Data Handling

Steps:

1. On a graph paper, draw two mutually perpendicular lines OX and OY which intersect each other at point O. The line OX is taken horizontal and is called the x-axis, whereas the line OY is taken vertical and is called the y-axis.

2. On the x-axis, starting from O, mark points at equal distances. At these points write the names of the different fruits i.e. Apple, Orange, Banana, Grapes, Guava, Pineapple and Papaya.

3. Along y-axis, mark the heights of the bars (rectangles) in proportion to the given data (number of people). Here, take 1 cm on the y-axis = 5 people.

The required bar graph is as given below:

Steps:

1. On a graph paper, draw two mutually perpendicular lines OX and OY which intersect each other at point O. The line OX is taken horizontal and is called the x-axis, whereas the line OY is taken vertical and is called the y-axis.

2. On the x-axis, starting from O, mark points at equal distances. At these points write the names of the different items i.e. Grill, Toaster, Oven, Blender and Coffee maker.

3. Along y-axis, mark the heights of the bars (rectangles) in proportion to the given data (number of items sold in Store A and Store B). Here, take 1 cm on the y-axis = 5 items.

4. For each item, draw two bars side by side — one for Store A and the other for Store B — and shade them with two different colours/designs. Show the meaning of the two shadings in an index.

The required double bar graph is as given below:

(i) The bar graph shows the expenditure (in ₹) of a monthly salary under different heads, namely Food, Clothing, Education, Misc and Savings.

Hence, the bar graph is representing the expenditure of monthly salary in different heads.

(ii) From the bar graph,

Expenditure on Food = ₹ 2,100

Expenditure on Clothing = ₹ 600

Expenditure on Education = ₹ 1,200

Expenditure on Misc = ₹ 1,500

Expenditure on Savings = ₹ 1,000

The tallest bar is that of Food.

Hence, the expenditure is maximum on food.

(iii) The shortest bar is that of Clothing.

Hence, the expenditure is minimum on clothing.

(iv) Expenditure on education = ₹ 1,200 and expenditure on clothing = ₹ 600.

⇒ 2 × Expenditure on clothing = 2 × 600 = ₹ 1,200 = Expenditure on education

Hence, the given statement is true.

From the double bar graph,

Mode of transport	Number of boys	Number of girls
School bus	75	135
Walking	120	60
Bicycle	240	180
Other vehicles	150	90

(i) The double bar graph compares, for each mode of transport, the number of boys with the number of girls.

Hence, the double bar graph is representing the number of boys and girls using different modes of transport for going to school.

(ii) The number of girls is more than the number of boys only where the girls' bar is taller than the boys' bar, that is for School bus (135 > 75).

Hence, girls use the school bus more than the boys.

(iii) Among the boys' bars, the tallest bar is that of Bicycle (240).

Hence, boys use the bicycle the most.

(iv) For Walking, number of boys = 120 and number of girls = 60.

⇒ Half the number of boys = $\frac{1}{2} \times 120$ = 60 = Number of girls

Hence, in walking the number of girls is half the number of boys.

Here, the maximum value = 22 and the minimum value = 1.

∴ Range = 22 - 1 = 21

Taking the class intervals 0-5, 5-10, 10-15, 15-20 and 20-25, each of size 5, the frequency distribution table of the given data is:

Class interval	Tally marks	Frequency
0-5	II	2
5-10	IIII II	7
10-15	IIII III	8
15-20	IIII I	6
20-25	II	2
Total		25

Hence, the required frequency distribution table is as given above.

Here, the maximum value = 48 and the minimum value = 0.

∴ Range = 48 - 0 = 48

Taking the class intervals 0-10, 10-20, 20-30, 30-40 and 40-50, each of size 10, the frequency distribution table of the given data is:

Classes	Tally marks	Frequency
0-10	IIII	5
10-20	IIII	5
20-30	IIII II	7
30-40	IIII IIII	10
40-50	IIII III	8
Total		35

The students obtaining below 20 marks are the students of the classes 0-10 and 10-20.

∴ Number of students obtaining below 20 marks = 5 + 5 = 10

Hence, 10 students obtained below 20 marks.

Steps:

1. Since the scale on the x-axis starts at 50, a break (kink) is shown near the origin along the x-axis to indicate that the graph is drawn to scale beginning at 50 and not at the origin itself.

2. Take 1 cm on the x-axis = 10 units (electricity bill in ₹).

3. Take 1 cm on the y-axis = 1 house.

4. Construct rectangles corresponding to the given data. As the data is grouped, the rectangles are drawn adjacent to one another with no gap in between.

The required histogram is as given below:

(i) The highest rectangle corresponds to the maximum number of houses (i.e. 8), which lies in the class interval 80-90.

Hence, the group 80-90 has the maximum number of houses.

(ii) The houses paying less than ₹ 100 are the houses of the classes 50-60, 60-70, 70-80, 80-90 and 90-100.

∴ Number of houses paying less than ₹ 100 = 2 + 6 + 5 + 8 + 5 = 26

Hence, 26 houses pay less than ₹ 100.

(iii) The houses paying ₹ 100 or more are the houses of the classes 100-110, 110-120 and 120-130.

∴ Number of houses paying ₹ 100 or more = 7 + 3 + 4 = 14

Hence, 14 houses pay ₹ 100 or more.

Steps:

1. Since the scale on the x-axis starts at 50, a break (kink) is shown near the origin along the x-axis to indicate that the graph is drawn to scale beginning at 50 and not at the origin itself.

2. Take 1 cm on the x-axis = 5 units (weight in kg).

3. Take 1 cm on the y-axis = 1 patient.

4. Construct rectangles corresponding to the given data. As the data is grouped, the rectangles are drawn adjacent to one another with no gap in between.

The required histogram is as given below:

Steps:

1. Take 1 cm on the x-axis = 10 units (age in years).

2. Take 1 cm on the y-axis = 5 patients.

3. Construct rectangles corresponding to the given data. As the data is grouped, the rectangles are drawn adjacent to one another with no gap in between.

The required histogram is as given below:

From the histogram,

Wages (in ₹)	Number of workers
375-400	6
400-425	18
425-450	10
450-475	20
475-500	4

(i) The height of the rectangle over the class 400-425 is 18.

Hence, the frequency of the class 400-425 is 18.

(ii) The shortest rectangle is the one over the class 475-500, whose height is 4.

Hence, the class 475-500 has the minimum frequency.

(iii) The workers getting more than ₹ 425 are the workers of the classes 425-450, 450-475 and 475-500.

∴ Number of workers getting more than ₹ 425 = 10 + 20 + 4 = 34

Hence, 34 workers get more than ₹ 425.

(iv) The workers getting less than ₹ 475 are the workers of the classes 375-400, 400-425, 425-450 and 450-475.

∴ Number of workers getting less than ₹ 475 = 6 + 18 + 10 + 20 = 54

Hence, 54 workers get less than ₹ 475.

(v) The workers whose daily wages are more than or equal to ₹ 400 but less than ₹ 450 are the workers of the classes 400-425 and 425-450.

∴ Required number of workers = 18 + 10 = 28

Hence, 28 workers have daily wages more than or equal to ₹ 400 but less than ₹ 450.

From the histogram,

Hours of T.V. watched per day	Number of students
1-2	4
2-3	8
3-4	22
4-5	32
5-6	8
6-7	6

(i) The highest rectangle corresponds to the maximum number of students (i.e. 32), which lies in the class interval 4-5.

Hence, the maximum number of students watched T.V. for 4-5 hours.

(ii) The students watching T.V. for less than 4 hours are the students of the classes 1-2, 2-3 and 3-4.

∴ Required number of students = 4 + 8 + 22 = 34

Hence, 34 students watched T.V. for less than 4 hours.

(iii) The students spending more than 5 hours are the students of the classes 5-6 and 6-7.

∴ Required number of students = 8 + 6 = 14

Hence, 14 students spent more than 5 hours in watching T.V.

(iv) The students spending more than 2 hours but less than 4 hours are the students of the classes 2-3 and 3-4.

∴ Required number of students = 8 + 22 = 30

Hence, 30 students spent more than 2 hours but less than 4 hours in watching T.V.

To construct a pie chart, the angles for various sectors can be calculated as:

Total number of animals = 40 + 10 + 30 + 15 + 25 = 120

Animals	Number of animals	Angle
Deer	40	(40/120) × 360° = 120°
Tiger	10	(10/120) × 360° = 30°
Elephant	30	(30/120) × 360° = 90°
Giraffe	15	(15/120) × 360° = 45°
Reptiles	25	(25/120) × 360° = 75°

Steps:

1. Draw a circle of any radius.

2. Draw a horizontal radius of the circle.

3. The angle of sector for deer is 120°. Using protractor, draw an angle of 120° on the horizontal radius.

4. Continue marking the remaining sectors moving in clockwise direction till all the sectors are marked.

5. Shade the sectors with different colours/designs and write the information in the sectors.

Thus the required pie chart is obtained.

To construct a pie chart, the angles for various sectors can be calculated as:

Total monthly expenditure = ₹ (12,500 + 5,000 + 7,500 + 10,000 + 5,000 + 10,000) = ₹ 50,000

Items	Expenditure (in ₹)	Angle
Food	12,500	(12,500/50,000) × 360° = 90°
House rent	5,000	(5,000/50,000) × 360° = 36°
Education	7,500	(7,500/50,000) × 360° = 54°
Savings	10,000	(10,000/50,000) × 360° = 72°
Health	5,000	(5,000/50,000) × 360° = 36°
Others	10,000	(10,000/50,000) × 360° = 72°

Steps:

1. Draw a circle of any radius.

2. Draw a horizontal radius of the circle.

3. The angle of sector for food is 90°. Using protractor, draw an angle of 90° on the horizontal radius.

4. Continue marking the remaining sectors moving in clockwise direction till all the sectors are marked.

5. Shade the sectors with different colours/designs and write the information in the sectors.

Thus the required pie chart is obtained.

To construct a pie chart, the angles for various sectors can be calculated as:

Total of all the percentages = 25 + 20 + 20 + 10 + 15 + 10 = 100

Items	Expenditure (in %)	Angle
Paper cost	25%	(25/100) × 360° = 90°
Printing cost	20%	(20/100) × 360° = 72°
Binding	20%	(20/100) × 360° = 72°
Royalty	10%	(10/100) × 360° = 36°
Transportation cost	15%	(15/100) × 360° = 54°
Promotion cost	10%	(10/100) × 360° = 36°

Steps:

1. Draw a circle of any radius.

2. Draw a horizontal radius of the circle.

3. The angle of sector for paper cost is 90°. Using, protractor draw an angle of 90° on the horizontal radius.

4. Continue marking the remaining sectors moving in clockwise direction till all the sectors are marked.

5. Shade the sectors with different colours/designs and write the information in the sectors.

Thus the required pie chart is obtained.

To construct a pie chart, the angles for various sectors can be calculated as:

Total number of students = 400 + 300 + 500 + 250 + 350 = 1800

Stream	Number of students	Angle
Science	400	(400/1800) × 360° = 80°
Arts	300	(300/1800) × 360° = 60°
Commerce	500	(500/1800) × 360° = 100°
Law	250	(250/1800) × 360° = 50°
Management	350	(350/1800) × 360° = 70°

Steps:

1. Draw a circle of any radius.

2. Draw a horizontal radius of the circle.

3. The angle of sector for science is 80°. Using protractor, draw an angle of 80° on the horizontal radius.

4. Continue marking the remaining sectors moving in clockwise direction till all the sectors are marked.

5. Shade the sectors with different colours/designs and write the information in the sectors.

Thus the required pie chart is obtained.

To find the answers of the questions given above, we have to find the central angles of the sectors of the different sports.

From the pie chart, the central angles of the sectors are:

Sports	Central angle
Cricket	90°
Hockey	75°
Football	40°
Tennis	50°
Badminton	60°
Athletics	30°
Others	15°

(i) The central angle of the sector for cricket = 90°

⇒ Percent spent on cricket = $\frac{90^{\circ}}{360^{\circ}}$ × 100%

= $\frac{1}{4}$ × 100%

= 25%

Hence, 25% of the total expenditure is spent on cricket.

(ii) The central angle of the sector for hockey = 75°

The central angle of the sector for tennis = 50°

∴ Difference in the central angles = 75° - 50° = 25°

⇒ Required percent = $\frac{25^{\circ}}{360^{\circ}}$ × 100%

= $\frac{2500}{360}$%

= 6.94%

Hence, 6.94% (approx.) more is spent on hockey than on tennis.

(iii) The central angle of the sector for badminton = 60° and the total amount spent = ₹ 1,80,00,000

⇒ Amount spent on badminton = $\frac{60^{\circ}}{360^{\circ}}$ × 1,80,00,000

= $\frac{1}{6}$ × 1,80,00,000

= 30,00,000

Hence, the amount spent on badminton is ₹ 30,00,000.

(iv) The central angle of the sector for cricket = 90°

The central angle of the sector for hockey = 75°

∴ Total central angle for cricket and hockey = 90° + 75° = 165°

The total amount spent = ₹ 2,40,00,000

⇒ Required amount = $\frac{165^{\circ}}{360^{\circ}}$ × 2,40,00,000

= $\frac{11}{24}$ × 2,40,00,000

= 1,10,00,000

Hence, the amount spent on cricket and hockey together is ₹ 1,10,00,000.

To find the answers of the questions given above, we have to find the number of students enrolled in the different classes.

Given, the total number of students enrolled from VI to X = 1440

Class	Central angle	Number of students
VI	80°	(80°/360°) × 1440 = 320
VII	70°	(70°/360°) × 1440 = 280
VIII	85°	(85°/360°) × 1440 = 340
IX	75°	(75°/360°) × 1440 = 300
X	50°	(50°/360°) × 1440 = 200

(i) Number of students enrolled in class VIII = 340

Hence, 340 students are enrolled in class VIII.

(ii) Number of students enrolled in class IX = 300

Number of students enrolled in class X = 200

∴ Required number of students = 300 - 200 = 100

Hence, 100 students are more in class IX than in class X.

(iii) Number of students enrolled in class VII = 280

Number of students enrolled in class VIII = 340

∴ Sum of students enrolled in VII and VIII = 280 + 340 = 620

Hence, 620 students are enrolled in classes VII and VIII together.

(iv) Number of students enrolled in class VI = 320

Number of students enrolled in class X = 200

⇒ Required ratio = 320 : 200

⇒ Required ratio = 8 : 5

Hence, the ratio of students enrolled in VI to students enrolled in X is 8 : 5.

(i) When we spin the wheel shown in the figure, the pointer can stop in any one of the sectors marked as A, A, A, B, C and D.

Hence, the possible outcomes are A, A, A, B, C, D.

(ii) The bag contains 5 identical balls marked W (white), R (red), B (blue), G (green) and Y (yellow). The ball drawn can be any one of these.

Hence, the possible outcomes are W, R, B, G, Y.

When a die is rolled once, the possible outcomes are 1, 2, 3, 4, 5 or 6.

∴ Total number of outcomes = 6

(i) The event is getting an even number and the outcomes 2, 4, 6 are even numbers.

So, the number of favourable outcomes = 3

⇒ P(getting an even number) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{3}{6}$

= $\frac{1}{2}$

Hence, the probability of getting an even number is $\mathbf{\frac{1}{2}}$.

(ii) The event is getting a multiple of 3 and the outcomes 3, 6 are multiples of 3.

So, the number of favourable outcomes = 2

⇒ P(getting a multiple of 3) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{2}{6}$

= $\frac{1}{3}$

Hence, the probability of getting a multiple of 3 is $\mathbf{\frac{1}{3}}$.

(iii) The event is not getting a multiple of 3 and the outcomes 1, 2, 4, 5 are not multiples of 3.

So, the number of favourable outcomes = 4

⇒ P(not getting a multiple of 3) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{6}$

= $\frac{2}{3}$

Hence, the probability of not getting a multiple of 3 is $\mathbf{\frac{2}{3}}$.

When two coins are tossed together, the possible outcomes are HH, HT, TH and TT.

∴ Total number of outcomes = 4

(i) The event is getting two tails and the outcome TT is favourable to the event.

So, the number of favourable outcomes = 1

⇒ P(getting two tails) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{1}{4}$

Hence, the probability of getting two tails is $\mathbf{\frac{1}{4}}$.

(ii) The event is getting atleast one tail, means getting one tail or two tails, and the outcomes HT, TH, TT are favourable to the event.

So, the number of favourable outcomes = 3

⇒ P(getting atleast one tail) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{3}{4}$

Hence, the probability of getting atleast one tail is $\mathbf{\frac{3}{4}}$.

(iii) The event is getting no tail and the outcome HH is favourable to the event.

So, the number of favourable outcomes = 1

⇒ P(getting no tail) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{1}{4}$

Hence, the probability of getting no tail is $\mathbf{\frac{1}{4}}$.

When three coins are tossed together, the possible outcomes are HHH, HHT, HTH, HTT, TTT, TTH, THT and THH.

∴ Total number of outcomes = 8

(i) The event is getting atleast two heads, means getting two heads or three heads, and the outcomes HHH, HHT, HTH, THH are favourable to the event.

So, the number of favourable outcomes = 4

⇒ P(getting atleast two heads) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{8}$

= $\frac{1}{2}$

Hence, the probability of getting atleast two heads is $\mathbf{\frac{1}{2}}$.

(ii) The event is getting atleast one tail, means getting one tail or two tails or three tails, and the outcomes HHT, HTH, HTT, TTT, TTH, THT, THH are favourable to the event.

So, the number of favourable outcomes = 7

⇒ P(getting atleast one tail) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{7}{8}$

Hence, the probability of getting atleast one tail is $\mathbf{\frac{7}{8}}$.

(iii) The event is getting atmost one tail, means getting no tail or one tail, and the outcomes HHH, HHT, HTH, THH are favourable to the event.

So, the number of favourable outcomes = 4

⇒ P(getting atmost one tail) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{8}$

= $\frac{1}{2}$

Hence, the probability of getting atmost one tail is $\mathbf{\frac{1}{2}}$.

Total number of screws in the box = 600

∴ Total number of outcomes = 600

Number of rusted screws = $\frac{1}{10}$ × 600 = 60

∴ Number of screws which are not rusted = 600 - 60 = 540

(i) The event is getting a rusted screw.

The number of rusted screws in the box is 60.

So, the number of favourable outcomes = 60

⇒ P(getting a rusted screw) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{60}{600}$

= $\frac{1}{10}$

Hence, the probability of getting a rusted screw is $\mathbf{\frac{1}{10}}$.

(ii) The event is getting a screw which is not rusted.

The number of screws which are not rusted in the box is 540.

So, the number of favourable outcomes = 540

⇒ P(getting a screw which is not rusted) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{540}{600}$

= $\frac{9}{10}$

Hence, the probability of getting a screw which is not rusted is $\mathbf{\frac{9}{10}}$.

In the word 'TRIANGLE' there are 8 letters T, R, I, A, N, G, L and E.

∴ Total number of outcomes = 8

Out of these 8 letters, there are three vowels I, A and E.

So, the number of favourable outcomes = 3

⇒ P(getting a vowel) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{3}{8}$

Hence, the probability that the letter chosen is a vowel is $\mathbf{\frac{3}{8}}$.

Total number of balls in the bag = 5 + 6 + 4 = 15

∴ Total number of outcomes = 15

(i) The event is getting a white ball.

The number of white balls in the bag is 4.

So, the number of favourable outcomes = 4

⇒ P(getting a white ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{15}$

Hence, the probability of getting a white ball is $\mathbf{\frac{4}{15}}$.

(ii) The event is getting a ball which is not black.

The number of balls which are not black in the bag = 5 + 4 = 9

So, the number of favourable outcomes = 9

⇒ P(getting a ball which is not black) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{9}{15}$

= $\frac{3}{5}$

Hence, the probability of getting a ball which is not black is $\mathbf{\frac{3}{5}}$.

(iii) The event is getting a red or a black ball.

The number of red or black balls in the bag = 5 + 6 = 11

So, the number of favourable outcomes = 11

⇒ P(getting a red or a black ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{11}{15}$

Hence, the probability of getting a red or a black ball is $\mathbf{\frac{11}{15}}$.

(iv) The event is getting a ball which is neither red nor black, that is getting a white ball.

The number of white balls in the bag is 4.

So, the number of favourable outcomes = 4

⇒ P(getting a ball which is neither red nor black) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{15}$

Hence, the probability of getting a ball which is neither red nor black is $\mathbf{\frac{4}{15}}$.

The box contains 17 cards numbered 1, 2, 3, ....., 17 and one card is drawn.

∴ Total number of outcomes = 17

(i) The event is getting an odd number and the outcomes 1, 3, 5, 7, 9, 11, 13, 15, 17 are odd numbers.

So, the number of favourable outcomes = 9

⇒ P(getting an odd number) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{9}{17}$

Hence, the probability that the number on the card is odd is $\mathbf{\frac{9}{17}}$.

(ii) The event is getting an even number and the outcomes 2, 4, 6, 8, 10, 12, 14, 16 are even numbers.

So, the number of favourable outcomes = 8

⇒ P(getting an even number) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{8}{17}$

Hence, the probability that the number on the card is even is $\mathbf{\frac{8}{17}}$.

(iii) The event is getting a prime number and the outcomes 2, 3, 5, 7, 11, 13, 17 are prime numbers.

So, the number of favourable outcomes = 7

⇒ P(getting a prime number) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{7}{17}$

Hence, the probability that the number on the card is prime is $\mathbf{\frac{7}{17}}$.

(iv) The event is getting a number divisible by 3 and the outcomes 3, 6, 9, 12, 15 are divisible by 3.

So, the number of favourable outcomes = 5

⇒ P(getting a number divisible by 3) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{5}{17}$

Hence, the probability that the number on the card is divisible by 3 is $\mathbf{\frac{5}{17}}$.

(v) A number divisible by both 2 and 3 is divisible by 6.

The event is getting a number divisible by 2 and 3 both and the outcomes 6, 12 are divisible by 6.

So, the number of favourable outcomes = 2

⇒ P(getting a number divisible by 2 and 3 both) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{2}{17}$

Hence, the probability that the number on the card is divisible by 2 and 3 both is $\mathbf{\frac{2}{17}}$.

There are 52 playing cards and one card is drawn.

∴ Total number of outcomes = 52

(i) The event is getting an ace. There are 4 aces in a pack of 52 cards.

So, the number of favourable outcomes = 4

⇒ P(getting an ace) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{52}$

= $\frac{1}{13}$

Hence, the probability of getting an ace is $\mathbf{\frac{1}{13}}$.

(ii) The event is getting a red card. There are 26 red cards, that is 13 cards of heart and 13 cards of diamond.

So, the number of favourable outcomes = 26

⇒ P(getting a red card) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{26}{52}$

= $\frac{1}{2}$

Hence, the probability of getting a red card is $\mathbf{\frac{1}{2}}$.

(iii) The event is getting neither a king nor a queen. There are 4 kings and 4 queens in a pack of 52 cards.

∴ Number of cards which are neither a king nor a queen = 52 - (4 + 4) = 52 - 8 = 44

So, the number of favourable outcomes = 44

⇒ P(getting neither a king nor a queen) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{44}{52}$

= $\frac{11}{13}$

Hence, the probability of getting neither a king nor a queen is $\mathbf{\frac{11}{13}}$.

(iv) The event is getting a red face card. The face cards are kings, queens and jacks, so the red face cards are 2 kings, 2 queens and 2 jacks.

∴ Number of red face cards = 2 + 2 + 2 = 6

So, the number of favourable outcomes = 6

⇒ P(getting a red face card) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{6}{52}$

= $\frac{3}{26}$

Hence, the probability of getting a red face card is $\mathbf{\frac{3}{26}}$.

(v) The event is getting a card of spade or an ace. There are 13 cards of spade, which include the ace of spade, and 3 more aces of the other suits.

∴ Number of cards which are of spade or an ace = 13 + 3 = 16

So, the number of favourable outcomes = 16

⇒ P(getting a card of spade or an ace) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{16}{52}$

= $\frac{4}{13}$

Hence, the probability of getting a card of spade or an ace is $\mathbf{\frac{4}{13}}$.

(vi) The event is getting a non face card of red colour. There are 26 red cards, out of which 6 are face cards.

∴ Number of non face cards of red colour = 26 - 6 = 20

So, the number of favourable outcomes = 20

⇒ P(getting a non face card of red colour) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{20}{52}$

= $\frac{5}{13}$

Hence, the probability of getting a non face card of red colour is $\mathbf{\frac{5}{13}}$.

Total number of tickets = 5 + 995 = 1000

∴ Total number of outcomes = 1000

The event is winning a prize.

The number of prized tickets is 5.

So, the number of favourable outcomes = 5

⇒ P(winning a prize) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{5}{1000}$

= $\frac{1}{200}$

Hence, the probability of his winning a prize is $\mathbf{\frac{1}{200}}$.

(i) Bar graphs are visual representation of ungrouped data.

(ii) In a grouped frequency distribution, the difference between lower limit and upper limit of a class is called class size or class width.

(iii) The mid point of the class interval is called class mark.

(iv) Bar graphs of grouped data are called histograms.

(v) The circle graphs are commonly called pie charts or pie graphs.

(vi) An experiment which has more than one possible outcomes and it is not possible to predict the outcome in advance is called random experiment.

(vii) The outcomes which ensures the occurrence of an event are called favourable outcomes.

(viii) An event which never happens is called an impossible event.

(i) The data arranged in ascending or descending order of size is called data array.

Hence, the given statement is true.

(ii) In the class 10-20, the smaller number 10 is the lower limit and the larger number 20 is the upper limit.

Hence, the given statement is false.

(iii) Class size = Upper limit - Lower limit

⇒ Class size = 30 - 20

⇒ Class size = 10

Hence, the given statement is true.

(iv) Class mark = $\frac{\text{Upper limit + Lower limit}}{2}$

⇒ Class mark = $\frac{35 + 25}{2}$

⇒ Class mark = $\frac{60}{2}$

⇒ Class mark = 30

Hence, the given statement is true.

(v) Bar graphs are used for ungrouped data and there are gaps between the bars, whereas histograms are used for grouped data and there is no gap between adjacent rectangles.

Hence, the given statement is false.

(vi) In a histogram the breadth of a rectangle is meaningful as it represents the class size.

Hence, the given statement is false.

(vii) In histograms, there is no gap between two adjacent rectangles as there is no gap between the class intervals.

Hence, the given statement is true.

(viii) In a pie chart, the whole circle is divided into sectors and the size of each sector is proportional to the value of the item represented by it.

Hence, the given statement is true.

(ix) As the angle about the centre of a circle is 360°, in a pie chart angle of sector = $\frac{\text{value of item}}{\text{sum of values of all items}} \times 360^{\circ}$ and not 180°.

Hence, the given statement is false.

(x) On tossing a coin there is no reason for a head to occur in preference to a tail, so getting a head or a tail are equally likely events.

Hence, the given statement is true.

(xi) The number of outcomes favourable to an event E is greater than or equal to 0 and is less than or equal to the total number of outcomes, so 0 ≤ P(E) ≤ 1.

Hence, the given statement is true.

(xii) P(occurrence of an event) and P(non-occurrence of an event) are equal only when the number of favourable outcomes is half of the total number of outcomes, which is not true for every event.

For example, when a die is thrown, P(getting the number 5) = $\frac{1}{6}$ whereas P(not getting the number 5) = $\frac{5}{6}$.

Hence, the given statement is false.

Class size = Upper limit - Lower limit

Taking the class 10-20,

⇒ Class size = 20 - 10

⇒ Class size = 10

Hence, option 3 is the correct option.

From the table, the frequencies are 14, 11, 11, 10 and 4.

The maximum of these frequencies is 14, which corresponds to the class 10-20.

Hence, option 1 is the correct option.

From the table, the frequencies are 14, 11, 11, 10 and 4.

The minimum of these frequencies is 4, which corresponds to the class 50-60.

The upper limit of the class 50-60 is 60.

Hence, option 4 is the correct option.

Class mark = $\frac{\text{Upper limit + Lower limit}}{2}$

Taking the class 20-30,

⇒ Class mark = $\frac{30 + 20}{2}$

⇒ Class mark = $\frac{50}{2}$

⇒ Class mark = 25

The frequency of the class 20-30 is 11.

Hence, option 2 is the correct option.

The pie graph shown in the adjoining figure represents different subjects liked by the students of class VIII. Study the pie graph carefully and choose the correct answer from the given four options for questions 7 to 10.

From the pie graph, the percentages of students liking the different subjects are:

Subject	Percentage of students
Maths	40%
Science	30%
S. Science	20%
English	10%

The maximum of these percentages is 40%, which corresponds to Maths.

Hence, option 1 is the correct option.

From the pie graph, the percentages of students liking the different subjects are:

Subject	Percentage of students
Maths	40%
Science	30%
S. Science	20%
English	10%

The minimum of these percentages is 10%, which corresponds to English.

Hence, option 4 is the correct option.

From the pie graph, 20% of the students like S. Science and the total number of students is 200.

⇒ Number of students who like S. Science = $\frac{20}{100}$ × 200

⇒ Number of students who like S. Science = 40

Hence, option 3 is the correct option.

From the pie graph, 30% of the students like Science and the total number of students is 200.

⇒ Number of students who like Science = $\frac{30}{100}$ × 200

⇒ Number of students who like Science = 60

Hence, option 3 is the correct option.

When three coins are tossed together, the possible outcomes are HHH, HHT, HTH, HTT, TTT, TTH, THT and THH.

∴ Total number of outcomes = 8

The event is getting exactly 2 heads and the outcomes HHT, HTH, THH are favourable to the event.

So, the number of favourable outcomes = 3

⇒ P(getting exactly 2 heads) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{3}{8}$

Hence, option 3 is the correct option.

In the word 'FATHER' there are 6 letters F, A, T, H, E and R.

∴ Total number of outcomes = 6

Out of these 6 letters, there are two vowels A and E, so there are four consonants F, T, H and R.

So, the number of favourable outcomes = 4

⇒ P(selecting a consonant) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{6}$

= $\frac{2}{3}$

Hence, option 4 is the correct option.

When a pair of coins is tossed, the possible outcomes are HH, HT, TH and TT.

∴ Total number of outcomes = 4

As only two coins are tossed, the number of heads cannot be more than 2 and so no outcome is favourable to the event.

So, the number of favourable outcomes = 0

⇒ P(getting more than 2 heads) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{0}{4}$

= 0

Thus, getting more than 2 heads is an impossible event and the probability of an impossible event is 0.

Hence, option 4 is the correct option.

Total number of balls in the bag = 20

∴ Total number of outcomes = 20

The event is getting a red ball.

As all the 20 balls in the bag are red, every outcome is favourable to the event.

So, the number of favourable outcomes = 20

⇒ P(getting a red ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{20}{20}$

= 1

Thus, getting a red ball is a sure event and the probability of a sure event is 1.

Hence, option 2 is the correct option.

Total number of balls in the bag = 4 + 5 + 3 = 12

∴ Total number of outcomes = 12

The event is getting a non red ball.

The number of non red balls in the bag = 5 + 3 = 8

So, the number of favourable outcomes = 8

⇒ P(getting a non red ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{8}{12}$

= $\frac{2}{3}$

Hence, option 2 is the correct option.

When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT.

∴ Total number of outcomes = 4

The event is getting different results on both the coins and the outcomes HT, TH are favourable to the event.

So, the number of favourable outcomes = 2

⇒ P(getting different results) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{2}{4}$

= $\frac{1}{2}$

So, Statement I is true.

The number of outcomes favourable to an event E is greater than or equal to 0 and is less than or equal to the total number of outcomes, so 0 ≤ P(E) ≤ 1.

So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

By definition, P(E) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes of the experiment}}$

So, Statement II is true.

There are 52 playing cards and one card is drawn.

∴ Total number of outcomes = 52

The event is getting a card of honour. The aces together with the face cards, that is 4 aces, 4 kings, 4 queens and 4 jacks, are called cards of honour.

∴ Number of cards of honour = 4 + 4 + 4 + 4 = 16

So, the number of favourable outcomes = 16

⇒ P(getting a card of honour) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{16}{52}$

= $\frac{4}{13}$

So, Statement I is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

The numbers appearing on the die are 1, 2, 3, 4, 5 and 6. Taking the observations of the given data one at a time and marking a tally mark against the corresponding number, the frequency distribution table of the given data is:

Number on the die (Variate)	Tally marks	Frequency (No. of students)
1	IIII I	6
2	IIII II	7
3	IIII I	6
4	IIII III	8
5	IIII II	7
6	IIII I	6
Total		40

Hence, the required frequency distribution table is as given above.

Here, maximum = 76 and minimum = 30.

∴ Range = 76 - 30 = 46

The classes are to be of size 5. Since we want to include 30 in the first class, 30 is the lower limit of the first class, and since we want to include 76 in the last class, 80 is the upper limit of the last class.

So, let us form 10 classes each of size 5.

The grouped frequency distribution table of the given data is:

Weight (in grams) (Class interval)	Tally marks	Number of oranges (Frequency)
30 - 35	III	3
35 - 40	II	2
40 - 45	III	3
45 - 50	III	3
50 - 55	IIII	5
55 - 60	IIII	4
60 - 65	IIII	5
65 - 70	II	2
70 - 75	IIII I	6
75 - 80	II	2
Total		35

(i) The classes are 30 - 35, 35 - 40, 40 - 45, 45 - 50, 50 - 55, 55 - 60, 60 - 65, 65 - 70, 70 - 75 and 75 - 80.

Hence, there are 10 classes in the frequency distribution table.

(ii) From the table, the frequencies are 3, 2, 3, 3, 5, 4, 5, 2, 6 and 2.

The lowest of these frequencies is 2, which corresponds to the classes 35 - 40, 65 - 70 and 75 - 80.

Hence, the weight groups 35 - 40, 65 - 70 and 75 - 80 have the lowest frequency.

(iii) From the table, the frequencies are 3, 2, 3, 3, 5, 4, 5, 2, 6 and 2.

The highest of these frequencies is 6, which corresponds to the class 70 - 75.

Hence, the weight group 70 - 75 has the highest frequency.

Steps:

1. Take 1 cm on x-axis = 10 units (marks obtained).

2. Take 1 cm on y-axis = 2 students (frequency).

3. Construct rectangles corresponding to the given data. As the data is grouped, the rectangles are drawn adjacent to one another with no gap in between.

The required histogram is as given below:

To construct a pie chart, the angles for various sectors can be calculated as:

Total number of students = 360 + 240 + 80 + 120 = 800

Mode of transport	Number of students	Angle
Bicycle	360	(360/800) × 360° = 162°
School bus	240	(240/800) × 360° = 108°
Car	80	(80/800) × 360° = 36°
Walking	120	(120/800) × 360° = 54°

Steps:

1. Draw a circle of any radius.

2. Draw a horizontal radius of the circle.

3. The angle of sector for bicycle is 162°. Using protractor draw an angle of 162° on the horizontal radius.

4. Continue marking the remaining sectors moving in clockwise direction till all the sectors are marked.

5. Shade the sectors with different colours/designs and write the information in the sectors.

Thus the required pie chart is obtained.

To find the answers of the questions given above, we have to find the percentage of viewers watching the different types of channels.

From the pie chart, the percentages of viewers are:

Type of programme	Percentage of viewers
Entertainment	50%
Sports	25%
News	15%
Informative	10%

(i) The largest of these percentages is 50%, which corresponds to Entertainment.

Hence, entertainment programmes are viewed the most.

(ii) The smallest of these percentages is 10%, which corresponds to Informative.

Hence, informative programmes are viewed the least.

(iii) Percentage of viewers watching sports channels = 25%

Considering the remaining types of programmes two at a time,

Percentage of viewers watching news and informative programmes = 15% + 10% = 25%

Hence, news and informative programmes together have the number of viewers equal to those watching sports channels.

When the wheel shown in the figure is spun, the pointer may stop in any one of the 8 sectors, out of which 5 sectors are green (G) and 3 sectors are red (R). Hence, the possible outcomes are G, G, G, G, G, R, R, R.

∴ Total number of outcomes = 8

(i) The outcomes of getting a green sector are the outcomes marked G and the outcomes of not getting a green sector are the outcomes marked R.

Hence, the outcomes of getting a green sector are G, G, G, G, G and the outcomes of not getting a green sector are R, R, R.

(ii) The event is getting a green sector and the outcomes G, G, G, G, G are favourable to the event.

So, the number of favourable outcomes = 5

⇒ P(getting a green sector) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{5}{8}$

Hence, the probability of getting a green sector is $\mathbf{\frac{5}{8}}$.

(iii) The event is not getting a green sector and the outcomes R, R, R are favourable to the event.

So, the number of favourable outcomes = 3

⇒ P(not getting a green sector) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{3}{8}$

Hence, the probability of not getting a green sector is $\mathbf{\frac{3}{8}}$.

Total number of balls in the bag = 4 + 2 = 6

∴ Total number of outcomes = 6

(i) The event is getting a red ball.

The number of red balls in the bag is 4.

So, the number of favourable outcomes = 4

⇒ P(getting a red ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{4}{6}$

= $\frac{2}{3}$

Hence, the probability of getting a red ball is $\mathbf{\frac{2}{3}}$.

(ii) The event is not getting a red ball, that is getting a yellow ball.

The number of yellow balls in the bag is 2.

So, the number of favourable outcomes = 2

⇒ P(not getting a red ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{2}{6}$

= $\frac{1}{3}$

Hence, the probability of not getting a red ball is $\mathbf{\frac{1}{3}}$.

When three coins are tossed together, the possible outcomes are HHH, HHT, HTH, HTT, TTT, TTH, THT and THH.

∴ Total number of outcomes = 8

(i) The event is getting atmost 2 heads, means getting no head or one head or two heads, and the outcomes HHT, HTH, HTT, TTT, TTH, THT, THH are favourable to the event.

So, the number of favourable outcomes = 7

⇒ P(getting atmost 2 heads) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{7}{8}$

Hence, the probability of getting atmost 2 heads is $\mathbf{\frac{7}{8}}$.

(ii) The event is getting 3 heads and the outcome HHH is favourable to the event.

So, the number of favourable outcomes = 1

⇒ P(getting 3 heads) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{1}{8}$

Hence, the probability of getting 3 heads is $\mathbf{\frac{1}{8}}$.

In the word 'RECTANGLE' there are 9 letters R, E, C, T, A, N, G, L and E.

∴ Total number of outcomes = 9

Out of these 9 letters, there are three vowels E, A and E, so there are six consonants R, C, T, N, G and L.

(i) The event is getting a consonant and the letters R, C, T, N, G, L are favourable to the event.

So, the number of favourable outcomes = 6

⇒ P(getting a consonant) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{6}{9}$

= $\frac{2}{3}$

Hence, the probability that the letter chosen is a consonant is $\mathbf{\frac{2}{3}}$.

(ii) The event is not getting a consonant, that is getting a vowel, and the letters E, A, E are favourable to the event.

So, the number of favourable outcomes = 3

⇒ P(not getting a consonant) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{3}{9}$

= $\frac{1}{3}$

Hence, the probability that the letter chosen is not a consonant is $\mathbf{\frac{1}{3}}$.

(i) Total number of balls in the bag = 12

∴ Total number of outcomes = 12

The event is getting a black ball.

The number of black balls in the bag is $x$.

So, the number of favourable outcomes = $x$

⇒ P(getting a black ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{x}{12}$

Hence, the probability of getting a black ball is $\mathbf{\frac{x}{12}}$.

(ii) When 6 more black balls are put in the bag,

Total number of balls in the bag = 12 + 6 = 18

∴ Total number of outcomes = 18

The number of black balls in the bag = $x$ + 6

So, the number of favourable outcomes = $x$ + 6

⇒ P(getting a black ball) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{x + 6}{18}$

Given, this probability is double of that in part (i).

$$\begin{aligned}\Rightarrow \frac{x + 6}{18} = 2 \times \frac{x}{12} \\[1em] \Rightarrow \frac{x + 6}{18} = \frac{x}{6} \\[1em] \Rightarrow 6(x + 6) = 18x \\[1em] \Rightarrow 6x + 36 = 18x \\[1em] \Rightarrow 36 = 18x - 6x \\[1em] \Rightarrow 36 = 12x \\[1em] \Rightarrow x = \frac{36}{12} \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, the value of x is 3.

The year 2010 is not a leap year, so it has 365 days. Ankita's birthday is fixed on one of these 365 days and Nagma's birthday can be any one of these 365 days.

∴ Total number of outcomes = 365

(i) The event is that Nagma's birthday is the same as Ankita's birthday, which can happen on only one of the 365 days.

So, the number of favourable outcomes = 1

⇒ P(they have the same birthday) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{1}{365}$

Hence, the probability that they have the same birthday is $\mathbf{\frac{1}{365}}$.

(ii) The event is that Nagma's birthday is different from Ankita's birthday, which can happen on any one of the remaining 365 - 1 = 364 days.

So, the number of favourable outcomes = 364

⇒ P(they have different birthdays) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{364}{365}$

Hence, the probability that they have different birthdays is $\mathbf{\frac{364}{365}}$.