KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 pdf, free KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7

Karnataka Board Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7

Question 1.
Find the cube root by prime factorisation,
i) 1728
ii) 3375
iii) 10648
iv) 46656
v) 15625
Answer:
i) 1728 = 2 × 2 × 2 × 6 × 6 × 6 = 12 × 12 × 12
1728−−−−√3=12
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 1

ii) 3375 = 5 × 5 × 5 × 3 × 3 × 3 = 15 × 15 × 15
3375−−−−√3=15
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 2

iii) 10648 = 2 × 2 × 2 × 11 × 11 × 11
= 2 × 11 × 2 × 11 × 2 × 11
= 22 × 22 × 22
10648 = 223
∴ 10648−−−−−√3=22
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 3

iv) 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= 4 × 4 × 4 × 9 × 9 × 9
= 4 × 9 × 9 × 4 × 9 × 4
46656 = 36 × 36 × 36 = 363
∴ 46656−−−−−√3=36
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 4

v) 15625 = 5 × 5 × 5 × 5 × 5 × 5
= 25 × 25 × 25
15625 = 253
∴ 15625−−−−−√3=25
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 5

2. Find the cube root of the following by looking at the last digit and using estimation.

Question (i)
91125
Answer:
Unit digit of 91125 in 5. Therefore the units digit in its cube root is 5.
Let us split 91125 as 91 and 125. We find that 43 = 64 < 91 < 125 = 53
Hence 403 = 64000 <91125 < 125000 = 503
∴Cube root of 91125 lies between 40 and 50 and units digit is 5 the only such number is 45.
∴ 91125−−−−−√3=45

Question (ii)
166375
Answer:
Units digit of 166375 is 5. Therefore the units digit of its cube root is 5.
Let us split 166375 as 166 and 375.
53 = 125 < 166 < 216 = 63
Hence 503 = 125000, < 166375 < 216000 = 603
∴ 166375−−−−−−√3 lies between 50 and 60. .
Since the units digit is 5 the only such number in 55.
∴ 166375−−−−−−√3=55

Question (iii)
704969
Answer:
The units digit of 704969 is 9. Therefore the units digits of its cube root are 9.
Let us split 704969 as 704 and 969
83 = 512 < 704 < 729 – 93
Hence = 803= 512000 < 704969 < 72900 = 903
∴ 704969−−−−−−√3 lies between 80 and 90.
Since the units digits is 9 the only such number is 89.
∴ 704969−−−−−−√3=89

3. Find the nearest integer to the cube root of each of the following.
(i) 331776
(ii) 46656
(iii) 373248

Question (i)
331776
Answer:
603 = 216000 < 331776 < 343000 = 703
Hence 331776−−−−−−√3 lies between 60 and 70.
We do not know whether 331776 in a perfect cube or not.
However we may sharpen the bound.
683 = 314432, 693 = 328509
Hence 331776−−−−−−√3 lies between 69 and 70
331776 – 328509 = 3267
343000 – 331776 = 11224
331776 in nearer to 693
∴ The closest integer to 331776−−−−−−√3 is 69.

Question (ii)
46656
Answer:
303 = 2700 < 46656 < 64000 – 40<sub>3</sub>
46656−−−−−√3 lies between 30 and 40 we do not know whether 46656 in a perfect cube or not. However we may sharper the bound
353 = 42, 875, 363 = 46656
∴ 46656−−−−−√3=36

Question (iii)
373248
Answer:
703 = 343000 < 373248 < 512000 – 803
373248−−−−−−√3 lies between 70 and 80. We do not know whether 373248 is a perfect cube or not.
However we may sharpen the bound. 713 = 357911, 723 = 373248
∴ 373248−−−−−−√3=72

All Chapter KSEEB Solutions For Class 8 maths

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