Download App

MATHS 2 Marks Questions

Data Handling · Gujarat Board (GSEB) STD 7 (GSEB - English Medium). 20 questions.

Questions

2 Marks Questions

Take a timed test

20 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
Discuss with your friends and give
two situations where mode would be an appropriate representative value to use.
Answer
Case I We can understand it by the following cases Let the numbers be $2,3,4,5,0,1,3,3,4$ and 3.
Arranging the numbers in ascending order, we have
$
0,1,2,3,3,3,3,4,4,5
$
Here, 3 occurs most frequently ( 4 times).
$\therefore$ Mode $=3$
Case II Let the numbers be $41,39,48,52,46,62,54$, $40,96,52,98,40,42,52$ and 60.
Arranging the data in descending order, we have $98,96,62,60,54,52,52,52,48,46,42,41,40,40,39$
Here, 52 occurs most frequently ( 3 times).
$\therefore$ Mode $=52$
Hence, it is clear that mode would be an appropriate representative value to use.
View full question & answer→
Question 22 Marks
Discuss with your friends and give
two situations where mean would be an appropriate representative value to use, and
Answer
We can understand it by the following cases
Case I Let the heights (in cm ) of six boys in a group be
$
\begin{array}{l}
150,154,152,170,164 \text { and } 156 . \\
\therefore\text { Mean }=\frac{\text { Sum of the given observations }}{\text { Number of observations }} \\
=\frac{150+154+152+170+164+156}{6} \\
=\frac{946}{6} \\
=157.67 cm
\end{array}
$
Case II We know that first six whole numbers are. 0 , $1,2,3,4$ and 5 .
$\therefore$ Mean of first six whole numbers
$
\begin{array}{l}
=\frac{\text { Sum of the numbers }}{\text { Number of terms }} \\
=\frac{0+1+2+3+4+5}{6} \\
=\frac{15}{6}=2.5
\end{array}
$
Hence, it is clear that mean is the appropriate representative value.
View full question & answer→
Question 32 Marks
Find atleast 5 numbers between $\frac{1}{2}$ and $\frac{1}{3}$.
Answer
Given numbers are $\frac{1}{2}$ and $\frac{1}{3}$.
1 st number $=\frac{\text { Sum of given numbers }}{\text { Number of digits }}=\frac{\frac{1}{2}+\frac{1}{3}}{2}=\frac{\frac{3+2}{6}}{2}$
$=\frac{5}{6 \times 2}=\frac{5}{12} \quad[\because$ LCM of 2 and $3=6]$
$\therefore \quad \frac{1}{2}>\frac{5}{12}>\frac{1}{3}$
2 nd number $=\frac{\text { Given number }+1 \text { st number }}{2}$
$
=\frac{\frac{1}{2}+\frac{5}{12}}{2}=\frac{\frac{6+5}{12}}{2}=\frac{11}{12 \times 2}=\frac{11}{24}
$
$
\therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{5}{12}>\frac{1}{3}
$
Similarly, 3rd number $=\frac{\frac{5}{12}+\frac{1}{3}}{2}=\frac{\frac{5+4}{12}}{2}=\frac{9}{12 \times 2}=\frac{9}{24}$
$
\begin{array}{l}
\therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{5}{12}>\frac{9}{24}>\frac{1}{3} \\
\text { 4th number }=\frac{\frac{5}{12}+\frac{11}{24}}{2}=\frac{\frac{10+11}{24}}{2}=\frac{21}{24 \times 2}=\frac{21}{48} \\
\therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{21}{48}>\frac{5}{12}>\frac{9}{24}>\frac{1}{3} \\
\text { 5th number }=\frac{\frac{5}{12}+\frac{9}{24}}{2}=\frac{\frac{10+9}{24}}{2}=\frac{10+9}{24 \times 2}=\frac{19}{48} \\
\therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{21}{48}>\frac{5}{12}>\frac{19}{48}>\frac{9}{24}>\frac{1}{3}
\end{array}
$
Hence, 5 numbers between $\frac{1}{2}$ and $\frac{1}{3}$ are
$
\frac{11}{24}, \frac{21}{48}, \frac{5}{12}, \frac{19}{48} \text { and } \frac{9}{24} .
$
View full question & answer→
Question 42 Marks
Consider this data collected from a survey of a colony
Favourite
sport		CricketBasket
ball		SwimmingHockeyAthletics
Watching1240470510430250
participating620320320250105
(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Answer
View full question & answer→
Question 62 Marks
Find the mode and median of the following data
13, 16, 12, 14, 19, 12, 14, 13, 14.
Answer
(i) On arranging the data in ascending order, we get
$
12,12,13,13,14,14,14,16,19 .
$
Here, 14 occurs more frequently i.e. 3 times.
$
\therefore \text { Mode }=14
$
(ii) Here, ascending order of the given data is
$
12,12,13,13,14,14,14,16,19
$
Since, the middle observation of the data is 14 .
$\therefore$ Median of data $=14$
Hence, mode $=14$ and median $=14$
View full question & answer→
Question 72 Marks
The enrolment in a school during six consecutive years was as follows
$ 1555,1670,1750,2013,2540,2820 . $
Find the mean enrolment of the school for this period.
Answer
Given, the enrolment of a school during six consecutive years
$
1555,1670,1750,2013,2540,2820
$
Now, sum of the enrolments
$
\begin{array}{l}
=1555+1670+1750+2013+2540+2820 \\
=12348
\end{array}
$
Number of years $=6$
$\therefore$ Mean enrolment of the school for this period
$
\begin{array}{l}
=\frac{\text { Sum of the enrolments }}{\text { Number of years }} \\
=\frac{12348}{6}=2058
\end{array}
$
Hence, the mean enrolment of the school for this period is 2058.
View full question & answer→
Question 82 Marks
A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.
Answer
Here, the sum of total runs =58+76+40+35+46+45+0+100
=400
Number of innings = 8
$\therefore$ Mean score $=\frac{\text { Total runs }}{\text { Number of innings }}$
=$\frac{400}{8}$=50
Hence, the mean score is 50 runs.
View full question & answer→
Question 92 Marks
Find the mean of the first five whole numbers.
Answer
We know that whole numbers are those which start from zero (0).
So, first five whole numbers are 0, 1, 2, 3 and 4.
$\begin{aligned} \therefore \quad \text { Mean } & =\frac{\text { Sum of numbers }}{\text { Number of terms }} \\ & =\frac{0+1+2+3+4}{5} \\ & =\frac{10}{5}=2\end{aligned}$
Hence, the mean of first five whole numbers is 2.
View full question & answer→
Question 102 Marks
Find the range of heights of any ten students of your class.
Answer
Let the heights (in cm ) of 10 students of your Class VII be
$150,151,147,149,150,152,149,143,146$ and 152.
Rearranging the heights in ascending order, we get
$143,146,147,149,149,150,150,151,152,152$
Now, range of the height of students
$
\begin{array}{l}
=\text { Highest observation }- \text { Lowest observation } \\
=152-143=9 cm
\end{array}
$
Hence, the range of height of 10 students is 9 cm .
View full question & answer→
Question 112 Marks
Find the mean of the first ten even natural numbers.
Answer
First ten even natural numbers
$
\begin{aligned}
=2,4,6,8,10,12,14,16,18,20 \\
\text { Mean } =\frac{\text { Sum of all observations }}{\text { Number of observations }}
\end{aligned}
$
$\begin{array}{l}\text { Sum of all observations }=2+4+6+8+10+12 \\ \begin{aligned} +14+16+18+20=110\end{aligned} \\ \therefore \text { Mean }=\frac{110}{10}=11\end{array}$
View full question & answer→
Question 122 Marks
The mean of 3 numbers is 40. All the three numbers are different natural numbers. If two number are 19 and 49. Then, what could be the highest possible value of remaining number?
Answer
Here, total numbers are 3 and mean of 3 numbers is 40 .
$
\begin{aligned}
\text { Sum of three numbers } & =\text { Total numbers } \times \text { Mean } \\
& =3 \times 40=120
\end{aligned}
$
If two numbers are 19 and 49 , then the highest number will be $120-19-49=52$.
View full question & answer→
Question 132 Marks
The height of 6 girls in a group are $142 cm, 150 cm$, $146 cm, 154 cm, x cm$ and $148 cm$ and their average height is 147 . Then, find the value of $x$.
Answer
Given, heights (in cm ) of 6 girls in a group are 142, 150, $146,154, x$ and 148.
$
\begin{aligned}
\text { Sum of heights } & =142+150+146+154+x+148 \\
& =740+x
\end{aligned}
$
$\begin{array}{l}\text { Mean/Average }=\frac{\text { Sum of observations }}{\text { Number of observations }} \\ \Rightarrow \quad 147=\frac{740+x}{6} \Rightarrow 147 \times 6=740+x\end{array}$
$\Rightarrow \quad 882=740+x \Rightarrow x=882-740=142$
View full question & answer→
Question 142 Marks
A car seller collects the following data of cars sold in his shop
Colour of carsNumber of cars sold
Red15
Black20
White17
Silver12
Others9
(i) Which colour of the car is most liked?
(ii) Which measure of central tendency was used in (i)?
Answer
(i) 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.
(ii) Mode concept used in (i).
View full question & answer→
Question 152 Marks
Find the mode and median of the following data
$
13, 16, 12, 14, 19, 12, 14, 13, 14 .
$
Answer
(i) On arranging the data in ascending order, we get
$
12,12,13,13,14,14,14,16,19
$
Here, 14 occurs more frequently i.e. 3 times.
$
\therefore \quad \text { Mode }=14
$
(ii) Here, ascending order of the given data is
$
12,12,13,13,14,14,14,16,19
$
Since, the middle observation of the data is 14 .
$\therefore$ Median of data $=14$
Hence, mode $=14$ and median $=14$
View full question & answer→
Question 162 Marks
Find the mode, mean and median of the following data
$
4,5,5,3,9,2,6,4,19
$
Answer
For median, we have to arrange the given data in ascending order, $2,3,4,4,5,5,6,9,19$
$
\begin{aligned}
\therefore \text { Mode } & =\text { Heighest frequency observation }=4,5 \\
\text { Median } & =\text { Middle observation }=5 \\
\text { Mean } & =\frac{\text { Sum of all observations }}{\text { Number of observations }}
\end{aligned}
$
Sum of all observations
$
\begin{array}{l}
\quad=2+3+4+4+5+5+6+9+19=57 \\
\therefore \text { Mean }=\frac{57}{9}=6.33
\end{array}
$
View full question & answer→
Question 172 Marks
Find the range of the following heights (in cm ) of 7 students of Class VII.
$
140,142,141,139,138,146,136 .
$
Answer
Given, heights of Class VII students are
$
140,142,141,139,138,146 \text { and } 136 \text {. }
$
Ascending order of the given heights
$
\begin{array}{l}
136,138,139,140,141,142,146 \\
\text { Range }=\text { Maximum value }- \text { Minimum value } \\
=146-136=10
\end{array}
$
View full question & answer→
Question 182 Marks
Find the range of the data $12,4,6,7,9,14,26,36$.
Answer
Given data, $12,4,6,7,9,14,26,36$
Ascending order of the given data
$
\begin{aligned}
& 4,6,7,9,12,14,26,36 \\
\text { Range } & =\text { Maximum value }- \text { Minimum value } \\
& =36-4=32
\end{aligned}
$
View full question & answer→
Question 192 Marks
The runs scored in a cricket match by 11 players is as follows
$
0,9,65,72,49,51,23,24,8,4,26 .
$
Find the mean runs scored of this data.
Answer
As per the given information, runs scored by 11 players in a cricket are $0,9,65,72,49,51,23,24,8,4$ and 26.
$
\text { Mean }=\frac{\text { Sum of all observations }}{\text { Number of observations }}
$
Sum of all observations
$
\begin{aligned}
& =0+9+65+72+49+51+23+24+8+4+26 \\
& =331 \\
\therefore \text { Mean } & =\frac{331}{11}=30.09
\end{aligned}
$
View full question & answer→
Question 202 Marks
Find a number between $\frac{1}{3}$ and $\frac{1}{4}$.
Answer
We know that the arithmetic mean of two given numbers is always between them.
Number between $\frac{1}{3}$ and $\frac{1}{4}$
$
=\frac{\frac{1}{3}+\frac{1}{4}}{2}=\frac{\frac{4+3}{12}}{2}=\frac{\frac{7}{12}}{2}=\frac{7}{12} \times \frac{1}{2}=\frac{7}{24}
$
Hence, the number between $\frac{1}{3}$ and $\frac{1}{4}$ is $\frac{7}{24}$.
View full question & answer→
Generate Paper FreeAll Data Handling topics
2 Marks Questions - MATHS STD 7 Questions - Vidyadip