Maths RD Sharma Solutions Class 8

RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.1

Other Exercises

• RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.1
• RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.2

Question 1.
Plot the points (5,0 ), (5,1), (5, 8). Do they lie on a line ? What is your observation ?
Solution:
Draw XOX’ and YOY’ the co-ordinates axis on the graph.
Take 1 cm = 1 unit
Point A (5, 0), B (5, 1) and C (5, 8) have been plotted on the graph. By joining A, B and C, we see that these points lie on the same line which is 5 units from y-axis.

Question 2.
Plot the points (2, 8), (7, 8) and (12, 8). Join these points in pairs. Do they lie on a line ? What do you observe ?
Solution:
Draw XOX’ and YOY’, the co-ordinate axis on the graph.
Take 0.5 cm = 1 unit.
Now points A (2, 8), B (7, 8), and C (12, 8) have been plotted on the graph. By joining them, we see that these points lie on the same line which is at a distance of 8 unit from x-axis.

Question 3.
Locate the points :
(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2,3), (2,4)
(iii) (1,3), (2,3), (3,3), (4,3)
(iv) (1,4), (2,4), (3,4), (4,4).
Solution:
The points given in (i) and (ii) are locates in first graph.

(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2, 3), (2, 4)
Points of (iii) and (iv) are located in the adjoining graph.
(iii) (1, 3), (2, 3), (3, 3), (4, 3)
(iv) (1,4), (2, 4), (3, 4), (4,4)

Question 4.
Find the coordinates of points A, B, C, D in the figure

Solution:
Draw perpendicular from A, B, C and D on x-axis and also on y-axis.
A is 1 unit from y-axis and 1 unit from x-axis.
∴ Co-ordinates of A are (1,1)
B is 1 unit fr onr y-axis and 4 units from x-axis
∴ Co-ordinates of B are (1,4)
C is 4 units from y-axis and 6 units from x-axis
∴ Co-ordinates of C are (4, 6)
D is 5 units fromy-axis and 3 units from x-axis
∴ Co-ordinates of D are (5,3)

Question 5.
Find the coordinates of points P, Q, R and S in Fig.

Solution:
Through P, Q, R and S, draw perpendiculars on x-axis and also on y-axis.
(i) P is 10 units form >’-axis and 70 units from y-axis
∴ Co-ordinates of P are (10, 70)
(ii) Q is 12 unit from x-axis and 80 units from y-axis
∴ Co-ordinates of Q are (12, 80)
(iii) R is 16 units from x-axis and 100 units from y-axis
∴ Co-ordinates of R are (16, 100)
(iv) S is 20 units from x-axis and 120 units from y-axis
∴ Co-ordinates of S are (20, 120)

Question 6.
Write the coordinates of each of the vertices of each polygon in the figure.

Solution:
(i) In figure OXYZ
Co-ordinates of O are (0, 0) ∵It is the origin
Co-ordinates of X are (0, 2) ∵ It lies on y – axis
Co-ordinates of Y are (2, 2)
Co-ordinates of Z are (2, 0) ∵It lies on x-axis
(ii) In figure ABCD, draw perpendicular from A, B, C, D on x-axis and y-axis.
A is 4 unit from y-axis and 5 units from x- axis
∴ Co-ordinates of A are (4, 5)
B is 7 units from y-axis and 5 units from x – axis
∴ Co-ordinates of B are (7, 5)
C is 6 units from y-axis and 3 units from x- axis
∴ Co-ordinates of C are (6, 3)
D is 3 units from y-axis as well x-axis
∴ Co-ordinates of D are (3, 3)
(iii) In figure PQR, perpendiculars for P, Q, R are drawn on x-axis and also on y-axis.
∴ Co-ordinates of P are (7, 4), of Q are (9, 5) and of R are (9, 3).

Question 7.
Decide which of the following statements is true and which is false. Give reasons for your answer.
(i) A point whose x-coordinate is zero, will lie on they-axis.
(ii) A point whose y-coordinate is zero, will lie on x-axis.
(iii) The coordinates of the origin are (0, 0).
(iv) Points whose x and y coordinates are equal, lie on a line passing through the origin.
Solution:
(i) Correct: ∵ every point on y-axis, its x = 0
(ii) Correct: ∵ every point on x-axis, its y = 0
(iii) Correct
(iv) Correct

Hope given RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.1 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4

Other Exercises

• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Question 1.
Find the cube roots of each of the following integers :
(i) -125
(ii) -5832
(iii) -2744000
(iv) -753571
Solution:

Question 2.
Show that :

Solution:

Question 3.
Find the cube root of each of the following numbers :
(i) 8 x 125
(ii) -1728 x 216
(iii) -27 x 2744
(iv) -729 x -15625
Solution:

Question 4.
Evaluate :

Solution:

Question 5.
Find the cube root of each of the following rational numbers.

Solution:

Question 6.
Find the cube root of each of the following rational numbers :
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331
Solution:

Question 7.
Evaluate each of the following :


Solution:

Question 8.
Show that :

Solution:

Question 9.
Fill in the Blanks :

Solution:

Question 10.
The volume of a cubical box is 474.552 cubic metres. Find the length of each side of the box.
Solution:

Question 11.
Three numbers are to one another 2:3: 4. The sum of their cubes is 0.334125. Find the numbers.
Solution:

Question 12.
Find side of a cube whose volume is

Solution:

Question 13.
Evaluate :

Solution:

Question 14.
Find the cube root of the numbers : 2460375,20346417,210644875,57066625 using the fact that
(i) 2460375 = 3375 x 729
(i) 20346417 = 9261 x 2197
(iii) 210644875 = 42875 x 4913
(iv) 57066625 = 166375 x 343
Solution:

Question 15.
Find the units digit of the cube root of the following numbers ?
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616
Solution:
(i) 226981
In it unit digit is 1
∴The units digit of its cube root will be = 1
(∵ 1 x 1 x 1 = 1)
∴Tens digit of the cube root will be = 6
(ii) 13824
∵ The units digit of 13824 = 4
(∵ 4 X 4 X 4 = 64)
∴Units digit of the cube root of it = 4
(iii) 571787
∵ The units digit of 571787 is 7
∴The units digit of its cube root = 3
(∵ 3 x 3 x 3 = 27)
(iv) 175616
∵ The units digit of 175616 is 6
∴The units digit of its cube root = 6
(∵ 6 x 6 x 6=216)

Question 16.
Find the tens digit of the cube root of each of the numbers in Question No. 15.
Solution:
(i) In 226981
∵ Units digit is 1
∴Units digit of its cube root = 1
We have 226
(Leaving three digits number 981)
6³ = 216 and 7³ = 343
∴6³ ∠226 ∠ T
∴The ten’s digit of cube root will be 6
(ii) In 13824
Leaving three digits number 824, we have 13
∵ (2)³ = 8, (3)³ = 27
∴2³ ∠13 ∠3′
∴Tens digit of cube root will be 2
(iii) In 571787
Leaving three digits number 787, we have 571
8³ = 512, 9³ = 729
∴ 8³ ∠571 ∠9³
Tens digit of the cube root will be = 8
(iv) In 175616
Leaving three digit number 616, we have 175
∵ 5³ = 125, 6³ = 216
∴5³ ∠175 ∠6³
∴Tens digit of the cube root will be = 5

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3

Other Exercises

• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Question 1.
Find the cube roots of the following numbers by successive subtraction of numbers : 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397,
(i) 64
(ii) 512
(iii) 1728
Solution:
(i) 64
64 – 1 = 63
63 – 7 = 56
56 – 19 = 37
37 – 37 = 0
∴ 64 = (4)³
∴ Cube root of 64 = 4

(ii) 512
512 -1 =511
511- 7 = 504
504 – 19 = 485
485 – 37 = 448
448 – 61 = 387
387 – 91 =296
296 – 127 = 169
169 – 169 = 0
∴ 512 = (8)³
∴ Cube root of 512 = 8

(iii) 1728
1728 – 1= 1727
1727 -7 = 1720
1720 -19 = 1701
1701 -37= 1664
1664 – 61 = 1603
1603 – 91 = 1512
1512 -127= 1385 .
1385 – 169= 1216
1216 – 217 = 999
999 – 271 =728
728 – 331 = 397
397 – 397=0
∴ 1728 = (12)³
∴ Cube root of 1728 = 1²

Question 2.
Using the method of successive subtraction, examine whether or not the following numbers are perfect cubes :
(i) 130
(ii) 345
(iii) 792
(iv) 1331
Solution:
(i) 130
130 – 1 = 129
129 -7 = 122
122 -19 = 103
103 -37 = 66
66 – 61 = 5
We see that 5 is left
∴ 130 is not a perfect cube.

(ii) 345
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
81 – 61 =220
220 – 91 = 129
129 – 127 = 2
We see that 2 is left
∴ 345 is not a perfect cube.

(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
∴ We see 280 is left as 280 <217
∴ 792 is not a perfect cube.

(iv) 1331
1331 – 1 = 1330
1330 -7 = 1323
1323 – 19 = 1304
1304 – 37 = 1267
1267 – 61 = 1206
1206 – 91 = 1115
1115 – 127 = 988
988 – 169 = 819
819 – 217 = 602
602 – 271 = 331
331 – 331 =0
∴ 1331 is a perfect cube

Question 3.
Find the smallest number that must be subtracted from those of the numbers in question 2, which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots ?
Solution:
We have examined in Question 2, the numbers 130, 345 and 792 are not perfect cubes. Therefore
(i) 130
130 – 1 = 129
129 -7= 122
122 -19 = 103
103 – 37 = 66
66 – 61 = 5
Here 5 is left
∴ 5 < 91 5 is to be subtracted to get a perfect cube.
Cube root of 130 – 5 = 125 is 5

(ii) 345
345 – 1 = 344
344 -7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 =220
220 – 91 = 129
129 – 127 = 2
Here 2 is left ∵ 2 < 169
∴ Cube root of 345 – 2 = 343 is 7
∴ 2 is to be subtracted to get a perfect cube.

(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 =667
667 – 91 = 576 5
76 – 127 = 449
449 – 169 = 280
280 – 217 = 63
∴ 63 <217
∴ 63 is to be subtracted
∴ Cube root of 792 – 63 = 729 is 9

Question 4.
Find the cube root of each of the following natural numbers :
(i) 343
(ii) 2744
(iii) 4913
(iv) 1728
(v) 35937
(vi) 17576
(vii) 134217728
(viii) 48228544
(ix) 74088000
(x) 157464
(xi) 1157625
(xii) 33698267
Solution:

Question 5.
Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.
Solution:

Grouping the factors in triplets of equal factors, we see that 2, 3 x 3 and 5 x 5 are left
∴ In order to complete the triplets, we have to multiply it by 2, 3 and 5.
∴ The smallest number to be multiplied = 2×2 x 3 x 5 = 60
Now product = 3600 x 60 = 216000 and cube root of 216000

Question 6.
Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.
Solution:

Grouping the factors in triplets of equal factors, we see that 41 x 41 is left
∴ In order to complete the triplet, we have to multiply it by 41
∴ Smallest number to be multiplied = 41
∴ Product = 210125 x 41 = 8615125
∴ Cube root of 8615125

Question 7.
What is the smallest number by which 8192 must be divided so that quotient is a perfect cube ? Also, find the cube root of the quotient so obtained.
Solution:

Grouping the factors in triplets of equal factors, we see that 2 is left
∴ Dividing by 2, we get the quotient a perfect cube
∴ Perfect cube = 8192 + 2 = 4096

Question 8.
Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.
Solution:
Ratio in numbers =1:2:3
Let first number = x
Then second number = 2x
and third number = 3x
∴ Sum of cubes of there numbers = (x)³ + (2x)³+(3x)³

Question 9.
The volume of a cube is 9261000 m³. Find the side of the cube.
Solution:

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2

Other Exercises

• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Question 1.
Find the cubes of:
(i) -11
(ii) -12
(iii) – 21
Solution:
(i) (-11)³=(-11)³=(11 x 11 x 11) =-1331
(ii) (-12)³=(-12)³=(12 x 12 x 12) =  -1728
(iii) (-21)³=(-21)³=(21 x 21 x 21) = -9261

Question 2.
Which of the following numbers are cubes of negative integers.
(i) -64
(ii) -1056
(iii) -2197
(iv) -2744
(v)  -42875
Solution:


∴ All factors of 64 can be grouped in triplets of the equal factors completely.
∴ -64 is a perfect cube of negative integer.

All the factors of 1056 can be grouped in triplets of equal factors grouped completely
∴ 1058 is not a perfect cube of negative integer.

All the factors of -2197 can be grouped in triplets of equal factors completely
∴ 2197 is a perfect cube of negative integer,

All the factors of -2744 can be grouped in triplets of equal factors completely
∴ 2744 is a perfect cube of negative integer

All the factors of -42875 can be grouped in triplets of equal factors completely
∴ 42875 is a perfect cube of negative integer.

Question 3.
Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer :
(i) -5832 (ii) -2744000
Solution:


Grouping the factors in triplets of equal factors, we see that no factor is left
∴ -5832 is a perfect cube
Now taking one factor from each triplet we find that
-5832 is a cube of – (2 x 3 x 3) = -18
∴ Cube root of-5832 = -18

Grouping the factors in tuplets of equal factors, we see that no factor is left. Therefore it is a perfect cube.
Now taking one factor from each triplet, we find that.
-2744000 is a cube of – (2 x 2 x 5 x 7) ie. -140
∴ Cube root of -2744000 = -140

Question 4.
Find the cube of :

Solution:

Question 5.
Which of the following numbers are cubes of rational numbers :

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

Other Exercises

• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Question 1.
Find the cubes of the following numbers:
(i) 7
(ii) 12
(iii) 16
(iv) 21
(v) 40
(vi) 55
(vii) 100
(viii) 302
(ix) 301
Solution:
(i) (7)³ = 7 x 7 x 7 = 343
(ii) (12)³ = 12 x-12 x 12 = 1728
(iii) (16)³ = 16 x 16 x 16 = 4096
(iv) (21)³ = 21 x 21 x 21 = 441 x 21 =9261
(v) (40)³ = 40 x 40 x 40 = 64000
(vi) (55)³ = 55 x 55 x 55 = 3025 x 55 = 166375
(vii) (100)³ = 100 x 100 x 100 =1000000
(viii)(302)³ = 302 x 302 x 302 = 91204 x 302 = 27543608
(ix) (301)³ = 301 x 301 x 301 = 90601 x 301 =27270901

Question 2.
Write the cubes of all natural numbers between 1 and 10 and verify the following statements :
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
Solution:
Cubes of first 10 natural numbers :
(1)³ = 1 x 1 x 1 = 1
(2)³ = 2 x 2 x 2 = 8
(3)³ = 3 x 3 x 3 = 27
(4)³= 4 x 4 x 4 = 64
(5)³ = 5 x 5 x 5 = 125
(6)³ = 6 x 6 x 6 = 216
(7)³ = 7 x 7 x 7 = 343
(8)³ = 8 x 8 x 8 = 512
(9)³ = 9 x 9 x 9= 729
(10)³ = 10 x 10 x 10= 1000
We see that the cubes of odd numbers is also odd and cubes of even numbers is also even.

Question 3.
Observe the following pattern :

Write the next three rows and calculate the value of 1³ + 2³ + 3³ +…. + 9³ + 10³ by the above pattern.
Solution:
We see the pattern

Question 4.
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings :
The cube of a natural number which is a multiple of 3 is a multiple of 27′
Solution:
5 natural numbers which are multiples of 3
3,6,9,12,15.
(3)³ = 3 x 3 x 3 = 27
Which is multiple of 27
(6)³ = 6 x 6 x 6 = 216 ÷ 27 = 8
Which is multiple of 27
(9)³ = 9 x 9 x 9 = 729 + 27 = 27
Which is multiple of 27
(12)³= 12 x 12 x 12 = 1728 ÷ 27 = 64
Which is multiple of 27
(15)³ = 15 x 15 x 15 = 3375 ÷ 27 = 125
Which is multiple of 27
Hence, cube of multiple of 3 is a multiple of 27

Question 5.
Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g.,4, 7, 10, …………) and verify the following :
‘The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’.
Solution:
3n + 1
Let n = 1, 2, 3, 4, 5, then
If n = 1, then 3n +1= 3 x 1+1= 3+1= 4
If n = 2, then 3n +1=3 x 2+1=6+1=7
If n = 3, then 3n + 1= 3 x 3 + 1= 9 + 1 = 10
If n = 4, then 3n + 1= 3 x 4+1 = 12 + 1= 13
If n = 5, then 3n +1=3 x 5 + 1 = 15 +1 = 16
Now
(4)³ = 4 x 4 x 4 = 64
Which is \(\frac {64 }{ 3 }\)=21, Remainder = 1
(7)³ = 7 x 7 x 7 = 343
Which is \(\frac {343 }{ 3 }\) =114, Remainder = 1
(10)³ = 10 x 10 x 10 = 1000 ÷ 3 = 333, Remainder = 1
(13)³ = 13 x 13 x 13 = 2197 ÷ 3 = 732, Remainder = 1
(16)³ = 16 x 16 x 16 = 4096 ÷ 3 = 1365, Remainder = 1
Hence cube of natural number of the form, 3n + 1, is a natural of the form 3n + 1

Question 6.
Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,……… ) and verify the following :
‘The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’.
Solution:
Natural numbers of the form 3n + 2, when n
is a natural number i.e. 1, 2, 3, 4, 5,………….
If n = 1, then 3n + 2 = 3 x 1+2 = 3+ 2 = 5
If n = 2, then 3n + 2 = 3 x 2 + 2 = 6 + 2 = 8
If n = 3, then 3n + 2 = 3 x 3 + 2 = 9 + 2 = 11
If n = 4, then 3n + 2 = 3 x 4 + 2 = 12 + 2 = 14
and if n = 5, then 3n + 2 = 3 x 5 + 2 = 15 + 2= 17
Now (5)³ = 5 x 5 x 5 = 125
125 + 3 = 41, Remainder = 2
(8)² = 8 x 8 x 8 = 512 512 -s- 3 = 170, Remainder = 2
(11)³ = 11 x 11 x 11 = 1331
1331 + 3 = 443, Remainder = 2
(14)³ = 14 x 14 x 14 = 2744
2744 + 3 = 914, Remainder = 2
(17)³ = 17 x 17 x 17 = 4913
4913 = 3 = 1637, Remainder = 2
We see the cube of the natural number of the
form 3n + 2 is also a natural number of the
form 3n + 2.

Question 7.
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following :
‘The cube of a multiple of 7 is a multiple of 7^3′.
Solution:
5 natural numbers which are multiple of 7,are 7, 14, 21, 28, 35
(7)³ = (7)³ which is multiple of 7³
(14)³ = (2 x 7)3 = 2³ x 7³, which is multiple of 7³
(21)³ = (3 x 7)3 = 33 x 7³_, which is multiple of 7³
(28)³ = (4 x 7)³ = 4³ x 7³, which is multiple of 7³ (35)³ = (5 x 7)³ = 5³ x 73 which is multiple of 7³
Hence proved.

Question 8.
Which of the following are perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1000
(v) 1728
(vi) 3087
(vii) 4608
(viii) 106480
(ix) 166375
(x) 456533
Solution:
(i) 64 = 2 x 2 x 2 x 2 x 2 x 2

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 64 is a perfect cube
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, we see that no factor is left
216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, we see that two factors 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1000 = 2 x 2 x 2 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 1000 is a perfect cube.
(v) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of the equal factors, we see that no factor is left
∴ 1728 is a perfect cube,
(vi) 3087 = 3 x 3 x 7 x 7 x 7

Grouping the factors in triplets of the equal factors, we see that two factor 3×3 are left
∴ 3087 is not a perfect cube.
(vii) 4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left
∴ 4609 is not a perfect cube.
(viii) 106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that factors 2, 5 are left
∴ 106480 is not a perfect cube.
(ix) 166375 = 5 x 5 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 166375 is a perfect cube.
(x) 456533 = 7 x 7 x 7 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 456533 is a perfect cube.

Question 9.
Which of the following are cubes of even natural numbers ?
216, 512, 729,1000, 3375, 13824
Solution:
We know that the cube of an even natural number is also an even natural number
∴ 216, 512, 1000, 13824 are even natural numbers.
∴ These can be the cubes of even natural number.

Question 10.
Which of the following are cubes of odd natural numbers ?
125, 343, 1728, 4096, 32768, 6859
Solution:
We know that the cube of an odd natural number is also an odd natural number,
∴ 125, 343, 6859 are the odd natural numbers
∴ These can be the cubes of odd natural numbers.

Question 11.
What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes ?
(i) 675
(ii) 1323
(iii) 2560
(iv) 7803
(v) 107311
(vi) 35721
Solution:
(i) 675 = 3 x 3 x 3 x 5 X 5

Grouping the factors in triplet of equal factors, 5 x 5 are left without triplet
So, by multiplying by 5, the triplet will be completed.
∴ Least number to be multiplied = 5
(ii) 1323 = 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors. We find that 7 x 7 has been left
So, multiplying by 7, we get a triplet
∴ The least number to be multiplied = 7
(iii) 2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5

Grouping the factors in triplet of equal factors, 5 is left.
∴ To complete a triplet 5 x 5 is to multiplied
∴ Least number to be multiplied = 5 x 5 = 25
(iv) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplet of equal factors, we find the 17 x 17 are left
So, to complete the triplet, we have to multiply by 17
∴ Least number to be multiplied = 17
(v) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplet of equal factors, factor 3 is left
So, to complete the triplet 3 x 3 is to be multiplied
∴ Least number to be multiplied = 3 x 3 = 9
(vi) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors, we find that 7 x 7 is left
So, in order to complete the triplets, we have to multiplied by 7
∴ Least number to be multiplied = 7

Question 12.
By which smallest number must the following numbers be divided so that the quotient is a perfect cube ?
(i) 675
(ii) 8640
(iii) 1600
(iv) 8788
(v) 7803
(vi) 107811
(vii) 35721
(viii) 243000
Solution:
(i) 675 = 3 x 3 x 3 x 5 x 5

Grouping the factors in triplet of equal factors, 5 x 5 is left
5 x 5 is to be divided so that the quotient will be a perfect cube.
∴ The least number to be divided = 5 x 5 = 25
(ii) 8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, 5 is left
∴ In order to get a perfect cube, 5 is to divided
∴ Least number to be divided = 5
(iii) 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

Grouping the factors in triplets of equal factors, we find that 5 x 5 is left
∴ In order to get a perfect cube 5 x 5 = 25 is to be divided.
∴ Least number to be divide = 25
(iv) 8788 = 2 x 2 x 13 x 13 x 13

Grouping the factors in triplets of equal factors, we find that 2 x 2 has been left
∴ In order to get a perfect cube, 2 x 2 is to be divided
∴ Least number to be divided = 4
(v) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplets of equal factors, we see that 17 x 17 has been left.
So, in order to get a perfect cube, 17 x 17 is be divided
∴ Least number to be divided = 17 x 17 = 289
(vi) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, 3 is left
∴ In order to get a perfect cube, 3 is to be divided
∴ Least number to be divided = 3
(vii) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplets of equal factors, we see that 7 x 7 is left
So, in order to get a perfect cube, 7 x 7 = 49 is to be divided
∴ Least number to be divided = 49
(viii) 243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, 3 x 3 is left
∴ By dividing 3 x 3, we get a perfect cube
∴ Least number to be divided = 3 x 3=9

Question 13.
Prove that if a number is trebled then its cube is 27 times the cube of the given number.
Solution:
Let x be the number, then trebled number of x = 3x
Cubing, we get:
(3x)³ = (3)³ x³ = 27x³
27x³ is 27 times the cube of x i.e., of x³

Question 14.
What happenes to the cube of a number if the number is multiplied by
(i) 3 ?
(ii) 4 ?
(iii) 5 ?
Solution:
number (x)³ = x³
(i) If x is multiplied by 3, then the cube of
∴ (3x)³ = (3)³ x x³ = 27x³
∴ The cube of the resulting number is 27 times of cube of the given number
(ii) If x is multiplied by 4, then the cube of
(4x)³ = (4)³ x x³ = 64x³
∴ The cube of the resulting number is 64 times of the cube of the given number
(ii) If x is multiplied by 5, then the cube of
(5x)³ = (5)³ x x³ = 125x³
∴ The cube of the resulting number is 125 times of the cube of the given number

Question 15.
Find the volume of a cube, one face of which has an area of 64 m².
Solution:
Area of one face of a cube = 64 m²
∴ Side (edge) of cube = √64
= √64 = 8 m
∴ Volume of the cube = (side)³ = (8 m)³ = 512 m³

Question 16.
Find the volume of a cube whose surface area is 384 m².
Solution:
Surface area of a cube = 384 m² Let side = a
Then 6a² = 384 ⇒ a² = \(\frac {384 }{ 6 }\)= 64 = (8)²
∴ a = 8 m
Now volume = a³ = (8)³ m³ = 512 m³

Question 17.
Evaluate the following :

Solution:

Question 18.
Write the units digit of the cube of each of the following numbers :
31,109,388,833,4276,5922,77774,44447, 125125125.
Solution:
We know that if unit digit of a number n is
= 1, then units digit of its cube = 1
= 2, then units digit of its cube = 8
= 3, then units digit of its cube = 7
= 4, then units digit of its cube = 4
= 5, then units digit of its cube = 5
= 6, then units digit of its cube = 6
= 7, then the units digit of its cube = 3
= 8, then units digit of its cube = 2
= 9, then units digit of its cube = 9
= 0, then units digit of its cube = 0
Now units digit of the cube of 31 = 1
Units digit of the cube of 109 = 9
Units digits of the cube of 388 = 2
Units digits of the cube of 833 = 7
Units digits of the cube of 4276 = 6
Units digit of the cube of 5922 = 8
Units digit of the cube of 77774 = 4
Units digit of tl. cube of 44447 = 3
Units digit of the cube of 125125125 = 5

Question 19.
Find the cubes of the following numbers by column method :
(i) 35
(ii) 56
(iii) 72
Solution:

Question 20.
Which of the following numbers are not perfect cubes ?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
Solution:
(i) 64 = 2 x 2 x 2 X 2 x 2 x 2

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 64 is a perfect cube.
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors m triplets, of equal factors, we see that no factor is left.
∴ 1728 is a perfect cube.

Question 21.
For each of the non-perfect cubes, in Q. 20, find the smallest number by which it must be
(a) multiplied so that the product is a perfect cube.
(b) divided so that the quotient is a perfect cube.
Solution:
In qustion 20, 243 is not a perfect cube and 243 = 3 x 3 x 3 x 3 x 3
Grouping the factors in triplets, of equal factors, we see that 3 x 3 is left.
(a) In order to make it a perfect cube, 3 is to be multiplied which makes a triplet.
(b) In order to make it a perfect cube, 3 x 3 or 9 is to be divided.

Question 22.
By taking three different values of n verify the truth of the following statements :
(i) If n is even, then n³ is also even.
(ii) If n is odd, then n³ is also odd.
(iii) If n leaves remainder 1 when divided by 3, then it3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p + 2 then n³ also a number of the same type.
Solution:
(i) n is even number.
Let n = 2, 4, 6 then
(a) n³ = (2)³ = 2 x 2 x 2 = 8, which is an even number.
(b) (n)³= (4)³ = 4 x 4 x 4 = 64, which is an even number.
(c) (n)³ = (6)³ = 6 x 6 x 6 = 216, which is an even number.

(ii) n is odd number.
Letx = 3, 5, 7
(a) (n)³ = (3)³ = 3 x 3 x 3 = 27, which is an odd number.
(b) (n)³ = (5)³ = 5 x 5 x 5 = 125, which is an odd number.
(c) (n)³ = (7)³ = 7 x 7 x 7 = 343, which is an odd number.

(iii) If n leaves remainder 1 when divided by 3, then n³ is also leaves 1 as remainder,
Let n = 4, 7, 10 If n = 4,
then «³ = (4)³ = 4 x 4 x 4 = 64
= 64 + 3 = 21, remainder = 1
If n = 7, then
n³ = (7)³ = 7 x 7 x 7 = 343
343 + 3 = 114, remainder = 1
If n – 10, then
(n)³ = (10)³ = 10 x 10 x 10 = 1000
1000 + 3 = 333, remainder = 1

(iv) If the natural number is of the form 3p + 2, then n³ is also of the same type
Let p =’1, 2, 3, then
(a) If p = 1, then
n = 3p + 2 = 3 x 1+2=3+2=5
∴ n³ = (5)³ = 5 x 5 x 5 = 125
125 = 3 x 41 + 2 = 3p +2

(b) If p = 2, then
n = 3p + 2 = 3 x 2 + 2 = 6 + 2 = 8
∴ n³ = (8)³ = 8 x 8 x 8 = 512
∴ 512 = 3 x 170 + 2 = 3p + 2

(c) If p = 3, then
n = 3p + 2 = 3 x 3 + 2 = 9 + 2 = 11
∴ n³ = (11)³ = 11 x 11 x 11 = 1331
and 1331 =3 x 443 + 2 = 3p + 2
Hence proved.

Question 23.
Write true (T) or false (F) for the following statements :
(i) 0 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a³ is always greater than a² > b²
(vi) If a and b are integers such that a² > b² , then a³ > b³.
(x) If a² ends in an even number of zeros, then a ends in an odd number of zeros.
Solution:

∵ 5 is left.
(iii)True : A number ending three zeros can be a perfect cube.
(iv) False : v (4)³ = 4 x 4 x 4 = 64, which ends with 4.
(v) False : If n is a proper fraction, it is not possible.
(vi) False : It is not true if a and b are proper fraction.
(vii) True.
(viii) False : as a² ends in 9, then a³ does not necessarily ends in 7. It ends in 3 also.
(ix) False : it is not necessarily that a³ ends in 25 it can end also in 75.
(x) False : If a² ends with even zeros, then a³ will ends with odd zeros but of multiple of 3.

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1 are helpful to complete your math homework.

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