THEORY OF INDICES
1. am × an = am + n
2. (am)n = amn
am
= am−n
3.
an
4. (ab)m = ambm
5. a0 = 1
6. ap/q = qth root of ap =
7. a1 p = pth root of a
m
amb m
⎛ ab ⎞
8. ⎜
⎟ =
cm
⎝ c ⎠
9. a∞ = ∞
10. a-∞ = 0
q
ap
1.Find the square root of 6561 (Factor method)
3
3
3
3
3
3
3
6561
2187
729
243
81
27
9
3
6561 = (3×3)×(3×3)×(3×3)×(3×3)
=(9×9)×(9×9)
= 81×81
6561 = 81
2. Find the least number with which you multiply 882, so that the product may be
a perfect square.
First find the factors of 882.
882 = 2 × 3 × 3 × 7 × 7
Now, 882 has factors as shown above, ‘3’ repeated twice, ‘7’ repeated twice
and ‘2’ only once. So when one more factor ‘2’ is used, then it becomes a perfect
square.
882 × 2 = (2 × 2) × (3 × 3) × (7 × 7)
The least number required is ‘2’
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2 882
3 441
3 147
7 49
7
3. Find the cube root of 2985984 (Factor method)
2985984 = 23 × 23 × 23 × 23 × 33 × 33
3
2985984 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
2985984
1492992
746496
373248
186624
93312
46656
23328
11664
5832
2916
1458
729
243
81
27
9
3
3
1
2
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
4. Find the value of
4 − 0.04
4 + 0.04
81
=
0.09
5.
4 − 0.04
4 + 0.04
6. Simplify
4 − 0.2 3.8 38 19
=
=
=
= 0.9
4 + 0.2 4.2 42 21
=
81
=
0.09
4
−
3
3
4
9
90
=
= 30
0.3
3
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4
−
3
3
2
3
4−3
1
=
−
=
=
4
2
3
2 3
2 3
8. Find the value of
410
1
16
410
1
=
16
6561
=
16
6561
=
16
1
81
= 20
4
4
6.Find the least number with which you multiply 882. So that the product may be
a perfect square.
First find the
2
3
3
7
factors of 882.
882
441
147
49
7
Now, 882 has factors as shown above.
‘3’ repeated twice, ‘7’ repeated twice and ‘2’ only
once. So, when one more factor ‘2’ is used, then
it becomes a perfect square.
∴882×2=(2×2)×(3×3)×(7×7)
∴The least number required is ‘2’.
7.
Find the cube root of 2985984 (Factor method)
2985984
1492992
746496
373248
186624
93312
46656
23328
11664
5832
2916
1458
729
243
81
27
9
3
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
∴2985984 = 23×23×23×23×33×33
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∴ 3 2985984 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
8.Find the value of
9.
81
=
0.09
10. Simplify
4 − 0.04
4 + 0.04
4 − 0.04
4 + 0.04
81
0.09
4
3
-
3
4
4 − 0.2
3.8
38 19
=
=
=
= 0.9 (approx)
4 + 0.2
4.242 42 21
=
9
90
=
= 30
0.3
3
=
4−3
4
3
2
3
1
-
=
=
=
−
3
4
2
2 3
3
2 3
11. Find the least number with which 1728 may be added so that the
resulting number is a perfect square.
42
4 1728
16
82 128
164
Note:
Take the square root of 1728 by long division method. It comes
to 41.+ something. As shown, if 128 is made 164, we get the
square root as an integer. The difference between 164 and 128
i.e., 36 must be added to 1728, so that 1764 is a perfect
square 1764 =42
Theory of Indices
Problems:
1. A certain number of persons agree to subscribe as many rupees each as there
are subscribers.
subscriber?
Let the number of subscribers be x, say since each subscriber agrees to
subscribe x rupees.
The whole subscription is Rs.2582449.
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The total subscription = no. of persons × subscription per person
= x × x = x2
given x2 = 2582449
∴x = 1607
2. Simplify:
3
192a 3 b 4
Use the 2 formulas
(abc)m
(a )
m n
3
192a3b 4 = 192a3b 4
= amb m c m
= amn
(
( ) (b )
= (2 × 3)
.ab
= (192 ) 3 . a3
1
3
1
4 3
4
3
= 2 2.3 3 .ab 3
= 4a. 3b 4
= 4a. 3b 4
x 9 y 12
3
)
1
3
1
6
1
4
( )
1
3
3
3. Simplify
Sol.
3
x 9 y 12 = x 9 y 12
(
( ) (y )
= x9
= x3y 4
)
1
3
1
3
1
12 3
4. Find the number whose square is equal to the difference between the squares of
75.12 and 60.12
Sol.
Let ‘x’ be the number required
∴x2 = (75.12)2 – (60.12)2
∴x =
= (75.12 + 60.12) (75.12 – 60.12)
= 135.24 ×15 = 2028.60
2028 .60 = 45.0399
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5. Reduce to an equivalent fraction write rational denominator
Sol.
3 5+ 3
5− 3
(3 5 + 3 )( 3 + 3 )
( 5 + 3 )( 5 − 3 )
=
=
=
15 + 3 15 + 15 + 3
5−3
18 + 4 15
= 9 + 2 15
2
7
9
21
×2
×
55 10
44
14
6. Find the value of
Sol.
14
7. An army general trying to draw
many men as true are rows, found that he had 31 men over. Find the number of
men in the front row.
Let ‘a’ be the number of men in the front row.
a2 + 31 = 161610
No. of men in the front row = 127
a2 = 161610 – 31 = 16129
∴a = 127
8. A man plants his orchid with 5625 trees and arranges them so that there are as
many rows as there are trees in a row. How many rows are there?
Sol.
Let ‘x’ be the number of rows and let the number of trees in a row be ‘x’ say
2
x = 5625
∴x = 75
∴There are 75 trees in a row and 75 rows are arranged.
21
7
9
×2
×
=
44
55 10
637 117
9
×
×
= 5.26 approx.
44
55
10
his 16,160 men in rows so that there are as
1. am × an = am + n
2. (am)n = amn
am
= am−n
3.
an
4. (ab)m = ambm
5. a0 = 1
6. ap/q = qth root of ap =
7. a1 p = pth root of a
m
amb m
⎛ ab ⎞
8. ⎜
⎟ =
cm
⎝ c ⎠
9. a∞ = ∞
10. a-∞ = 0
q
ap
1.Find the square root of 6561 (Factor method)
3
3
3
3
3
3
3
6561
2187
729
243
81
27
9
3
6561 = (3×3)×(3×3)×(3×3)×(3×3)
=(9×9)×(9×9)
= 81×81
6561 = 81
2. Find the least number with which you multiply 882, so that the product may be
a perfect square.
First find the factors of 882.
882 = 2 × 3 × 3 × 7 × 7
Now, 882 has factors as shown above, ‘3’ repeated twice, ‘7’ repeated twice
and ‘2’ only once. So when one more factor ‘2’ is used, then it becomes a perfect
square.
882 × 2 = (2 × 2) × (3 × 3) × (7 × 7)
The least number required is ‘2’
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2 882
3 441
3 147
7 49
7
3. Find the cube root of 2985984 (Factor method)
2985984 = 23 × 23 × 23 × 23 × 33 × 33
3
2985984 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
2985984
1492992
746496
373248
186624
93312
46656
23328
11664
5832
2916
1458
729
243
81
27
9
3
3
1
2
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
4. Find the value of
4 − 0.04
4 + 0.04
81
=
0.09
5.
4 − 0.04
4 + 0.04
6. Simplify
4 − 0.2 3.8 38 19
=
=
=
= 0.9
4 + 0.2 4.2 42 21
=
81
=
0.09
4
−
3
3
4
9
90
=
= 30
0.3
3
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4
−
3
3
2
3
4−3
1
=
−
=
=
4
2
3
2 3
2 3
8. Find the value of
410
1
16
410
1
=
16
6561
=
16
6561
=
16
1
81
= 20
4
4
6.Find the least number with which you multiply 882. So that the product may be
a perfect square.
First find the
2
3
3
7
factors of 882.
882
441
147
49
7
Now, 882 has factors as shown above.
‘3’ repeated twice, ‘7’ repeated twice and ‘2’ only
once. So, when one more factor ‘2’ is used, then
it becomes a perfect square.
∴882×2=(2×2)×(3×3)×(7×7)
∴The least number required is ‘2’.
7.
Find the cube root of 2985984 (Factor method)
2985984
1492992
746496
373248
186624
93312
46656
23328
11664
5832
2916
1458
729
243
81
27
9
3
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
∴2985984 = 23×23×23×23×33×33
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∴ 3 2985984 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144
8.Find the value of
9.
81
=
0.09
10. Simplify
4 − 0.04
4 + 0.04
4 − 0.04
4 + 0.04
81
0.09
4
3
-
3
4
4 − 0.2
3.8
38 19
=
=
=
= 0.9 (approx)
4 + 0.2
4.242 42 21
=
9
90
=
= 30
0.3
3
=
4−3
4
3
2
3
1
-
=
=
=
−
3
4
2
2 3
3
2 3
11. Find the least number with which 1728 may be added so that the
resulting number is a perfect square.
42
4 1728
16
82 128
164
Note:
Take the square root of 1728 by long division method. It comes
to 41.+ something. As shown, if 128 is made 164, we get the
square root as an integer. The difference between 164 and 128
i.e., 36 must be added to 1728, so that 1764 is a perfect
square 1764 =42
Theory of Indices
Problems:
1. A certain number of persons agree to subscribe as many rupees each as there
are subscribers.
subscriber?
Let the number of subscribers be x, say since each subscriber agrees to
subscribe x rupees.
The whole subscription is Rs.2582449.
© Entranceguru.com Private Limited
Basic concepts.doc
-9-
www.tnpscquestionpapers.com
Takes you to places where you belong.
The total subscription = no. of persons × subscription per person
= x × x = x2
given x2 = 2582449
∴x = 1607
2. Simplify:
3
192a 3 b 4
Use the 2 formulas
(abc)m
(a )
m n
3
192a3b 4 = 192a3b 4
= amb m c m
= amn
(
( ) (b )
= (2 × 3)
.ab
= (192 ) 3 . a3
1
3
1
4 3
4
3
= 2 2.3 3 .ab 3
= 4a. 3b 4
= 4a. 3b 4
x 9 y 12
3
)
1
3
1
6
1
4
( )
1
3
3
3. Simplify
Sol.
3
x 9 y 12 = x 9 y 12
(
( ) (y )
= x9
= x3y 4
)
1
3
1
3
1
12 3
4. Find the number whose square is equal to the difference between the squares of
75.12 and 60.12
Sol.
Let ‘x’ be the number required
∴x2 = (75.12)2 – (60.12)2
∴x =
= (75.12 + 60.12) (75.12 – 60.12)
= 135.24 ×15 = 2028.60
2028 .60 = 45.0399
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5. Reduce to an equivalent fraction write rational denominator
Sol.
3 5+ 3
5− 3
(3 5 + 3 )( 3 + 3 )
( 5 + 3 )( 5 − 3 )
=
=
=
15 + 3 15 + 15 + 3
5−3
18 + 4 15
= 9 + 2 15
2
7
9
21
×2
×
55 10
44
14
6. Find the value of
Sol.
14
7. An army general trying to draw
many men as true are rows, found that he had 31 men over. Find the number of
men in the front row.
Let ‘a’ be the number of men in the front row.
a2 + 31 = 161610
No. of men in the front row = 127
a2 = 161610 – 31 = 16129
∴a = 127
8. A man plants his orchid with 5625 trees and arranges them so that there are as
many rows as there are trees in a row. How many rows are there?
Sol.
Let ‘x’ be the number of rows and let the number of trees in a row be ‘x’ say
2
x = 5625
∴x = 75
∴There are 75 trees in a row and 75 rows are arranged.
21
7
9
×2
×
=
44
55 10
637 117
9
×
×
= 5.26 approx.
44
55
10
his 16,160 men in rows so that there are as
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