre and the same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q and then with Q as centre the same radius cut off the arc again at R.
4. With Q and R as centres and radius more than half of RQ draw arcs cutting each other at L.
5. Join OL and produce it to B. Then \(\angle AOB\) = 150°.
6. Take a point M on the arc where OL intersects the arc.
7. With M and Q as centres and radius more than half of MQ draw arcs cutting each other at N.
8. Join ON and produce it to E. \(\angle AOE\) = 135°.

Question (iv)
180°
Solution:
To Construct Angle of 180°.

Steps of Construction:
1. Draw a line AB and mark a point C on it.
2. Taking C as centre and with any suitable radius, draw an are PQ cutting AB at P and Q.

3. Here \(\angle ACB\) = 180° (It is a straight line).

Question (v)
120°
Solution:
To Construct Angle of 120°:

Steps of Construction:
1. Draw a line segment OA.

2. With O as centre and any suitable radius draw an arc cutting OA at point M.
3. With M as centre and same radius draw an arc which cuts the arc at N and then with N as centre and the same radius cut off the arc again at Q.
4. Join OQ and produce it to B.
Then \(\angle AOB\) = 120°.

Question (vi)
75°.
Solution:
To Construct Angle of 75°.

Steps of Construction:
1. Draw a line segment OA.
2. With O as centre and any suitable radius draw an arc cutting OA at C.
3. With C as centre and same radius cut off the arc at P and then with P as centre and the same radius cut off the arc again at Q.
4. With P and Q as centres and radius more than half of PQ draw arcs cutting each other at R.
5. Join OR and produce it to B. \(\angle AOB\) = 90°.
6. Bisect \(\angle AOB\).
7. OD is the bisector of \(\angle AOB\).
\(\angle BOD\) = \(\angle DOA\) = 45°.
8. Again draw OE bisector of \(\angle DOB\).
Thus angle \(\angle EOA\) = 75°.

6. Draw an angle of 30° by protractor and bisect it by a ruler and compasses.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the centre of the protractor at O.
3. Starting with 0° mark point B at 30°.
4. Join OB. Then \(\angle AOB\) = 30°.
5. With O as centre and using compasses draw an arc that cuts both rays of \(\angle AOB\), with the point intersection as C and D.
6. With C as centre, draw an arc whose radius is more than half of the length of CD.
7. With D as centre and same radius as in step 6 draw another arc which cuts the first arc at point E.
8. Join OE, it bisect \(\angle AOE\).

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 15 Probability MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
When a balanced die is thrown, the probability of getting 3 is …………….. .
A. \(\frac{1}{3}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{6}\)
Answer:
D. \(\frac{1}{6}\)

Question 2.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a king is …………………. .
A. \(\frac{1}{52}\)
B. \(\frac{1}{26}\)
C. \(\frac{1}{13}\)
D. 1
Answer:
C. \(\frac{1}{13}\)

Question 3.
A card is drawn at random from a well shuffled pack of cards. The probability of that card being a card other than picture cards is ……………….. .
A. \(\frac{4}{13}\)
B. \(\frac{10}{13}\)
C. \(\frac{3}{13}\)
D. \(\frac{1}{13}\)
Answer:
B. \(\frac{10}{13}\)

Question 4.
When an unbiased coin is tossed thrice, the probability of receiving three heads is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{2}\)
D. \(\frac{3}{8}\)
Answer:
A. \(\frac{1}{8}\)

Question 5.
When three unbiased coins are tossed simultaneously, the probability of receiving exactly one tail is ………………… .
A. \(\frac{1}{8}\)
B. \(\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{3}{8}\)
Answer:
D. \(\frac{3}{8}\)

Question 6.
When a balanced die is thrown, the probability of receiving an even number is ………………… .
A. \(\frac{1}{6}\)
B. \(\frac{5}{6}\)
C. \(\frac{1}{2}\)
D. \(\frac{1}{4}\)
Answer:
C. \(\frac{1}{2}\)

Question 7.
When a balanced die is thrown, the probability of receiving a prime number is ……………….. .
A. \(\frac{2}{3}\)
B. \(\frac{3}{4}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{2}\)
Answer:
D. \(\frac{1}{2}\)

Question 8.
When two balanced dice are thrown simultaneously, the probability of getting the total of numbers on dice as 9 is ………………. .
A. \(\frac{1}{9}\)
B. \(\frac{1}{6}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{12}\)
Answer:
A. \(\frac{1}{9}\)

Question 9.
Out of 100 days, the forecast predicted by the wheather department proved to be true on 20 days. Chosen any one day from these 100 days, the probability that the forecast proved to be false is ………………… .
A. \(\frac{1}{3}\)
B. \(\frac{1}{4}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
Answer:
D. \(\frac{4}{5}\)

Question 10.
The probability of a month of January having 5 Sundays is ………………….. .
A. \(\frac{2}{7}\)
B. \(\frac{3}{7}\)
C. \(\frac{5}{7}\)
D. \(\frac{1}{7}\)
Answer:
B. \(\frac{3}{7}\)