Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 23 - Line Graphs And Linear Graphs

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

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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

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(i) Both boys and girls achieve their maximum height in 18th years.
(ii) Boys grows faster than girls in puberty.

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(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.

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.

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.

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.

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

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.

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 100⁰ c.
(iv) The temperature after 25 second was 18⁰ c.
(v)  The temperature after 1.5 min will be 100⁰ c as it is uniform after 50 seconds.

Answer:

(i) On Tuesday, Friday, sunday the forecast temperature was same as the actual temperature.
(ii) Maximum forecast temperature was 35⁰ c during the week.
(iii) Minimum actual temperature was 15⁰ c during the week.
(iv) On Thursday, the actual temperature differ the most from the forecast temperature.

Answer:

(i) The car travels 180 km in $4\frac{1}{2}$ 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.

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.

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.

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

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No, the drawn graph is not linear.

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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 

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

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$$\begin{gathered} \mathrm{Using} SI=\frac{P\times R\times T}{100} \\ \mathrm{Where} \mathrm{SI}= \mathrm{simple} \mathrm{interest} \\ \mathrm{P}=\mathrm{principal} \\ \mathrm{R}= \mathrm{rate} \mathrm{of} \mathrm{interest} \\ \mathrm{T}=\mathrm{time} \end{gathered}$$

We have,

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We have,
speed $=\frac{\mathrm{distance}}{\mathrm{time}}$


We have the following table for Distance-Time graph
 

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 (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.

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.