NCERT Solutions Class 7 Maths Chapter 3

Q.1 Find the range of heights of any ten students of your class.

Ans.

$\begin{array}{l}\text{Let the heights (in cm) of any ten students be}\\ \text{124, 126, 133,135, 136, 139, 140, 142, 144, 147}\\ \text{Highest value among these observations}=\text{147}\text{cm}\\ \text{Lowest value among these observations}=\text{124 cm}\\ \text{Range}=\text{Highest value-Lowest value}\\ =(\text{147}-\text{124})\text{cm}\\ =\boxed{\text{23 cm}}\end{array}$

Q.2 Organise the following marks in a class assessment, inat a bular form.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}\mathrm{Which}\mathrm{number}\mathrm{is}\mathrm{the}\mathrm{greatest}?\\ \left(\mathrm{ii}\right)\mathrm{Which}\mathrm{number}\mathrm{is}\mathrm{the}\mathrm{lowest}?\\ \left(\mathrm{iii}\right)\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{data}?\\ \left(\mathrm{iv}\right)\mathrm{Find}\mathrm{the}\mathrm{arithmetic}\mathrm{mean}.\end{array}$

Ans.

\begin{array}{l}\text{(i)}\text{9}\\ \text{(ii)}1\\ (iii)\text{Range=9-1}=\boxed{\text{8}}\\ (iv)\text{Arithmetic}\text{mean=}\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{100}{20}=\boxed{5}\end{array}

Q.3 Find the mean of the first five whole numbers.

Ans.

$\begin{array}{l}\text{First five whole numbers are 0, 1, 2, 3 and 4}\\ \text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ \text{So, Mean =}\frac{0+1+2+3+4}{5}=\frac{10}{2}=\boxed{5}\\ \text{Thus, mean of first five whole number is 5.}\end{array}$

Q.4 A cricket scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.

Ans.

$\begin{array}{l}\text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ \text{So},\text{Mean}=\frac{58+76+40+35+46+45+0+100}{8}\\ =\frac{400}{8}\\ =\boxed{50}\\ \text{Thus, the mean score is 50}\end{array}$

Q.5 Following table shows the points of each player scored in four game.

Now answer the following questions:
(i) find the mean to determine A’ save range number of points scored per game.
(ii) To find mean number of points per game of C,
would you divide the total points by 3 or by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who is the best performer?

Ans.

\begin{array}{l}\text{(i)}\\ \text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ \text{So,}\text{Mean}=\frac{14+16+10+10}{4}\\ =\frac{50}{4}=\boxed{12.5}\\ \text{Thus, A}’\text{s average number of points scored per game is 12}\text{.5}\\ \text{(ii)}\\ \text{Since C played only 3 games, so to find the mean number of}\\ \text{points per game for C},\text{divide the total points by 3}\text{.}\\ \text{(iii)}\\ \text{Mean of B =}\frac{0+8+6+4}{4}=\frac{18}{4}=4.5\\ \text{(iv)}\\ \text{Since, the mean of A}\text{}\text{is better and bigger than B and C, so A is}\\ \text{the best performer}\text{.}\end{array}

Q.6 The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75.
Find the :
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.

Ans.

\begin{array}{l}\text{(i) Clearly, the from the given data, the highest and the lowest}\\ \text{marks are 95 and 39 respectively}\text{.}\\ \text{(ii) Range}=\text{Highest}-\text{Lowest}\\ =95-39=56\\ \text{(iii)}\text{}\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{85+76+90+58+39+48+56+95+81+75}{10}\\ =\boxed{70.3}\end{array}

Q.7 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.

Ans.

\begin{array}{c}\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{1555+1670+1750+2013+2540+2820}{6}\\ =\frac{12348}{6}\\ =\boxed{2058}\end{array}

Q.8 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.

Ans.

\begin{array}{c}\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{1555+1670+1750+2013+2540+2820}{6}\\ =\frac{12348}{6}\\ =\boxed{2058}\end{array}

Q.9 The rain fall (in mm) in a city on 7 days of a certain week was recorded as follows:

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.

Ans.

\begin{array}{l}\text{(i) Range}=\text{Highest-Lowest}\\ =20.\text{5 mm}–0.\text{0 mm}=\boxed{20.5}\text{mm}\\ \text{(ii)}\text{Mean rainfall}=\frac{0.0+12.2+2.1+0.0+20.5+5.5+1.0}{7}\\ =\frac{41.3}{7}=5.9\text{mm}\\ \text{(iii)}\text{For five days}\text{(Monday, Wednesday, Thursday, Saturday}\\ \text{and Sunday), rainfall was less than the mean rainfall}\text{.}\end{array}

Q.10 The heights of 10 girls were measured in cm and theresults are as follows:135,150,139,128,151,132,146,149,143,141iWhat is the height of the tallest girl?iiWhat is the height of the shortest girl?iiiWhat is the range of the data?ivWhat is the mean height of the girls?vHow many girls have heights more than the mean

Ans.

\begin{array}{l}\text{(i) The height of the tallest girl is 151 cm}\text{.}\\ \text{(ii) The height of the shortest girl is 128}\text{.}\\ \text{(iii) Range}=\text{Highest}-\text{Lowest}=\left(151-128\right)\text{cm}=\boxed{23\text{cm}}\\ \text{(iv) Mean height}\\ =\frac{135+150+139+128+151+132+146+149+143+141}{10}\\ =\frac{1414}{10}\\ =\boxed{141.4\text{cm}}\\ \text{Thus, five girls have more height than the mean height}\text{.}\end{array}

Q.11 The scores in mathmatics test (out of 25) of 15 students is as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data, are they same?

Ans.

$\begin{array}{l}\text{Arranging the score in ascending order.}\\ \text{5,9,10,12,15,16,19,20,20,20,20,23,24,25,25}\\ \text{Since, mode is that value of observation which occurs for}\\ \text{the most number of time and median is the middle observation.}\\ \text{There are 15 values, so median would be the 8th observation}\\ \text{and 20 occurs 4 times.}\\ \text{So,}\boxed{\text{Mode}=\text{20\hspace{0.17em}and Median}=\text{20}}\\ \text{Yes both are same.}\end{array}$

Q.12 The runs scored in a cricket match by 11 players isas follows:6,15,120,50,100,80,10,15,8,10,15Find the mean, mode and median of this data.Are the three same?

Ans.

\begin{array}{l}\text{Arranging in ascending order to get}\\ \text{6,8,10,10,15,15,15,50,80,100,120}\\ \text{Mean}=\frac{6+8+10+10+15+15+15+50+80+100+120}{11}\\ =\frac{429}{11}=\boxed{39}\\ \text{Mode is that value of observation which occurs for the most}\\ \text{number of time and median is the middle observation}\text{.}\\ \text{There are 11 values, so median would be the 6th observation}\\ \text{and 15 occurs 3 times}\text{.}\\ \text{So,}\boxed{\text{Mode}=\text{15}\text{and Median}=\text{15}}\\ \text{No, these three are not same}\text{.}\end{array}

Q.13 The weights inkg. of 15 students of a class are :38,42,35,37,45,50,32,43,43,40,36,38,43,38,47i Find the mode and median of this data.ii Is there more than one mode?

Ans.

\begin{array}{l}\text{Arranging the weight in ascending order}\text{.}\\ \text{32,35,36,37,38,38,38,40,42,43,43,43,45,47,50}\\ \text{Mode of given data is that value of observation which occurs for}\\ \text{the most number of time and median is the middle observation}\text{.}\\ \text{There are 15 values, so median would be the 8th observation}\\ \text{So, Median}=\boxed{\text{40}}\\ \text{Also 38 and 43 both occurs 3 times}\text{.}\\ \text{So,}\boxed{\text{Mode}=\text{38}\text{and 43}}\\ \text{Yes there are two mode for given data}\text{.}\end{array}

Q.14 Find the mode and median of the data:
13, 16, 12, 14, 19, 12, 14, 13, 14

Ans.

\begin{array}{l}\text{Arrange in ascending order to get:}\\ \text{12,12,13,13,14,14,1416,19}\\ \text{Mode of given data is that value of observation which occurs for}\\ \text{the most number of time and median is the middle observation}\text{.}\\ \text{There are 9 values, so median would be the 5th observation}\\ \text{So, Median}=\boxed{\text{14}}\\ \text{Also 14 occurs 3 times}\text{.}\\ \text{So,}\boxed{\text{Mode}=\text{14}}\end{array}

Q.15

\begin{array}{l}Tellwhetherthestatementistrueorfalse:\\ (i)Themodeisalwaysoneofthenumbersinadata.\\ \left(ii\right)Themeancanbeoneofthenumbersinadata.\\ \left(iii\right)Themedianisalwaysoneofthenumbersinadata.\\ \left(iv\right)Adataalwayshasamode.\\ \left(v\right)Thedata6,4,3,8,9,12,13,9hasmean9.\end{array}

Ans.

\begin{array}{l}\text{(i)}\boxed{\text{True}}\text{, because mode is the value that occurs most of}\\ \text{the time and it belongs to the given data}\text{.}\\ \text{(ii)}\boxed{\text{False}}\text{, Mean may or may not belong to the data}\text{.}\\ \text{(iii)}\boxed{\text{True}}\text{, because median is the middle observation}\\ \text{of given data}\text{.}\\ \text{(iv)}\boxed{\text{False}}\\ \text{The given data is: 6,4,3,8,9,12,13,9}\text{.}\\ \text{So, Mean}=\frac{6+4+3+8+9+12+13+9}{8}\\ =\frac{64}{8}\\ =8\end{array}

Q.16 Use the bar graph (Fig below) to a answer the following question
(a) Which is the most popular pet?
(b) How many children have dog as a pet?

Ans.

(a) The bar showing cats is the tallest, so cat is the most popular pet.
(b) From the bar graph, the number of children having dog as a pet is 8.

Q.17

\begin{array}{l}\left(\text{i}\right)\text{About how many books were sold in 1989? 1990? 1992?}\\ \left(\text{ii}\right)\text{Inwhich year were about 475 books sold?}\\ \text{About 225 books sold?}\\ \left(\text{iii}\right)\text{In which years were fewer than 25}0\text{books sold}?\\ \left(\text{iv}\right)\text{Can you explain how you would estimate the number}\\ \text{of books sold in 1989?}\end{array}

Ans.

\begin{array}{l}\text{(i) In 1989, 175 books were sold in 1990, 475 books were sold}\text{.}\\ \text{In 1992, 225 books were sold}\text{.}\\ \text{(ii) From the graph, it can be concluded that 475 books were}\\ \text{sold in the year 1990 and 225 books were sold in the year 1992}\text{.}\\ \text{(iii) From the graph, it can be concluded that in the years}\\ \text{1989 and 1992, the number of books sold were less than 250}\text{.}\\ \text{(iv) From the graph, it can be concluded that the number of}\\ \text{books sold in the year 1989 is about 1 and}\frac{3}{4}\text{th}\text{part of 1 cm}\\ \text{The scale is taken as 1 cm = 100 books}\\ \text{So, 100+}\frac{3}{4}\times 100=100+75=175\\ \text{Therefore, about 175 books were sold in the year 1989}\text{.}\end{array}

Q.18 Number of children in six different classes are given below.
Represent the data on a bar graph.

(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? and the minimum?
(ii) Find the ratio of students of class sixth to the students of class eight.

Ans.

\begin{array}{l}\text{(a)}\text{Choose a scale as 1 unit = 10 children, because we can}\\ \text{show a clear difference between the number of students of}\\ \text{class 7th and class 9th}\text{.}\\ \text{(b)}\\ \text{(i)}\\ \text{The bar representing the number of children of class fifth is}\\ \text{tallest, there are maximum number of children in class fifth}\\ \text{and the bar representing the number of children of class tenth is}\\ \text{smallest, there are minimum number of children in class tenth}\text{.}\\ \text{(ii)}\\ \text{The number of student in class sixth is 120 and the number of}\\ \text{student in class eight is 100}\text{.}\\ \text{So, the ratio ratio is =}\frac{120}{100}=\frac{20\times 6}{20\times 5}=\boxed{\frac{6}{5}}\end{array}

Q.19 The performance of student sin 1^st Term and 2^nd Term is given.
Draw a double bar graph choosing appropriate
scale and answer the following:

(i) In which subject, has the child improved h is performance the most?
(ii) in which subject is the improvement the least?
(iii) Has the performance gone down in any subject?

Ans.

\begin{array}{l}\text{(i) There was a maximum increase in the marks obtained in}\\ \text{Maths}\text{. Thus, the child improved his performance the most}\\ \text{in Maths}\text{.}\\ \text{(ii) From the graph, we observe that the improvment was least}\\ \text{in S}\text{. Science}\text{.}\\ \text{(iii) From the graph, we observe that the performance in Hindi}\\ \text{has grown down}\text{.}\end{array}

Q.20 Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the graph?
(ii) Which sport is most popular?
(iii) Which is more preferred watching or participating sport?

Ans.

(i)

\begin{array}{l}\text{(ii) From the bar graph, the bar representing the number}\\ \text{of people who like watching and participating in cricket}\\ \text{is the tallest among all the bars}\text{.Thus, cricket is the most}\\ \text{popular sport}\text{.}\\ \text{(iii) The bars showing watchig spot are longer than the bars}\\ \text{showing participating in spot}\text{. Thus, watching different types}\\ \text{of sports is more preffered than participating in the sports}\text{.}\end{array}

Q.21 Take the data giving the mininum and the maximum temperature of various cities given in the beginning of this chapter.

Plot a double bar graph using thedata and answer the following:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.

Ans.

\text{A double bar graph for given data is shown as below:}

\begin{array}{l}\text{(i) From the graph, we observe that Jammu has the largest}\\ \text{difference in its minimum and maximum temprature on the}\\ \text{given date}\text{.}\\ \text{(ii) From graph, we observe that Jammu is the hottest city}\\ \text{and Banglore is the coldest city}\text{.}\\ \text{(iii) Banglore and Jaipur, Banglore and Ahemdabad}\text{.}\\ \text{For Banglore, the maximum temprature was 28}°C\text{, while}\\ \text{temprature of both cities was 29}°C.\\ \text{(iv) From graph, we observe that the city which has the}\\ \text{least difference between its minimum and maximum}\\ \text{temprature is Mumbai}\text{.}\end{array}

Q.22 Tell whether the following is cretain to happen, impossible, can happed but not certain.
(i) You are older today than yesterday.
(ii) Atossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrowl will be a coludy day.

Ans.

(i) Certain
(ii) Can happen but not certain
(iii) Impossible
(iv) Can happen but not certain
(v) Can happen but not certain

Q.23 There are 6 marbles in a box with numbers from 1 to 6 marked on each o0f them.
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with umber 5?

Ans.

\begin{array}{l}\text{(i)}\text{Since,}\boxed{\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}}\\ \text{So,P(apperance of 2)}=\boxed{\frac{1}{6}}\\ \text{(ii) P(apperance of 2)}=\boxed{\frac{1}{6}}\end{array}

Q.24 A coin is flipped to decide which team starts the game.
what is the probability that your team will start?

Ans.

\begin{array}{l}\text{A coin has two faces namely}-\text{Head and Tail}\text{.}\\ \text{One team can opt either Head or Tail}\text{.}\\ \text{Since,}\boxed{\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Number of favourable outcomes}}}\\ \text{So, Probability(our team start first)}=\boxed{\frac{1}{2}}\end{array}

Q.25 A box contains pairs of socks of two colours (black and white) I have picked out a white sock.
I pick out one more with my eyes closed. What is the probability that it will make a pair?

Ans.

\begin{array}{l}\text{While closing the eyes, one can draw either a black sock}\\ \text{or a white sock}\text{.}\\ \text{Therefore, there are two possible cases}\text{.}\\ \text{Since,}\boxed{\text{Probability=}\frac{\text{Number of favourable outcomes}}{\text{Number of favourable outcomes}}}\\ \text{So, P(a pair of white socks will be formed) =}\boxed{\frac{1}{2}}\end{array}