GSEB Solutions for Class 7 Mathematics – Power and Exponent (English Medium)
GSEB SolutionsMathsScience
Exercise
Solution 1:
Solution 2:
Solution 3(1):
[(35)3 ÷ (32)2] ÷ 38
= [35×3 ÷ 32×2]÷ 38
= [315 ÷ 34]÷38
= [315-4]÷ 38
= 311-8
= 33
= 27
Solution 3(2):
Solution 3(3):
Solution 3(4):
Solution 3(5):
Solution 3(6):
Solution 4(1):
Solution 4(2):
Solution 4(3):
Solution 4(4):
Solution 4(5):
Solution 5:
Solution 6:
Solution 7:
Practice – 1
Solution 1:
Solution 2(1):
Solution 2(2):
Solution 2(3):
Solution 2(4):
Solution 2(5):
Solution 3(1):
Solution 3(2):
Solution 3(3):
Solution 3(4):
y12 ÷ (y6 × y3 × y)
= y12 ÷ (y6+3+1)
= y12 ÷ (y10)
= y12-10
= y2
Solution 3(5):
x10 ÷ (x2 × x3 ÷ x)
= x10 ÷ (x2+3 ÷ x)
= x10 ÷ (x5 ÷ x)
= x10 ÷ ( x5-1)
= x10 ÷ x4
= x10-4
= x6
Solution 3(6):
(y6 ÷ y4) × (y2 × y3)
= (y6-4) × (y2+3)
= y2 × y5
= y2+5
= y7
Solution 4:
x × x2 × x3
= x1+2+3
= x6
Substituting x = 2, we have
x6 = 26
∴ x6 = 64
∴ x× x2 × x3 = 64
x8 ÷ x7 × x2
= x8-7 × x2
= x1 × x2
= x1+2
= x3
Substituting x = 2, we have
x3 = 23 = 8
∴ x8 ÷ x7 × x2 = 8
Solution 5:
x15 ÷ (x17 ÷ x6)
= x15 ÷ (x17-6)
= x15 ÷ (x11)
= x15 – 11
= x4
Substituting x = (-3), we have
x4 = (-3)4 = (81)
(When a negative number is raised to an even power, the result is positive)
∴ x15 ÷ (x17 ÷ x6) = 81
(x6 × x4) ÷ (x2 × x3)
= (x6+4) ÷ (x2+3)
= (x10) ÷ (x5)
= x10 – 5
= x5
Substituting x = (-3), we have
x5 = (-3)5 = (-243)
(When a negative number is raised to an odd power, the result is negative)
∴ (x6 × x4) ÷ (x2 × x3) = (-243)
Practice – 2
Solution 1:
Solution 2(1):
Solution 2(2):
26 × 22 × 25 ÷ 83
= 26+2+5 ÷ (23)3
= 213 ÷ 29
= 213 – 9
= 24
= 16
Solution 2(3):
Solution 2(4):
Solution 2(5):