Maharashtra Board Class 7 Math Solution Chapter 6 – Indices
Balbharati Maharashtra Board Class 7 Math Solution Chapter 6: Indices. Marathi or English Medium Students of Class 7 get here Indices full Exercise Solution.
Std |
Maharashtra Class 7 |
Subject |
Math Solution |
Chapter |
Indices |
Practice Set 26
1.) Complete the table below.
ANS:
I)163
Here,
Base = 16
Index = 3
Multiplication form = 16 x 16 x 16
Value = 4096
Ii(-8)2
Here,
Base = -8
Index = 2
Multiplication form = -8 x -8
When base contain negative sign and index is even number then answer we get is always positive.
Value = 64
iii) (3/7)4
Here,
Base = 3/7
Index = 4
Multiplication form = 3/7 x 3/7 x 3/7 x 3/7
Value = 81/2401
iv)(-13)4
Here,
Base = -13
Index = 4
Multiplication form = -13 x -13 x -13 x -13
When base contain negative sign and index is even number then answer we get is always positive.
Value = 28,561
2.) Find the value.
(i) 210
ANS:
Here,
Base = 2
Index = 10
We multiply 2 by 10 times.
Multiplication form = 2 x 2 x 2 x 2 x2 x 2 x 2 x 2 x 2 x 2
Value of 210 = 1024
(ii) 53
ANS:
Here,
Base = 5
Index = 3
We multiply 5 by 3 times.
Multiplication form = 5 x 5 x 5
Value of 53 = 125
(iii) (-7)4
ANS:
Here,
Base = -7
Index = 4
We multiply -7 by 4 times.
Multiplication form = -7 x -7 x -7 x -7
When base contain negative sign and index is even number then answer we get is always positive.
Value of (-7)4 = 2401
(iv) (-6)3
ANS:
Here,
Base = -6
Index = 3
We multiply -6 by 3 times.
Multiplication form = -6 x -6 x -6
When base contain negative sign and index is odd number then answer we get is always negative.
Value of (-6)3 = -216
(v) 93
ANS:
Here,
Base = 9
Index = 3
We multiply 9 by 3 times.
Multiplication form = 9 x 9 x 9
Value of 93 = 729
(vi) 81
ANS:
Here,
Base = 8
Index = 1
We multiply 8 by 1 times. i.e. 8 itself.
Value of 81 = 8
(vii) (4/ 5)3
ANS:
Here,
Base = 4/5
Index = 3
We multiply 4/5 by 3 times.
Multiplication form = 4/5 x 4/5 x 4/5
Value of (4/5)3 = 64/125
(viii) (– 1/ 2) 4
ANS:
Here,
Base = -1/2
Index = 4
We multiply -1/2 by 4 times.
Multiplication form = -1/2 x -1/2 x -1/2 x -1/2
When base contain negative sign and index is even number then answer we get is always positive.
Value of (-1/2)4 = 1/16
Practice Set 27
(1) Simplify.
(i) 74 × 72
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them multiplication operation then we add their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am × an = am+n
Here,
Base = 7
Index of indices = 4,2
74 × 72 =7 4 + 2
= 76
Multiplication form = 7 x 7 x 7 x 7 x 7 x 7
Value of indices = 1, 17,649
(ii) (-11)5 × (-11)2
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them multiplication operation then we add their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am × an = am+n
Here,
Base = -11
Index of indices = 5,2
-115 × -112 =-11 5 + 2
= -117
Multiplication form = -11 x -11 x -11 x -11 x -11 x -11 x -11
When base contain negative sign and index is odd number then answer we get is always negative.
Value of indices = – 1, 94, 87,171
(iii) (6/ 7)3 x (6 /7)5
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them multiplication operation then we add their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am × an = am+n
Here,
Base = 6/7
Index of indices = 3,5
(6/5)3 × (6/5)5 =(6/5)3 + 5
= (6/5)8
Multiplication form = (6/5) x (6/5) x (6/5) x (6/5) x (6/5) x (6/5) x (6/5) x (6/5)
Value of indices =1679616 /390625
(iv) (− 3/2)5× (− 3/2)3
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them multiplication operation then we add their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am × an = am+n
Here,
Base = – 3/2
Index of indices = 5,3
(- 3/2)5 × ((- 3/2)3 =(- 3/2)5 + 3
= (- 3/2)8
Multiplication form = (- 3/2) x (- 3/2)x(- 3/2)x(- 3/2) x (- 3/2)x (- 3/2)x (- 3/2)x (- 3/2)
When base contain negative sign and index is even number then answer we get is always positive.
Value of indices = 6561/256
(v) a16 × a7
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them multiplication operation then we add their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am × an = am+n
Here,
Base = a
Index of indices = 16,7
a16 × a7 =a 16 + 7
= a23
(vi) (P/ 5) 3 × (P/ 5)7
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them multiplication operation then we add their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am × an = am+n
Here,
Base = p/5
Index of indices = 3,7
(p/5)3 × p/5)7 =(p/5) 3 + 7
= (p/5)10
Practice Set 28
1.) Simplify.
(i) a6 ÷ a4
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = a
Index of indices = 6,4
a6 / a4 =a 6 – 4
a6 ÷ a4= a2
(ii) m5 ÷ m8
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = m
Index of indices = 5,8
m5 / m8 =m 5 – 8
m5 ÷ m8= m(-3)
(iii) p3 ÷ p13
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = p
Index of indices = 3,13
P3 / p13 =p 3 – 13
P3 ÷ p13= p(-10)
(iv) x10 ÷ x10
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = x
Index of indices = 10,10
X10 / x10 = x 10 – 10
X10 ÷ x10= x0
Any index having power 0 then their value is always 1.
= 1
2.) Find the value.
(i) (-7)12 ÷ (-7)12
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = -7
Index of indices = 12,12
(-7)12 / (-7)12 = (-7) 12 – 12
(-7)12 ÷ (-7)12= (-7)0
Any index having power 0 then their value is always 1.
= 1
(ii) 75 ÷ 73
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = 7
Index of indices = 5,3
(7)5 / (7)3 = (7) 5 – 3
(7)5 ÷ (7)3= (7)2
= 49
(iii) (4/ 5)3 ÷( 4 /5)2
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = (4/ 5)
Index of indices = 3,2
(4/ 5)3 / (4/ 5)2 = (4/ 5)3 – 2
(4/ 5)3 ÷ (4/ 5)2 = (4/ 5)1
Any indices having power 1 have value itself.
= 4/5
(iv) 47÷ 45
ANS:
We know the properties of indices,
When the base of indices are same and their index are different and in between them division operation then we subtract their indexes.
Let,
a = Base of indices.
m,n = Index of indices
am / an = am-n
Here,
Base = 4
Index of indices = 7,5
(4)7 / (4)5 = (4) 7 – 5
(4)7 ÷ (4)5= (4)2
= 16
Practice Set 29
1.) Simplify.
(i) [(15 /12)3] 4
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base =(15 /12)
Index = 3,4
[(15 /12)3] 4 = (15 /12)3x 4
[(15 /12)3] 4 = (15 /12)12
(ii) [(3)4] 2
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = 3
Index = 4, 2
[(3)4] 2 = (3)4x 2
[(3)4] 2 = (3)8
(iii) [(1/ 7) -3] 4
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = (1/ 7)
Index = -3,4
[(1/ 7) -3] 4 = (1/ 7)-3x 4
When there is multiplication between two anyone have negative sign then answer contain negative sign.
[(1/ 7) -3] 4 = (1/ 7)-12
(iv)[(2 /5) -2]– 3
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = (2/ 5)
Index = -2,-3
[(2 /5) -2]– 3= (2/ 5)-2x -3
When there is multiplication between two and both have negative sign then answer contain positive sign.
[(2 /5) -2]– 3 = (2/ 5)6
(v) [(6)5] 4
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = 6
Index = 5, 4
[(6)5] 4 = (6)5x 4
[(6)5] 4 = (6)20
(vi) [(6/7) 5] 2
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base =(6/7)
Index = 5,2
[(6/7) 5] 2 = (6/7)5 x 2
[(6/7) 5] 2 = (6/7)10
(vii) [(2/3) -4]5
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = (2/3)
Index = -4,5
[(2/3) -4] 5 = (2/3)-4x 5
When there is multiplication between two anyone have negative sign then answer contain negative sign.
[(2/3) -4] 5 = (2/ 3)-20
(viii) [(5 /8) 3] -2
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = (5/8)
Index = 3,-2
[(5/8) 3]-2 = (5/8)3x -2
When there is multiplication between two anyone have negative sign then answer contain negative sign.
[(5/8) 3]-2 = (5/8)-6
(ix) [(3/4) 6] 1
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = (3/4)
Index = 6,1
[(3/4) 6]1 = (3/4)6 x1
[(3/4) 6]1 = (3/4)6
(x) [(2/5) -3] 2
ANS:
We know the properties of indices,
When there is common base of indices and having more than 1 index or power then we have to multiply all the powers taking common base.
Let,
Here
Base = a
Index = m,n
(am)n = am × n = amn
Base = (2/5)
Index = -3,2
[(2/5) -3] 2 = (2/5)-3x 2
When there is multiplication between two anyone have negative sign then answer contain negative sign.
[(2/5) -3] 2 = (2/ 5)-6
- Write the following numbers using positive indices.
(i) [2/7] -2
ANS:
Here we have write numbers using positive indices.
We have to change sign of index (power) of indices.
[2/7] -2 positive integer of that is [2/7] 2
(ii) [11/ 3] -5
ANS:
Here we have write numbers using positive indices.
We have to change sign of index (power) of indices.
[11/3] -5 positive integer of that is [11/3] 5
(iii) [1/ 6] -3
ANS:
Here we have write numbers using positive indices.
We have to change sign of index (power) of indices.
[1/6] -3 positive integer of that is [1/6] 3
(iv) (y)-4
ANS:
Here we have write numbers using positive indices.
We have to change sign of index (power) of indices.
(y)-4positive integer of that is (y)4
Practice Set 30
¤ Find the square root.
(i) 625
ANS:
Here we have to find square root to find square root we are use factors method.
Here we make pair of two common factors and taking one from them and multiply them.
625 = 5 x 5
Square root of 625 is 25.
(ii) 1225
ANS:
Here we have to find square root to find square root we are use factors method.
Here we make pair of two common factors and taking one from them and multiply them.
1225 = 5 x 7
Square root of 1225 is 35.
(iii) 289
ANS:
Here we have to find square root to find square root we are use factors method.
Here we make pair of two common factors and taking one from them and multiply them.
289 = 17 x 17
Square root of 289 is 17.
(iv) 4096
ANS:
Here we have to find square root to find square root we are use factors method.
Here we make pair of two common factors and taking one from them and multiply them.
4096 = 4 x 4 x 4
Square root of 4096 is 64.
(v) 1089
ANS:
Here we have to find square root to find square root we are use factors method.
Here we make pair of two common factors and taking one from them and multiply them.
1089 = 3x 11
Square root of 1089 is 33.