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

ii) 3375 = 5 × 5 × 5 × 3 × 3 × 3 = 15 × 15 × 15

3375−−−−√3=15

iii) 10648 = 2 × 2 × 2 × 11 × 11 × 11

= 2 × 11 × 2 × 11 × 2 × 11

= 22 × 22 × 22

10648 = 22^{3}

∴ 10648−−−−−√3=22

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 = 36^{3}

∴ 46656−−−−−√3=36

v) 15625 = 5 × 5 × 5 × 5 × 5 × 5

= 25 × 25 × 25

15625 = 25^{3}

∴ 15625−−−−−√3=25

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 = 5^{3}

Hence 40^{3} = 64000 <91125 < 125000 = 50^{3}

∴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.

5^{3} = 125 < 166 < 216 = 6^{3}

Hence 50^{3} = 125000, < 166375 < 216000 = 60^{3}

∴ 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

8^{3} = 512 < 704 < 729 – 9^{3}

Hence = 80^{3}= 512000 < 704969 < 72900 = 90^{3}

∴ 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:

60^{3} = 216000 < 331776 < 343000 = 70^{3}

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.

68^{3} = 314432, 69^{3} = 328509

Hence 331776−−−−−−√3 lies between 69 and 70

331776 – 328509 = 3267

343000 – 331776 = 11224

331776 in nearer to 69^{3}

∴ The closest integer to 331776−−−−−−√3 is 69.

Question (ii)

46656

Answer:

30^{3} = 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

35^{3} = 42, 875, 36^{3} = 46656

∴ 46656−−−−−√3=36

Question (iii)

373248

Answer:

70^{3} = 343000 < 373248 < 512000 – 80^{3}

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. 71^{3} = 357911, 72^{3} = 373248

∴ 373248−−−−−−√3=72

**All Chapter KSEEB Solutions For Class 8 maths**

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