Rs Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 23 Line Graphs And Linear Graphs are provided here with simple step-by-step explanations. These solutions for Line Graphs And Linear Graphs are extremely popular among Class 8 students for Maths Line Graphs And Linear Graphs Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 23 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Page No 261:

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#### Page No 261:

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From the graph, we can see that outout for the inputs of 3 and 8 are 8 and 23 respectively.

#### Page No 262:

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ans

#### Page No 262:

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distance of Ajeets's school from her home is given by

speed = $\frac{\mathrm{distance}}{\mathrm{time}}$

30 = $\frac{\mathrm{distance}}{1/2}$

distance = 15 km

Distance-Time graph is given by

#### Page No 262:

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ans

#### Page No 262:

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(i) Both boys and girls achieve their maximum height in 18th years.

(ii) Boys grows faster than girls in puberty.

#### Page No 262:

#### Answer:

(i) The x-axis shows the time y-axis shows the distance of the car from city P.

(ii) The car begins its journey from city P at 8:00 a.m.

(iii) The car covers 50km in first hour.

(iv) The car covers

(a) 100 km in second hour.

(b) 50km in third hour.

(v) No, because the distance covered per hour is not uniform.

(vi) Yes, the car stopped during the interval 11 a.m. to 12 a.m. ,as indicated by horizontal line segment during this period.

(vii) At 2 p.m ,the car reached at the city Q.

#### Page No 263:

#### Answer:

(i) The scale taken at time-axis is 1 unit = 15 minutes.

(ii) The person travels for 3 hours 30 minutes.

(iii) The place of merchant is 22 km from the town.

(iv) Yes, the person stopped from 10 a.m. to 11 a.m. as indicated by the horizonatal line in the graph.

(v) He ride fastest during 8 a.m. to 9 a.m.

#### Page No 263:

#### Answer:

(i) He starts his journey at 5.30 a.m. and end at 6 p.m.

(ii) Total duration of the journey was 12 hr 30 min.

(iii) Forward.

(iv) He did not move for 6 hours.

(v) He had the fastest speed from 2 p.m. to 5.30 p.m.

#### Page No 264:

#### Answer:

(i) The sales in

(a) 2013 was Rs 7 cr

(b) 2015 was Rs10 cr

(c) 2016 was Rs 8 cr

(ii) The difference between the sales in 2012 and 2016 is Rs 4 cr.

(iii) The greatest difference was in 2015.

#### Page No 265:

#### Answer:

(i) Amit make least progress from 25 min to 40 min, as indicated by the line in the graph.

(ii) Total distance covered = 2km

Total time = 55 min = $\frac{55}{60}$ hr

Average speed = $\frac{\mathrm{Total}\mathrm{distance}\mathrm{covered}}{\mathrm{Total}\mathrm{time}}$ = $\frac{2}{55}\times 60=2.18$ km/hr

#### Page No 265:

#### Answer:

(i) Plant A was

(a) 7 cm high after 2 weeks.

(b) 9 cm high after 3 weeks.

(ii) Plant B was

(a) 7 cm high after 2 weeks.

(b) 10 cm high after 3 weeks.

(iii) Plant A grow 2 cm during the 3rd week.

(iv) Plant B grow 3 cm in the given period.

(v) Plant A grow most in 2nd week.

(vi) Plant B grow least in 1st week.

(vii) At the end of the 2nd week both plants were of same height can be seen in the graph.

#### Page No 266:

#### Answer:

(i) The ice block had no change for 20 seconds.

(ii) There was a change in the ttemperature for 30 seconds

(iii) After 50 second, the temperture became constant at 100^{0} c.

(iv) The temperature after 25 second was 18^{0} c.

(v) The temperature after 1.5 min will be 100^{0} c as it is uniform after 50 seconds.

#### Page No 267:

#### Answer:

(i) On Tuesday, Friday, sunday the forecast temperature was same as the actual temperature.

(ii) Maximum forecast temperature was 35^{0} c during the week.

(iii) Minimum actual temperature was 15^{0} c during the week.

(iv) On Thursday, the actual temperature differ the most from the forecast temperature.

#### Page No 267:

#### Answer:

(i) The car travels 180 km in $4\frac{1}{2}\phantom{\rule{0ex}{0ex}}$ hours.

(ii) The car takes 5 hours to reach R.

(iii) The car takes 2 hours to cover 180 km.

(iv) Q is 120 km far from the starting point.

(v) The car reach the place S after 6 hours.

#### Page No 268:

#### Answer:

(i) Cyclist II rest at 8.45 a.m. for 15 min.

(ii) Cyclist II was faster after the rest.

(iii) The two cyclist met at 9 a.m.

(iv) Cyclist II travelled 10 km when he met cyclist I.

(v) Cyclic I was 10 km far from town A.

#### Page No 268:

#### Answer:

(i) The car covers 20 km during the period from 7.30 a.m. to 8 a.m.

(ii) The car covered 100 km at 7.30 a.m.

(iii) The car covered 140 km by 8.30 a.m.

#### Page No 268:

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Yes, we can see that the drawn graph is linear.

#### Page No 269:

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No, the drawn graph is not linear.

#### Page No 269:

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We have p = 4q

(i) When q = 6

then p = $4\times 4=16$

(ii) when p = 20

then, 20 = 4q

5 = q

And the graph of p = 4q is given by

q |
1 | 2 | 3 | 4 | 5 | 6 |

p |
4 | 8 | 12 | 16 | 20 | 24 |

#### Page No 269:

#### Answer:

We have the following table for y = 2x+1

x | 0 | 1 | 2 | 3 | 4 |

y | 1 | 3 | 5 | 7 | 9 |

#### Page No 269:

#### Answer:

$\mathrm{Using}SI=\frac{P\times R\times T}{100}\phantom{\rule{0ex}{0ex}}\mathrm{Where}\mathrm{SI}=\mathrm{simple}\mathrm{interest}\phantom{\rule{0ex}{0ex}}\mathrm{P}=\mathrm{principal}\phantom{\rule{0ex}{0ex}}\mathrm{R}=\mathrm{rate}\mathrm{of}\mathrm{interest}\phantom{\rule{0ex}{0ex}}\mathrm{T}=\mathrm{time}$

We have,

P | 500 | 1000 | 1500 | 2000 |

SI | 50 | 100 | 150 | 200 |

(i) Given sum deposited i.e P = 250 , R = 10%, Time = 1 year

using formula we have,

$SI=\frac{250\times 10\times 1}{100}\phantom{\rule{0ex}{0ex}}SI=25\mathrm{Rs}$

Hence, the annual interest for an investment of 250 rs is 25 rs.

(ii) Given annual simple interest = 70 rs , R = 10%, T=1 year

using formula we have,

$70=\frac{P\times 10\times 1}{100}\phantom{\rule{0ex}{0ex}}700=P\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{he}\mathrm{has}\mathrm{to}\mathrm{invest}700\mathrm{rs}\mathrm{in}\mathrm{order}\mathrm{to}\mathrm{get}\mathrm{an}\mathrm{annual}\mathrm{simple}\mathrm{interest}\mathrm{of}70\mathrm{rs}$

#### Page No 269:

#### Answer:

We have,

speed $=\frac{\mathrm{distance}}{\mathrm{time}}$

We have the following table for Distance-Time graph

Time(hrs) | 1 | 2 | 3 | 4 |

Distance(km) | 30 | 60 | 90 | 120 |

(i) Given speed = 30 km/hr , distance = 75 km

using formula we have

$30=\frac{75}{\mathrm{time}\mathrm{taken}}\phantom{\rule{0ex}{0ex}}\mathrm{time}\mathrm{taken}=2.5\mathrm{hr}$

Hence, the time taken by Sajal to ride 75 km is 2.5 hr.

(ii) Given speed = 30 km/hr , time = $3\frac{1}{2}$ hr

using formula we have,

$30=\frac{\mathrm{distance}}{7/2}\phantom{\rule{0ex}{0ex}}\mathrm{distance}=105\mathrm{km}$

Hence, the distance covered by Sajal in $3\frac{1}{2}$ hours is 105 km.

#### Page No 269:

#### Answer:

(a) Graph shows the uniform speed.

(b) Moves with uniform speed and then comes to rest.

(c) Moves with non uniform speed and then slowly comes to rest.

#### Page No 269:

#### Answer:

(i) Yes, temperature is directly proportional to time.

(ii) Yes, temperature decreases with time.

(iii) No, tempertaure can not change for a particular time.

(iv) Yes, temperature is constant.

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