Rs Aggarwal 2017 Solutions for Class 8 Math Chapter 1 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 8 students for Math Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Page No 242:

#### Question 1:

The marks of a student in different subjects are given below:

Subject |
English | Hindi | Mathematics | Science | Social science |

Marks |
72 | 63 | 80 | 56 | 65 |

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the names of the subjects at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 1 mark

Step 4: Then the height of the various bars are:

Hindi = 72

English = 63

Mathematics = 80

Science = 56

Social science = 65

Step 5: On the x-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 242:

#### Question 2:

The following table shows the yearwise strength of a school

Year |
2010−11 | 2011−12 | 2012−13 | 2013−14 | 2014−15 |

No. of students |
950 | 1125 | 1400 | 1750 | 1900 |

Represent the above data by a bar graph.

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the time intervals in years at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 50 students

Step 4: Then the height of the various bars are:

$\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{year}2010-11=\left(\frac{1}{50}\times 950\right)=19\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{year}2011-12=\left(\frac{1}{50}\times 1125\right)=22.5\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{year}2012-13=\left(\frac{1}{50}\times 1400\right)=28\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{year}2013-14=\left(\frac{1}{50}\times 1750\right)=35\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{year}2014-15=\left(\frac{1}{50}\times 1900\right)=38\mathrm{small}\mathrm{divisions}$

Step 5: On the x-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 242:

#### Question 3:

The following table shows the favourite sports of 300 students of a school.

Sports |
Cricket | Badminton | Football | Tennis | Swimming |

No. of students |
85 | 60 | 55 | 35 | 65 |

Represent the above data by a bar graph.

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the names of the sports at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 1 students

Step 4: Then the height of the various bars are:

Cricket = 85

Football = 60

Tennis = 55

Badminton = 35

Swimming = 65

Step 5: On the x-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 242:

#### Question 4:

The air distances of four cities from Delhi (in km) are given below:

City |
Mumbai | Kolkata | Hyderabad | Chennai |

Distance from Delhi(in km) |
1100 | 1340 | 1220 | 1700 |

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y-*axis, respectively.

Step 2: Along OX, write the time interval in years at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 10 km

Step 4: Then the height of the various bars are:

$\mathrm{Distance}\mathrm{of}\mathrm{Mumbai}\mathrm{from}\mathrm{Delhi}=\left(\frac{1}{10}\times 1100\right)=110\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{of}\mathrm{Kolkata}\mathrm{from}\mathrm{Delhi}=\left(\frac{1}{10}\times 1340\right)=134\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{of}\mathrm{Hyderabad}\mathrm{from}\mathrm{Delhi}=\left(\frac{1}{10}\times 1220\right)=122\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{of}\mathrm{Chennai}\mathrm{from}\mathrm{Delhi}=\left(\frac{1}{10}\times 1700\right)=170\mathrm{small}\mathrm{divisions}$

Step 5: On the x-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 242:

#### Question 5:

The following table shows the life expectancy (average age to which people live) in various

countries in a particular year.

Country | India | Japan | Britain | Ethiopia | Cambodia |

Lifeexpectancy (in years) | 65 | 80 | 75 | 50 | 45 |

Represent the above data by a bar graph.

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis respectively.

Step 2: Along OX, write the names of the countries at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 1 year

Step 4: Then the height of the various bars are:

India : 65

Japan : 80

Britain : 75

Ethiopia : 50

Cambodia : 45

Step 5: On the *x*-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 242:

#### Question 6:

Various modes of transport used by 2150 students of a school are given below:

School bus | Private bus | Scooter | Bicycle | By foot |

750 | 460 | 130 | 370 | 440 |

Draw a bar graph to represent the above data.

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the names of the states of India at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 10 student

Step 4: Then the height of the various bars are:

$\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{used}\mathrm{School}\mathrm{bus}=\left(\frac{1}{10}\times 750\right)=75\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{used}\mathrm{Private}\mathrm{bus}=\left(\frac{1}{10}\times 460\right)=46\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{used}\mathrm{Scooter}=\left(\frac{1}{10}\times 130\right)=13\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{used}\mathrm{Bicycle}=\left(\frac{1}{10}\times 370\right)=37\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{went}\mathrm{on}\mathrm{their}\mathrm{foot}=\left(\frac{1}{10}\times 440\right)=44\mathrm{small}\mathrm{divisions}$

Step 5: On the x-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 242:

#### Question 7:

The following table shows the number of motorcycles produced by a company during six

consecutive years.

2009 | 2010 | 2011 | 2012 | 2013 | 2014 |

14000 | 16000 | 18500 | 21000 | 24000 | 30000 |

Draw a bar graph to represent the above data

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the years of census at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 500 motorcycles

Step 4: Then the height of the various bars are:

$\mathrm{Number}\mathrm{of}\mathrm{motorcycles}\mathrm{in}2009=\left(\frac{1}{500}\times 14000\right)=28\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{motorcycles}\mathrm{in}2010=\left(\frac{1}{500}\times 16000\right)=32\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{motorcycles}\mathrm{in}2011=\left(\frac{1}{500}\times 18500\right)=37\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{motorcycles}\mathrm{in}2012=\left(\frac{1}{500}\times 21000\right)=42\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{motorcycles}\mathrm{in}2013=\left(\frac{1}{500}\times 24000\right)=48\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Number}\mathrm{of}\mathrm{motorcycles}\mathrm{in}2014=\left(\frac{1}{500}\times 30000\right)=60\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Step 5: On the *x*-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 243:

#### Question 8:

The present population (in lakhs) of six Indian states is given below:

State | Population(in lakhs) |

Bihar | 1040 |

Chhattisgarh | 260 |

Jharkhand | 330 |

Madhya Pradesh | 720 |

West Bengal | 910 |

Rajasthan | 690 |

Represent the above data by a bar graph.

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the years at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 10 lakhs

Step 4: Then the height of the various bars are:

$\mathrm{Population}\mathrm{of}\mathrm{Bihar}=\left(\frac{1}{10}\times 1040\right)=104\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Population}\mathrm{of}\mathrm{Chhattisgarh}=\left(\frac{1}{10}\times 260\right)=26\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Population}\mathrm{of}\mathrm{Jharkhand}=\left(\frac{1}{10}\times 330\right)=33\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Population}\mathrm{of}\mathrm{Madhya}\mathrm{Pradesh}=\left(\frac{1}{10}\times 720\right)=72\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Population}\mathrm{of}\mathrm{West}\mathrm{Bengal}=\left(\frac{1}{10}\times 910\right)=91\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Population}\mathrm{of}\mathrm{Rajasthan}=\left(\frac{1}{10}\times 690\right)=69\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Step 5: On the *x*-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 243:

#### Question 9:

There are 1120 creatures in a zoo as per list given below:

Beast animals | Other land animals | Water animals | Birds | Reptiles |

220 | 410 | 165 | 190 | 135 |

Represent the above data by a bar graph.

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the names of the places at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 10 animals

Step 4: Then the height of the various bars are:

$\mathrm{Beast}\mathrm{animals}\mathrm{in}\mathrm{zoo}=\left(\frac{1}{10}\times 220\right)=22\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Other}\mathrm{land}\mathrm{animals}\mathrm{in}\mathrm{zoo}=\left(\frac{1}{10}\times 410\right)=41\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Water}\mathrm{animals}\mathrm{in}\mathrm{zoo}=\left(\frac{1}{10}\times 165\right)=16.5\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Birds}\mathrm{in}\mathrm{zoo}=\left(\frac{1}{10}\times 190\right)=19\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}\mathrm{Reptiles}\mathrm{in}\mathrm{zoo}=\left(\frac{1}{10}\times 135\right)=13.5\mathrm{small}\mathrm{divisions}\phantom{\rule{0ex}{0ex}}$

Step 5: On the *x*-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 243:

#### Question 10:

The following table shows the export earnings of India (in thousand crore rupees) during five consecutive years.

Year |
2010−11 | 2011−12 | 2012−13 | 2013−14 | 2014−15 |

Export(in thousand crores rupees) |
540 | 600 | 750 | 950 | 110 |

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the names of the countries at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 10 thousand crore rupees.

Step 4: Then the height of the various bars are:

Export in 2010−11 : $\frac{540}{10}=54\mathrm{small}\mathrm{divisions}$

Export in 2011−12 : $\frac{600}{10}=60\mathrm{small}\mathrm{divisions}$

Export in 2012−13 : $\frac{750}{10}=75\mathrm{small}\mathrm{divisions}$

Export in 2013−14 : $\frac{950}{10}=95\mathrm{small}\mathrm{divisions}$

Export in 2013−15 : $\frac{1100}{10}=110\mathrm{small}\mathrm{divisions}$

Step 5: On the *x*-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 243:

#### Question 11:

The following data shows India's total population (in millions) from 1961 to 201l.

Year | 1961 | 1971 | 1981 | 1991 | 2001 | 2011 |

Population (in millions) | 360 | 420 | 540 | 680 | 1020 | 1200 |

Represent the above data by a bar graph

#### Answer:

The following steps are followed while drawing the bar graph:

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY, representing the *x*-axis and *y*-axis, respectively.

Step 2: Along OX, write the names of the soap brands at points taken at uniform gaps.

Step 3: Choose the scale: 1 small division = 10 millions people

Step 4: Then the height of the various bars are:

Population in 1961 = $\left(\frac{1}{10}\times 360\right)=36$ small divisions

Population in 1971 = $\left(\frac{1}{10}\times 420\right)=42$ small divisions

Population in 1981 = $\left(\frac{1}{10}\times 540\right)=54$ small divisions

Population in 1991 = $\left(\frac{1}{10}\times 680\right)=68$ small divisions

Population in 2001= $\left(\frac{1}{10}\times 1020\right)=102$ small divisions

Population in 2011 = $\left(\frac{1}{10}\times 1200\right)=120$ small divisions

Step 5: On the *x*-axis, draw bars of equal width and of heights obtained in step 4 at the points marked in step 2.

The completed bar graph is as shown below:

#### Page No 243:

#### Question 12:

In a survey of 100 families of a village, the number of members in each family was recorded, as shown by the bar graph given below:

Read the bar graph carefully and answer the following questions.

(I) What does the bar graph show?

(il) How many families have less than fivemembers?

(Iii) How many families have more than three members?

(iv) How many families have two children?

#### Answer:

(i) The given bar graph shows the number of members in 100 families of a village.

(ii) Number of families having 2 members = 5

Number of families having 3 members = 30

Number of families having 4 members = 55

∴Total number of families haing less than five members = 5 + 30 + 55 = 90

(iii) Number of families having 4 members = 55

Number of families having 5 members = 10

∴Total number of families haing more than three members = 55 + 10 = 65

(iv) Number of families having 2 children = Number of families having 4 members = 55

#### Page No 244:

#### Question 13:

Look at the bar graph given below:

Read the above bar graph carefully and answer the questions given below.

(i) What does the bar graph show?

(tt) In which subject is the student very poor?

(Iii) If maximum marks in each subject be 100, what is the average of his marks?

(Iv) On the basis of marks obtained, fmd the subject in which the student has special interest.

#### Answer:

(i) The given bar graph shows the marks scored by a student in five different subjects in his exams.

(ii) It is clear from the graph that the bar of the minimum height corresponds to English. So, the student is poor in English.

(iii) Average marks scored by the student $=\frac{\left(55+80+30+60+45\right)}{5}=\frac{270}{5}=54$

(iv) It is clear from the graph that the bar of the maximum height corresponds to Mmathematics. So, the student has special interest in mathematics.

#### Page No 244:

#### Question 14:

Given below is a bar graph showing the heights of five mountain peaks.

Read the bar graph carefully and answer the following questions.

(i) Name the highest peak and mention its height.

(ii) What is the ratio between the heights of the lowest and the highest peaks?

(iii) Arrange the heights of the given peaks in an ascending order.

(iv) Which peaks differ in height by 600 metres?

#### Answer:

(i) It is clear from the graph that the bar of the maximum height corresponds to Mount Everest. So, the highest peak is Mount Everest.

The height of Mount Everest is 8800 m.

(ii) It is clear from the graph that the bar of the maximum height corresponds to Mount Everest. So, the highest peak is Mount Everest.

The height of Mount Everest is 8800 m.

It is clear from the graph that the bar of the minimum height corresponds to Annapurna. So, the lowest peak is Annapurna.

The height of Annapurna is 6000 m.

Required ratio = $\frac{8800}{6000}=\frac{22}{15}$ or 22 : 15

(iii) The heights of the given peaks 8800 m, 6000 m, 7500 m, 8200 m and 8000 m.

The heights of the given peaks in an ascending order is given by

6000 m, 7500 m, 8000 m, 8200 m and 8800 m.

(iv)The height of Mount Everest is 8800 m.

The height of Kanchenjunga is 8200 m.

The difference between the heights of these two peaks is 8800 − 8200 = 600 m.

The two peaks differ in height by 600 metres are Mount Everest and Kanchenjunga.

View NCERT Solutions for all chapters of Class 8