Std 7 Mathematics · Chapter 6

6. Indices

Step-by-step solutions to Practice Sets 26 to 30. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

Practice Set 26

1Complete the table below.

Sr. No.	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)	(37)⁴	37	4	37 × 37 × 37 × 37	812401
(v)	(−13)⁴	−13	4	(−13) × (−13) × (−13) × (−13)	28561

Row (i) is given in the textbook. In (iii) and (v) the base is negative and the index is even, so the value is positive.

2Find the value. (i) 2¹⁰ (ii) 5³ (iii) (−7)⁴ (iv) (−6)³ (v) 9³ (vi) 8¹ (vii) (45)³ (viii) (−12)⁴

(i) 2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

(ii) 5³ = 5 × 5 × 5 = 125

(iii) (−7)⁴ = (−7) × (−7) × (−7) × (−7) = 49 × 49 = 2401

(iv) (−6)³ = (−6) × (−6) × (−6) = 36 × (−6) = −216

(v) 9³ = 9 × 9 × 9 = 729

(vi) 8¹ = 8

(vii) (45)³ = 4 × 4 × 45 × 5 × 5 = 64125

(viii) (−12)⁴ = (−1) × (−1) × (−1) × (−1)2 × 2 × 2 × 2 = 116

A negative base with an even index gives a positive value; with an odd index it gives a negative value — compare (iii) and (iv).

Practice Set 27

1Simplify.

Rule: a^m × aⁿ = a^{m + n} (same base — add the indices).

(i) 7⁴ × 7² = 7^{4 + 2} = 7⁶

(ii) (−11)⁵ × (−11)² = (−11)^{5 + 2} = (−11)⁷

(iii) (67)³ × (67)⁵ = (67)^{3 + 5} = (67)⁸

(iv) (−32)⁵ × (−32)³ = (−32)^{5 + 3} = (−32)⁸

(v) a¹⁶ × a⁷ = a^{16 + 7} = a²³

(vi) (P5)³ × (P5)⁷ = (P5)^{3 + 7} = (P5)¹⁰

Practice Set 28

1Simplify. (i) a⁶ ÷ a⁴ (ii) m⁵ ÷ m⁸ (iii) p³ ÷ p¹³ (iv) x¹⁰ ÷ x¹⁰

Rule: a^m ÷ aⁿ = a^{m − n}, and a^−m = 1a^m, a⁰ = 1.

(i) a⁶ ÷ a⁴ = a^{6 − 4} = a²

(ii) m⁵ ÷ m⁸ = m^{5 − 8} = m⁻³ = 1m³

(iii) p³ ÷ p¹³ = p^{3 − 13} = p⁻¹⁰ = 1p¹⁰

(iv) x¹⁰ ÷ x¹⁰ = x^{10 − 10} = x⁰ = 1

2Find the value. (i) (−7)¹² ÷ (−7)¹² (ii) 7⁵ ÷ 7³ (iii) (45)³ ÷ (45)² (iv) 4⁷ ÷ 4⁵

(i) (−7)¹² ÷ (−7)¹² = (−7)^{12 − 12} = (−7)⁰ = 1

(ii) 7⁵ ÷ 7³ = 7^{5 − 3} = 7² = 49

(iii) (45)³ ÷ (45)² = (45)^{3 − 2} = (45)¹ = 45

(iv) 4⁷ ÷ 4⁵ = 4^{7 − 5} = 4² = 16

Practice Set 29

1Simplify.

Rules used: (a^m)ⁿ = a^mn, a^−m = 1a^m and (ab)^−m = (ba)^m.

(i) [(1512)³]⁴ = (1512)^{3 × 4} = (1512)¹² (15/12 = 5/4, so this also equals (5/4)¹²)

(ii) (3⁴)⁻² = 3^{4 × (−2)} = 3⁻⁸ = 13⁸

(iii) [(17)⁻³]⁴ = (17)^{(−3) × 4} = (17)⁻¹² = 7¹²

(iv) [(25)⁻²]⁻³ = (25)^{(−2) × (−3)} = (25)⁶

(v) (6⁵)⁴ = 6^{5 × 4} = 6²⁰

(vi) [(67)⁵]² = (67)^{5 × 2} = (67)¹⁰

(vii) [(23)⁻⁴]⁵ = (23)^{(−4) × 5} = (23)⁻²⁰ = (32)²⁰

(viii) [(58)³]⁻² = (58)^{3 × (−2)} = (58)⁻⁶ = (85)⁶

(ix) [(34)⁶]¹ = (34)^{6 × 1} = (34)⁶

(x) [(25)⁻³]² = (25)^{(−3) × 2} = (25)⁻⁶ = (52)⁶

2Write the following numbers using positive indices. (i) (27)⁻² (ii) (113)⁻⁵ (iii) (16)⁻³ (iv) (y)⁻⁴

(i) (27)⁻² = (72)²

(ii) (113)⁻⁵ = (311)⁵

(iii) (16)⁻³ = (61)³ = 6³

(iv) (y)⁻⁴ = 1y⁴

Practice Set 30

1Find the square root. (i) 625 (ii) 1225 (iii) 289 (iv) 4096 (v) 1089

Method: factorise into primes, make pairs of equal factors, and take one factor from each pair.

(i) 625

5	625
5	125
5	25
5	5
	1

625 = (5 × 5) × (5 × 5)
√625 = 5 × 5 = 25

(ii) 1225

5	1225
5	245
7	49
7	7
	1

1225 = (5 × 5) × (7 × 7)
√1225 = 5 × 7 = 35

(iii) 289

17	289
17	17
	1

289 = (17 × 17)
√289 = 17 = 17

(iv) 4096

2	4096
2	2048
2	1024
2	512
2	256
2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

4096 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2)
√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

(v) 1089

3	1089
3	363
11	121
11	11
	1

1089 = (3 × 3) × (11 × 11)
√1089 = 3 × 11 = 33

Check by squaring: 25² = 625, 35² = 1225, 17² = 289, 64² = 4096, 33² = 1089.

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