ML Aggarwal Exponents and Powers Exe-2 Class 8 Ch-2 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-2 Questions for Exponents and Powers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.
ML Aggarwal Exponents and Powers Exe-2 Class 8 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 8th |
| Chapter-2 | Exponents and Powers |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-2 Questions |
| Edition | 2023-2024 |
Exponents and Powers Exe-2
ML Aggarwal Class 8 ICSE Maths Solutions
Page-40
Question 1. Evaluate:
(i) (3/5)-2
(ii) (-3)-3
(iii) (2/7)-4
Answer:
(i) (3/5)-2
By using the formula, (a-n = 1/an)
(3/5)-2 = (5/3)2
= (5/3) × (5/3)
= 25/9
(ii) (-3)-3
By using the formula, (a-n = 1/an)
(-3)-3 = (-1/3)3
= (-1/3) × (-1/3) × (-1/3)
= – 1/27
(iii) (2/7)-4
By using the formula, (a-n = 1/an)
(2/7)-4 = (7/2)4
= (7/2) × (7/2) × (7/2) × (7/2)
= 2401/16
Question 2. Simplify:
(i) [(2)-1 + (4)-1 + (3)-1]-1
(ii) [(4)-1 – (5)-1]2 × (5/8)-1
(iii) [40 + 42 – 23] × 3-2
(iv) [(5)2 – (1/4)-2] × (3/4)-2
Answer :
(i) [(2)-1 + (4)-1 + (3)-1]-1
[(2)-1 + (4)-1 + (3)-1]-1 = [(1/2) + (1/4) + (1/3)]-1
= [(6+3+4)/12]-1
= [13/12]-1
= 12/13
(ii) [(4)-1 – (5)-1]2 × (5/8)-1
[(4)-1 – (5)-1]2 × (5/8)-1 = [(1/4) – (1/5)]2 × (8/5)1
= [(5-4)/20]2 × (8/5)
= [1/20]2 × (8/5)
= (1/20) × (1/20) × (8/5)
= 1/250
(iii) [40 + 42 – 23] × 3-2
[40 + 42 – 23] × 3-2 = [1 + 16 – 8] × (1/32)
= 9 × (1/9)
= 1
(iv) [(5)2 – (1/4)-2] × (3/4)-2
[(5)2 – (1/4)-2] × (3/4)-2 = [25 – (4)2] × (4/3)2
= [25 – 16] × (16/9)
= 9 × (16/9)
= 16
Question 3. Find the multiplicative inverse of the following:
(i) (81/16)-3/4
(ii) {(-3/2)-4}1/2
(iii) (5/7)-2 × (5/7)4 ÷ (5/7)3
Answer:
(i) (81/16)-3/4
(81/16)-3/4 = (16/81)3/4
= (24/34)3/4
= (2/3)4×3/4
= (2/3)3
= (2/3) × (2/3) × (2/3)
= 8/27
So, the multiplicative inverse of 8/27 is 27/8.
(ii) {(-3/2)-4}1/2
{(-3/2)-4}1/2 = (-3/2)-4×1/2
= (-3/2)-2
= (2/-3)2
= (2/-3) × (2/-3)
= 4/9
So, the multiplicative inverse of 4/9 is 9/4.
(iii) (5/7)-2 × (5/7)4 ÷ (5/7)3
(5/7)-2 × (5/7)4 ÷ (5/7)3 = (7/5)2 × (5/7)4 ÷ (5/7)3
= (7/5)2 × (5/7)4-3
= (7/5)2 × (5/7)
= (7/5) × (7/5) × (5/7)
= 7/5
So, the multiplicative inverse of 7/5 is 5/7.
Question 4.
(i) Express 16-2 as a power with base 2.
(ii) Express 125-4 as a power with base 5.
Answer:
(i) Express 16-2 as a power with base 2.
(16)-2 = (24)-2
= 2-8
= 1/28
(ii) Express 125-4 as a power with base 5.
(125)-4 = (53)-4
= 5-12
= 1/512
Question 5. Simplify and write in exponential form with positive exponent:
(i) [{(5/7)2}-1]-3
(ii) (2/7)2 × (7/2)-3 ÷ {(7/5)-2}-4
(iii) (4/5)2 × 54 × (2/5)-2 ÷ (5/2)-3
(iv) [8-1 × 53]/2-4
Answer:
(i) [{(5/7)2}-1]-3
[{(5/7)2}-1]-3 = {(5/7)2}-1×-3
= (5/7)2×3
= (5/7)6
(ii) (2/7)2 × (7/2)-3 ÷ {(7/5)-2}-4
(2/7)2 × (7/2)-3 ÷ {(7/5)-2}-4 = (2/7)2 × (2/7)3 ÷ (7/5)8
= (2/7)2 × (2/7)3 × (5/7)8
= (22/72) × (23/73) × (58/78)
= (22+3/72+3) × (58/78)
= (25/75) × (58/78)
= (25 × 58)/75+8
= (25 × 58)/713
(iii) (4/5)2 × 54 × (2/5)-2 ÷ (5/2)-3
(4/5)2 × 54 × (2/5)-2 ÷ (5/2)-3 = ((22)2/52) × 54 × (2-2/5-2) × (2-3/5-3)
= (24 × 54 × 2-2 × 2-3) / (52 × 5-2 × 5-3)
= 24-2-3 × 54-2+2+3
= 2-1 × 57
= 57/21
(iv) [8-1 × 53]/2-4
[8-1 × 53]/2-4 = [(23)-1 × 53]/2-4
= [2-3 × 53]/2-4
= 2-3+4 × 53
= 21 × 53
= 2 × 5 × 5 × 5
= 250
Question 6. Simplify and write the following in exponential form:
(i) ((-2)3)2 + 5-3 ÷ 5-5 – (-1/2)0
(ii) 3-5 × 32 ÷ 3-6 + (22 × 3)2 + (2/3)-1 + 2-1 + (1/19)-1
Answer:
(i) ((-2)3)2 + 5-3 ÷ 5-5 – (-1/2)0
((-2)3)2 + 5-3 ÷ 5-5 – (-1/2)0 = (-2)6 + 1/53 ÷ 1/55 – 1
= 64 + 1/53 × 55 – 1
= 64 + 55-3 – 1
= 64 + 52 – 1
= 64 + 25 – 1
= 88
(ii) 3-5 × 32 ÷ 3-6 + (22 × 3)2 + (2/3)-1 + 2-1 + (1/19)-1
3-5 × 32 ÷ 3-6 + (22 × 3)2 + (2/3)-1 + 2-1 + (1/19)-1 = 3-5+2+6 + (24 × 32) + 3/2 + ½ + 191
= 33 + (16×9) + 4/2 + 19
= 27 + 144 + 2 + 19
= 192
Question 7. Simplify and write in exponential form with negative exponent:
(i) 53 × (4/5)3
(ii) [(3/7)-2]-3
(iii) (5/9)-2 × (5/3)2 ÷ (1/5)-2
(iv) 2-1 [(5/3)4 + (3/5)-2] ÷ (17/9)
(v) (-7)3 × (1/-7)-9 ÷ (-7)10
Answer:
(i) 53 × (4/5)3
53 × (4/5)3 = 53 × (43/53)
= 53-3 × 43
= 50 × 43
= 1 × 43
= (1/4)-3
(ii) [(3/7)-2]-3
[(3/7)-2]-3 = (3/7)-2×-3
= (3/7)6
= (7/3)-6
(iii) (5/9)-2 × (5/3)2 ÷ (1/5)-2
(5/9)-2 × (5/3)2 ÷ (1/5)-2 = (5-2/9-2) × (52/32) ÷ (1/5-2)
= (5-2 × 52 × 5-2)/ [(32)-2 × 32]
= (5-2+2-2)/(3-4+2)
= 5-2/3-2
= (5/3)-2
(iv) 2-1 [(5/3)4 + (3/5)-2] ÷ (17/9)
2-1 [(5/3)4 + (3/5)-2] ÷ (17/9) = 2-1 [(54/34) + (3-2/5-2)] ÷ (17/9)
= 2-1 [(54/34) + (52/32)] ÷ (17/9)
= 2-1 [(625/81) + (25/9)] ÷ (17/9)
= ½ [(625 + 225)/81] × (9/17)
= ½ × (850/81) × 9/17
= 25/9
= (5/3)2
= (3/5)-2
(v) (-7)3 × (1/-7)-9 ÷ (-7)10
(-7)3 × (1/-7)-9 ÷ (-7)10 = (-7)3 × (-7)9 ÷ (-7)10
= (-7)3+9-10
= (-7)2
= (1/-7)-2
Question 8. Simplify:
(i) (49 × z-3) / (7-3 × 10 × z-5) (z ≠ 0)
(ii) (93 × 27 × t4) / (32 × 34 × t2)
(iii) [(3-2)2 × (52)-3 × (-t-3)2] / [(3-2)5 × (53)-2 × (t-4)3]
(iv) (2-5 × 15-5 × 500) / (5-6 × 6-5)
Answer:
(i) (49 × z-3) / (7-3 × 10 × z-5) (z ≠ 0)
(49 × z-3) / (7-3 × 10 × z-5) = (72 × z-3) / (7-3 × 10 × z-5)
= (72+3 × z-3+5) / 10
= (75 × z2)/10
(ii) (93 × 27 × t4) / (32 × 34 × t2)
(93 × 27 × t4) / (32 × 34 × t2) = [(32)3 × (3)3 × t4] / (32 × 34 × t2)
= (36 × 33 × t4) / (32 × 34 × t2)
= 36+3-2-4 × t4-2
= 33 × t2
= 27t2
(iii) [(3-2)2 × (52)-3 × (t-3)2] / [(3-2)5 × (53)-2 × (t-4)3]
[(3-2)2 × (52)-3 × (t-3)2] / [(3-2)5 × (53)-2 × (t-4)3] = [3-4 × 5-6 × t-6] / [3-10 × 5-6 × t-12]
= 3-4+10 × 5-6+6 × t-6+12
= 36 × 50 × t6
= 36 × 1 × t6
= 729t6
(iv) (2-5 × 15-5 × 500) / (5-6 × 6-5)
(2-5 × 15-5 × 500) / (5-6 × 6-5) = (2-5 × (3×5)-5 × 22 × 53) / (5-6 × (2×3)-5)
= (2-5 × 3-5 × 5-5 × 22 × 53) / (5-6 × 2-5 × 3-5)
= 2-5+2+5 × 3-5+5 × 5-5+3+6
= 22 × 30 × 54
= 4 × 1 × 625
= 2500
Exponents and Powers Exe-2
ML Aggarwal Class 8 ICSE Maths Solutions
Page-41
Question 9. By what number should (3/-2)-3 be divided to get (2/3)2?
Answer:
The required number is (3/-2)-3 ÷ (2/3)2
(3/-2)-3 ÷ (2/3)2 = (3-3)/(-2)-3 × (32)/(22)
= [3-3 × 32]/[-2-3 × -22]
= [3-3+2]/(-2)-3+2
= 3-1/-2-1
= -21/31
= -2/3
Question 10. Find the value of m for which 9m ÷ 3-2 = 94.
Answer:
9m ÷ 3-2 = 94
(32)m ÷ 3-2 = (32)4
32m ÷ 3-2 = 38
32m+2 = 38
Now by comparing the powers,
2m + 2 = 8
2m = 8 – 2
= 6
m = 6/2
= 3
Question 11. If (-5/7)-4 × (-5/7)12 = {(-5/7)3} × (-5/7)-1, find the value of x.
Answer:
(-5/7)-4 × (-5/7)12 = {(-5/7)3} × (-5/7)-1
(-5/7)-4 × (-5/7)12 = (-5/7)3x × (-5/7)-1
(-5/7)-4+12 = (-5/7)3x-1
(-5/7)8 = (-5/7)3x-1
Now by comparing the powers, we get
8 = 3x – 1
3x = 8 + 1
3x = 9
x = 9/3
= 3
Question 12. Find x, if (-2/3)-13 × (3/-2)8 = (-2/3)-2x+1
Answer:
(-2/3)-13 × (3/-2)8 = (-2/3)-2x+1
(-2/3)-13 × (-2/3)-8 = (-2/3)-2x+1
(-2/3)-13-8 = (-2/3)-2x+1
(-2/3)-21 = (-2/3)-2x+1
Now by comparing the powers,
-21 = -2x +1
2x = 1 + 21
2x = 22
x = 22/2
= 11
(ML Aggarwal Exponents and Powers Exe-2 Class 8)
Question 13.
(i) If 52x-1 = 1/(125)x-3, find x.
(ii) If (9n × 35 × 273)/(3×814) = 27, find n.
Answer:
(i) If 52x-1 = 1/(125)x-3, find x.
52x-1 = 1/(53)x-3
52x-1 = 1/53x-9
52x-1 = 5-3x+9
Now by comparing the powers,
2x – 1 = -3x + 9
2x + 3x = 9 + 1
5x = 10
x = 10/5
= 2
(ii) If (9n × 35 × 273)/(3×814) = 27
[(32)n × 35 × (33)3]/(3×(34)4) = 33 [32n × 35 × 39] = (3×316) = 33
32n+5+9-1-16 = 33
Now by comparing the powers,
2n + 5 + 9 – 1 – 16 = 3
2n – 3 = 3
2n = 3 + 3
2n = 6
n = 6/2
= 3
— : End of ML Aggarwal Exponents and Powers Exe-2 Class 8 ICSE Maths Solutions :–
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