Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 3 - Data Handling

Answer:

Let x, y, z be three observations.

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\\ & =& \frac{x+y+z}{3}\end{array}$

Hence, the correct answer is option (b).

Answer:

The data is 33, 38, 48, 33, 34, 34, 33 and 24.
Since, 33 occurs most frequently i.e., three times.
∴ Mode = 33

Hence, the correct answer is option (c).

Answer:

Mean and mode affected by observations on the ends.

Hence, the correct answer is option (a).

Answer:

Data is 21, 6, 17, 18, 12, 8, 4, 13 Highest observation is 21 and lowest observation is 4.
∴ Range = 21 – 4 = 17

Hence, the correct answer is option (a).

Answer:

Arranging the data in ascending order, we have
3, 3, 4, 4, 5, 6, 7
Observations are odd in numbers.
∴ Medium is the middle most observation i.e., 4.

Hence, the correct answer is option (c).

Answer:

Since mode is the measure of central tendency which is use to find that event which occurs most often.

Hence, the correct answer is option (b).

Answer:

Since in option (c), only two cards are placed face down i.e., probability of picking up 2 aces in this is certain.

Hence, the correct answer is option (c).

Answer:

Set (d) has 6 cards.
We have to pick 2 aces.
∴ Probability (picking up 2 aces) = $\frac{2}{6}$

Hence, the correct answer is option (b).

Answer:

Since, Range = Highest observation – Lowest observation

Hence, the correct answer is option (c).

Answer:

Probability (picking up student who is participating in a quiz) = $\frac{2}{5}$

∴ Required percentage $=\frac{2}{5}\times 100=40\%$

Hence, the correct answer is option (b).

Answer:

Lowest observation = -17
Highest observation – 20
∴ Range = 20 – (-17) = 20 + 17 = 37

Hence, the correct answer is option (b).

Answer:

As we know that when a coin is tossed, it has two possible outcomes, head or tail.

Hence, the correct answer is option (d).

Answer:

Lowest number = 19
Let highest number be x and middle number be y,
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\\ & =& \frac{x+y+z}{3}\end{array}$

Given that mean is 40.
x + y + 19 = 120
⇒ x + y = 120 – 19 = 101

Since 19 < x and 19 < y and x > y

∴ Maximum value of x could be 81 as 81 + 20 = 101 and 81 > 19,81 > 20, 20 > 19

Hence, the correct answer is option (a).

Answer:

Mean of the first three examinations scores = $\frac{97+73+88}{3}=\frac{258}{3}=86$

Mean of the first four examinations scores = $\frac{97+73+88+80}{4}=\frac{338}{4}=84.5$


Thus, the average score of Khilona will be decreased by 86 – 84.5 = 1.5

Hence, the correct answer is option (d).

Answer:

Since the mode is the measure of central tendency which is use to find that event which occurs most often.

Hence, the correct answer is option (c).

Answer:

(a) Data is 1, 2, 3, 4, 4, 5, 6
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{1+2+3+4+4+5+6}{7}\\ & =& \frac{25}{7}\\ & =& 3.57\end{array}$
Mode = 4 (∵ 4 occurs two times)
∴ Mean ≠ Mode
(b) Data is 1, 2, 2, 2, 3, 3, 5
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{1+2+2+2+3+3+5}{7}\\ & =& \frac{17}{7}\\ & =& 2.42\end{array}$
Mode = 2 (∵ 2 occurs three times)
∴ Mean ≠ Mode

(c) Data is 2, 2, 3, 3, 3, 7, 8
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{2+2+3+3+3+7+8}{7}\\ & =& \frac{28}{7}\\ & =& 4\end{array}$
Mode = 3 (∵ 3 occurs three times)
∴ Mean ≠ Mode
(d) Data is 3, 3, 4, 4, 4, 4, 6
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{3+3+4+4+4+4+6}{7}\\ & =& \frac{28}{7}\\ & =& 4\end{array}$
Mode = 4 (∵ 4 occurs four times)
Since, observation are odd in number
∴ Median = Middle most observation i.e., 4
Hence mean = median = mode
Hence, the correct answer is option (d).

Answer:

Range

Answer:

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\end{array}$

Answer:

Mode

Answer:

Median

Answer:

Central tendency

Answer:

1

Answer:

Fill in the blanks to make the statements true.
The probability of an event which is impossible to happen is _________.

Answer:

Total outcomes when a die is thrown are 1, 2, 3, 4, 5, 6 i.e., 6 in number.
All are less than 7.
∴ Probability (getting a number less than 7) $=\frac{6}{6}=1$

Answer:

When a die is thrown then possible outcomes are 1, 2, 3, 4, 5, 6 i.e., 6 in number.

Answer:

A double bar graph

Answer:

Bar graph

Answer:

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\\ 5& =& \frac{8+4+x+6+2+7}{6}\\ 30& =& 27+x\\ x& =& 3\end{array}$

Answer:

Minimum, maximum : Since, median is the middle most observation.

Answer:

Odd

Answer:

Since maximum number of students i.e., 25 lies in interval 52-55.

Answer:

When a die is thrown all possible outcome are less than or equal to 6. There is no number which is greater than 6.
∴ Probability of getting a number greater than 6 is 0.

Thus, the statement is false.

Answer:

When a coin is tossed, there are two possible outcomes i.e., Head or Tail.

Thus, the statement is true.

Answer:

Since extreme observations affect mean and mode not the median.

Thus, the statement is false.

Answer:

Since measures of central tendency always lie between maximum and minimum values of data.

Thus, the statement is false.

Answer:

Since the maximum sum of the numbers on both dice is 12.

Thus, the statement is true.

Answer:

Total parts = 4
Shaded parts = 2
∴ Probability (spining arrow stopping in shaded region) $=\frac{2}{4}=\frac{1}{2}$

Thus, the statement is true.

Answer:

Total outcomes = 15
Number of occurrence of head = 7 .
∴ The percentage of the chance of occurrence of a head $=\frac{7}{15}\times 100=46.67\%$

Thus, the statement is false.

Answer:

Mean, Median and Mode may be the same for some data like normal distribution.

Thus, the statement is true.

Answer:

Since, probability of getting an ace out of a deck of cards is $\frac{4}{52}=\frac{1}{13}$ which is less than 1.

Thus, the statement is false.

Answer:

Mean of the data may or may not from the given data.

Thus, the statement is false.

Answer:

Median of the data may or may not be from the given data.

Thus, the statement is true.

Answer:

Mode of the data is always from the given data.

Thus, the statement is true.

Answer:

Mean of the data cannot be lesser than each of the observations.

Thus, the statement is false.

Answer:

Mean may be a fraction.

Thus, the statement is false.

Answer:

Since the range is the difference between the highest observation and lowest observation.
∴ It may or may not be from the data.


Thus, the statement is false.

Answer:

Data is 12, 13, 14, 15, 16
Since, every observation occurs one time.
∴ Every observation can taken as mode.

Thus, the statement is true.

Answer:

Data is 2, -5, 4, 3, 7, 6
Range = Highest value – Lowest value = 7 – (-5) = 7 + 5 = 12
After subtracting 2 from each value, data becomes 0, -7, 2, 1, 5, 4
Now, range = Highest value – Lowest value = 5 – (-7) – 5 + 7 = 12
Hence, in both cases range is same.


Thus, the statement is false.

Answer:

Data is 3, 7, 1,-2, 2, 6, -3, -5
Range = Highest value – Lowest value = 7 – (-5) = 12
After adding 8 to each value, data becomes 11, 15, 9, 6, 10, 14, 5, 3
∴ Range = Highest value – Lowest value = 15 – 3 = 12
Hence, in both cases range is same.

Thus, the statement is false.

Answer:

Data is 5, 10, 10, 12, 13
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\\ & =& \frac{5+10+10+12+13}{5}\\ & =& \frac{50}{5}\\ & =& 10\end{array}$

Since, 10 occurs most frequently i.e., two times.
∴ Mode = 10
Observations are odd in number.
∴ Median = Middle most observation i.e., 3rd observation
∴ Median = 10
Mean = Median = Mode = 10
Yes, these three are equal.

Answer:

First ten even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{2+4+6+8+10+12+14+16+18+20}{10}\\ & =& \frac{110}{11}\\ & =& 11\end{array}$

Answer:

Mode is the height that occurs most frequently in the set of heights of 50 children.

Answer:

(a) Black colour of car is most liked because number of cars sold of black car is greatest.

(b) Mode is used in (a) part.

Answer:

Arranging the data in ascending order, we get 17, 18, 23, 25, 29, 31, 35, 35, 35, 37, 38, 42
(a) Range – Highest value – Lowest value = 42 – 17 = 25

(b)
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{17+18+23+25+29+31+35+35+35+37+38+42}{12}\\ & =& \frac{365}{12}\\ & =& 30.41\end{array}$
(c) Observations are even in number.
∴ Median = Mean of two middle most observations
$$\begin{gathered} =\frac{31+35}{2} \\ =\frac{66}{2} \\ =33 \end{gathered}$$

(d) Since, 35 occurs most frequently i.e., 3 times.
∴ Mode = 35.

Answer:

Arranging the data in ascending order, we get
54, 57, 59, 60, 62, 62, 62, 65, 70, 75, 78, 84
$\begin{array}{rcl}\left(\mathrm{a}\right) \mathrm{Mean}& =& \frac{54+57+59+60+62+62+62+65+70+75+78+84}{12}\\ & =& \frac{788}{12}\\ & =& 65.6 \mathrm{kg}\end{array}$

(b) Since, mean = 65.6 kg
∴ 4 people have weight above mean weight.

(c) Range = Highest weight – Lowest weight
= (84 – 54) kg = 30 kg

Answer:

Total cards = 5
(a) Here, all cards are vowel.
∴ Chance of drawing out a vowel is 1.
(b) Chance of drawing out a card marked A or I is $\frac{2}{5}$.
(c) Since, one card has marked U.
∴ Chance of drawing out a card marked U is $\frac{1}{5}$.
(d) Since, there is no consonant.
∴ Chance of drawing out a consonant is 0.

Answer:

Arranging the data in a ascending order, we get
2, 2, 3, 3, 4, 4, 4, 5, 6
Since, observations are odd in number.
∴ Median = Middle most observation = 4
Since, 4 occurs most frequently i.e., 3 times
∴ Mode = 4
Now, mean of median and mode is given by

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Median}+\mathrm{Mode}}{2}\\ & =& \frac{4+4}{2}\\ & =& 4\end{array}$

Answer:

The data is 5, 7, 7, 8,x,5, 4, 3, 1, 2.

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{5+7+7+8+x+5+4+3+1+2}{10}\\ 4.5& =& \frac{x+42}{10}\\ 45-42& =& x\\ x& =& 3\end{array}$
Now, arranging the data in ascending order, we get
1, 2, 3, 3, 4, 5, 5, 7, 7, 8
Since, observations are even in number.
∴ Median = Mean of two middle most observations
$$\begin{gathered} =\frac{4+5}{2} \\ =4.5 \end{gathered}$$

Answer:

It is certain that sun setting every evening
∴ Probability of sun setting tomorrow is 1.

Answer:

Colour Y has more chance to show up with arrow than the others because Y has more region than both colours B and R.

Answer:

Total number of students = 3 + 4 = 7
Total number of boys = 4
∴ Probability (student chosen is a boy) = $\frac{4}{7}$

Answer:

Total letters in the word MEDIAN = 6

D occurs only at once.
∴ Probability (Drawn slip bears the letter D) = $\frac{1}{6}$

Answer:

(a) Since, no number is less than 1 on a die.
∴Getting a number less than 1 on throwing a die is impossible to happen.

(b) When a coin is tossed two outcomes are possible i.e., Head or Tail.
∴ Getting head when a coin is tossed may or may not happen.

(c) A team winning the match may or may not happen because a team can lose the match.

(d) Christmas will be on 25 December is certain to happen.

(e) Since, moon always revolves around the earth.
∴Today moon will not revolve around the earth is impossible to happen.

(f) A ball thrown up in the air will fall down after some time is certain to happen due to gravity of earth.

Answer:

Arranging the data in ascending order, we get
1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\\ & =& \frac{5+3+4+1+2+6+4+2+2+3+1+5+6+1+2}{15}\\ & =& \frac{47}{15}\\ & =& 3.13\end{array}$
Since observations are odd in number.

∴ Median = Middle most observation i.e., 8th observation = 3

Since 2 occurs most frequently i.e., 4 times
∴ Mode = 2

Answer:

First six multiples of 4 are 4, 8, 12, 16, 20, 24.
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\\ & =& \frac{84}{6}\\ & =& 14\end{array}$
 

Answer:

First nine even natural numbers are
2, 4, 6, 8, 10, 12, 14, 16, 18
Since, observations are odd in number.
∴ Median = Middle most observation i.e., 5th observation = 10

Answer:

Mean of 3 numbers = 10
Mean of 4 numbers = 12
$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\\ & \Rightarrow & 10=\frac{\mathrm{Sum} \mathrm{of} 3 \mathrm{observations}}{3}\\ & & \end{array}$
⇒ Sum of 3 observations = 30          .....(1)

$\begin{array}{rcl}& \Rightarrow & 12=\frac{\mathrm{Sum} \mathrm{of} 4 \mathrm{observations}}{4}\\ & & \end{array}$
⇒ Sum of 4 observations = 48          .....(2)

From (1) and (2), we get

Sum of 7 observations = 30 + 48 = 78

$\begin{array}{rcl}\therefore \mathrm{Required} \mathrm{mean}& =& \frac{\mathrm{Sum} \mathrm{of} 7 \mathrm{observations}}{7}\\ & =& \frac{78}{7}\\ & =& 11.14\end{array}$

Answer:

Arranging the data in ascending order, we get
2, 3, 4, 4, 6, 7, 8, 8, 8, 8, 10, 11, 15

Since 8 occurs most frequently in the data
i.e., 4 times.

∴ Mode = 8

Answer:

Heights in ascending order: 128, 129, 136, 138, 140, 140, 142, 142, 143, 144, 144, 146, 150, 152, 154

(a) Height of the tallest boy is 154 cm.
(b) Height of the shortest boy is 128 cm.
(c) Range = Highest value − lowest value = (154 – 128) cm = 26 cm
(d) Since, observations are odd in number
∴ Median height = Middle most height = 142 cm

Answer:

The data is 16, 15, 16, 16, 8, 15, 17
(a) Since, 16 occurs most frequently in the data i.e., 3 times.
∴ Mode = 16
15 occurs two times, therefore if 15 will be put in the data, mode would be changed.
Hence, 8 or 17 or 16 can be put in the data so that mode remain the same.
(b) 15 occurs two times, therefore mode will be 15 if 15 occurs atleast 4 times.
∴ Atleast two times 15 must be put into the data to change the mode to 15.
(c) 17 occurs only once in the data.
∴ Atleast 3 times 17 must be put in to the data to change the mode to 17

Answer:

(a) For group A;
Both 7 and 10 occurs two times.
∴ Mode = 7 year and 10 year.
Range = Highest age – Lowest age
= (10 – 7) year = 3 year
For group B;
12 occurs most frequently i.e., 3 times.
∴ Mode = 12
Range – Highest age – Lowest age
= (12 – 7) year = 5 year

(b) If both groups A and B are combined, then arranging the data in ascending order, we get
7, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 12
In this data both 7 and 12 occurs most frequently i.e., 3 times.
∴ Mode = 7 year and 12 year
Range = Highest age – Lowest age
= (12 – 7) year = 5 year

Answer:

(a) The bar graph depicts the information about production of motor bikes by XYZ automobiles Ltd. during January to June.
(b) The motor bikes produced in first three months are = 600 + 800 + 700 = 2100
(c) Production in January = 600
Production in May = 900
∴ Increase in production in May over the production in January = 900 – 600 = 300
(d) In the month of June the production was minimum and it was 500.
(e) 
$\begin{array}{rcl}\mathrm{Average} \mathrm{production}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{production} \mathrm{in} \mathrm{every} \mathrm{month}}{\mathrm{Number} \mathrm{of} \mathrm{months}}\\ & =& \frac{600+800+700+1100+900+500}{6}\\ & =& \frac{4600}{6}\\ & =& 766.66\\ & & ~767\end{array}$

Answer:

(a) Number of students scored marks below 40 i.e., 30-39 are 4. 
Thus, 4 students have failed.
(b) Number of students got marks from 50 to 59 is 7 and number of students got marks from 60 to 69 is 11.
∴ 7 + 11 = 18 students got marks from 50 to 69.
(c) 4 students scored marks from 90 and above.
(d) Number of students got marks from 80 to 89 is 6. Number of students got marks from 90 to 92 is 4.
∴ 6 + 4 = 10 students are merit holders.
(e) Total number of students in the class = 4 + 2 +7 + 11 + 8 + 6 + 4 = 42

Answer:

(a) The bar graph represents the information about the production of rice by a country during the years 2005 to 2009.

(b) In 2006, the production was least.

(c) In 2007, there was a maximum rise in the production
∴ After 2006, production has a maximum rise.

(d)
$\begin{array}{rcl}\mathrm{Average} \mathrm{production}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{producation} \mathrm{in} \mathrm{every} \mathrm{year}}{\mathrm{Number} \mathrm{of} \mathrm{years}}\\ & =& \frac{\left(50+40+70+50+60\right)}{5} \mathrm{million} \mathrm{years}\\ & =& \frac{270}{5}\mathrm{million} \mathrm{tonnes}\\ & =& 54 \mathrm{million} \mathrm{tonnes}\end{array}$

(e) Rice production in 2006 = 40 million tonnes
Rice production in 2008 = 50 million tonnes
∴ Difference = (50 – 40) million tonnes = 10 million tonnes

Answer:

(a) The bar graph depicts the information about marks obtained by a student in 5 different subjects.
(b) Since student get the highest marks in Maths i.e., 82
∴ The student is very good at Maths.
(c)
$\begin{array}{rcl}\mathrm{Average} \mathrm{marks}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{marks} \mathrm{in} \mathrm{every} \mathrm{subject}}{\mathrm{Numbner} \mathrm{of} \mathrm{subjects}}\\ & =& \frac{64+75+82+71+49}{5}\\ & =& \frac{341}{5}\\ & =& 68.2\end{array}$

(d) The student got 75 marks in Hindi and 82 marks in Maths.
∴ In Hindi and Maths, the student got distinction.

(e) Total marks obtained by the student = 64 + 75 + 82 + 71 + 49 = 341
∴ Percentage marks = $\frac{341}{500}\times 100$ = 68.2%

Answer:

(a) Total number of newspapers read in Hindi, Punjabi, Urdu, Marathi and Tamil = 800 + 400 + 200 + 300 + 100 = 1800
(b) Number of newspapers read in Hindi = 800
Number of newspapers read in English = 500
∴ 800 – 500 = 300 excess newspapers read in Hindi than in English.
(c) The least number of newspapers are read in Tamil language.
(d) Total number of newspapers in the town = 200 + 100 + 500 + 800 + 300 + 400 = 2300

Answer:

(a) Cricket has heighest value for Class VIII students in the graph.
∴ Cricket is liked the most by Class VIII students.

(b) Number of students like Hockey of Class VII = 7
Number of students like Tennis of Class VII = 10
∴ 7 + 10 = 17 students of Class VII like Hockey and Tennis.

(c) Total number of students in Class VII = 7 + 16 + 18 + 10 + 14 = 65

(d) For Cricket only, the number of students of Class VII is less than that of Class VIII.

(e) For Hockey Football, Tennis and Badminton i.e., 4 sports, students of Class VIII are less than that of Class VII.

(f) Number of students like Badminton in Class VII = 14
Number of students like Tennis in Class VIII = 7
∴ Required ratio = $\frac{14}{7}=\frac{2}{1} \mathrm{or} 2 : 1$$\frac{14}{7}=\frac{2}{1} \mathrm{or} 2 :\hspace{0.17em}1$

Answer:

(a) The double bar graph represents the information about the comparison sales of two brands A and B during January to June.
(b) Sales of Brand A in the month of March is less than sales in the month of February.
∴ In the month of March, sales of Brand A decreased as compared to the previous month.
(c) Sales of Brand A in the month of June = ₹ 57 lakh
Sales of Brand B in the month of June = ₹ 54 lakh.
Difference in sales = ₹ 57 lakh – ₹ 54 lakh = ₹ 3 lakh
$$\begin{gathered} \mathrm{Average} \mathrm{sales} \mathrm{of} \mathrm{brands} \mathrm{B}=₹\left(\frac{36+38+43+35+45+54}{6}\right) \mathrm{lakhs} \\ =₹\left(\frac{251}{6}\right) \mathrm{lakhs} \\ =₹41.83 \mathrm{lakhs} \end{gathered}$$
(e) In the month of April and June, the sales of Brand B was less than that of Brand A.
(f) Sales of Brand A in the month of January = ₹ 31 lakh
Sales of Brand B in the month of January = ₹ 36 lakh
∴ Required ratio = $\frac{31}{36}$ = 31 : 36

Answer:

(a) The double bar graph compares the information about minimum temperature (°C) of year 2008 and year 2009 during the months November to February.
(b) Minimum temperature in the year 2008 for the month of November = 18°C
Minimum temperature in the year 2009 for the month of November = 15°C
∴ Required ratio = $\frac{18}{15}$ = 6:5
(c) For the month of November and February, i.e., 2 months the minimum temperature in 2008 was greater than that of 2009
(d) Average minimum temperature for 2008 
 $$\begin{gathered} =\frac{18°\mathrm{C}+11°\mathrm{C}+4°\mathrm{C}+12°\mathrm{C} }{4} \\ =\frac{ 45°\mathrm{C}}{4} \\ =11.25°\mathrm{C} \end{gathered}$$

(e) In the month of February, the variation in two temperatures is maximum.

Answer:

Answer:

(a) The double bar graph represents the information about the number of girls and boys in different sections of Class VII
(b) Total number of boys in all sections of Class VIII = 15 + 30 + 20 + 20 + 25 =110
(c) In sections A and D, the number of girls is greater than the number of boys.
(d) In section B, the number of boys is the maximum.
(e) In section C, the number of girls is the least.

Answer:

(a)


(b) Number of readers on Monday = 400 + 150 = 550
Number of readers on Tuesday = 600 + 100 = 700
Number of readers on Wednesday = 350 + 200 = 550
Number of readers on Thursday = 550 + 300 = 850
Number of readers on Friday = 500 + 250 = 750
Number of readers on Saturday = 350 + 200 = 550
Hence, on Thursday, the number of readers in the library was maximum.


(c) Total number of magazine readers = 150 + 100 + 200 + 300 + 250 + 200 = 1200
Mean number of magazine readers = $\frac{1200}{6}$
= 200
 

Answer:

(a)

The double bar graph gives information about total number of students and number of students present from Class VI to Class X.

(b) Class VIII has the maximum number of students.

(c) Difference of total students and number of students present in Class VI = 90 – 81 = 9
Difference of total students and number of students present in Class VII = 82 – 76 = 6
Difference of total students and number of students present in Class VIII = 95 – 91 = 4
Difference of total students and number of students present in Class IX = 70 – 65 = 5
Difference of total students and number of students present in Class X = 63 – 62 = 1

Hence, in Class X, the difference between the total number of students and the number of students present is minimum.

(d) Total number of students in Class IX = 70
Number of present students in Class IX = 65
∴ Required ratio = $\frac{65}{70}$ = 13 : 14

(e) Total number of students in Class VI = 90
Number of students present in Class VI = 81
∴ Number of students absent in Class VI = 90 – 81 = 9
∴ Percentage of students who were absent =$\frac{9}{90}\times 100$ = 10%

Answer:

(a)


(b) On Saturday, the sales was maximum.

(c) Total sale during the week = 50 + 45 + 30 + 55 + 27 + 60 = 267

(d) Maximum sale = 60
Minimum sale = 27
∴ Required ratio =$\frac{27}{60}=\frac{9}{20}$ = 9:20

(e) Total sales during the week = 50 + 45 + 30 + 55 + 27 + 60 = 267
∴ Average sale = $\frac{267}{6}$ = 44.5
(f) Average sale is 44.5.
∴ 4 days of the week i.e., Monday, Tuesday, Thursday and Saturday has the sale above the average sale.

Answer:

(a)

(b) Average height of one storey for three tallest buildings = (3.9 + 4.15 + 4.15) m = 12.23 m = 4.06 m
(c) Mumbai has the largest percentage of skys crappers
Total number of sky scrappers = 10
Total number sky scrappers in Mumbai = 9
∴ Percentage of skyscrappers in Mumbai =$\left(\frac{9}{10}\times 100\right)\%$ =90%
(d) Range = Highest value – Lowest value
                 = 249 m – 156 m = 93 m

(e) Arranging the height (inm) in ascending order, we get
156, 170, 170, 170, 180, 181, 184, 193, 249, 249
Since, observations are even in number.
∴ Median = Mean of two middle most observations
                = (180 + 181) m
                =$\frac{361}{2}$ m
               =180.5 m

(f)

Answer:

(a)


(b) Total marks obtained by Soni = 86 + 89 + 90 + 82 + 75 + 82 = 504
Total maximum marks = 6 × 100=600
∴ Percentage of marks obtained by Soni = $\left(\frac{504}{600}\times 100\right)\%$ = 84%

(c) Total marks obtained by Kunal = 72 + 81 + 92 +96 + 64 + 85 = 490
Total maximum marks = 6 × 100 = 600
∴ Percentage of marks obtained by Kunal = $\left(\frac{490}{600}\times 100\right)\%$ = 81.6%
(d)
$\begin{array}{rcl}\mathrm{Required} \mathrm{ratio}& =& \frac{\mathrm{Percentage} \mathrm{marks} \mathrm{of} \mathrm{Kunal}}{\mathrm{Percentage} \mathrm{marks} \mathrm{of} \mathrm{Soni}}\\ & =& \frac{81.6}{84}\\ & =& \frac{34}{35}\\ & =& 34:35\end{array}$


(e) In English, Hindi and S. Science i.e., 3 subjects Soni get more marks than Kunal.

(f) Soni got more marks in S. Science and difference of marks = 75 – 64 = 11

(g) Difference of marks in English = 86 – 72 = 14
Difference of marks in Hindi = 89 – 81 = 8
Difference of marks in Maths = 92 – 90 = 2
Difference of marks in Science = 96 – 82 – 14
Difference of marks in S. Science = 75 – 64 = 11
Difference of marks in Sanskrit = 85 – 82 = 3
Hence, in English and Science, the difference of marks was maximum by 14 marks.

Answer:

(a)


(b) Total number of girls in Class VII = 15 + 24 + 10 + 19 + 27 + 21 = 116
Total number of boys in Class VII = 12 + 16 + 8 + 17 + 11 + 30 = 94
∴ Total number of students in Class VII = 116 + 94 = 210

(c) Electronics club is the most preferred by boys.

(d) Yoga club is the least preferred by girls.

(e) Difference between the number of boys and girls for Music club 15 – 12 = 3
Difference between the number of boys and girls for Dance club = 24 – 16 = 8
Difference between the number of boys and girls for Yoga club = 10 – 8 = 2
Difference between the number of boys and girls for Dramatics club = 19 – 17 = 2
Difference between the number of boys and girls for Fine Arts = 27 – 11 = 16
Difference between the number of boys and girls for the Electronics club = 30 – 21 = 9
Hence, for the Dramatics club and Yoga club, the difference between the number of boys and girls is the least.

(f) For the Fine Arts club, the difference between the number of boys and girls is the maximum.

Answer:

(a)


(b) Total output in year 2007 = 2800+ 4500 + 4800 + 4800 + 5200 = 22100
Total output in year 2008 = 2700+ 3200 + 6000 + 5000 + 4200 = 21100
Thus, in year 2007, the total output was the maximum.

(c) Total production for the year 2007 = 22100
∴ Mean production = $\frac{22100}{5}$ = 4420

(d) Difference between the production for February = 2800 – 2700 = 100
Difference between the production for May = 4500 – 3200 = 1300
Difference between the production for August = 6000 – 4800 = 1200
Difference between the production for October = 5000 – 4800 = 200
Difference between the production for December = 5200 – 4200 = 1000
Thus, in the month of May the difference between the production for two years was the maximum.

(e) In the month of August for the year 2008, the production was the maximum.

(f) In the month of February for the year 2007, the production was the least.

Answer:

(a)

(b) Population growth in town A = 3200 – 2900 = 300
Population growth in town B = 7500 – 6400 = 1100
Population growth in town C = 9200 – 8300 = 900
Population growth in town D = 6300 – 4600 = 1700
Hence, in town the population growth was maximum.

(c) In town A, the population growth was the least.

Answer:

(a)


(b) The hill station Mussoorie was visited by the maximum number of tourists in 2008.

(c) The hill station Manali was visited by the least number of tourists in 2009.

(d) In the hill stations Nainital, Manali, Mussoorie and Kullu, there was an increase in the number of tourists in the year 2009.

Answer:

(a)


(b) Butterscotch flavour is liked the most by the boys.

(c) Total number of girls = 8 + 12 + 7 + 9 + 10 = 46

(d) Number of girls, like chocolate flavour = 12
Number of boys, like chocolate flavour = 9
Thus, 12 + 9 = 21 children like the chocolate flavour of ice cream.

(e) Total number of children who like strawberry flavour = 3 + 7 = 10
Total number of children who like vanilla flavour = 4 + 8 = 12
∴ Required ratio = $\frac{10}{12}=\frac{5}{6}$ = 5 : 6