RS Aggarwal Maths Solutions Class 7

RS Aggarwal Class 7 Solutions Chapter 23 Probability Ex 23

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 23 Probability Ex 23.

Question 1.
Solution:
(i) Here, total number of trials = 300
Number of heads got = 136.
P(E) = \(\frac { 136 }{ 300 }\) = \(\frac { 34 }{ 75 }\)
(ii) Total number of trials = 300
Number of tails got = 164
P(E) = \(\frac { 164 }{ 300 }\) = \(\frac { 41 }{ 75 }\)

Question 2.
Solution:
Number times, the two coins were tossed = 200
Number of times got two heads = 58
Number of times got one head = 83
and number of times got no head = 59
(i) Probability of getting 2 heads : P(E) = \(\frac { 58 }{ 200 }\) = \(\frac { 29 }{ 100 }\)
(ii) Probability of getting one head : P(E) = \(\frac { 83 }{ 200 }\)
(iii) Probability of getting no head : P(E) = \(\frac { 59 }{ 200 }\)

Question 3.
Solution:
Number of times, a dice was thrown = 100
(i) Number of times got 3 = 18
Probability will be
P(E) = \(\frac { 18 }{ 100 }\) = \(\frac { 9 }{ 50 }\)
(ii) Number of times got 6 = 9
Probability will be
P(E) = \(\frac { 9 }{ 100 }\)
iii) Number of times got 4 = 15
Probability will be
P(E) = \(\frac { 15 }{ 100 }\) = \(\frac { 3 }{ 20 }\)
(iv) Number of times got 1 = 21
Probability will be
P(E) = \(\frac { 21 }{ 100 }\)

Question 4.
Solution:
Total number of ladies = 100
Number of ladies also like coffee = 36.
Number of ladies who dislike coffee = 64
(i) Probability of lady who like coffee :
P(E) = \(\frac { 36 }{ 100 }\) = \(\frac { 9 }{ 25 }\)
(ii) Probability of lady who dislikes coffee:
P(E) = \(\frac { 64 }{ 100 }\) = \(\frac { 16 }{ 25 }\)

 

Hope given RS Aggarwal Solutions Class 7 Chapter 23 Probability Ex 23 are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 22 Bar Graphs Ex 22.

Question 1.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph representing x-axis and y-axis.
(ii) Along OX, mark subjects and along y-axis, mark, number of marks
(iii) Take one division = 10 marks.
Now we shall draw bar graph as shown.

Question 2.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph paper. These two lines represent x-axis and y-axis respectively.
(ii) Along OX, write sports and along OY, number of students choosing on division equal to 10 students.
(iii) Now draw the bars of various heights according to the no. of students as shown on the graph.

Question 3.
Solution:
(i) Draw a horizontal line OX and a vertical line OY. These represent x-axis and y-axis respectively on the graph paper.
(ii) Along OX, write years and along OY, no. of students choosing one division = 200 students.
(iii) Draw bars of various heights according to the number of students given.
This is the required bar graph as shown.

Question 4.
Solution:
(i) Draw a horizontal line OX and a vertical line OY which represent x-axis and y-axis respectively on the graph.
(ii) Along OX, write years and along OY, no. of scooters
(iii) Choose 1 division = 300
(iv) Now draw bars of different heights according to given data as shown on the graph.

Question 5.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along OX write countries and along OY, take Birth rate per thousand.
(iii) Choose 1 division = 10.
(iv) Now draw bars of different heights according to the given data as shown on the graph.

Question 6.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis on the graph paper.
(ii) Along x-axis write states and along y-axis population in lakhs.
(iii) Choose one division = 200 (Lakhs)
(iv) Draw bars of different heights according to the data given as shown on the graph.

Question 7.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis Interest in thousand cores rupees.
(iii) Choose one division = 20 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 8.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write city and along y-axis the distance (in km).
(iii) Choose one division = 200 km.
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 9.
Solution:
(i) Draw a horizontal line OX and a vertical line OY represent x-axis and y-axis respectively on the graph paper.
(i) Along x-axis write countries and along y-axis life expectancy (in years)
(ii) Choose one division = 10 years
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 10.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis imports in thousand crores rupees.
(iii) Choose one division = 50 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 11.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write months and along y-axis average rainfall in cm.
(iii) Choose one small division = 5cm.
(iv) Draw different bars of different heights according to the given data as shown on the graph.

Question 12.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write Brand and along y-axis, percentage of buyers.
(iii) Choose one division = 5% of buyers.
(iv) Draw bars of different heights according to the given data, as shown on the graph.

Question 13.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write week and along y-axis. Rate per 10gm in rupees.
(iii) Choose and division = 1000
(iv) Draw bars of different heights according to the data given as shown on the graph.

Question 14.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write mode of transport and on the y-axis is number of students.
(iii) Choose one division =100 students
(iv) Draw bars of different heights according to given data as shown on the graph.

Question 15.
Solution:
(i) The bar graph shows the marks obtained by a student in different subjects.
(ii) The student is very good in Mathematics.
(iii) The student is very poor in Hindi.
(iv) Average marks

Question 16.
Solution:
(i) The bar graph shows the number of members in each of the 85 families.
(ii) 40 families have 3 number each.
(iii) Number of people living alone is nil.
(iv) The families having 3 members each is most common.

Question 17.
Solution:
(i) The highest peak is Mount Everest whose heighest is 8800 m.
(ii) The required ratio between the highest peak and the next heighest peak = 8800 : 8200 or 44 : 41
(iii) Arranging the heights of peaks in descending order are : 8800 m, 8200 m, 8000 m, 7500 m, 6000 m

Question 18.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale : 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are :
Mon. 350 and 200; Tues. 400 and 450; Wed. 500 and 300; Thurs. 450 and 250; Fri. 550 and 100 and Sat. 450 and 50.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) The number of readers in the library was maximum on TUESDAY.
(iii) Total Number of magazine readers in a week = 200 + 450 + 300 + 250 + 100 + 50 = 1350
Mean Number of readers per day = \(\frac { 1350 }{ 6 }\) = 225

Question 19.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 4
(d) The heights of various pairs of bars in terms of the number of small divisions are:
VI 95 and 92; VII 90 and 85; VIII 82 and 78; IX 75 and 69; X 68 and 62.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) For class VII, total Number of Students = 90
Number of students present = 85

Question 20.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
January 2 and 1.5 ; February 3.25 and 3; March 4 and 3.5; April 4.5 and 6; May 7.75 and 5.5; June 8 and 6.5.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) June
(iii) January

Question 21.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y’-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
Town A 640000 and 750000; Town B 830000 and 920000; Town C 460000 and 630000; Town D 290000 and 320000
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) Town B
(iii) Town D

 

Hope given RS Aggarwal Solutions Class 7 Chapter 22 Bar Graphs Ex 22 are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21A
• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B
• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C

Question 1.
Solution:
(i) Arranging in ascending order :
4, 6, 7, 8, 8, 8, 8, 10, 11, 15
We see that 8 occurs maximum times
Mode = 8
(ii) Arranging in ascending order :
18, 21, 23, 27, 27, 27, 27, 27, 36, 39, 40
We see that 27 occurs maximum times
Mode = 27

Question 2.
Solution:
Arranging in ascending order :
28, 31, 32, 32, 32, 32, 34, 36, 38, 40, 41.
We see that 32 occurs maximum times
Mode = 32 years

Question 3.
Solution:
We prepare the table as given below:

Here, number of terms = 45, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 45 + 1 }{ 2 }\) = \(\frac { 46 }{ 2 }\) th term
= 23th term = 450
Now, mode = 3(median) – 2(mean)
= 3 x 450 – 2 x 470
= 1350 – 940
= 410

Question 4.
Solution:
We prepare the table as given below:

Here, number of terms (N) = 41, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 41 + 1 }{ 2 }\) th term
= \(\frac { 42 }{ 2 }\) = 21 th term = 22 {value of 18 to 29 = 22}
Mode = 3 (median) – 2 (mean)
= 3 x 22 – 2 x 21.92 = 66 – 43.84 = 22.16

Question 5.
Solution:
We prepare the table as given below:

Hope given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21A
• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B
• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C

Question 1.
Solution:
(i) Arranging in ascending order.
2, 2, 3, 5, 7, 9, 9, 10, 11
Here number of terms = 9 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 9 + 1 }{ 2 }\) th term
= 5th term = 7
Hence median = 7
(ii) Arranging in ascending order,
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms (n) = 9 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 9 + 1 }{ 2 }\) th term
= 5th term = 16
(iii) Arranging in ascending order,
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25 Here number of terms (n) = 11 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 11 + 1 }{ 2 }\) th term
= 6th term = 16

Question 2.
Solution:
(i) Arranging in ascending order,
9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms = 8 which is even

Question 3.
Solution:
First 15 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.
Here, number of terms (n) = 15 which is odd
Median = \(\frac { n + 1 }{ 2 }\) = \(\frac { 15 + 1 }{ 2 }\) th term
= 8th term = 15

Question 4.
Solution:
First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Here, number of terms = 10 which is even

Question 5.
Solution:
First 50 whole numbers
0, 1, 2, 3, 4, …, 49
Here, number of terms = 50, which is even

Question 6.
Solution:
Arranging in ascending order,
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40 .
Here, number of terms = 15 which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 15 + 1 }{ 2 }\) = \(\frac { 16 }{ 2 }\) th
= 8th term = 23

Question 7.
Solution:
Arranging is ascending order,
31, 34, 36, 37, 40, 43, 46, 50, 52, 53
Here, number of terms = 10, which is even

Question 8.
Solution:
Preparing the cumulative frequency table

Here, number of terms (N) = 41, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term
= \(\frac { 41 + 1 }{ 2 }\) = \(\frac { 42 }{ 2 }\) th = 21 th term = 50kg (value of 20 to 28 = 50)
Hence median = 50kg

Question 9.
Solution:
Arranging in order and preparing the cumulative frequency table.

Here, number of terms (N) = 37 which is odd.
Median = \(\frac { n + 1 }{ 2 }\) th term
= \(\frac { 37 + 1 }{ 2 }\) = \(\frac { 38 }{ 2 }\) th term
= 19th term = 22 (Value of 18 to 21 = 22)
Hence median = 22

Question 10.
Solution:
Arranging in order and then preparing its cumulative frequency table :

Here, number of terms (N) = 50, which is even

Hence median = 154.5 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21A
• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B
• RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C

Question 1.
Solution:
(i) Data : A collection of numerical figures giving some particular type of information is called data
(ii) Raw data : Data obtained in the original form is called raw data.
(iii) Array : Arranging the numerical figures of a data in ascending or descending order is called an array.
(iv) Tabulation of data : Arranging the data in a systematic form in the form of a table is called tabulation of the data.
(v) Observations : Each numerical figure in a data is called an observation.
(vi) Frequency of an observation : The number of times a particular observation occurs is called its frequency.
(vii) Statistics : It is the subject that deals with the collection presentation analysis and interpretation of numerical data.

Question 2.
Solution:
Arranging the given data in ascending order is as given below :
1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6 and 6 its frequency table will be as under:

Question 3.
Solution:
Arranging the given data in ascending order,
260, 260, 300, 300, 300, 300, 360, 360, 360, 360, 360, 360, 400, 400, 400.
and its frequency table will be as under.

Question 4.
Solution:
Arranging the given data in ascending order we find
5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 10, 10 and its frequency table will be as under.

Question 5.
Solution:
(i) Data means information in the form of numerical figures.
(ii) Data obtained in the original form is called raw data.
(iii) Arranging the numerical figures in ascending or descending order is called an array.
(iv) The number of times a particular observation occurs is called its frequency.
(v) Arranging the data in the form of a table is called tabulation of data.

Question 6.
Solution:
First five natural numbers are 1, 2, 3, 4, 5

Question 7.
Solution:
First six odd natural numbers are 1, 3, 5, 7, 9, 11

Question 8.
Solution:
First seven even natural numbers are 2, 4, 6, 8, 10, 12, 14

Question 9.
Solution:
First five prime numbers are 2, 3, 5, 7, 11

Question 10.
Solution:
First six multiples of 5 are 5, 10, 15, 20, 25, 30

Question 11.
Solution:

Question 12.
Solution:

Mean = Rs. 159

Question 13.
Solution:

Mean = Rs. 318

Question 14.
Solution:

Question 15.
Solution:

 

Hope given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21A are helpful to complete your math homework.

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