Selina Concise Class 9 Maths Chapter 7 Indices Exercise 7A Solutions
EXERCISE – 7A
Selina Concise Class 9 Maths
(1) Evaluate:
(i) 33 × (243)-213 × 9-113
Solution:
(3)3 × (3×3×3×3×3)-2/3 × (3×3)-1/3
= (3)3 × (35)-2/3 × (32)-1/3
= (3)3 × (3)-10/3 × (3)-2/3
= (3)3+(-10/3)+(-2/3)
= (3)3-10/3-2/3
= (3)9-10-2/3
= (3)9-12/3
= (3)-3/3
= (3)-1
= 1/3
(ii) 5-4 × (125)5/3 ÷ (25)-1/2
Solution:
5-4 × (125)5/3 ÷ (25)-1/2
= 5-4 × (5×5×5)5/3 ÷ (5×5)-1/2
= 5-4 × (53)5/3 ÷ (52)-1/2
= 5-4 × (5)15/3 ÷ (5)-2/2
= 5-4 × (5)15/3 ÷ (5)-1
= 5-4 × (5)5 ÷ (5)-1
= (5)-4 × (5)5 ÷ 1/5
= (5)-4 × (5)5 × (5)1
= (5)-4+5+1
= (5)-4+6
= (5)2
= 25
(iii) (27/125)2/3 × (9/25)-3/2
Solution:
(27/125)2/3 × (9/25)-3/2
= (33/53)2/3 × (32/52)-3/2
= (3/5)3×2/3 × (3/5)2×-3/2
= (3/5)2 × (3/5)-3
= (3/5)2-3
= (3/5)-1
= (5/3)
(iv) 70 × (25)-3/2 – 5-3
Solution:
70 × (25)-3/2 – 5-3
= 1 × (52)-3/2 – 5-3 {a0 = 1}
= (5)2×(-3/2) – 5-3
= (5)-3 – (5)-3
= 1/53 – 1/53
= 1/125 – 1/125
= 0
(v) (16/81)-3/4 × (49/9)3/2 ÷ (343/216)2/3
Solution:
(16/81)-3/4 × (49/9)3/2 ÷ (343/216)2/3
= (24/34)-3/4 × (72/32)3/2 ÷ (73/63)2/3
= (2/3)4×-3/4 × (7/3)2×3/2 ÷ (7/6)3×2/3
= (2/3)-3 × (7/3)3 ÷ (7/6)2
= (2/3)-3 × (7/3)3 × (6/7)2
= (2/3)-3 × 7×7×7/3×3×3 × 6×6/7×7
= (3/2)+3 × 7/3 × 4
= 3×3×3/2×2×2 × 7/3 × 4
= 3×3/2×2 × 7 × 2
= 3×3/2 × 7
= 9/2 × 7
= 63/2
= 31.5
(Q2) Simplify:
(i) (8x3 ÷ 125 y3)2/3
Solution:
(8x3 ÷ 125y3)2/3
= [(2x)3/(5y)3]2/3
= [2x/5y]3×2/3
= [2x/5y]2
= [4x2/25y2]
(ii) (a + b)-1. (a-1 + b-1)
Solution:
(a + b)-1 (a-1 + b-1)
= 1/(a + b) . (1/a + 1/b)
= 1/(a + b) {b + a/ab}
= 1/(a + b) {(a+b)/ab}
= 1/ab
(iii) 5n+3 – 6×5n+1/9×5n – 5n×22
Solution:
5n+3 – 6×5n+1/9×5n – 5n×22
= 5n×53-6×5n×51/9×5n-5n×22
= 5n [53 – 6×51]/5n [9-22]
= [125 – 30]/[9-4]
= 95/5
= 19
(iv) (3x2)-3 × (x9)2/3
Solution:
(3x2)-3 × (x9)2/3
= 1/(3x2)3 × x9×2/3
= 1/27x6 × x6
= 1/27
(Q3) Evaluate:
(i) √1/4 + (0.01)-1/2 – (27)2/3
Solution:
√1/4 + (0.01)-1/2 – (27)2/3
= 1/2 + (1/100)-1/2 – (33)2/3
= 1/2 + (100/1)1/2 – (3)3×2/3
= 1/2 + (102)1/2 – (3)2
= 1/2 + (10)2×1/2 – (3)2
= 1/2 + (10) – 9
= 1/2 + 1
= 1+2/2
= 3/2
(ii) (27/8)2/3 – (1/4)-2 + 50
Solution:
(27/8)2/3 – (1/4)-2 + 50
= (33/23)2/3 – (4/1)2 + 1 (a0 = 1)
= (3/2)3×2/3 – (4)2 + 1
= (3/2)2 – (4)2 + 1
= 9/4 – 16 + 1
= 9/4 – 15
= 9-60/4
= -51/4
(Q4) Simplify each of the following and express with positive index:
(i) (3-4/2-8)1/4
Solution:
(3-4/2-8)1/4
= (3)-4×1/4/(2)-8×1/4
= (3)-1/(2)-2
= (2)2/(3)1
= 4/3
(ii) (27-3/9-3)1/5
Solution:
(27-3/9-3)1/5
= ((33)-3)/(32)-3)1/5
= [(33)/(32)]-3/5
= [3×3×3/3×3]-3/5
= [3]-3/5
= 1/(3)3/5
(iii) (32)-2/5 ÷ (125)-2/3
Solution:
(32)-2/5 ÷ (125)-2/3
= (25)-2/5 ÷ (53)-2/3
= (2)5×-2/5 ÷ (5)3×-2/3
= (2)-2 ÷ (5)-2
= ½2 ÷ 1/52
= 1/22 × 52/1
= 1/4 × 25
= 25/4
(iv) [1 – {1 – (1 – n)-1}-1]-1
Solution:
[1 – {1 – (1 – n)-1}-1]-1
= [1 – {1 – 1/(1-n)}-1]-1
= [1 – {1-n-1/(1-n)}-1]-1
= [1 – {-n/1-n}-1]-1
= [1 – [1-n/-n]1]-1
= [-n – 1 + n]-1
= [1/-n]-1
= n
(Q5) If 2160 = 2a.3b.5c, find a, b and c. Hence calculate the value of 3a × 2-b × 5-c
Solution:
Given: 2160 = 2a.3b.5c
Find a, b, c and 3a × 2-b × 5-c
2160 = 2a × 3b × 5c
2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2a × 3b × 5c
24 × 33 × 51 = 2a × 3b × 5c
∴ 24 = 2a, 33 = 3b, 51 = 5c
∴ a = 4, b = 3, c = 1
Now, 3a × 2-b × 5-c = 34 × 2-3 × 5-1
= 3×3×3×3× 1/23 × 1/5
= 81/2×2×2×5
= 81/40
= 2 1/40
(Q6) If 1960 = 2a.5b.7c, Calculate the value of 2-a, 7b, 5-c
Solution:
1960 = 2a.5b.7c
2×2×2×5×7×7 = 2a.5b.7c
23 × 51 × 72 = 2a × 5b × 7c
23 = 2a, 51 = 5b , 72 = 7c
∴ a = 3, b = 1, c = 2
Now, 2-a × 7b × 5-c = 2-3 × 71 × 5-2
= 1/23 × 7 × 1/52
= 1/2×2×2 × 7 × 1/5×5
= 7/2×2×2×5×5
= 7/8×25
= 7/200
(Q7) Simplify:
(i) 83a/25 × 22a/4×211a × 2-2a
Solution:
83a × 25 × 22a/4×211a×2-2a
= (23)3a × 25 × 22a/4×211a × 2-2a
= (2)9a × 25 × 22a/4×2(11a-2a)
= (2)9a × 2×2×2×2×2×22a/4×2(9a)
= 2×2×2×2×2×22a/2×2
= 23 × 22a
= 8.22a
= 23 × 22a
= 22a+3
(ii) 3 × 27n+1 + 9×33n-1/8×33n – 5×27n
Solution:
3×27n+1 + 9 × 33n-1/8×33n – 5 × 27n
= 3 × 27n × 27 + 9 × 33n.3-1/8×(33)n – 5×(33)n
= 27n×81+9×27n.3-1/8×27n-5×27n
= 27n [81+9×3-1]/27n [8-5]
= 81+9×1/3/3
= 81+3/3
= 84/3
= 28
(Q8) Show that:
(am/a-n)m-n × (an/a-1)n-1 × (a1/a-m)1-m = 1
(Q9) If a = xm+n . xl, b = xn+1, xm and c = xl+m . xn
Prove that: am+n . bn-l . cl-m = 1
Solution:
Q10.)
Solution –
ii.)
Solution –
Here is your solution of Selina Concise Class 9 Maths Chapter 7 Exercise 7A
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