Mathematics Solutions Solutions for Class 7 Maths Chapter 6 Indices are provided here with simple step-by-step explanations. These solutions for Indices are extremely popular among Class 7 students for Maths Indices Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Complete the table below.

 Sr. No. Indices (Numbers in index      form) Base Index Multiplication form Value (i) 34 3 4 3$×$3$×$3$×$3 81 (ii) 163 (iii) ($-$8) 2 (iv) $\frac{3}{7}×\frac{3}{7}×\frac{3}{7}×\frac{3}{7}$ $\frac{81}{2401}$ (v) ($-$13)4

 Sr. No. (Numbers in index form) Base Index Multiplication form Value (i) 34 3 4 3 $×$ 3 $×$ 3 $×$ 3 81 (ii) 163 16 3 16 $×$ 16 $×$ 16 4096 (iii) (−8)2 (−8) 2 (−8) $×$ (−8) 64 (iv) ${\left(\frac{3}{7}\right)}^{4}$ $\frac{3}{7}$ 4 $\frac{3}{7}×\frac{3}{7}×\frac{3}{7}×\frac{3}{7}$ $\frac{81}{2401}$ (v) (−13)4 −13 4 (−13) $×$ (−13) $×$ (−13) $×$ (−13) 28561

#### Question 2:

Find the value.

(i) 210 (ii) 53    (iii) ($-$7)4    (iv) ($-$6)    (v) 93    (vi) 81  (vii) ${\left(\frac{4}{5}\right)}^{3}$ (viii) ${\left(-\frac{1}{2}\right)}^{4}$

#### Question 1:

Simplify.

(i) 74 × 72 (ii) ($-$11)5 × ($-$11)2 (iii)

(iv) (v) (vi)

It is known that, am × an = am+n, where m and n are integers and a is a non-zero rational number.

#### Question 1:

Simplify.
(i) a6 ÷ a4    (ii) m5 ÷ m8 (iii) p3 ÷ p13 (iv) x10 ÷ x10

It is known that, am ÷ an = amn, where m and n are integers and a is non-zero rational number.

#### Question 2:

Find the value

(i) (    (ii) ${7}^{5}$    (iii) ${\left(\frac{4}{5}\right)}^{3}÷{\left(\frac{4}{5}\right)}^{2}$   (iv) ${4}^{7}÷{4}^{5}$

It is known that, am ÷ an = amn, where m and n are integers and a is a non-zero rational number.

#### Question 1:

Simplify

(i)  ${\left[{\left(\frac{15}{12}\right)}^{3}\right]}^{4}$   (ii) ${\left({3}^{4}\right)}^{-2}$          (iii) ${\left({\left(\frac{1}{7}\right)}^{-3}\right)}^{4}$    (iv)  ${\left({\left(\frac{2}{5}\right)}^{-2}\right)}^{-3}$    (v) ${\left({6}^{5}\right)}^{4}$

(vi)  ${\left[{\left(\frac{6}{7}\right)}^{5}\right]}^{2}$   (vii) ${\left[{\left(\frac{2}{3}\right)}^{-4}\right]}^{5}$  (viii)  ${\left[{\left(\frac{5}{8}\right)}^{3}\right]}^{-2}$   (ix) ${\left[{\left(\frac{3}{4}\right)}^{6}\right]}^{1}$   (x) ${\left[{\left(\frac{2}{5}\right)}^{-3}\right]}^{2}$

It is known that, (am)n = amn, where m and n are integers and a is a non-zero rational number.

#### Question 2:

Write the following numbers using positive indices.

(i) ${\left(\frac{2}{7}\right)}^{-2}$   (ii) ${\left(\frac{11}{3}\right)}^{-5}$  (iii) ${\left(\frac{1}{6}\right)}^{-3}$   (iv) ${\left(y\right)}^{-4}$

It is known that, ${a}^{-m}=\frac{1}{{a}^{m}}$ where m is an integer and a is a non-zero rational number.

#### Question 1:

Find the square root.
(i) 625   (ii) 1225   (iii) 289   (iv) 4096   (v) 1089

(i) The prime factorization of 625 is,
625 = 5 × 5 ×  5 × 5
To find the square root, we will take one number from each pair and multiply.
$\sqrt{625}=5×5=25\phantom{\rule{0ex}{0ex}}\therefore \sqrt{625}=25$
(ii) The prime factorization of 1225 is,
1225 = 5 × 5 × 7 × 7
To find the square root, we will take one number from each pair and multiply.
$\sqrt{1225}=5×7=35\phantom{\rule{0ex}{0ex}}\therefore \sqrt{1225}=35$
(iii) The prime factorisation of 289 is,
289 = 17 × 17
To find the square root, we will take one number from each pair and multiply.
$\sqrt{289}=17×17=17\phantom{\rule{0ex}{0ex}}\therefore \sqrt{289}=17$
(iv) The prime factorization of 4096 is,
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ​× 2
To find the square root, we will take one number from each pair and multiply.

(v) The prime factorizationn of 1089 is,
1089 = 3 × 3 × 11 × 11
To find the square root, we will take one number from each pair and multipy.
$\sqrt{1089}=3×11=33\phantom{\rule{0ex}{0ex}}\therefore \sqrt{1089}=33$

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