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Joy of Mathematics Class 8 Solutions Chapter 5 Cubes and Cube Roots
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 5 Cubes and Cube Roots. Here students can easily find step by step solutions of all the problems for Cubes and Cube Roots. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 5
Cubes and Cube Roots Exercise 5 Solution
Question no – (1)
Solution :
(a) (52)3
= 52 × 52 × 52
= 140,608
(b) (73)3
= 73 × 73 × 73
= 389,017
(c) (- 30)3
= – 3- × (- 30) × (- 30)
= – 27000
(d) (1 1/4)3
= (5/4)3
= 5/4 × 5/4 × 5/4
= 125/64
(e) (0.003)3
= 0.003 × 0.003 × 0.003
= 27/100000000
(f) (- 15)3
= – 15 × (- 15) × – 15)
= – 3375
(g) – 8/11
= – 8/11 × (- 8/11) × (- 8/11)
= – 512/1331
(h) 0.064
= 0.064 × 0.064 × 0.064
= 0.000262144
Question no – (2)
Solution :
Cube of all natural numbers between 12 and 17 are
(13)3 = 13 × 13 × 13 = 2197
(14)3 = 14 × 14 × 14 = 2744
(15)3 = 15 × 15 × 15 = 3375
(16)3 = 16 × 16 × 16 = 4096
Question no – (3)
Solution :
(a) ∛625
=> √625 = ∛5 × 5 × 5× 5
= 5∛5
∴ This is not perfect cube.
(b) ∛1458
=> ∛1458 = ∛2 × 3 × 3 × 3 × 3 × 3 × 3
= 3 × 3 ∛2
= 9 ∛2
∴ This is not perfect cube.
(c) ∛16875
=> ∛16875 = ∛5 × 5 × 5 × 5 × 3 × 3 × 3
= 5 × 3 × ∛5
= 15 ∛5
∴ This is not perfect cube.
(d) ∛12167
=> ∛12167 = ∛23 × 23 × 23
= 23
∴ This is perfect cube.
(e) ∛1000
=> ∛1000 = ∛2 × 2 × 2 × 5 × 5 × 5
= 2 × 5
= 10
∴ This is perfect cube.
(f) ∛6859
=> ∛6859 = ∛19 × 19 × 19
= 19
∴ This is perfect cube.
(g) ∛13824
=> ∛13824 = ∛3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 3 × 2 × 2 × 2
= 24
∴ This is a perfect cube.
(h) ∛46875
=> ∛46875 = ∛5 × 5 × 5 × 5 × 5 × 3
= 5 ∛5 × 5 × 3
∴ This is not perfect cube.
Question no – (4)
Solution :
(a) 5 natural number which is multiple of 3 is a multiple of 27, are, 27, 729, 1728, 3375
(b) 5 natural number which is multiple of 2 is a multiple of 8, are 8, 64, 512.
Question no – (5)
Solution :
The side of a cube (a) = 17.5 cm
∴ Volume of the cube = (a)3 cm3
= (17.5)3 cm3
= (17.5 × 17.5 × 17.5) cm3
= 5359.375 cm3
Question no – (6)
Solution :
69120
=> ∛68120
=> ∛5 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
If we multiple 5 × 5 = 25, then 69120 obtain a perfect cube.
∴ 69120 × 25 = 1728000
=> √5 × 5 × 5 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
=> 5 × 2 × 2 × 2 × 3
= 120
Question no – (7)
Solution :
81920
=> ∛5 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
=> 81920 must be perfect cube if we divided by 5 × 2 × 2 = 20
Question no – (8)
Solution :
(a) 1728
=> ∛2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 2 × 3 × 2
= 12
(b) 3375
= ∛5 × 5 × 5 × 3 × 3 × 3
= 5 × 3
= 15
(c) 17576
= ∛2 × 2 × 2 × 13 × 13 × 13
= 2 × 13
= 26
(d) 0.000343
= 343/1000000
= ∛7 × 7 × 7/∛2 × 2 × 2 × 2 × 2 × 2 ×5 × 5 × 5 × 5 × 5 × 5
= 7/2 × 2 × 5 × 5
= 7/100
= 0.07
(e) 2 93/125
= 343/125
= ∛7 × 7 × 7/∛5 × 5 × 5
= 7/5
(f) 0.000064
= 64/1000000
= ∛2 × 2 × 2 × 2 × 2 × 2/100
= 4/100
= 0.04
(g) 50653
= ∛37 × 37 × 37
= 37
(h) – 1 91/125
= 216/125
= ∛2 × 2 × 2 × 3 × 3 × 3/∛5 × 5 × 5
= 2 × 3/5
= 6/5
Question no – (9)
Solution :
(a) ∛216 × ∛64
= ∛2 × 2 × 2 × 3 × 3 × 3 × ∛2 × 2 × 2 × 2 ×2 × 2
= 2 × 3 × 2 × 2
= 24
(b) ∛729 × ∛(- 27)
= ∛3 × 3 × 3 × 3 × 3 × 3 × ∛(- 3 × 3 × 3)
= 3 × 3 × (- 3)
= – 27
(c) ∛3 – 67/64
= ∛192 – 67/64
= ∛125/64
= ∛5 × 5 × 5/2 × 2 × 2 × 2 × 2 ×2
= 5/ 2 × 2
= 5/4
(d) ∛512 + ∛(- 1728)
=> √2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 + √(- 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3)
= (2 × 2 × 2) + (- 2 × 2 × 3)
= 8 – 12
= – 4
(e) ∛3375 × ∛(- 216)
= ∛5 × 5 × 5 × 3 × 3 × 3 × ∛(- 2) × (- 2) × (- 2) × (- 3) × (- 3) × (- 3)
= 5 × 3 × (- 2) × – 3
= – 90
(f) ∛2744 × ∛343
= ∛2 × 2 × 2 × 7 × 7 × 7 × √7 × 7 × 7
= 2 × 7 × 7
= 14 × 7
= 98
Question no – (10)
Solution :
Let, the length of the cubical box = x cm
∴ Then the volume of the cubical box = x3 cm3
∴ x3 = 1953.125
=> x = ∛5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5/ 10× 10 × 10
= 5 × 5 × 5/10
= 125/10
= 12.5
∴ Length of the cubical = 12.5
Question no – (11)
Solution :
Let, the length of the cubical box = x m
∴ Volume of the cubical box = x3 m3
∴ x3 = 0.001331
=> x = ∛0.001331
= ∛1331/1000000
= ∛11 × 11 × 11/10 × 10 × 10 × 10 × 10 × 1 0
= 11/10 × 10
= 11/100
= 0.11
∴ The length of the cubical box = 0.11 m
∴ Total surface area of the box is
= 6 × (0.11)2 m
= 6 × 0.11/100 × 0.11/100
= 66 × 11/10000
= 726/10000
= 0.0726 m
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