Class 8 Maths Chapter 15 INTRODUCTION TO GRAPHS NCERT SOLUTIONS

AMEND EDUCATION ACADEMY
B6 97 SECTOR 8 ROHINI

15 - Introduction to Graphs

Exercise 15.1


Question 1- The following graph shows the temperature of a patient in a hospital, recorded every hour.

(a) What was the patient’s temperature at 1 p.m.?

(b) When was the patient’s temperature 38.5°C?

(c) The patient’s temperature was the same two times during the period given. What were these two times?

(d) What was the temperature at 1.30 p.m? How did you arrive at your answer?

(e) During which periods did the patient’s temperature show an upward trend?

Answer - (a) At 1 p.m., the patient’s temperature was 36.5°C.

(b) The patient’s temperature was 38.5°C at 12 noon.

(c) The patient’s temperature was same at 1 p.m. and 2 p.m.

(d) The graph between the times 1 p.m. and 2 p.m. is parallel to the x-axis. The temperature at 1 p.m. and 2 p.m. is 36.5°C. So, the temperature at 1:30 p.m. is 36.5°C.

(e) During the following periods, the patient’s temperature showed an upward trend.

9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 2 p.m. to 3 p.m.

Question 2-  The following line graph shows the yearly sales figure for a manufacturing company.

(a) What were the sales in (i) 2002 (ii) 2006?

(b) What were the sales in (i) 2003 (ii) 2005?

(c) Compute the difference between the sales in 2002 and 2006.

(d) In which year was there the greatest difference between the sales as compared to its previous year?

Answer -  (a)

(i) In 2002, the sales were Rs 4 crores.

(ii) In 2006, the sales were Rs 8 crores.

(b)

(i) In 2003, the sales were Rs 7 crores.

(ii) In 2005, the sales were Rs 10 crores.

(c)

(i) In 2002, the sales were Rs 4 crores and in 2006, the sales were Rs 8 crores.

Difference between the sales in 2002 and 2006

= Rs (8 − 4) crores = Rs 4 crores

(d) Difference between the sales of the year 2006 and 2005

= Rs (10 − 8) crores = Rs 2 crores

Difference between the sales of the year 2005 and 2004

= Rs (10 − 6) crores = Rs 4 crores

Difference between the sales of the year 2004 and 2003

= Rs (7 − 6) crore = Rs 1 crore

Difference between the sales of the year 2003 and 2002

= Rs (7 − 4) crores = Rs 3 crores

Hence, the difference was the maximum in the year 2005 as compared to its previous year 2004.

Question 3-  For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.

(a) How high was Plant A after (i) 2 weeks (ii) 3weeks?

(b) How high was Plant B after (i) 2 weeks (ii) 3weeks?

(c) How much did Plant A grow during the 3^rd week?

(d) How much did Plant B grow from the end of the 2^nd week to the end of the 3^rd week?

(e) During which week did Plant A grow most?

(f) During which week did Plant B grow least?

(g) Were the two plants of the same height during any week shown here? Specify.

Answer - (a)

(i) After 2 weeks, the height of plant A was 7 cm.

(ii) After 3 weeks, the height of plant A was 9 cm.

(b)

(i) After 2 weeks, the height of plant B was 7 cm.

(ii) After 3 weeks, the height of plant B was 10 cm.

(c) Growth of plant A during 3^rd week = 9 cm − 7 cm = 2 cm

(d) Growth of plant B from the end of the 2^nd week to the end of the 3^rd week

= 10 cm − 7 cm = 3 cm

(e) Growth of plant A during 1^st week = 2 cm − 0 cm = 2 cm

Growth of plant A during 2^nd week = 7 cm − 2 cm = 5 cm

Growth of plant A during 3^rd week = 9 cm − 7 cm = 2 cm

Therefore, plant A grew the most, i.e. 5 cm, during the 2^nd week.

(f) Growth of plant B during 1^st week = 1 cm − 0 cm = 1 cm

Growth of plant B during 2^nd week = 7 cm − 1 cm = 6 cm

Growth of plant B during 3^rd week = 10 cm − 7 cm = 3 cm

Therefore, plant B grew the least, i.e. 1 cm, during the 1^st week.

(g) At the end of the 2^nd week, the heights of both plants were same.

Question 4- The following graph shows the temperature forecast and the actual temperature for each day of a week.

(a) On which days was the forecast temperature the same as the actual temperature?

(b) What was the maximum forecast temperature during the week?

(c) What was the minimum actual temperature during the week?

(d) On which day did the actual temperature differ the most from the forecast temperature?

Answer - (a) The forecast temperature was same as the actual temperature on Tuesday, Friday, and Sunday.

(b) The maximum forecast temperature during the week was 35°C.

(c) The minimum actual temperature during the week was 15°C.

(d) The actual temperature differs the most from the forecast temperature on Thursday.

Question 5 - Use the tables below to draw linear graphs.

(a) The number of days a hill side city received snow in different years.

Year	2003	2004	2005	2006
Days	8	10	5	12

(b) Population (in thousands) of men and women in a village in different years.

Year	2003	2004	2005	2006	2007
Number of men	12	12.5	13	13.2	13.5
Number of women	11.3	11.9	13	13.6	12.8

Answer - 

(a) By taking the years on x-axis and the number of days on y-axis and taking scale as 1 unit = 2 days on y-axis and 2 unit = 1 year on x-axis, the linear graph of the given information can be drawn as follows.

(b) By taking the years on x-axis and population on y-axis and scale as 1 unit = 0.5 thousand on y-axis and 2 unit = 1 year on x-axis, the linear graph of the given information can be drawn as follows.

Question 6- 

A courier-person cycles from a town to a neighboring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph.

(a) What is the scale taken for the time axis?

(b) How much time did the person take for the travel?

(c) How far is the place of the merchant from the town?

(d) Did the person stop on his way? Explain.

(e) During which period did he ride fastest?

Answer - 

(a) Scale taken for the time axis is 4 units = 1 hour

(b) The person travelled during the time 8 a.m. − 11:30 a.m.

Therefore, the person took hours to travel.

(c) The merchant is 22 km far from the town.

(d) Yes, the person stopped on his way from 10 a.m. to 10: 30 a.m. This is indicated by the horizontal part of the graph.

(e) From the graph, it can be observed that during 8 a.m. to 9 a.m., the person travelled the maximum distance. Thus, the person’s ride was the fastest between 8 a.m. and 9 a.m.

Question 7- 

Can there be a time temperature graph as follows? Justify you’re answer:

(i)	(ii)
(iii) Answer - (i) This can be a time−temperature graph, as the temperature can increase with the increase in time. (ii) This can be a time−temperature graph, as the temperature can decrease with the decrease in time. (iii) This cannot be a time−temperature graph since different temperatures at the same time are not possible. (iv) This can be a time−temperature graph, as same temperature at different times is possible. Exercise 15.2 Question 1- Plot the following points on a graph sheet. Verify if they lie on a line (a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5) (b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4) (c) K(2, 3), L(5, 3), M(5, 5), N(2, 5) Answer - (a) We can plot the given points and join the consecutive points on a graph paper as follows. From the graph, it can be observed that the points A, B, C, and D lie on the same line. (b) We can plot the given points and join the consecutive points on a graph paper as follows. Hence, points P, Q, R, and S lie on the same line. (c) We can plot the given points and join the consecutive points on a graph paper as follows. Hence, points K, L, M, and N are not lying on the same line. Question 2- Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis. Answer -   From the graph, it can be observed that the line joining the points (2, 3) and (3, 2) meets the x-axis at the point (5, 0) and the y-axis at the point (0, 5). Question 3- Write the coordinates of the vertices of each of these adjoining figures. Answer -  The coordinates of the vertices in the given figure are as follows. O (0, 0), A (2, 0), B (2, 3), C (0, 3) P (4, 3), Q (6, 1), R (6, 5), S (4, 7) K (10, 5), L (7, 7), M (10, 8) Question 4- State whether True or False. Correct those are false. (i) A point whose x coordinate is zero and y-coordinate is non-zero will lie on the y-axis. (ii) A point whose y coordinate is zero and x-coordinate is 5 will lie on y-axis. (iii) The coordinates of the origin are (0, 0). Answer -  (i) True (ii) False The point whose y-coordinate is zero and x-coordinate is 5 will lie on x-axis. (iii) True Exercise -15.3 Question 1-  Draw the graphs for the following tables of values, with suitable scales on the axes. (a) Cost of apples Number of apples 1 2 3 4 5 Cost (in Rs) 5 10 15 20 25 (b) Distance travelled by a car Time (in hours) 6 a.m. 7 a.m. 8 a.m. 9 a.m. Distance (in km) 40 80 120 160 (i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m.? (ii) What was the time when the car had covered a distance of 100 km since its start? (c) Interest on deposits for a year: Deposit (in Rs) 1000 2000 3000 4000 5000 Simple interest (in Rs) 80 160 240 320 400 (i) Does the graph pass through the origin? (ii) Use the graph to find the interest on Rs 2500 for a year: (iii) To get an interest of Rs 280 per year, how much money should be deposited? Answer - (a) Taking a suitable scale (for x-axis, 1 unit = 1 apple and for y-axis, 1 unit = Rs 5), we can mark the number of apples on x-axis and the cost of apples on y-axis. A graph of the given data is as follows. (b) Taking a suitable scale (for x-axis, 2 units = 1 hour and for y-axis, 2 units = 40 km), we can represent the time on x-axis and the distance covered by the car on y-axis. A graph of the given data is as follows. (i) During the period 7:30 a.m. to 8 a.m., the car covered a distance of 20 km. (ii) The car covered a distance of 100 km at 7:30 a.m. since its start. (c) Taking a suitable scale, For x-axis, 1 unit = Rs 1000 and for y-axis, 1 unit = Rs 80 We can represent the deposit on x-axis and the interest earned on that deposit on y-axis. A graph of the given data is obtained as follows. From the graph, the following points can be observed. (i) Yes. The graph passes through the origin. (ii) The interest earned in a year on a deposit of Rs 2500 is Rs 200. (iii) To get an interest of Rs 280 per year, Rs 3500 should be deposited. Question 2- Draw a graph for the following. (i) Side of square (in cm) 2 3 3.5 5 6 Perimeter (in cm) 8 12 14 20 24 Is it a linear graph? (ii) Side of square (in cm) 2 3 4 5 6 Area (in cm2) 4 9 16 25 36 Is it a linear graph? Answer - (i) Choosing a suitable scale, For x-axis, 1 unit = 1 cm and for y-axis, 1 unit = 4 cm We can represent the side of a square on x-axis and the perimeter of that square on y-axis. A graph of the given data is drawn as follows. It is a linear graph. (ii)Choosing a suitable scale, For x-axis, 1 unit = 1 cm and for y-axis, 1 unit = 4 cm2 We can represent the side of a square on the x-axis and the area of that square on y-axis. A graph of the given data is as follows. It is not a linear graph.

Class 8 Maths Chapter 7 Cube and Cube Roots NCERT SOLUTIONS

AMEND EDUCATION ACADEMY 9999908238
B6 97 SECTOR 8 ROHINI

CHAPTER    7- Cubes and Cube Roots NCERT SOLUTIONS

Exercise 7.1

Question 1- Which of the following numbers are notperfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Answer - (i) The prime factorisation of 216 is as follows.

2	216
2	108
2	54
3	27
3	9
3	3
	1

216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 216 is a perfect cube.

(ii)The prime factorisation of 128 is as follows.

2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, each prime factor is not appearing as many times as a perfect multiple of 3. One 2 is remaining after grouping the triplets of 2. Therefore, 128 is not a perfect cube.

(iii) The prime factorisation of 1000 is as follows.

2	1000
2	500
2	250
5	125
5	25
5	5
	1

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.

(iv)The prime factorisation of 100 is as follows.

2	100
2	50
5	25
5	5
	1

100 = 2 × 2 × 5 × 5

Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.

(v)The prime factorisation of 46656 is as follows.

2	46656
2	23328
2	11664
2	5832
2	2916
2	1458
3	729
3	243
3	81
3	27
3	9
3	3
	1

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 46656 is a perfect cube.

Question 2- Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Answer - (i) 243 = 3 × 3 × 3 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.

In that case, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube.

Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.

Then, we obtain

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) 72 = 2 × 2 × 2 × 3 × 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.

Then, we obtain

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) 675 = 3 × 3 × 3 × 5 × 5

Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required.

Then, we obtain

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) 100 = 2 × 2 × 5 × 5

Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.

Then, we obtain

100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube

Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

Question 3- Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

(ii) 128

(iii) 135

(iv) 192

(v) 704

Answer - (i) 81 = 3 × 3 × 3 × 3

Here, one 3 is left which is not in a triplet.

If we divide 81 by 3, then it will become a perfect cube.

Thus, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, one 2 is left which is not in a triplet.

If we divide 128 by 2, then it will become a perfect cube.

Thus, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

(iii) 135 = 3 × 3 × 3 × 5

Here, one 5 is left which is not in a triplet.

If we divide 135 by 5, then it will become a perfect cube.

Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.

Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

(iv) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, one 3 is left which is not in a triplet.

If we divide 192 by 3, then it will become a perfect cube.

Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

(v) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Here, one 11 is left which is not in a triplet.

If we divide 704 by 11, then it will become a perfect cube.

Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.

Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

Question 4- Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Answer - Here, some cuboids of size 5 × 2 × 5 are given.

When these cuboids are arranged to form a cube, the side of this cube so formed will be a common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.

LCM of 5, 2, and 5 = 10

Let us try to make a cube of 10 cm side.

For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width, and 2 along with its height.

Total cuboids required according to this arrangement = 2 × 5 × 2 = 20

With the help of 20 cuboids of such measures, a cube is formed as follows.

Alternatively

Volume of the cube of sides 5 cm, 2 cm, 5 cm

= 5 cm × 2 cm × 5 cm = (5 × 5 × 2) cm³

Here, two 5s and one 2 are left which are not in a triplet.

If we multiply this expression by 2 × 2 × 5 = 20, then it will become a perfect cube.

Thus, (5 × 5 × 2 × 2 × 2 × 5) = (5 × 5 × 5 × 2 × 2 × 2) = 1000 is a perfect cube. Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.

Exercise 7.2

Question 1- Find the cube root of each of the following numbers by prime factorisation method.

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

Answer - i) Prime factorisation of 

∴ 

(ii) Prime factorisation of 

∴ 

(iii) Prime factorisation of 

∴ 

(iv) Prime factorisation of 

∴ 

(v) Prime factorisation of 

∴ 

(vi) Prime factorisation of 

∴ 

(vii) Prime factorisation of 

∴ 

(viii) Prime factorisation of 

∴ 

(ix) Prime factorisation of 

∴ 

(x)Prime factorisation of 

∴ 

Question 2-  State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeroes.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

Answer - For finding the cube of any number, the number is first multiplied with itself and this product is again multiplied with this number.

(i) False. When we find out the cube of an odd number, we will find an odd number as the result because the unit place digit of an odd number is odd and we are multiplying three odd numbers. Therefore, the product will be again an odd number.

For example, the cube of 3 (i.e., an odd number) is 27, which is again an odd number.

(ii) True. Perfect cube will end with a certain number of zeroes that are always a perfect multiple of 3.

Foe example, the cube of 10 is 1000 and there are 3 zeroes at the end of it.

The cube of 100 is 1000000 and there are 6 zeroes at the end of it.

(iii) False. It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.

For example, the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625. However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is 42875 which does not end with 25.

(iv) False. There are many cubes which will end with 8. The cubes of all the numbers having their unit place digit as 2 will end with 8.

The cube of 12 is 1728 and the cube of 22 is 10648.

(v) False. The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it.

(vi) False. The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.

(vii)True, as the cube of 1 and 2 are 1 and 8 respectively.

Question 3-  You are told that 1331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768

Answer - Firstly, we will make groups of three digits starting from the rightmost digit of the number as.

There are 2 groups, 1 and 331, in it.

Considering 331,

The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place digit of the required cube root can be taken as 1.

Taking the other group i.e., 1,

The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit of our cube root will be taken as the unit place of the smaller number whose cube is near to the number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.

Hence, 

The cube root of 4913 has to be calculated.

We will make groups of three digits starting from the rightmost digit of 4913, as. The groups are 4 and 913.

Considering the group 913,

The number 913 ends with 3. We know that if the digit 3 is at the end of a perfect cube number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place digit of the required cube root is taken as 7.

Taking the other group i.e., 4,

We know that, 1³ = 1 and 2³ = 8

Also, 1 < 4 < 8

Therefore, 1 will be taken at the tens place of the required cube root.

Thus, 

The cube root of 12167 has to be calculated.

We will make groups of three digits starting from the rightmost digit of the number 12167, as. The groups are 12 and 167.

Considering the group 167,

167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required cube root can be taken as 3.

Taking the other group i.e., 12,

We know that, 2³ = 8 and 3³ = 27

Also, 8 < 12 < 27

2 is smaller between 2 and 3. Therefore, 2 will be taken at the tens place of the required cube root.

Thus, 

The cube root of 32768 has to be calculated.

We will make groups of three digits starting from the rightmost digit of the number 32768, as .

Considering the group 768,

768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required cube root will be taken as 2.

Taking the other group i.e., 32,

We know that, 3³ = 27 and 4³ = 64

Also, 27 < 32 < 64

3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube root.

Thus,