Unit: 3: Data Handling
In Questions 1 to 16, there are four options, out of which only one is correct. Write the correct answer.
Let $x\text{,}y\text{,}z$ be three observations. The mean of these observations is
a. $\frac{x\times y\times z}{3}$
b. $\frac{x+y+z}{3}$
c. $\frac{xyz}{3}$
d. $\frac{x\times y+z}{3}$
(b)
Here, $x\text{,}y$ and $z$ are three observations.
We know that, $\text{Mean}=\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$$=\frac{x+y+z}{3}$
The number of trees in different parks of a city are $33\text{,}38\text{,}48\text{,}33\text{,}34\text{,}34\text{,}33\text{and}24.$ The mode of this data is
a. $24$
b. $34$
c. $33$
d. $48$
(c)
The given data is: $33\text{,}38\text{,}48\text{,}33\text{,}34\text{,}34\text{,}33,\text{}24$
On arranging the given data in ascending order, we get: $24\text{,}33\text{,}33\text{,}33\text{,}38\text{,}34\text{,}34\text{,}48$
Here, $33$ occurs most frequently.
Mode is the observation that occurs most frequently in the data.
Therefore, mode of the given data $=33$
Which measures of central tendency get affected if the extreme observations on both the ends of a data arranged in descending order are removed?
a. Mean and mode
b. Mean and Median
c. Mode and Median
d. Mean, Median and Mode
(a)
Mean is defined as follows: $\text{Mean}=\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$
So, if we remove the extreme values, sum of observations and total number of observations will change.
Hence, mean will also change.
Mode is that observation which occurs the most.
So, if extreme values are removed, mode can be affected. Median is the mid value when the given data is arranged in ascending or descending order. So, if extreme values are removed then the mid value remains same.
Hence, median will not change.
The range of the data: $21\text{,}6\text{,}17\text{,}18\text{,}12\text{,}8\text{,}4\text{,}13$ is
a. $17$
b. $12$
c. $8$
d. $15$
(a)
Highest observation $=21$
Lowest observation $=4$
Range $=$ Highest observation $$ Lowest observation $=214=17$
The median of the data: $3\text{,}4\text{,}5\text{,}6\text{,}7\text{,}3\text{,}4$ is
a. $5$
b. $3$
c. $4$
d. $6$
(c)
We know that, median is the middle most observation.
For finding the median of the data, arrange the data in ascending order.
Ascending order: $3\text{,}3\text{,}4\text{,}4\text{,}5\text{,}6\text{,}7$
$\text{numberofobservations}=7\text{}\left(\text{odd}\right)$
$\therefore $ Median
$=$ Value of $\left(\frac{n+1}{2}\right)\text{th}$ observation
$=$ Value of $\left(\frac{7+1}{2}\right)\text{th}$ observation
$={4}^{\text{th}}$ observation $=4$
Out of $5$ brands of chocolates in a shop, a boy has to purchase the brand which is most liked by children. What measure of central tendency would be most appropriate if the data is provided to him?
a. Mean
b. Mode
c. Median
d. Any of the three
(b)
Mode is the most appropriate central tendency because it is the observation that occurs most frequently.
Here, by the measurement of mode, we can find out the chocolates which is most liked by children.
There are $2$ aces in each of the given set of cards placed face down. From which set are you certain to pick the two aces in the first go?
a.
b.
c.
d.
(c)
From third set, we are certain to pick the two aces in the first go because it has only $2$ cards and it is given that every set has $2$ aces.
In the previous question, what is the probability of picking up an ace from set (d)?
a. $\frac{1}{6}$
b. $\frac{2}{6}$
c. $\frac{3}{6}$
d. $\frac{4}{6}$
(b)
Probability $=\frac{\text{Numberofpossibleoutcomes}}{\text{Totalnumberofoutcomes}}$
Total no. of cards in set $\left(d\right)=6$
No. of possible outcomes $=2$ [given, $2$ aces in every set]
So, probability of getting an ace from set (d) $=\frac{2}{6}$
The difference between the highest and the lowest observations in a data is its
a. frequency
b. width
c. range
d. mode
(c)
The difference between the highest and the lowest observations in a data is its range.
In a school, only $2$ out of $5$ students can participate in a quiz. What is the chance that a student picked at random makes it to the competition?
a. $20\%$
b. $40\%$
c. $50\%$
d. $30\%$
(b)
Total number of outcomes $=$ Total number of students $=5$
Number of possible outcomes $=$ Students participating in a quiz $=2$
Probability $=\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\frac{\left(\text{Numberofpossibleoutcomes}\right)}{\left(\text{Totalnumberofoutcomes}\right)}$
$=\frac{2}{5}$
To find percentage, multiply it by hundred.
Probability $=\frac{2}{5}\times 100=40\%$
Some integers are marked on a board. What is the range of these integers?
a. $31$
b. $37$
c. $20$
d. $3$
(b)
Here, highest observation $=+20$
and lowest observation $=17$
Range $=$ Highest observation $$ Lowest observation
$\begin{array}{l}=+20\left(17\right)\\ =20+17\\ =37\end{array}$
On tossing a coin, the outcome is
a. only head
b. only tail
c. neither head nor tail
d. either head or tail
(d)
When we toss a coin, the outcome is either head or tail.
The mean of three numbers is $40.$ All the three numbers are different natural numbers. If lowest is $19\text{,}$ what could be highest possible number of remaining two numbers?
a. $81$
b. $40$
c. $100$
d. $71$
(a)
Mean of $3$ numbers $=40$ and the lowest number $=19$
Let the other two numbers be $x$ and $y\text{,}$ respectively.
$\text{Mean}=\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$
$\begin{array}{l}\Rightarrow 40=\frac{19+x+y}{3}\\ \Rightarrow 3\times 40=19+x+y\\ \Rightarrow 120=19+x+y\\ \Rightarrow x+y=12019\\ \Rightarrow x+y=101\text{}\mathrm{...}\text{(i)}\end{array}$
Since $19$ is the lowest number, so for highest possible value of remaining two numbers, one must be $20.$
Let $x=20$
From eq. (i), we get
$\begin{array}{l}\Rightarrow 20+y=101\\ \Rightarrow y=10120=81\end{array}$
So, highest possible number $=81$
Khilona earned scores of $97\text{,}73\text{and}88$ respectively in her first three examinations. If she scored $80$ in the fourth examination, then her average score will be
a. increased by $1$
b. increased by $1.5$
c. decreased by $1$
d. decreased by $1.5$
(d)
Average score $=\frac{\left(\text{sumofscoresinallexams}\right)}{(\text{total}\text{numberofexams})}$
Average score in first three examinations $=\frac{97+73+88}{3}=\frac{258}{3}=86$
Also, average score in four examinations $=\frac{97+73+88+80}{4}=\frac{338}{4}=84.5$
$84.5<86$
Decrease in average score $=8684.5=1.5$
Hence, average score is decreased by $\mathrm{1.5.}$
Which measure of central tendency best represents the data of the most popular politician after a debate?
a. Mean
b. Median
c. Mode
d. Any of the above
(c)
Mode is the most frequent observation in a data. So, this measure of central tendency best represents the data of most popular politician after a debate.
Which of the following has the same mean, median and mode?
a. $6\text{,}2\text{,}5\text{,}4\text{,}3\text{,}4\text{,}1$
b. $4\text{,}2\text{,}2\text{,}1\text{,}3\text{,}2\text{,}3$
c. $2\text{,}3\text{,}7\text{,}3\text{,}8\text{,}3\text{,}2$
d. $4\text{,}3\text{,}4\text{,}3\text{,}4\text{,}6\text{,}4$
(d)
We have to find out mean, median and mode of the data given for each option.
For option (a)
Data (in ascending order) $\to 1\text{,}2\text{,}3\text{,}4\text{,}4\text{,}5\text{,}6$
Here, $n=7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{odd}\right)$
Median $=$ Value of $\left(\frac{n+1}{2}\right)\text{th}$ observation
$=$ Value of $\left(\frac{8}{2}\right)\text{th}$ observation $=4$
$\text{Mean}=\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$
$\begin{array}{l}=\frac{1+2+3+4+4+5+6}{7}\\ =\frac{25}{7}\\ =3.57\end{array}$
Mode $=$ most frequent observation $=4$
Hence, Mean $\ne $ Median $=$ Mode
For option (b)
Data (in ascending order) $\to 1\text{,}2\text{,}2\text{,}2\text{,}3\text{,}3\text{,}4$
Here, $n=7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{odd}\right)$
Median $=$ Value of $\left(\frac{n+1}{2}\right)\text{th}$ observation
$=$ Value of $\left(\frac{8}{2}\right)\text{th}$ observation $=2$
$\text{Mean}=\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$
$\begin{array}{l}=\frac{1+2+2+2+3+3+4}{7}\\ =\frac{17}{7}\\ =2.428\end{array}$
Mode $=$ most frequent observation $=2$
Hence, Mean $\ne $ Median $=$ Mode
For option (c)
Data (in ascending order) $\to 2\text{,}2\text{,}3\text{,}3\text{,}3\text{,}7\text{,}8$
Here, $n=7\left(\text{odd}\right)$
Median $=$ Value of $\left(\frac{n+1}{2}\right)\text{th}$ observation
$=$ Value of $\left(\frac{8}{2}\right)\text{th}$ observation $=3$
$\text{Mean}=\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$
$\begin{array}{l}=\frac{2+2+3+3+3+7+8}{7}\\ =\frac{28}{7}\\ =4\end{array}$
Mode $=$ most frequent observation $=3$
Hence, Mean $\ne $ Median $=$ Mode
For option (d)
Data (in ascending order) $\to 3\text{,}3\text{,}4\text{,}4\text{,}4\text{,}4\text{,}6$
Here, $n=7\left(\text{odd}\right)$
Median $=$ Value of $\left(\frac{n+1}{2}\right)\text{th}$ observation
$=$ Value of $\left(\frac{8}{2}\right)\text{th}$ observation $=4$
$\text{Mean}=\frac{\text{Sumofobservations}}{\text{Numberofobservations}}$
$\begin{array}{l}=\frac{3+3+4+4+4+4+6}{7}\\ =\frac{28}{7}\\ =4\end{array}$
Mode $=$ most frequent observation $=4$
Hence, Mean $=$ Median $=$ Mode
In Questions 17 to 31, fill in the blanks to make the statements true.
The difference between the highest and the lowest observations of a data is called _________.
The difference between the highest and the lowest observations of a data is called range.
The mean of a data is defined as _________.
$\text{Mean}=\frac{\text{Sumofallobservations}}{\text{Numberofobservations}}$
In a set of observations, the observation that occurs the most often is called _________.
In a set of observations, the observation that occurs the most often is called mode.
If a given data, arranged in ascending or descending order, the middle most observation is called _________.
If a given data, arranged in ascending or descending order, the middle most observation is called median.
Mean, Median, Mode are the measures of _________.
Mean, Median, Mode are the measures of central tendency.
The probability of an event which is certain to happen is _________.
The probability of an event which is certain to happen is 1
The probability of an event which is impossible to happen is _________.
The probability of an event which is impossible to happen is 0.
When a die is thrown, the probability of getting a number less than $7$ is _________.
When we throw a die, $6$ outcomes are possible, i.e. $1\text{,}2\text{,}3\text{,}4\text{,}5\text{,}6.$
Total possible outcomes $=6$
Number of favourable outcomes (less than $7$ ) $=6$
[all outcomes are less than $7$ ]
$\begin{array}{l}\text{Probabilityofgettinganumberlessthan}7\\ =\frac{\text{favourableoutcomes}}{\text{totalpossibleoutcomes}}\end{array}$
$=\frac{6}{6}=1$
In throwing a die, the number of possible outcomes is _________.
When we throw a die, $6$ outcomes are possible. They are $1\text{,}2\text{,}3\text{,}4\text{,}5\text{and}6.$
In throwing a die, the number of possible outcomes is 6.
_________ can be used to compare two collections of data.
A double bar graph can be used to compare two collections of data.
The representation of data with bars of uniform width is called _________.
The representation of data with bars of uniform width is called a bar graph.
If the arithmetic mean of $8\text{,}4\text{,}x\text{,}6\text{,}2\text{,}7\text{is}5\text{,}$ then the value of $x$ is _________.
We know that,
$\text{Mean}=\frac{\text{Sumofallobservations}}{\text{totalno}.\text{ofobservations}}$
$\begin{array}{l}\Rightarrow 5=\frac{8+4+x+6+2+7}{6}\\ \Rightarrow 30=27+x\\ \Rightarrow 3027=x\\ \Rightarrow x=3\end{array}$
Hence, the value of $x=3$
The median of any data lies between the _________ and _________ observations.
The median of any data lies between the minimum and maximum observations.
Median is one of the observations in the data if number of observations is _________.
Median is one of the observations in the data if number of observations is odd.
Rohit collected the data regarding weights of students of his class and prepared the following table:
Weight (in kg) 
$4447$ 
$4851$ 
$5255$ 
$5660$ 
Number of Students 
$3$ 
$5$ 
$25$ 
$7$ 
A student is to be selected randomly from his class for some competition. The probability of selection of the student is highest whose weight is in the interval _________.
We know that,
Probability $=\frac{\text{no}.\text{offavourableoutcomes}}{\text{totalno}.\text{ofpossibleoutcomes}}$
$\therefore $ To make the probability highest, we have to take the interval where number of students, i.e. possible outcomes are highest.
Here, probability is highest whose weight is in the interval $5255$.
In Questions 32 to 49, state whether the statements are True or False.
If a die is thrown, the probability of getting a number greater than $6$ is $1.$
False
As we know, a die has six numbers on it, i.e. $1\text{to}6.$ So, it is impossible to get a number greater than $6.$ Hence, if a die is thrown, the probability of getting a number greater than $6$ is $0.$
When a coin is tossed, there are $2$ possible outcomes.
True
If a coin is tossed, then
Maximum possible outcomes $=2\text{,}$ i.e. head or tail.
If the extreme observations on both the ends of a data arranged in ascending order are removed, the median gets affected.
False
If the extreme observations on both the ends of a data arranged in ascending order are removed, then the mean and mode gets affected but median remains same.
The measures of central tendency may not lie between the maximum and minimum values of data.
False
The measures of central tendency lie between the maximum and minimum values of the data
It is impossible to get a sum of $14$ of the numbers on both dice when a pair of dice is thrown together.
True
When a die is thrown, total possible outcomes are $6\text{,i}\text{.e}\text{.}1\text{,}2\text{,}3\text{,}4\text{,}5\text{,}6.$
So, when a pair of dice is thrown together, maximum sum will be $12\text{,}$ if and only if both dice get $6$ together.
So, the pair will be $\left(6\text{,}6\right)$ and the sum is $12.$
$\therefore $ It is impossible to get a sum of $14$ on both dice, when a pair of dice is thrown together.
The probability of the spinning arrow stopping in the shaded region (Fig. 3.4) is $\frac{1}{2}\text{.}$
True
Favourable outcomes
$=$ Number of shaded regions $=2$
Total number of possible outcomes
$=$ Total number of regions $=4$
Probability
$=\frac{\text{favourableoutcomes}}{\left(\text{totalno}.\text{ofpossibleoutcomes}\right)}$
$=\frac{2}{4}=\frac{1}{2}$
A coin is tossed $15$ times and the outcomes are recorded as follows:
H T T H T H H H T T H T H T T.
The chance of occurence of a head is $50$ per cent.
False
Number of times in which head occurs $=7$
Total number of times, the coin is tossed $=15$
$\therefore $ Probability of getting a head $=\frac{7}{15}=\frac{7}{15}\times 100\%=46.67\%$,
which is not equal to $50$ per cent.
Mean, Median and Mode may be the same for some data.
True
Mean, median and mode can be the same for some data.
See Q. No. 16.
The probability of getting an ace out of a deck of cards is greater than $1.$
False
Probability of an event can never be
greater than $1.$
For any event E,
$0\le \text{P}\left(\text{E}\right)\le 1$.
Mean of the data is always from the given data.
False
It is not compulsory that mean of the data is always from the given data. It may or may not be the observation from given data.
Median of the data may or may not be from the given data.
True
Case 1: $2\text{,}4\text{,}6\text{,}8\text{,}10$
Here, $n=5\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{odd}\right)$
Median $=$ Value of $\left(\frac{n+1}{2}\right)\text{th}$ observation
$=$ Value of $\left(\frac{5+1}{2}\right)\text{th}$ observation
$=$ Value of $3\text{rd}$ observation
$=6$
This value belongs to the given data.
Case 2: $4\text{,}6\text{,}8\text{,}8\text{,}12\text{,}14\text{,}15\text{,}16$
Here, $n=8\left(\text{even}\right)$
$\begin{array}{l}\text{Median}\\ =\frac{\text{Valueof}\left(\frac{n}{2}\right)\text{thobservations}+\text{Valueof}\left(\frac{n+1}{2}\right)\text{thobservations}}{2}\\ =\frac{\text{Valueof}4\text{thobservations}+\text{Valueof}5\text{thobservations}}{2}\\ =\frac{8+12}{2}\\ =\frac{20}{2}\\ =10\end{array}$
This value does not belong to the given data.
Mode of the data is always from the given data.
True
Mode of the data is always from the given data as it is the most frequent observation in the data.
Mean of the observations can be lesser than each of the observations.
False
Mean is the average value of all the observations. Some of the observations are less than it and some of observations are greater than it.
Mean can never be a fraction.
False
e.g., mean of $\frac{1}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\&}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{6}$
$\begin{array}{l}=\left(\frac{\text{Sumof}\frac{1}{4}\text{and}\frac{1}{6}}{n}\right)\\ =\frac{\left(\frac{1}{4}+\frac{1}{6}\right)}{2}\text{}\left[\text{no}\text{.ofobservations}\left(n\right)=2\right]\end{array}$
$\begin{array}{l}=\frac{\frac{6+4}{24}}{2}\\ =\frac{10}{24}\times \frac{1}{2}\\ =\frac{5}{24}\end{array}$
Range of the data is always from the data.
False
It is not necessary as range is the difference of highest observation and lowest observation.
The data $12\text{,}13\text{,}14\text{,}15\text{,}16$ has every observation as mode.
True
Given data is $12\text{,}13\text{,}14\text{,}15\text{,}16.$ Here, each observation has same frequency, so every observation is a mode.
The range of the data $2\text{,}5\text{,}4\text{,}3\text{,}7\text{,}6$ would change if $2$ was subtracted from each value in the data.
False
Range before subtraction
$=\text{Highestobservation}\text{Lowestobservation}$
$\begin{array}{l}=7(5)\\ =7+5\\ =12\end{array}$
Data after subtraction is:
$\begin{array}{l}22\text{,}52\text{,}42\text{,}32\text{,}72\text{,}62\text{,}\\ \text{i}\text{.e}\text{.}0\text{,}7\text{,}2\text{,}1\text{,}5\text{,}4\end{array}$
Range after subtraction $=5\left(7\right)=5+7=12$
Range of the data in both the cases is same.
The range of the data $3\text{,}7\text{,}1\text{,}2\text{,}2\text{,}6\text{,}3\text{,}5$ would change if $8$ was added to each value in the data.
False
Range before adding 8
$=$ Maximum observation $$ Minimum observation
$=7(5)=7+5=12$
Data after adding 8:
$3+8\text{,}7+8\text{,}1+8\text{,}2+8\text{,}2+8\text{,}$
$\begin{array}{l}6+8,3+8\text{,}5+8\\ \text{i}\text{.e}\text{.}11\text{,}15\text{,}9\text{,}6\text{,}10\text{,}14\text{,}5\text{,}3\end{array}$
Range after adding $8$:
$=$ Maximum observation $$ Minimum observation
$=153=12$
So, the range is same.
Calculate the Mean, Median and Mode of the following data:
$5\text{,}10\text{,}10\text{,}12\text{,}13.$
Are these three equal?
Given data is $5\text{,}10\text{,}10\text{,}12\text{,}13.$
Sum of all observations $=5+10+10+12+13=50$
No. of observations $=5$
Mean $={\scriptscriptstyle \frac{\text{sumofallobservations}}{\text{totalobservations}}}$
$=\frac{50}{5}=10$
Given data is in ascending order.
So, median $=$ value of $\left(\frac{n+1}{2}\right)\text{th}$ observation
$=$ value of $\left(\frac{5+1}{2}\right)\text{th}$ observation
$=$ value of $3\text{rd}$ observation
$=10$
Mode $=$ most frequent data $=10$
Hence, mean $=$ median $=$ mode.
Find the mean of the first ten even natural numbers.
First ten even natural numbers $=2\text{,}4\text{,}6\text{,}8\text{,}10\text{,}12\text{,}$
$14\text{,}16\text{,}18\text{,}20.$
Sum of all observations
$=2+4+6+8+10+12+14+16+18+20=110$
No. of observations $=10$
Mean $=\frac{\text{sumofallobservations}}{\text{totalobservations}}$
$\begin{array}{l}=\frac{110}{10}\\ =11\end{array}$
A data constitutes of heights (in cm) of $50$ children. What do you understand by mode for the data?
Mode is the observation that occurs most frequently in a set of observation. Here, for the given data, mode is the height that appears most frequently.
A car seller collects the following data of cars sold in his shop.
Colour of Car 
Number of Cars Sold 
Red 
$15$ 
Black 
$20$ 
White 
$17$ 
Silver 
$12$ 
Others 
$9$ 
a. Which colour of the car is most liked?
b. Which measure of central tendency was used in (a)?
a. Red colour of the car liked by people $=15$
Black colour of the car liked by people $=20$
White colour of the car liked by people $=17$
Silver colour of the car liked by people $=12$
Other colour of the car liked by people $=9$
Hence, black colour of the car is the most liked.
b. Mode is used in (a).
The marks in a subject for $12$ students are as follows:
$31\text{,}37\text{,}35\text{,}38\text{,}42\text{,}23\text{,}17\text{,}18\text{,}35\text{,}25\text{,}35\text{,}29$
For the given data, find the
a. Range
b. Mean
c. Median
d. Mode
Given data is: $31\text{,}37\text{,}35\text{,}38\text{,}42\text{,}23\text{,}17\text{,}18\text{,}35\text{,}25\text{,}35\text{,}29.$
The given data in ascending order:
$17\text{,}18\text{,}23\text{,}25\text{,}29\text{,}31\text{,}35\text{,}35\text{,}35\text{,}37\text{,}38\text{,}42$
a. Range $=$ highest observation $$ lowest observation
$=4217=25$
b. Mean $=\frac{\text{sumofallobservations}}{\left(\text{totalno}.\text{ofobservations}\right)}$
$=\frac{17+18+23+25+29+31+35+35+35+37+38+42}{12}$$\begin{array}{l}=\frac{365}{12}\\ =30.41\end{array}$
c. Here, $n=12\text{}(\text{even})$
$\begin{array}{l}\text{Median}\\ =\frac{\text{Valueof}\left(\frac{n}{2}\right)\text{thobservation}+\text{Valueof}\left(\frac{n+1}{2}\right)\text{thobservation}}{2}\\ =\frac{\text{Valueof}6\text{thobservation}+\text{Valueof}7\text{thobservation}}{2}\\ =\frac{31+35}{2}\\ =\frac{66}{2}\\ =33\end{array}$
d. Mode $=$ Most frequent observation $=35$
The following are weights (in kg) of $12$ people.
$70\text{,}62\text{,}54\text{,}57\text{,}62\text{,}84\text{,}75\text{,}59\text{,}62\text{,}65\text{,}78\text{,}60$
a. Find the mean of the weights of the people.
b. How many people weigh above the mean weight?
c. Find the range of the given data.
a. The weights of $12$ persons are: $70\text{,}62\text{,}54\text{,}57\text{,}62\text{,}84\text{,}75\text{,}59\text{,}62\text{,}65\text{,}78\text{,and}60.$
Sum of weights of $12$ people $70+62+54+57+62+84+75+59+62+65+78+60=788$
No. of observation (persons) $=12$
Mean $=\frac{\text{sumofallobservations}}{\text{no}.\text{ofobservations}}$
$\begin{array}{l}=\frac{788}{12}\\ =65.66\end{array}$
b. Weights above $65.66$ are $70\text{,}84\text{,}75\text{and}78\text{,}$ i.e. $4$ persons.
c. Range
$=$ Maximum observation $$ Minimum observation
$=8454=30$
Following cards are put facing down:
$\overline{)\text{A}}\text{}\overline{)\text{E}}\text{}\overline{)\text{I}}\text{}\overline{)\text{O}}\text{}\overline{)\text{U}}$
What is the chance of drawing out
a. a vowel
b. A or I
c. a card marked U
d. a consonant
a. We can clearly see that all the $5$ letters are vowels, i.e. A, E, I, O, U
Hence, it is certain to draw a vowel.
So, probability $=1$
b.
probability
$=\frac{\text{no}.\text{ofcardsmarkedAorl}}{\text{totalno}.\text{ofcards}}$
$=\frac{2}{5}$
c.
Probability
$=\frac{\text{no}.\text{ofcardsmarkedU}}{\text{totalno}.\text{ofcards}}$
$=\frac{1}{5}$
d.
probability
$=\frac{\text{no}.\text{ofcardsmarkedwithaconstant}}{\text{totalno}.\text{ofcards}}$
$\begin{array}{l}=\frac{0}{5}\\ =0\end{array}$
Hence, it is not possible to draw a consonant.
For the data given below, calculate the mean of its median and mode.
$6\text{,}2\text{,}5\text{,}4\text{,}3\text{,}4\text{,}4\text{,}2\text{,}3$
Given data in ascending order is: $2\text{,}2\text{,}3\text{,}3\text{,}4\text{,}4\text{,}4\text{,}5\text{,}6$
Here, $n=9\text{}(\text{odd})$
Median $=$ Value of $\left(\frac{n}{2}\right)\text{th}$ observations
$=$ Value of $\left(\frac{9+1}{2}\right)\text{th}$ observation
$=5\text{th}$ observation $=4$
Mode $=$ most frequent observation $=4$
Mean of median and mode
$=\left(\text{median}+\text{mode}\right)\xf72$
$=\left(4+4\right)\xf72$
$=8\xf72$
$=4$
Find the median of the given data if the mean is $\mathrm{4.5.}$
$5\text{,}7\text{,}7\text{,}8\text{,}x\text{,}5\text{,}4\text{,}3\text{,}1\text{,}2$
Given, mean $=4.5$
We know that,
mean $=\frac{\text{sumofallobservations}}{\left(\text{totalno}.\text{ofobservations}\right)}$
$4.5=\frac{5+7+7+8+x+5+4+3+1+2}{10}$
$\begin{array}{l}4.5\times 10=42+x=4542=x\\ \therefore x=3\end{array}$
Now arrange the data in ascending order. $1\text{,}2\text{,}3\text{,}3\text{,}4\text{,}5\text{,}5\text{,}7\text{,}7\text{,}8$
Here, $n=12\text{}(\text{even})$
$\begin{array}{l}\text{Median}\\ =\frac{\text{Valueof}\left(\frac{n}{2}\right)\text{thobservation}+\text{Valueof}\left(\frac{n}{2}+1\right)\text{thobservation}}{2}\\ =\frac{\text{Valueof}5\text{thobservation}+\text{Valueof}6\text{thobservations}}{2}\\ =\frac{4+5}{2}\\ =\frac{9}{2}\\ =4.5\end{array}$
What is the probability of the sun setting tomorrow?
Setting of the sun is a sure event. Hence, its probability is $1.$
When a spinner with three colours (Fig. 3.5) is rotated, which colour has more chance to show up with arrow than the others?
From the figure, area covered by the yellow colour is maximum out of the given three colours. Hence, chance of yellow colour to show up with arrow will be more.
What is the probability that a student chosen at random out of $3$ girls and $4$ boys is a boy?
Given, total children $=7=4$ boys and $3$ girls
So, favourable outcomes for a boy $=4$
Total number of possible outcomes $=7$
Probability
$=\frac{\text{favourableoutcomes}}{(\text{totalno}.\text{ofpossibleoutcomes})}$
$=\frac{4}{7}$
The letters written on paper slips of the word MEDIAN are put in a bag. If one slip is drawn randomly, what is the probability that it bears the letter D?
In the word ‘MEDIAN’, there is only one D.
So, favourable outcomes $=$ number of letter D $=1$
Total number of possible outcomes $=6$
Probability
$=\frac{\text{favourableoutcomes}}{\text{totalno}.\text{ofoutcomes}}$
$=\frac{1}{6}$
Classify the following events as certain to happen, impossible to happen, may or may not happen:
a. Getting a number less than $1$ on throwing a die.
b. Getting head when a coin is tossed.
c. A team winning the match.
d. Christmas will be on $25$ December.
e. Today moon will not revolve around the earth.
f. A ball thrown up in the air will fall down after some time.
a. Getting a number less than $1$ on throwing a die is impossible, as a die does not have a number less than $1$ on it.
b. Getting head, when a coin is tossed may or may not happen as a coin has head and tail on its two faces. So, we might get a head or a tail on tossing it.
c. A team may or may not win a match.
d. Christmas is certain to happen on $25\text{th}$ December.
e. It is impossible that moon will not revolve around the earth.
f. It is certain to happen that a ball thrown up in the air will fall down after sometime due to gravity.
A die was thrown $15$ times and the outcomes recorded were
$5\text{,}3\text{,}4\text{,}1\text{,}2\text{,}6\text{,}4\text{,}2\text{,}2\text{,}3\text{,}1\text{,}5\text{,}6\text{,}1\text{,}2$
Find the mean, median and mode of the data.
Given data is $5\text{,}3\text{,}4\text{,}1\text{,}2\text{,}6\text{,}4\text{,}2\text{,}2\text{,}3\text{,}1\text{,}5\text{,}6\text{,}1\text{,}2$
Arranging the data in ascending order, we have
$1\text{,}1\text{,}1\text{,}2\text{,}2\text{,}2\text{,}2\text{,}3\text{,}3\text{,}4\text{,}4\text{,}5\text{,}5\text{,}6\text{,}6.$
Mean $=\frac{\text{sumofallobservations}}{\text{totalno}.\text{ofobservations}}$
$\begin{array}{l}=\frac{1+1+1+2+2+2+2+3+3+4+4+5+5+6+6}{15}\\ =\frac{47}{15}\\ =3.13\end{array}$
Mode $=$ most frequent observation $=2$
Median $=$ value of $\left(\frac{n+1}{2}\right)\text{th}$ observation $=$ value of $\left(\frac{15+1}{2}\right)\text{th}$ observation
$=$ value of $8\text{th}$ observation $=3$
Find the mean of first six multiples of $4.$
First six multiplies of 4 are: $4\text{,}8\text{,}12\text{,}16\text{,}20\text{,}24$
Mean $=\frac{\text{sumofallobservations}}{\text{totalno}.\text{ofobservations}}$
$\begin{array}{l}=\frac{4+8+12+16+20+24}{6}\\ =\frac{84}{6}\\ =14\end{array}$
Hence, the mean of ${1}^{\text{st}}$ six multiplies of $4\text{is}14.$
Find the median of first nine even natural numbers.
First nine even natural numbers are
$2\text{,}4\text{,}6\text{,}8\text{,}10\text{,}12\text{,}14\text{,}16\text{,}18.$
Here, $n=9\text{}(\text{odd})$
Median $=$ value of $\left(\frac{n+1}{2}\right)\text{th}$ observation $=$ value of $\left(\frac{9+1}{2}\right)\text{th}$ observation
$=$ value of $5\text{th}$ observation $=10$
The mean of three numbers is $10.$ The mean of other four numbers is $12.$ Find the mean of all the numbers.
Mean of $3$ numbers $=\frac{\text{sumofthreenumbers}}{3}$
$10=\frac{\text{sumofthreenumbers}}{3}$
Hence, sum of three numbers $=30$
Mean of other $4$ numbers $=\frac{\text{sumofother}4\text{numbers}}{4}$
$12=\frac{\text{sumofotherfournumbers}}{4}$
Hence, sum of other $4$ numbers $=48$
Mean of all the numbers
$=\frac{\text{sumofallthenumbers}}{\text{totalnumbers}}$
$=\frac{\left[\text{sumoffirstthreenumbers}+\text{sumofotherfournumbers}\right]}{7}$$\begin{array}{l}=\frac{30+48}{7}\\ =\frac{78}{7}\\ =11.14\end{array}$
Find the mode of the given data:
$10\text{,}8\text{,}4\text{,}7\text{,}8\text{,}11\text{,}15\text{,}8\text{,}4\text{,}2\text{,}3\text{,}6\text{,}8$
We know that, mode is the most frequent observation in the data.
$\therefore $ Mode $=8$
Given below are heights of $15$ boys of a class measured in cm: $128\text{,}144\text{,}146\text{,}143\text{,}136\text{,}142\text{,}138\text{,}129\text{,}$ $140\text{,}152\text{,}144\text{,}140\text{,}150\text{,}142\text{,}154.$
Find
a. The height of the tallest boy.
b. The height of the shortest boy.
c. The range of the given data.
d. The median height of the boys.
Given, height (data) of $15$ boys of a class are $128\text{,}144\text{,}146\text{,}143\text{,}136\text{,}142\text{,}138\text{,}129\text{,}$ $140\text{,}152\text{,}144\text{,}140\text{,}150\text{,}142\text{,}154.$
Arranging the given data in ascending order, we have
$128\text{,}129\text{,}136\text{,}138\text{,}140\text{,}140\text{,}142\text{,}$
$142\text{,}143\text{,}144\text{,}144\text{,}146\text{,}150\text{,}152\text{,}154$
a. By observing the data, height of the tallest boy $=154\text{cm}$
b. By observing the data, height of the shortest boy $=128\text{cm}$
c. Here, highest observation $=154$ and lowest observation $=128$,
Range $=$ highest observation $$ lowest observation
$\begin{array}{l}=154128\\ =26\text{cm}\end{array}$
d. Median $=$ value of $\left(\frac{n+1}{2}\right)\text{th}$ observation $=$ value of $\left(\frac{15+1}{2}\right)\text{th}$ observation
$=$ value of $8\text{th}$ observation $=142\text{cm}$
Observe the data and answer the questions that follow:
$16\text{,}15\text{,}16\text{,}16\text{,}8\text{,}15\text{,}17$
a. Which data value can be put in the data so that the mode remains the same?
b. At least how many and which value(s) must be put in to change the mode to $15\text{?}$
c. What is the least number of data values that must be put in to change the mode to $17\text{?}$ Name them.
Given data; $16\text{,}15\text{,}16\text{,}16\text{,}8\text{,}15\text{,}17$
Arranging the given data in ascending order, we have $8\text{,}15\text{,}15\text{,}16\text{,}16\text{,}16\text{,}17$
a. As per the given data, $16$ is the mode of data, since it has highest frequency, i.e. $3.$
Now, if $15$ is added to the given data, mode will get changed to $15$ and $16\text{,}$ whereas if any other number, i.e. $8\text{,}16\text{or}17$ is added, mode will remain same.
b. At least two $15\text{'s}$ should be added to change the mode to $15.$ On adding two $15\text{'s}$ the frequency of $15$ will be maximum, i.e. $4.$
c. We will have to add at least three $17\text{'s}$ to change the mode to $17.$ On adding three $17\text{'s,}$ the frequency of $17$ will be maximum, i.e. $4.$
Age (in years) of $6$ children of two groups are recorded as below:
Age (in Years) 

Group A 
Group B 
$7$ $7$ $9$ $8$ $10$ $10$ 
$7$ $9$ $11$ $12$ $12$ $12$ 
a. Find the mode and range for each group.
b. Find the range and mode if the two groups are combined together.
Measures of central tendency are used to describe the middle of a data set. Mean, median, and mode are measures of central tendency.
From the given table, age of children in
group A: $7\text{yr,}7\text{yr,}9\text{yr,}8\text{yr,}10\text{yr,}10\text{yr}$
Age of children in group B: $7\text{yr,}9\text{yr,}11\text{yr,}12\text{yr,}12\text{yr,}12\text{yr}$
a. Mode in group A $=7\text{yrand}10\text{yr}\text{.}$ [ $\because 7\text{yrand}10\text{yr}$ occurs most frequent, i.e. $2$ times]
Range in group A
$=$ Maximum value $$ Minimum value $=107=3$
Mode in group B $=12\text{yr}$
[ $\because 12\text{yr}$ is the most frequent, i.e. $3$ ]
Range in group B $=$ Maximum value $$ Minimum value $=127=5$
b. If both groups are combined together $7\text{,}7\text{,}7\text{,}9\text{,}9\text{,}11\text{,}8\text{,}12\text{,}10\text{,}12\text{,}10\text{,}12.$
Mode $=7\text{and}12$ [ $\because 7\text{and}12$ occurs most frequent, i.e. $3$ times]
$\because $ Range $=$ Maximum value $$ Minimum value $=127=5$
Observe the given bar graph carefully and answer the questions that follow.
a. What information does the bar graph depict?
b. How many motor bikes were produced in the first three months?
c. Calculate the increase in production in May over the production in January.
b. In which month the production was minimum and what was it?
c. Calculate the average (mean) production of bikes in $6$ months.
a. The given bar graph shows the production of motor bikes by XYZ automobiles Ltd. during January to June.
b.
Total number of motor bikes
produced in first three months
$=$ Motor bikes produced in January $+$ Motor bikes produced in February $+$ Motor bikes produced in March $=600+800+700=2100$
c.
Increase in production in May
over the production in January
$=$ Production in May $$ Production in January $=900600=300$
d. By observing the graph, we can say that the production was minimum in the month of June, i.e. $500.$
e.
Average production
$=\frac{\text{Totalproduction}}{\text{no}.\text{ofmonths}}$
$\begin{array}{l}=\frac{600+800+700+1100+900+500}{6}\\ =\frac{4600}{6}\\ =767\text{bikes(approx}\text{.)}\end{array}$
The bar graph given below shows the marks of students of a class in a particular subject:
Study the bar graph and answer the following questions:
a. If $40$ is the pass mark, then how many students have failed?
b. How many students got marks from $50$ to $69\text{?}$
c. How many students scored $90$ marks and above?
d. If students who scored marks above $80$ are given merits then how many merit holders are there?
e. What is the strength of the class?
a. If $40$ is the passing marks, then students who got marks less than $40$ will be failed.
No. of students who failed $=4$
b.
No. of students who got marks
from $50\text{to}69$ $=$ (No. of students who got marks from $50\text{to}59$ )
+ (No. of students who got marks from $60$ to $69$ ) $=7+11=18$
c. No. of students who scored $90$ marks & above $=$ No. of students who scored marks $90\text{to}92=4$
d. No. of students who scored marks above $80=$ No. of students who score $80\text{to}89+$ No. of students who score $90\text{to}92=6+4=10$
Since students who scored marks above $80$ are given merits.
Number of students who are merit holders $=10$
e. Strength of the class $=$ Total no. of students who scored different marks
$=4+2+7+11+8+6+4=42$
Study the bar graph given below and answer the questions that follow.
a. What information does the above bar graph represent?
b. In which year was production the least?
c. After which year was the maximum rise in the production?
d. Find the average production of rice during the $5$ years.
e. Find difference of rice production between years $2006$ and $2008.$
After studying the bar graph, we have
Production of rice in $2005=50$ million tonnes
Production of rice in $2006=40$ million tonnes
Production of rice in $2007=70$ million tonnes
Production of rice in $2008=50$ million tonnes
Production of rice in $2009=60$ million tonnes
a. The bar graph shows the production of rice in million tonnes by a country during years $2005\text{to}2009.$
b. The production of rice was the least in $2006\text{,}$ i.e. $40$ million tonnes.
c. The maximum production of rice was in $2007.$ The production rose after $2006.$
d. Sum of productions $=50+40+70+50+60=270$
Average production $=$ $\frac{\text{sumofobservations}}{\text{no}.\text{ofobservations}}$
Average production $=\frac{270}{5}=54$ million tonnes
e. Production in $2006=40$ million tonnes and production in $2008=50$ million tonnes
Difference $=5040=10$ million tonnes
Study the bar graph given below and answer the questions that follow :
a. What information is depicted from the bar graph?
b. In which subject is the student very good?
c. Calculate the average marks of the student.
d. If $75$ and above marks denote a distinction, then name the subjects in which the student got distinction.
e. Calculate the percentage of marks the student got out of $500.$
a. The given bar graph shows marks obtained by a student in different subjects out of $100.$
b. Subject in which student is very good is Maths as he scored highest marks in it.
c. $\text{Averagemarks}=\frac{\text{sumofallmarksobtainedinvarioussubjects}}{\text{Totalsubjects}}$
$\begin{array}{l}=\frac{64+75+82+71+49}{5}\\ =\frac{341}{5}\\ =68.2\text{\%}\end{array}$
d. In Hindi & Maths, student got $75\text{}82$ marks, respectively. Since, the marks equal to $75$ or above denote a distinction. Hence, student got distinction in Hindi & Maths.
e. Percentage marks $=\frac{\text{Totalmarksscored}}{\text{Totalmarks}}\times 100\text{\%}$
$\begin{array}{l}=\frac{341}{500}\times 100\text{\%}\\ =68.2\text{\%}\end{array}$
The bar graph given below represents the circulation of newspapers (dailies) in a town in six languages (the figures are approximated to hundreds).
Study the bar graph and answer the following questions:
a. Find the total number of newspapers read in Hindi, Punjabi, Urdu, Marathi and Tamil.
b. Find the excess number of newspapers read in Hindi than those in English.
c. Name the language in which the least number of newspapers are read.
d. Write the total circulation of newspapers in the town.
Number of newspapers in Urdu $=200$
Number of newspapers in Tamil $=100$
Number of newspapers in English $=500$
Number of newspapers in Hindi $=800$
Number of newspapers in Marathi $=300$
Number of newspapers in Punjabi $=400$
a. Total no. of newspapers read in Hindi, Punjabi, Urdu, Marathi & Tamil $=800+400+200+300+100=1800$
b. Excess no. of newspapers read in Hindi than those English $=$ No. of newspapers read in Hindi $$ no. of newspapers read in English $=800500=300$
c. Out of all the newspapers, least no. of newspapers is in Tamil, i.e., $100$ newspapers are read.
d. Total circulation of newspapers in the town $=$ No. of newspapers in six different languages $=200+100+500+800+300+400=2300$
Study the double bar graphs given below and answer the following questions:
a. Which sport is liked the most by Class VIII students?
b. How many students of Class VII like Hockey and Tennis in all?
c. How many students are there in Class VII?
d. For which sport is the number of students of Class VII less than that of Class VIII?
e. For how many sports students of Class VIII are less than Class VII?
f. Find the ratio of students who like Badminton in Class VII to students who like Tennis in Class VIII.
a. By observing the graph, we can say that the height of the bar corresponding to cricket for class VIII student is largest.
Hence, cricket is liked the most by class VIII students.
b. Height of bar corresponding to hockey and tennis for class VIl are $7$ and $10$ respectively.
So, total students of class VII who like hockey and tennis $=7+10=17$
c. Total number of students in class VIl $=$ Sum of heights of all the bars for class VII $=7+16+18+10+14=65$
d. The sport for which number of students of class VII is less than that of class VIII will be that for which height of bar is less.
By observing the graph in case of cricket height of bar is less for class VII as compared to class VIII.
e. We can clearly see from the double bar graph for Hockey, Football, Tennis and Badminton, the number of students are less for class VIIl as compared to class VII.
f. Number of students who like badminton in class VII $=14$ and number of students who like tennis in class VII $=7$
$\therefore $ Required ratio $=14:7=2:1$
Study the double bar graph shown below and answer the questions that follow:
a. What information is represented by the above double bar graph?
b. In which month sales of Brand A decreased as compared to the previous month?
c. What is the difference in sales of both the Brands for the month of June?
d. Find the average sales of Brand B for the six months.
e. List all months for which the sales of Brand B was less than that of Brand A.
f. Find the ratio of sales of Brand A as compared to Brand B for the month of January.
a. The given double bar graph compares the sales of brands A and B during the months of January to June.
b. We can clearly see from the double bar graph that sales for brand A reduced in the month of March compared to that of February.
c. Sales of brand A in June $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}57\text{lakh}$ and sales of brandB in June $=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}54\text{lakh}$
Difference in sales $=5754=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{lakh}$
d. Average sales of brand B $=$ Total sales of brand B in $6$ months from January to June $\text{/}\text{\hspace{0.17em}}6$
$\begin{array}{l}=\frac{36+38+43+35+45+54}{6}\\ =\frac{251}{6}\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}41.83\text{lakhs}\end{array}$
e. We can clearly see from the double bar graph that sales of brand B is less than sales of brand A in the month of April and June.
f. Sales of brand A in January $=31$ and sales of brand B in January $=36$
$\therefore $ Required ratio $=\frac{31}{36}\text{or}31:36$
Study the double bar graph given below and answer the questions that follow:
a. What information is compared in the above given double bar graph?
b. Calculate the ratio of minimum temperatures in the year $2008$ to the year $2009$ for the month of November.
c. For how many months was the minimum temperature in the year $2008$ greater than that of year $2009\text{?}$ Name those months.
d. Find the average minimum temperature for the year $2008$ for the four months.
e. In which month is the variation in the two temperatures maximum?
a. The given double bar graph compares the minimum temperature during the month November to February for the years $2008$ and $2009.$
b. Minimum temperature of November in year $2008=18\xb0\text{C}$
Minimum temperature of November in year $2009=15\xb0\text{C}$
$\therefore $ Required ratio $=\frac{18}{15}=18:15=6:5$
c. We can clearly see from the double bar graph that the minimum temperature in the year $2008$ greater than that of the year $2009$ for the month of February and November.
d. Average minimum temperature for year $2008=$ $\frac{\text{totaltemperatureforyearin}\text{\hspace{0.17em}}4\text{months}}{4}\text{}$
$=\frac{18+11+4+12}{4}=\frac{45}{4}=11.25$
e. Difference of temperature for different months can be shown by following table:
Month 
Difference of temperature 
November 
$1815=3$ 
December 
$1211=1$ 
January 
$54=1$ 
February 
$128=4$ 
From the above table, it is clear that for the month of February variation in two temperatures is maximum.
The following table shows the average intake of nutrients in calories by rural and urban groups in a particular year. Using a suitable scale for the given data, draw a double bar graph to compare the data.
Foodstuff 
Rural 
Urban 
Pulses 
$35$ 
$49$ 
Leafy vegetables 
$14$ 
$21$ 
Other vegetables 
$51$ 
$89$ 
Fruits 
$35$ 
$66$ 
Milk 
$70$ 
$250$ 
Fish and flesh foods 
$10$ 
$22$ 
Fats and Oils 
$9$ 
$35$ 
Sugar/Jaggery 
$19$ 
$31$ 
Steps to construct the bar graphs are as follows:
Step I Firstly, we draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes.
Step Il Along the horizontal axis, we mark the foodstuff and along the vertical axis, we mark the intake of nutrients (calories).
Step Ill We choose a suitable scale to determine the heights of bars. Here, we choose the scale as $1$ small division to represent $20.$
Step IV First, we draw the bars for rural and then bars of urban for different foodstuff. Bars for rural and urban are shaded separately and the shading is shown at the top right corner of the graph.
Study the double bar graph and answer the questions that follow:
a. What information does the double bar graph represent?
b. Find the total number of boys in all sections of Class VII.
c. In which sections, the number of girls is greater than the number of boys?
d. In which section, the number of boys is the maximum?
e. In which section, the number of girls is the least?
a. The given graph shows the number of students (boys and girls) in different sections of class VII.
b. It is clear from the graph, total number of boys in all sections of class VIl $=$ Sum of heights of all the bars corresponding to boys in different sections $=15+30+20+20+25=110$
c. It is clear from the graph that in sections VII A and VII D, the number of girls are greater than the number of boys.
d. From the graph, it is clear that in section VII B, number of boys is maximum.
e. From the graph, it is clear that in section VII C, number of girls is minimum,
In a public library, the following observations were recorded by the librarian in a particular week:
Days 
Mon 
Tues 
Wed 
Thurs 
Fri 
Sat 
Newspaper Readers 
$400$ 
$600$ 
$350$ 
$550$ 
$500$ 
$350$ 
Magazine Readers 
$150$ 
$100$ 
$200$ 
$300$ 
$250$ 
$200$ 
a. Draw a double bar graph choosing an appropriate scale.
b. On which day, the number of readers in the library was maximum?
c. What is the mean number of magazine readers?
Step I We draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes.
Step Il Along the horizontal axis, we mark the days and along the vertical axis, we mark the readers.
Step Ill We choose a suitable scale to determine the heights of bars. Here, we choose the scale as $1$ small division to represent $50.$
Step IV First, we draw the bars for newspaper readers and then bars for magazine readers for different days. Bars for newspapers and magazine readers are shaded separately and the shading is shown in the top right corner of the graph paper.
b. Total no. of both readers on different days are:
Day 
Readers 
Mon 
$400+150=550$ 
Tue 
$600+100=700$ 
Wed 
$350+200=550$ 
Thur 
$550+300=850$ 
Fri 
$500+250=750$ 
Sat 
$350+200=550$ 
Hence, it is clear that the no. of readers was maximum on Thursday.
c. Mean of readers $=$ $\frac{\text{sumofallmagazinesreaderson}\text{\hspace{0.17em}}6\text{days}}{6}$
$\begin{array}{l}=\frac{150+100+200+300+250+200}{6}\\ =\frac{1200}{6}\\ =200\end{array}$
Observe the following data:
Government School, Chandpur 

Daily Attendance Date : $15.4.1994$ 

Class 
Total Students 
Number of Students Present on that Day 
VI 
$90$ 
$81$ 
VII 
$82$ 
$76$ 
VIII 
$95$ 
$91$ 
IX 
$70$ 
$65$ 
X 
$63$ 
$62$ 
a. Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
b. Which class has the maximum number of students?
c. In which class, the difference of total students and number of students present is minimum?
d. Find the ratio of number of students present to the total number of students of Class IX.
e. What per cent of Class VI students were absent?
a. A double bar graph is shown below:
From the bar graph, maximum no. of students was absent in Class VI on $\mathrm{15.4.2009}\text{,}$ whereas minimum no. of students was absent in class X.
b. Clearly, class VIII has maximum no. of students i.e, $95$
c. The difference of total no. of students & no. of students present is minimum for class X i.e, $6362=1$
d. No. of students present in class IX $=65$
Total no. of students in class IX $=70$
Hence, required ratio $=\frac{65}{70}=\frac{13}{14}\text{or}13:14$
e. Total no. of students in class VI $=90$
No. of students present in class VI $=81$
No. of absent students $=9081=9$
Percentage of absent students of class VI $=\left(\frac{\text{numberofabsentstudents}}{\text{totalnumberofstudents}}\times 100\right)\text{\%}$
$\begin{array}{l}=\left(\frac{9}{90}\times 100\right)\text{\%}\\ =10\text{\%}\end{array}$
Observe the given data:
Days of the Week 
Mon 
Tues 
Wed 
Thurs 
Fri 
Sat 
Number of Mobile Phone Sets Sold 
$50$ 
$45$ 
$30$ 
$55$ 
$27$ 
$60$ 
a. Draw a bar graph to represent the above given information.
b. On which day of the week was the sales maximum?
c. Find the total sales during the week.
d. Find the ratio of the minimum sale to the maximum sale.
e. Calculate the average sale during the week.
f. On how many days of the week was the sale above the average sales?
a. In order to construct a bar graph representing the above data, we follow the following steps:
Step I Take a graph paper and draw two mutually perpendicular lines OX and OY. Call OX as the horizontal axis and OY as the vertical axis.
Step Il Along OX, mark days and along OY, mark number of mobile phone sets sold.
Step Ill Along OX, choose the uniform (equal) width of the bars and the uniform gap between them, according to the space available for the graph,
Step IV Choose a suitable scale to determine the heights of the bars, according to the availability of space. Here, we choose $1$ small division to represent $5$ mobile sets.
b. It is clear from the graph that on Saturday the sales was maximum.
c. Total sale during the week $=$ sum of all the sales on each day $=50+45+30+55+27+60=267$
d. Minimum sale on Friday $=27$
Maximum sale on Saturday $=60$
Required ratio $=27:60=9:20$
e. Average sale $=\frac{\text{totalsale}}{6}$
$=\frac{267}{6}=44.5$
f. On Monday, Tuesday, Thursday & Saturday, i.e., on $4$ days, the sale was above the average sale.
Below is a list of $10$ tallest buildings in India.
This list ranks buildings in India that stand at least $150\text{m}\left(492\text{ft}\text{.}\right)$ tall, based on standard height measurement. This includes spires and architectural details but does not include antenna marks. Following data is given as per the available information till $2009$. Since new buildings are always under construction, go online to check new taller buildings.
Use the information given in the table about sky scrapers to answer the following questions:
Name 
City 
Height 
Floors 
Year 
Planet 
Mumbai 
$181\text{m}$ 
$51$ 
$2009$ 
UB Tower 
Bengaluru 
$184\text{m}$ 
$20$ 
$2006$ 
Ashok Towers 
Mumbai 
$193\text{m}$ 
$49$ 
$2009$ 
The Imperial I 
Mumbai 
$249\text{m}$ 
$60$ 
$2009$ 
The Imperial II 
Mumbai 
$249\text{m}$ 
$60$ 
$2009$ 
RNA Mirage Oberoi Woods 
Mumbai

$180\text{m}$

$40$

$2009$

Tower I Oberoi Woods 
Mumbai

$170\text{m}$

$40$

$2009$

Tower II Oberoi Woods 
Mumbai

$170\text{m}$

$40$

$2009$

Tower III 
Mumbai 
$170\text{m}$ 
$40$ 
$2009$ 
MVRDC 
Mumbai 
$156\text{m}$ 
$35$ 
$2002$ 
a. Find the height of each storey of the three tallest buildings and write them in the following table:
Building 
Height 
Number of Storeys 
Height of Each Storey 












b. The average height of one storey for the buildings given in (a) is ______________.
c. Which city in this list has the largest percentage of skyscrappers? What is the percentage?
d. What is the range of data?
e. Find the median of the data.
f. Draw a bar graph for given data.
a. Clearly, Imperial I, Imperial Il & Ashok Towers are $3$ tallest buildings.
Building 
Height 
Number of Storeys 
Height of Each Storey 
The Imperial I 
$249\text{m}$ 
$60$ 
$249/60=4.15$ 
The Imperial II 
$249\text{m}$ 
$60$ 
$249/60=4.15$ 
Ashok Towers 
$193\text{m}$ 
$49$ 
$193/49=3.94$ 
b. Average height of each storey of buildings in (a) $=$ sum of heights of a storey of each building $\text{/}3$
$\begin{array}{l}=\frac{4.15+4.15+3.94}{3}\\ =\frac{12.24}{3}\\ =4.08\end{array}$
c. We can clearly see from data, Mumbai has maximum no. of skyscrapers. It has $9$ skyscrapers.
Required percentage $=\frac{9}{10}\times 100=90\text{\%}$
d. Range of data $=$ maximum height $$ maximum height $=249156=93$
e. Median
$=\frac{\frac{n}{2}\text{thobservation}+\left(\frac{n}{2}+1\right)\text{thobservation}}{2}$
$=\frac{\left(\frac{10}{2}\right)\text{thobservation}+\left(\frac{10}{2}+1\right)\text{thobservation}}{2}$
$=\frac{5\text{thobservation}+6\text{thobservation}}{2}$
$\begin{array}{l}=\frac{180+181}{2}\\ =180.5\end{array}$
f. A bar graph is as shown below:
The marks out of $100$ obtained by Kunal and Soni in the Half Yearly Examination are given below:
Subjects 
English 
Hindi 
Maths 
Science 
S. Science 
Sanskrit 
Kunal 
$72$ 
$81$ 
$92$ 
$96$ 
$64$ 
$85$ 
Soni 
$86$ 
$89$ 
$90$ 
$82$ 
$75$ 
$82$ 
a. Draw a double bar graph by choosing appropriate scale.
b. Calculate the total percentage of marks obtained by Soni.
c. Calculate the total percentage of marks obtained by Kunal.
d. Compare the percentages of marks obtained by Kunal and Soni.
e. In how many subjects did Soni get more marks than Kunal? Which are those subjects?
f. Who got more marks in S. Science and what was the difference of marks?
g. In which subject the difference of marks was maximum and by how much?
a. Steps to construct the bar graphs are as follows :
Step I We draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes.
Step Il Along the horizontal axis, OX mark the subjects and along vertical axis, OY mark the marks obtained.
Step Ill We choose a suitable scale to determine the heights of bars. Here, we choose the scale as $1$ small division to represent $5$ marks.
Step IV First, we draw the bars for Kunal and then bars for Soni for different years. Bars for Kunal and Soni shaded separately and the shading is shown in the top right corner of the graph paper
b. Marks obtained by Soni $=\left(\frac{\text{totalmarksobtainedbySoniin}\text{}6\text{subjects}}{600}\right)\times 100$
$\begin{array}{l}=\frac{86+89+90+82+75+82}{600}\times 100\text{\%}\\ =\left(\frac{504}{600}\times 100\right)=84\text{\%}\end{array}$
c. Marks obtained by Kunal
$=\left(\frac{\text{totalmarksobtainedbyKunalin}6\text{subjects}}{600}\right)\times 100$
$\begin{array}{l}=\frac{72+81+92+96+64+85\text{}}{600}\times 100\text{\%}\\ =\left(\frac{490}{600}\times 100\right)=81.6\text{\%}\end{array}$
d. Ratio of percentage marks obtained by Soni & Kunal $=81.6:84=34:35$
e. In English, Hindi & S.Science, Soni get more marks than Kunal.
f. Marks obtained by Kunal & Soni is S.Science are $64\text{}75$ respectively. Therefore, Soni got more marks than Kunal by $11$ marks.
g. In English & Science the difference of marks was maximum $\left(504490\right)=14\text{marks}$
The students of Class VII have to choose one club from Music, Dance, Yoga, Dramatics, Fine arts and Electronics clubs. The data given below shows the choices made by girls and boys of the class. Study the table and answer the questions that follow:
Clubs 
Music 
Dance 
Yoga 
Dramatics 
Fine Arts 
Electronics 
Girls 
$15$ 
$24$ 
$10$ 
$19$ 
$27$ 
$21$ 
Boys 
$12$ 
$16$ 
$8$ 
$17$ 
$11$ 
$30$ 
a. Draw a double bar graph using appropriate scale to depict the above data.
b. How many students are there in Class VII?
c. Which is the most preferred club by boys?
d. Which is the least preferred club by girls?
e. For which club the difference between boys and girls is the least?
f. For which club is the difference between boys and girls the maximum?
a. Steps to construct the bar graph are as follows :
Step I We draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes.
Step Il Along the horizontal axis, OX mark the clubs and along the vertical axis, OY mark the number of boys and girls.
Step Ill We choose a suitable scale to determine the heights of bars. Here, we choose the scale as $1$ small division to represent $2.$
Step IV First, we draw the bars for girls and then bars for boys for different years. Bars for girls and boys are shaded separately and the shading is shown in the top right corner of the graph paper.
b. Total students in class VII $=15+12+24+16+10+8+19+17+27$$+11+21+30=210$
c. From the given data, we can say that most preferred club by boys is Electronics.
d. From the given data, we can say that least preferred club by girls is Yoga.
e. It is clear from the given data that in Yoga and Dramatics, the difference between boys and girls is the least, i.e. $\left(1917\right)=2$
f. It is clear from the given data that in Fine Arts the difference between boys and girls is maximum, i.e. $\left(2711\right)=16$
The data given below shows the production of motor bikes in a factory for some months of two consecutive years.
Months 
Feb 
May 
August 
October 
December 
$2008$ 
$2700$ 
$3200$ 
$6000$ 
$5000$ 
$4200$ 
$2007$ 
$2800$ 
$4500$ 
$4800$ 
$4800$ 
$5200$ 
Study the table given above and answer the following questions:
a. Draw a double bar graph using appropriate scale to depict the above information and compare them.
b. In which year was the total output the maximum?
c. Find the mean production for the year $2007.$
d. For which month was the difference between the production for the two years the maximum?
e. In which month for the year $2008\text{,}$ the production was the maximum?
f. In which month for the year $2007\text{,}$ the production was the least?
a. Steps to construct the bar graphs are as follows:
Step I We draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes.
Step Il Along the horizontal axis, OX mark the months and along the vertical axis, OY mark the production of motor bikes.
Step Ill We choose a suitable scale to determine the heights of bars. Here, we choose the scale as $1$ big division to represent $400.$
Step IV First, we draw the bars for Year $2008$ and then bars for Year $2007$ for different months. Bars for year $2008$ and year $2007$ months are shaded separately and the shading is shown in the top right corner of the graph paper.
b. total output in year $2008$ $=2700+3200+6000+5000+4200=21100$
Total output in year $2007$ $=2800+4500+4800+4800+5200=22100$
Total output in year $2007$ is more than that of year $2008.$
c. Mean production of the year $2007$ $=\frac{\text{totalproductioninyear}2007\text{for}5\text{months}}{5}$
$\begin{array}{l}=\frac{22100}{5}\\ =4420\end{array}$
d. It is clear from the given data in May the difference between the production for the two years in maximum, i.e. $1300.$
e. In August the production was maximum, i.e. $6000$ as compared to other months of year $2008.$
f. In February the production was minimum, i.e. $2800$ as compared to other months of year $2007.$
The table below compares the population (in hundreds) of $4$ towns over two years:
Towns 
A 
B 
C 
D 
$2007$ 
$2900$ 
$6400$ 
$8300$ 
$4600$ 
$2009$ 
$3200$ 
$7500$ 
$9200$ 
$6300$ 
Study the table and answer the following questions:
a. Draw a double bar graph using appropriate scale to depict the above information.
b. In which town was the population growth maximum?
c. In which town was the population growth least?
a. Steps to construct the bar graph are as follows :
Step I We draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes.
Step Il Along the horizontal axis, OX mark the towns and along the vertical axis, OY mark the population.
Step Ill We choose a suitable scale to determine the heights of bars. Here, we choose the scale as $1$ small division to represent $500\text{,}$
Step IV First, we draw the bars for year $2007$ and then bars for year $2009$ for different towns. Bars for year $2007$ and $2009$ are shaded separately and the shading is shown in the top right corner of the graph paper.
b. It is clear from the graph, the population growth of town D was maximum.
c. It is clear from the graph, the population growth of town A was minimum.
The table below gives the data of tourists visiting $5$ hill stations over two consecutive years. Study the table and answer the questions that follow:
Hill stations 
Nainital 
Shimla 
Manali 
Mussoorie 
Kullu 
$2008$ 
$4000$ 
$5200$ 
$3700$ 
$5800$ 
$3500$ 
$2009$ 
$4800$ 
$4500$ 
$4200$ 
$6200$ 
$4600$ 
a. Draw a double bar graph to depict the above information using appropriate scale.
b. Which hill station was visited by the maximum number of tourists in $2008\text{?}$
c. Which hill station was visited by the least number of tourists in $2009\text{?}$
d. In which hill stations was there increase in number of tourists in the year $2009\text{?}$
a. Steps to construct the bar graph as follows:
Step I We draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes.
Step Il Along the horizontal axis, OX mark the hill stations and along the vertical axis, OY mark the tourist visitors.
Step Ill We choose a suitable scale to determine the heights of bars. Here, we choose the scale as $1$ small division to represent $400$ tourists.
Step IV First, we draw the bars for year $2008$ and then bars for year $2009$ for different hill stations. Bars for years $2008$ and $2009$ are shaded separately and the shading is shown in the top right corner of the graph paper
b. It is clear from the given data that in year $2008$ tourists visit Mussoorie the most.
c. It is clear from the given data that in year $2009$ tourists visit Manali the least.
d. From the graph, we can say that in $2009\text{,}$ there is increase in tourist visitors in the places; Manali, Nainital, Mussoorie and Kullu.
The table below gives the flavours of ice cream liked by children (boys and girls) of a society.
Flavours 
Vanilla 
Chocolate 
Strawberry 
Mango 
Butterscotch 
Boys 
$4$ 
$9$ 
$3$ 
$8$ 
$13$ 
Girls 
$8$ 
$12$ 
$7$ 
$9$ 
$10$ 
Study the table and answer the following questions:
a. Draw a double bar graph using appropriate scale to represent the above information.
b. Which flavour is liked the most by the boys?
c. How many girls are there in all?
d. How many children like chocolate flavour of ice cream?
e. Find the ratio of children who like strawberry flavour to vanilla flavour of ice cream.
a. Bar graph:
b. On observing the bar graph, we can say that boys like butterscotch the most because the bar for butterscotch in case of boys is of highest length, i.e. $13.$
c. Total number of girls $=$ Sum of heights of bars corresponding to girls $=8+4+12+7+9+10=46$
d. Number of children who like chocolate flavour $=$ Sum of heights of bars for both boys and girls corresponding to chocolate $=9+12=21$
e. Total number of children who like strawberry $=3+7=10$
Total number of children who like vanilla $=4+8=12$
$\therefore $ Ratio of children who like strawberry flavour to vanilla flavour of icecream $=10:12$ $=5:6$