Bhagya

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. \(\frac{3}{7}\)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Answer:
The batswoman played 30 balls. Hence, the total number of trials = 30. If the event that she did not hit a boundary is denoted by A, then. the number of trials when event A occured is 30 – 6 = 24.
∴ p(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{24}{30}\)
= \(\frac{4}{5}\)
Thus, the probability that she did not hit a boundary is \(\frac{4}{5}\).

Question 2.
1500 families with 2 children were selected randomly, and the following data were s recorded:

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl. Also check whether the sum of these probabilities is 1.
Answer:
Here, the total number of families is 1500.
Hence, the total number of trials = 1500

(i) Let event A denote the event that the family chosen at random is having 2 girls.
Then, the number of trials when event A occured is 475.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{475}{1500}\)
= \(\frac{19}{60}\)

(ii) Let event B denote the event that the family chosen at random is having 1 girl.
Then, the number of trials when event B occured is 814.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{814}{1500}\)
= \(\frac{407}{750}\)

(iii) Let event C denote the event that the family chosen at random Is having no girl.
Then, the number of trials when event C occured is 211.
∴ p(C) = \(\frac{\text { No. of trials in which event } \mathrm{C} \text { occured }}{\text { The total number of trials }}\)
= \(\frac{211}{1500}\)
Now,
P(A) + P(B) + P(C) = \(\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\)
= \(\frac{475+814+211}{1500}\)
= \(\frac{1500}{1500}\)
= 1

Question 3.
Refer to sum no. 5 of “Sums to Enrich ‘Remember’” in chapter 14. Find the probability that a student of the class was born in August.
Answer:
From the Bar graph in the sum which is referred here, we get the following information:

Total number of students = 40 and the number of students born in August = 6.
Hence, if event A denotes the event that a student of the class is born in August, then the number of trials when event A occured is 6 and the total number of trials is 40.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{6}{40}\)
= \(\frac{3}{20}\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Answer:
Here, the total number of trials = 200. If event A denotes the event that 2 heads come up, then the number of trials when event A occured is 72.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{72}{200}\)
= \(\frac{9}{25}\)

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below :

Suppose a family is chosen. Find the probability that the family chosen is ( i ) earning ? 10000- ? 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not .own any vehicle.
(iv) earning ₹ 13000 – ₹ 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Answer:
Here, the total number of families is 2400. Hence, the total number of trials = 2400

(i) Let event A denote the event that the family is earning ₹ 10000 – ₹ 13000 per month and owning exactly 2 vehicles.
Then, the number of trials when event A occured = 29.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{29}{2400}\)

(ii) Let event B denote the event that the family is earning ₹ 16000 or more per month and owning exactly 1 vehicle.
Then, the number of trials when even B occured = 579.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{579}{2400}\)
= = \(\frac{193}{800}\)

(iii) Let event C denote the event that the family is earning less than ₹ 7000 per month and does not own any vehicle.
Then, the number of trials when event C occured = 10.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{10}{2400}\)
= = \(\frac{1}{240}\)

(iv) Let event D denote the event that the family is earning ? 13000 -? 16000 per month and is owning more than 2 vehicles. Then, the number of trials when event D occured = 25.
∴ P(D) = \(\frac{\text { No. of trials in which event D occured }}{\text { The total number of trials }}\)
= \(\frac{25}{2400}\)
= \(\frac{1}{96}\)

(v) Let event E denote the event that the family is owning not more than 1 vehicle, i.e., 1 vehicle or no vehicle.
Then, the number of trials when event E occured.
= 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
∴ P(E) = \(\frac{\text { No. of trials in which event E occured }}{\text { The total number of trials }}\)
= \(\frac{2062}{2400}\)
= \(\frac{1031}{1200}\)

Question 6.
Refer to table 7 of sum no. 7 in “Sums to Enrich ‘Remember’” in chapter 14.
(i) Find the probability that a student obtained less than 20 marks in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Answer:
According to the table referred here, the total number of students = 90.
Hence, the total number of trials = 90.
(i) According to the same table, the number of students who obtained less than 20 marks in the mathematics test is 7. So, if the event that a student obtained less than 20 marks in mathematics test is called event A, then the number of trials when event A occured is 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{90}\)

(ii) Let event B denote the event that a student obtained 60 or more marks. Then, , according to the same table, the number of trials when event B occured = 15 + 8 = 23.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{23}{90}\)

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

Opinion	Number of students
Like	135
Dislike	65

Find the probability that a student chosen at random
(i) Likes statistics,
(ii) Does not like it.
Answer:
Here, the total number of students = 200.
Hence, the total number of trials = 200.

(i) Let event A denote the event that a student likes statistics.
Then, the number of trials when event A occured = 135
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{135}{200}\)
= \(\frac{27}{40}\)

(ii) Let event B denote the event that a student does not like statistics. Then, the number of trials when event B occured = 65.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{65}{200}\)
= \(\frac{13}{40}\)

Question 8.
Refer to sum no. 2, Exercise 14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work ?
(ii) more than or equal to 7km from her place of work ?
(iii) within \(\frac{1}{2}\)km from her place to work?
Answer:
The total number of observations in the question referred here is 40.
Hence, the total number of trials = 40.

(i) Let event A denote the event that the distance between her residence and the place of work is less than 7 km. Then there are 9 such observations, viz., 5, 3, 2, 3, 6, 5, 6, 2, 3.
Hence, the number of trials when event A occured = 9.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{9}{40}\)

(ii) Let event B denote the event that the said distance is 7 km or more than 7 km. Then, all the remaining 31(40-9) observations refer to event B.
Hence, the number of trials when event B occured = 31
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{31}{40}\)

(iii) Let event C denote the event that the engineer lives within \(\frac{1}{2}\) km from her place of work. There is no observation which is \(\frac{1}{2}\) or less than \(\frac{1}{2}\).
Hence, the number of trials when event C occured = 0.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{0}{40}\)
= 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer:
Note: Students should do this Activity themselves.

Question 10.
Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her / him is divisible by 3 ? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Answer:
Note: Students should do this Activity themselves.

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg) :
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer:
The total number of bags = 11.
Hence, the total number of trials = 11.
Let event A denote the event that a bag contains more than 5 kg of flour.
There are 7 bags weighing more than 5 kg.
Their weights (in kg) are 5.05, 5.08, 5.03, 5.06, 5.08, 5.04 and 5.07. Hence, the number of trials when event A occured = 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{11}\)

Question 12.
In sum no. 5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Answer:
In sum no. 5, Exercise 14.2, total number of days is 30.
Hence, the total number of trials = 30.
In the table prepared there, we see that the frequency of class 0.12 – 0.16 is 2.
Hence, during 2 days the concentration of sulphur dioxide (in ppm) was in the interval 0.12 – 0.16.
Let event A denote the event that the concentration of sulphur dioxide (in ppm) is in the interval 0.12 – 0.16.
Hence, the number of trials when event A occured = 2.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{2}{30}\)
= \(\frac{1}{15}\)

Question 13.
In sum no. 1, Exercise 14.2, you were asked) to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Answer:
In sum no. 1, Exercise 14.2, the total number of students is 30.
Hence, the total number of trials = 30.
Let event A denote the event that a student has blood group AB. The number of students having blood group AB is 3.
Hence, the number of trials when event A occured = 3.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{3}{30}\)
= \(\frac{1}{10}\)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 14 Statistics 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.
The marks scored by Kavya in 10 tests of Mathematics are 35, 18, 41, 24, 45, 10, 28, 32, 40, 15. Then, the range of the data is …………….. .
A. 45
B. 10
C. 35
D. 28.8
Answer:
C. 35

Question 2.
The average of the observations 3, 4, 5, 8, 12, 10, 13, 16, 18, 11 is …………………. .
A. 100
B. 10
C. 18
D. 3
Answer:
B. 10

Question 3.
The mean of first five odd natural numbers is ……………….. .
A. 3
B. 5
C. 4
D. 25
Answer:
B. 5

Question 4.
The mean of first four even natural numbers is ……………….. .
A. 5
B. 10
C. 20
D. 4
Answer:
A. 5

Question 5.
The mean of first five prime numbers is
A. 28
B. 2.8
C. 5.6
D. 1.4
Answer:
C. 5.6

Question 6.
If the mean of 2x, 5, 3x, 12, 5x, 17 and 6 is 20, then x = ………………….. .
A. 10
B. 20
C. 15
D. 40
Answer:
A. 10

Question 7.
The mean of the following distribution is ………………. .

A. 3.9
B. 7.8
C. 78
D. 39
Answer:
A. 3.9

Question 8.
If the mean of 12, 13, x, 17, 18 and 20 is 16, then x = ………………. .
A. 8
B. 4
C. 16
D. 32
Answer:
C. 16

Question 9.
For a given frequency distribution, n = 20 and Σf_ix_i = 140, then X̄ = ………………… .
A. 20
B. 14
C. 7
D. 28
Answer:
C. 7

Question 10.
The mean of \(\frac{2}{5},\), \(\frac{5}{7},\), \(\frac{3}{5},\) and \(\frac{2}{7},\) is ……………… .
A. \(\frac{1}{2},\)
B. \(\frac{3}{5},\)
C. \(\frac{5}{7},\)
D. 2
Answer:
A. \(\frac{1}{2},\)

Question 11.
The median of 14, 6, 2, 13, 9, 15 and 12 is …………………. .
A. 12
B. 10
C. 2
D. 15
Answer:
A. 12

Question 12.
The median of 21, 17, 13, 33, 19, 23 is ………………… .
A. 21
B. 20
C. 33
D. 19
Answer:
B. 20

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches:
2, 3. 4, 5, 0. 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Answer:
Here, n = 10.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\frac{2+3+4+5+0+1+3+3+4+3}{10}\)
= \(\frac{28}{10}\)
= 2.8
Thus, the mean of the given scores is 2.8 goals.

Arranging the observations in the ascending order, we get:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Since n = 10 is an even number, \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6.

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
= \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
= \(\frac{3+3}{2}\) = 3
Thus, the median of the given scores is 3 goals.
In the given data, observation 3 occurs most frequently (4 times). Hence, the mode of the data is 3 goals.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Answer:
Here, n = 15.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\begin{gathered}
41+39+48+52+46+62+54+40 \\
+96+52+98+40+42+52+60 \\
\hline 15
\end{gathered}\)
= \(\frac{822}{15}\) = 54.8
Thus, the mean of the data is 54.8 marks.
Arranging the observations in the ascending order, we get:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Here, n = 15 is an odd number.
Median M = \(\left(\frac{n+1}{2}\right)\)th observation
= \(\left(\frac{15+1}{2}\right)\)th observation
= 8 th observation
= 52
Thus, the median of the data is 52 marks.
In the given data, observation 52 occurs most frequently (3 times). Hence, the mode of the data is 52 marks.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Answer:
Here, the median = 63 and n = 10.
∴ \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
∴ 63 = \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
∴ 63 = \(\frac{(x)+(x+2)}{2}\)
∴63 × 2 = x + x + 12
∴126 = 2x + 2
∴ 2x = 124
∴ x = 62

Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Answer:
Here, just by simple observation, it is clearly seen that observation 14 occurs most frequently, i.e., 4 times.
Hence, the mode of the data is 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table:

Answer:

Mean X̄ = \(\frac{\Sigma f_{i} x_{i}}{n}\)
= \(\) = \(\frac{3,05,000}{60}\) = 5083.33
Thus, the mean salary is ₹ 5083.33.

Question 6.
Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
Answer:
For the students studying in the same class, usually their level of knowledge and understanding would be more or less equal. There would be a few student having this level low and there would be a few students having this level high. Their level of knowledge and understanding would be reflected in the marks scored by them at an examination. Hence, the mean of marks scored by them at an examination is an appropriate measure of central tendency.

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
If we consider the monthly income of the people of certain region, the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %):

Causes	Female fatality rate (%)
1. Reproductive health conditions	31.8
2. Neuropsychiatric conditions	25.4
3. Injuries	12.4
4. Cardiovascular conditions	4.3
5. Respiratory conditions	4.1
6. Other causes	22.0

(i) Represent the information given above graphically.
Answer:

(ii) Which condition is the major cause of women’s ill health and death worldwide?
Answer:
‘Reproductive health conditions’ is the major cause of womens ill health and death worldwide.

(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Answer:
‘Malnutrition’ and ‘Lack of necessary medical facilities’ can be considered as two other factors which play a major role in female fatality.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:

Section	Number of girls per thousand bays
Scheduled Caste (SC)	940
Scheduled Tribe (ST)	970
Non-SC/ST	920
Backward districts	950
Non-backward districts	920
Rural	930
Urban	910

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Answer:

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) Draw a bar graph to represent the polling results.
Answer:
Seats won by different political parties

(ii) Which political party won the maximum number of seats?
Answer:
Political party: A won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	.5
163-171	4
172-180	2

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous.]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
Answer:
Making the class intervals continuous, we get the following table:

Length (in mm)	Number of leaves
117.5-126.5	3
126.5- 135.5	5
135.5-144.5	9
144.5-153.5	12
153.5- 162.5	5
162.5-171.5	4
171.5-180.5	2

(i) Length of leaves in millimetre

(ii) Yes. The given data can also be represented by ‘Frequency polygon’.

(iii) It is not correct to conclude that the maximum number of leaves are 153 mm long, because even if the frequency of class 145-153 is 12, we do not have the information about the length of each of those 12 leaves individually.

Question 5.
The following table gives the life times of 400 neon lamps:

Life time (in hours)	Number of lamps
300 – 400	14
400 – 500	56
500 – 600	60
600 – 700	86
700 – 800	74
800 – 900	62
900 – 1000	48

(i) Represent the given information with the help of a histogram.
Answer:

(ii) How many lamps have a life time of 700 hours or more than 700 hours ?
Answer:
The-frequencies of classes 700-800, 800-900 and 900-1000 are 74, 62 and 48 respectively.
Hence, the life time of 184 (74 + 62 + 48) lamps is 700 hours or more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Answer:
To draw the frequency polygons of both the sections, we find the class marks of each class and prepare the following tables:

Section A

Marks	Class mark	Frequency
0-10	5	3
10-20	15	9
20-30	25	17
30-40	35	12
40-50	45	9

Section B

Marks	Class mark	Frequency
0-10	5	5
10-20	15	19
20-30	25	15
30-40	35	10
40-50	45	1

Comparing the performance of both the sections from the frequency polygons, we observe that the performance of students of section A is better than the performance of students of section B.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Number of balls	Team A	Team B
1-6	2	5
7-12	1	6
13-18	8	2
19-24	9	10
25-30	4	5
31-36	5	6
37-42	6	3
43-48	10	4
49-54	6	8
55-60	2	10

Represent the data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Answer:

Number of runs made by Team A and Team B in first 60 balls.

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows:

Age (in years)	Number of children
1-2	5
2-3	3
3-5	6
5-7	12
7-10	9
10-15	10
15-17	4

Draw a histogram to represent the data above.
Answer:

Children of various age groups playing in a park

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of letters	Number of surnames
1-4	6
4-6	30
6-8	44
8-12	16
12-20	4

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Answer:

(i) Information regarding the number of surnames having given number of letters
Answer:

(ii) Write the class interval in which the maximum number of surnames lie.
Answer:
The maximum number of surnames lie in the class interval 6-8.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows : Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see the given figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.

Answer:
Outer faces to be polished:

• One face on back side of the bookshelf, measuring 110 cm × 85 cm.
• Two faces on the sides, each of those measuring 110 cm × 25 cm.
• The top and the base, each of those measuring 85 cm × 25 cm.
• Two vertical strips on the front side, each of those measuring 110 cm × 5 cm.
• Four horizontal strips on the front side, each of those measuring 75 cm × 5 cm.

Thus, total area of region to be polished
= [(110 × 85) + 2(110 × 25) + 2 (85 × 25) + 2(110 × 5) + 4(75 × 5)] cm²
= (9350 + 5500 + 4250 + 1100+ 1500) cm²
= 21700 cm²
20 paise per cm² = ₹ 0.20 per cm²
Cost of polishing 1 cm2 region = ₹ 0.20
∴ Cost of polishing 21700 cm² region
= ₹ (21700 × 0.20)
= ₹ 4340

Inner faces to be painted:

• Two faces on the sides each of those measuring 90 cm × 20 cm.
• Two faces each of two shelves, the top face and the bottom face, in all six face, each of those measuring 75 cm × 20 cm.
• Face on the back side, measuring 90 cm × 75 cm.

Thus, total area of the region to be painted
= [2 (90 × 20) + 6 (75 × 20) + (90 × 75)] cm²
= (3600 + 9000 + 6750) cm²
= 19350 cm²
10 paise per cm² = ₹0.10 per cm²
Cost of painting 1 cm² region = ₹ 0.10
∴ Cost of painting 19350 cm² region = ₹ (19350 × 0.10) = ₹ 1935
Then, the total expense of polishing and painting = ₹ 4340 + ₹ 1935 = ₹ 6275

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be ‘ painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Answer:
For each wooden sphere,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{21}{2}\) cm
Curved surface area of 1 sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm²
= 1386 cm²
For each cylindrical support, radius r = 1.5 cm and height h = 7 cm.
Area of top of cylindrical support
= πr²
= \(\frac{22}{7}\) × 1.5 × 1.5 cm²
= 7.07 cm² (approx.)
Hence, the area of each sphere to be painted silver = 1386 cm² – 7.07 cm² = 1378.93 cm²
∴ Total area of eight spheres to be painted silver = 1378.93 cm² × 8 = 11031.44 cm²
25 paise per cm² = ₹ 0.25 per cm²
Cost of painting silver in 1 cm² region = ₹ 0.25
∴ Cost of painting silver in 11031.44 cm² region
= ₹ (11031.44 x 0.25)
= ₹ 2757.86 (approx.)
Curved surface area of 1 cylindrical support
= 2πrh
= 2 × \(\frac{22}{7}\) × 1.5 × 7 cm
= 66 cm²
∴ Total area of eight cylindrical supports to be painted black = 66 cm² × 8 = 528 cm²
5 paise per cm² = ₹ 0.05 per cm²
Cost of painting black in 1 cm² region = ₹ 0.05
∴ Cost of painting black in 528 cm² region = ₹ (528 × 0.05)
= ₹ 26.40
Thus, the total cost of painting = ₹ 2757.86 + ₹ 26.40
= ₹ 2784.26 (approx.)

Question 3.
The diameter of a sphere is decreased by 25 %. By what per cent does its curved surface area decrease?
Answer:
Suppose, the initial diameter of the sphere is d units and radius is r units.
∴ d = 2r
Original curved surface area of the sphere
= 4πr²
= π (4r²)
= π (2r)²
= πd² unit²
Now, the diameter of the sphere is reduced by 25 %. Hence, the new diameter of the sphere is 0.75d units.
New curved surface area of the sphere
= π (diameter)
= π (0.75d)² unit²
= 0.5625 πd² unit²
∴ The decrease in the curved surface area of the sphere = πd² – 0.5625 πd²
= 0.4375 πd² unit²
∴Percentage decrease in the curved surface area of the sphere = \(\frac{0.4375 \pi d^{2}}{\pi d^{2}}\) × 100 = 43.75 %
Thus, when the diameter of a sphere is decreased by 25 %, its curved surface area decreases by 43.75 %.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan² 45° + cos² 30° – sin² 60°

(iii)

(iv)

(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
Solution:
(i) sin 60° cos 30° + sin 30° cos 60°
= \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\)

= \(\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\)
= \(\frac{3}{4}+\frac{1}{4}\) = 1.

(ii) 2 tan² 45° + cos² 30° – sin² 60° = 2 (tan 45°)² + (cos 30°)² – (sin 60°)²
= 2 (1)² + (\(\frac{\sqrt{3}}{2}\))² – (\(\frac{\sqrt{3}}{2}\))² = 2.

(iii)
= \(\frac{\frac{1}{\sqrt{2}}}{\left(\frac{2}{\sqrt{3}}\right)+(2)}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)

= \(\frac{1}{\sqrt{2}}: \frac{\sqrt{3}}{2+2 \sqrt{3}}=\frac{\sqrt{3}}{2 \sqrt{2}(\sqrt{3}+1)}\)

= \(\frac{\sqrt{3}(\sqrt{3}-1)}{2 \sqrt{2}(\sqrt{3}+1)(\sqrt{3}-1)}\)

= \(\frac{\sqrt{2} \times \sqrt{3} \times(\sqrt{3}-1)}{4(3-1)}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\).

(iv)

= \(\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}\)

= \(\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}}\)

= \(\frac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}\)

= \(\frac{27+16-24 \sqrt{3}}{27-16}\)

= \(\frac{43-24 \sqrt{3}}{11}\)

(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)

= \(\begin{array}{r}
5\left(\cos 60^{\circ}\right)^{2}+4\left(\sec 30^{\circ}\right)^{2} \\
\frac{-\left(\tan 45^{\circ}\right)^{2}}{\left(\sin 30^{\circ}\right)^{2}+\left(\cos 30^{\circ}\right)^{2}}
\end{array}\)

= \(\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\)

= \(\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{5}{4}+\frac{1}{3}-1}{\frac{1}{4}+\frac{3}{4}}\)

= \(\frac{5}{4}+\frac{16}{3}-1=\frac{15+64-12}{12}=\frac{67}{12}\).

Question 2.
Choose the correct option and justify your choice.

(i) \(\frac{2 \tan 30^{\circ}}{1+\tan 30^{\circ}}\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan 45^{\circ}}\)
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0.

(iii) sin 2A = 2 sin A is true when
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\)
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°.

Solution:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}\)

\(\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}\) = sin 60°.
So, correct anwer is (A).

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\frac{1-(1)^{2}}{1+(1)^{2}}\) = 0
So, correct anwer is (D).

(iii) Here when A = 0°
L.H.S. = sin 2A = sin 0° = 0
and R.H.S. = 2 sin A = 2 sin 0°
= 2 × 0 = 0
∴ Option (A) is correct.

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}\)

= \(\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}\)

= tan 60°
∴ Option (C) is correct.

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\); 0° ∠A + B ≤ 90°; A > B. find A and B.
Solution:
tan (A + B) = \(\sqrt{3}\). Given
tan (A + B) = tan 60°
⇒ A + B = 60° ……………..(1)
tan (A – B) = \(\frac{1}{\sqrt{3}}\) (Given)
or tan (A – B) = tan 30°
⇒ A – B = 30° …………….(2)
On adding (1) and (2),

A = 45°

Pu value of A = 45° in (1)
45° + B = 60°
B = 60° – 45°
B = 15°
Hence A = 45° and B = 15°.

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin q increases as q increases.
(iii) The value of cos q Increases as q increases
(iv) sin q = cos q for all value of q.
(v) cot A is not defined for A = 0°.
Solution:
(i) False.
When A = 60°, B = 30°
L.H.S. = sin (A + B) = sin (60° + 30°) = sin 90° = 1
R.H.S. = sin A + sin B
= sin 60° + sin 30°
= \(\frac{\sqrt{3}}{2}+\frac{1}{2}\) ≠ 1
i.e., L.H.S. ≠ R.H.S.

(ii) True, sin 30° = \(\frac{1}{2}\) = 0.5,
Note that sin 0° = 0,
sin 45° = \(\frac{1}{\sqrt{2}}\) = 0.7 (approx.)
sin 60° = \(\frac{\sqrt{3}}{2}\) = 0.87 (approx.)
and sin 90° = 1
i.e., value of sin θ increases as θ increases from 0° to 90°.

(iii) False.
Note that cos 0° = 1,
cos 30° = \(\frac{\sqrt{3}}{2}\) = 0.87(approx.)
cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.7.(approx.)
cos 60° = \(\frac{1}{2}\) = 0.5
and cos 90° = 0.
Hence, value of θ decreases as θ increases from 0° to 90°.

(iv) False
Since sin 30° = \(\frac{1}{2}\)
and cos 30° = \(\frac{\sqrt{3}}{2}\)
or sin 30° ≠ cos 30°
Only we have: sin 45° = cos 45°.
\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)

(v) True.
cot 0° = \(\frac{1}{\tan 0^{\circ}}=\frac{1}{0}\), or not defined.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 1 Rational Numbers Ex 1.1Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1

1. Represent these numbers on the number line:

Question (i).
\(\frac {7}{4}\)
Solution:
To represent \(\frac {7}{4}\), make 7 markings each of a distance equal to \(\frac {1}{4}\) on the right side of 0. The 7th point represents the rational number \(\frac {7}{4}\).

The point A is \(\frac {7}{4}\).

Question (ii).
\(\frac {-5}{6}\)
Solution:
To represent (\(\frac {-5}{6}\)) on the number line, make 5 markings each of a distance equal to on the left side of 0. The 5th point represents the rational number (\(\frac {-5}{6}\)).

The point B is (\(\frac {-5}{6}\))

2. Represent \(\frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11}\) on the number line.
Solution:
To represent the given rational numbers on a number line, make 11 markings each being equal to distance \(\frac {1}{11}\) on the left of 0.

Here, the point A is (\(\frac {-2}{11}\)).
the point B is (\(\frac {-5}{11}\)).
the point C is (\(\frac {-9}{11}\)).

3. Write five rational numbers which are smaller than 2.
Solution:
There are infinite rational numbers below 2, positive as well as negative.
Five of them are 1, \(\frac {1}{3}\), \(\frac {1}{4}\), 0, – 1.

4. Find ten rational numbers between \(\frac {-2}{5}\) and \(\frac {1}{2}\).
Solution:
First, convert \(\frac {-2}{5}\) and \(\frac {1}{2}\) having the same denominator, such that the difference between the numerators is more than 10.
\(\frac{-2}{5}=\frac{-2}{5} \times \frac{4}{4}=\frac{-8}{20}\);
\(\frac{1}{2}=\frac{1}{2} \times \frac{10}{10}=\frac{10}{20}\)
∴ The ten rational numbers between \(\frac {-8}{20}\) and \(\frac {10}{20}\) are
\(\frac{-7}{20}, \frac{-6}{20}, \frac{-5}{20}, \frac{-4}{20}, \frac{-3}{20}, \ldots, 0, \frac{1}{20}, \ldots, \frac{9}{20} .\)
(There can be many more such rational numbers.)

5. Find five rational numbers between

Question (i).
\(\frac {2}{3}\) and \(\frac {4}{5}\)
Solution:
First, convert \(\frac {2}{3}\) and \(\frac {4}{5}\) having the same denominator, such that the difference between the numerators is more than 5.
\(\frac{2}{3}=\frac{2}{3} \times \frac{20}{20}=\frac{40}{60}\);
\(\frac{4}{5}=\frac{4}{5} \times \frac{12}{12}=\frac{48}{60}\)
∴ The five rational numbers between \(\frac {2}{3}\) and \(\frac {4}{5}\) are \(\frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60}, \frac{46}{60}\).

Question (ii).
\(\frac {-3}{2}\) and \(\frac {5}{3}\)
Solution:
First, convert \(\frac {-3}{2}\) and \(\frac {5}{3}\) having the same denominator, such that the difference between the numerators is more than 5.
\(\frac{-3}{2}=\frac{-3}{2} \times \frac{3}{3}=\frac{-9}{6}\);
\(\frac{5}{3}=\frac{5}{3} \times \frac{2}{2}=\frac{10}{6}\)
∴ The five rational numbers between \(\frac {-3}{2}\) and \(\frac {5}{3}\) are \(\frac{-8}{6}, \frac{-7}{6}, 0, \frac{7}{6}, \frac{8}{6}\).

Question (iii).
\(\frac {1}{4}\) and \(\frac {1}{2}\)
Solution:
First, convert \(\frac {1}{4}\) and \(\frac {1}{2}\) having the same denominator, such that the difference between the numerators is more than 5.
\(\frac{1}{4}=\frac{1}{4} \times \frac{8}{8}=\frac{8}{32}\);
\(\frac{1}{2}=\frac{1}{2} \times \frac{16}{16}=\frac{16}{32}\)
∴ The five rational numbers between \(\frac {1}{4}\) and \(\frac {1}{2}\) are \(\frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}, \frac{14}{32}\).
(There can be many more such rational numbers.)
[Note : You can write rational numbers of your choice.]

6. Write five rational numbers greater than -2.
Solution:
There can be many rational numbers greater than – 2. Five of them are \(\frac{-3}{2}, \frac{-1}{4}, 0, \frac{1}{2}, \frac{1}{5}\).

7. Find ten rational numbers between \(\frac {3}{5}\) and \(\frac {3}{4}\).
Solution:
First, convert \(\frac {3}{5}\) and \(\frac {3}{4}\) having the same denominator, such that the difference between the numerators is more than 10.
\(\frac{3}{5}=\frac{3}{5} \times \frac{20}{20}=\frac{60}{100}\);
\(\frac{3}{4}=\frac{3}{4} \times \frac{25}{25}=\frac{75}{100}\)
∴ The ten rational numbers between \(\frac {3}{5}\) and \(\frac {3}{4}\) are \(\frac{61}{100}, \frac{62}{100}, \frac{63}{100}, \frac{64}{100}, \frac{65}{100}, \frac{66}{100}, \frac{67}{100}, \frac{68}{100},\)\(\frac{69}{100}, \frac{70}{100}\)