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Frank ICSE Mathematics Class 8 Solutions Chapter 4 Cubes and Cube Roots
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 4, Cubes and Cube Roots. Here students can easily find step by step solutions of all the problems for Cubes and Cube Roots, Exercise 4.1 and 4.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 4 solutions. Here in this post all the solutions are based on latest Syllabus.
Cubes and Cube Roots Exercise 4.1 Solution :
Question no – (1)
Solution :
(a) 37 – 73 = 34
So, units digit is 3
(b) 164 – 43 = 6
So, units digit is – 4
(c) 82 – 23 = (8)
So, units digit is – (8)
(d) 50 – 0
(e) 99 – 93 = 72
So, units digit is 9
(f) 53 – 33 = 2
So, unit digit is 7
(g) 78 – 83 = 51
So, unit digit is 2
Question no – (2)
Solution :
(b) 121000 is not a perfect cube
Question no – (3)
Solution :
(a) 173 – odd
(b) 243 – even
(c) 353 – odd
(d) 963 – even
Question no – (4)
Solution :
(a) This statement is – False
(b) This statement is – False
(c) This statement is – True
(d) This statement is – True
Question no – (5)
Solution :
Correct Option – (d)
= – 17
Question no – (6)
Solution :
= 8/11
Question no – (7)
Solution :
(b) (0.5)3 = 1/8
Question no – (8)
Solution :
(c) is false if a be negative integer
Question no – (9)
Solution :
(a) (- 15)3 – 3375
(b) (- 21)3 = – 9261
(c) (0.05)3 = 0.000125
(d) (1.2)3 = 1.728
Question no – (10)
Solution :
(a) (- 5/6)3 = – 125/216
(b) (11/12)3 = 1331/1728
(c) (1 1/7)3 = (8/7)3 = 512/343
(d) (- 2 1/4)3 = (- 9/4) = – 729/64
Question no – (12)
Solution :
(a) 63 + 31 + 35 + 37 + 39 + 41
(b) 73 + 43 + 45 + 47 + 49 + 51 + 53 + 55
(c) 83 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71
(d) 93 = 73 + 75 + 77 + 79 + 81 + 83 + 85 + 87 + 89
Cubes and Cube Roots Exercise 4.2 Solution :
Question no – (1)
Solution :
Given numbers are, 289, 537, 414, 729 here we have to find which of this is a perfect cube of a number.
(d) 729 is a perfect cube.
In the given numbers only 729 is the perfect cube number.
Question no – (3)
Solution :
The units digit of cube root of 4913 is 7
Given number is, 4913
So now,
∴ Units digit of cube root of 4913 is 7
Question no – (4)
Solution :
12 is the cube root of 1728
First Cube of 12
∴ (12)³
= 12 × 12 × 12
= 1728
Question no – (5)
Solution :
(a) -35937
∴ – 35937 = 3 × 3 × 3 × 11 × 11 × 11 × (-1) × (-1) × (-1)
∴ ∛-35937 = 3 × 11 × (-1)
= – 33
∴ The cube root of – 35937 is -33
(b) -5832
∴ 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × (-1) × (-1) × (-1)
∴ ∛5832 = 2 × 3 × 3 ×(-1)
= – 18
∴ The cube root of -5832 is -18
(c) 2744
∴ +2744 = 2 × 2 × 2 × 7 × 7 × 7
∴ ∛+2744 = 2 × 7
= 14
∴ The cube root of 2744 is 14
(d) 9261
∴ 9261 = 3 × 3 × 3 × 7 × 7 × 7
∴ ∛9261 = 3 × 7
= 21
∴ The cube root of 9261 is 21
(e) 10648
∴ 10648 = 2 × 2 × 2 × 11 × 11 × 11
∴ ∛10648 = 2 × 11
= 22
∴ The cube root of 10648 is 22
(f) 21952
∴ 21952 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7
∴ ∛21952 = 2 × 2 × 7
= 28
∴ The cube root of 21952 is 28
(g) 74088
∴ 74088 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7
∴ ∛74088 = 2 × 3 × 7
= 42
∴ The cube root of 74088 is 42
(h) 42875
∴ 42875 = 5 × 5 × 5 × 7 × 7 × 7
∴ ∛42875 = 5 × 7
= 35
∴ The cube root of 42875 is 35
(i) 166375
∴ 166375 = 5 × 5 × 5 × 11 × 11 × 11
∴ ∛166375 = 5 × 11
= 55
∴ The cube root of 166375 is 55
Question no – (6)
Solution :
(a) 1331/1728
= ∛1331/1728
= 11/12
(b) -343/3375
= ∛-343/3375
= -7/15
(c) 0.004096
= ∛0.004096
= 0.16
(d) -9.261
= ∛-9.261
= -2.1
Question no – (7)
Solution :
(a) 216 × 343
= ∛216 × 343
= 3√6 × 6 × 6 × 7 × 7 × 7
= 6 × 7
= 42
(b) 144 × 96
= ∛144 × 96
= ∛2 × 2 × 2 × 2 × 3 × 3 × 2 × 2 × 2 × 2 × 3
= 2 × 2 × 2 × 3
= 24
(c) 250 × 28 × 49
= ∛250 × 28 × 49
= ∛2 × 5 × 5 × 5 × 2 × 2 × 7 × 7 × 7
= 2 × 5 × 7
= 70
(d) -216 × 729
= ∛-216 × 729
= √(-6) × (-6) × (-6) × 9 × 9 × 9
= (-6) × 9
= -54
Question no – (8)
Solution :
(a) 6859 → 9 in unit place
Now, 23 > 6 > 13.
So, 1 in tens place
= ∛6859
= 19
(b) 29791 → 1 in unit place.
and, 43 > 29 > 33 → 3 in tens place
= ∛29791
= 31
(c) 158877 → 3 in unit place
and, 63 > 148 > 53 → 5 in tens place
∴ ∛148877
= 53
(d) 238328 → 2 in unit place
and, 73 > 238 > 63 → 6 in tens place
(e) 4913 -> 7 in unit place
and, 23 > 4 > 13 → 1 in tens place
∴ ∛4913
= 17
(f) 46656 → 6 in unit place.
and, 43 > 46 > 33 → 3 in tens place
∴ ∛46656
= 36
(g) 110592 -> 8 in unit place
and, 53 > 110 > 43 → 4 in tens place
∴ ∛110592
= 48
(h) 405224 -> 4 in unit place
and, 83 > 405 > 73 → 7 in tens place.
∴ ∛405224
= 74
(i) 421875 → 5 in unit place
and, 83 > 421 > 73 → 7 in tens place
∴ ∛421875
= 75
Question no – (9)
Solution :
In the given question :
What is the smallest number by which 4116 must be multiplied?
So that the product is a perfect cube.
the cube root of the perfect cube so obtained.
Clearly, we have to multiply 4116 with = 2 × 3 × 3 = 18 to get a perfect cube.
And, the cube root will be 2 × 3 × 7 = 72
Therefore, the cube root of the perfect cube 72
Question no – (10)
Solution :
From the above solution its Clear that we have to divided 3584 with 7 to get a perfect cube and the cube root is 2 × 2 × 2 = 8
Therefore, the cube root of the perfect cube so obtained is 8
Question no – (11)
Solution :
In the given question we get,
Cube volume is = 729 cm3
length of the edge of a cube = ?
Given volume of the cube is 729 cm3
So now,
Therefore, the length of the edge of the cube will be 9 cm.
Question no – (12)
Solution :
According to the question we get,
Three numbers are in the ratio = 2 : 3 : 4
The sum of their cubes is = 2673.
The numbers are = ?
Let, the numbers be 2x, 3x, 4x
∴ (2x)3 + (3x)3 + (4x)3 = 2673
or, 8x3 + 27x3 + 64x3 = 2673
or, 99x3 = 2673 or x3
= 2673/99 = 27
∴ x = 3
∴ The required numbers are –
= (2 × 3) = 6
= (3 × 3) = 9
= (4 × 3) = 12
Therefore, numbers are 6, 9 and 12
Revision Exercise Questions Solution :
Question no – (1)
Solution :
(a) 10648
Here we get,
= 10648 = 2 × 2 × 2 × 11 × 11 × 11
= 23 × 113
Thus, we can say 10648 is a perfect cube.
(b) 42875
Here we get,
= 42875 = 5 × 5 × 5 × 7 × 7 × 7
= 53 × 73
Therefore, 42875 also a perfect cube.
Question no – (2)
Solution :
Given 1125 by Prime factorisation method,
∴ 1125 = 5 × 5 × 5 × 3 × 3 × 3
Clearly, 1125 has to be multiplied by 3 to get a perfect cube.
Because the smallest pair is 3
Question no – (4)
Solution :
(a) ∛216 – ∛64/√625 – √196
= ∛216 – ∛64/√625 – √196
= 6 – 4/25 -14
= 2/11 …(Simplified)
(b) √729/∛729 – √64/∛64
= √729/∛729 – √64/∛64
= 27/9 – 8/4
= 3 – 2
= 1 …(Simplified)
Question no – (7)
Solution :
(a) 125 × 729
= ∛125×729
= √5 × 5 × 5 × 9 × 9 × 9
= 5 × 9
= 45
∴ Cube Root is 45
(b) 512/1331
= ∛512/1331
= ∛8 × 8 × 8/11 × 11 × 11
= 8/11
∴ Cube Root is 8/11
(c) 64/343
= ∛64/343
= ∛4 × 4 × 4/7 × 7 × 7
= 4/7
∴ Cube Root is 4/7
(d) 216 × 27
= ∛216 × 27
= ∛6 × 6 × 6 × 3 × 3 × 3
= 6 × 3
= 18
∴ Cube Root is 18
Question no – (8)
Solution :
Given number, 17576
17576 → the unit digit would be 6.
And, 23 < 17 < 33, so tens digit will be 2.
∴ ∛17576 = 26
Therefore, the Cube root of 17576 is 26
Next Chapter Solution :
👉 Chapter 5 👈
