Exercise 3 (A)
Question1
Find the square of the following numbers :
(i) 15
(ii) 48
(iii) \( \dfrac{6}{7} \)
(iv) \( \dfrac{21}{25} \)
(v) \( 6\dfrac{3}{8} \)
(vi) 0.9
(vii) 1.1
(viii) 0.018
Sol :
(i)
=15×15
=225
(ii)
=48×48
=2304
(iii)
$=\frac{6}{7} \times \frac{6}{7}$
$=\frac{36}{49}$
(iv)
$=\frac{21}{25} \times \frac{21}{25}$
$=\frac{441}{625}$
(v)
$=\frac{48+3}{8}$
$=\frac{51}{8} \times \frac{51}{8}$
$=\frac{2601}{64}$
$=40 \frac{41}{64}$
(vi)
=0.9×0.9
=0.81
(vii)
=1.1×1.1
=1.21
(viii)
=0.018×0.018
=0.000324
Question 2
Determine whether square of the following numbers will be even or odd.(Do not find the square)
Note : Square of even number are even
Square of odd number is odd
(i) 529
Sol :
529 is odd and we know square of odd number is odd
(ii) 30976
Sol :
30976 is even and we know square of even number is even
(iii) 893025
Sol :
893025 is odd and we know square of odd number is odd
(iv) 6990736
Sol :
6990736 is even and we know square of even number is even
Question3
Just by looking at the following numbers, give reason why they are not perfect squares.
Note : Any number ending with 2,3,7,8 and having odd number of zeros are not perfect square
(i) 4893
Sol : Not a perfect square
(ii) 65000
Sol : Not a perfect square
(iii) 4422
Sol : Perfect square
(iv) 89138
Sol : Perfect square
(v) 150087
Sol : Not a perfect square
Question 4
Using prime factorization method, find which of the following are perfect square numbers.
Note : For a number to be a perfect square it's prime factors should be to a power of 2 or any even number.
(i) 100
Sol :
$\begin{array}{l|l}
2&100 \\ \hline
2&50 \\ \hline
5&25 \\ \hline
5&5 \\ \hline
&1
\end{array}$
100=2×2×5×5 =22×52
Since the powers of the prime factors are even therefore 100 is a perfect square.
(ii) 284
Sol :
$\begin{array}{l|l}2&284\\ \hline 2&142 \\ \hline 71& 71 \\ \hline & 1\end{array}$
281=2×2×71 =22×711
Since the powers of the prime factors are not even (i.e. 711) therefore 284 is not a perfect square.
(iii) 784
Sol :
$\begin{array}{l|l}2&784 \\ \hline2&392 \\ \hline
2&196 \\ \hline
2&98 \\ \hline
7&49 \\ \hline
7&7 \\ \hline
&1\end{array}$
784=2×2×2×2×7×7 =24×72
Since the powers of the prime factors are even therefore 784 is a perfect square.
(iv) 1444
Sol :
$\begin{array}{l|l}2 & 1444 \\
\hline 2 & 722 \\
\hline 19 & 361 \\
\hline 19 & 19 \\
\hline & 1 \\
\end{array}$
1444=2×2×19×19 =22×192
Since the powers of the prime factors are even therefore 1444 is a perfect square.
(v) 768
Sol :
$\begin{array}{l|l}2 & 768 \\
\hline 2 & 384 \\
\hline 2 & 192 \\
\hline 2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1 \\
\end{array}$
768=2×2×2×2×2×2×2×2×3 =28×31
Since the powers of the prime factors are not even therefore 768 is not a perfect square.
(vi) 4225
Sol :
$\begin{array}{l|l}5 & 4225 \\
\hline 5 & 845 \\
\hline 13 & 169 \\
\hline 13 & 13 \\
\hline & 1 \\
\end{array}$
4225=5×5×13×13 =52×132
Since the powers of the prime factors are even therefore 4225 is a perfect square.
(vii) 3375
Sol :
$\begin{array}{l|l}3 & 3375 \\
\hline 3 & 1125 \\
\hline 3 & 375 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1 \\
\end{array}$
3375=3×3×3×5×5×5 =33×53
Since the powers of the prime factors are not even therefore 3375 is not a perfect square.
(viii) 15625
Sol :
$\begin{array}{l|l}5 & 15625 \\
\hline 5 & 3125 \\
\hline 5 & 625 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1 \\
\end{array}$
15625=5×5×5×5×5×5 =56
Since the powers of the prime factors are even therefore 15625 is a perfect square.
Question 5
Using prime factorization method, find the square root of the following numbers.
(i) 256
Sol :
$\begin{array}{l|l}
2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}$
√15625=√2×2×2×2×2×2×2×2 =24=16
(ii) 1936
Sol :
$\begin{array}{l|l}2 & 1936 \\
\hline 2 & 968 \\
\hline 2 & 484 \\
\hline 2 & 242 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1 \\
\end{array}$
√1936=√2×2×2×2×11×11=2×2×11=44
(iii) 3364
Sol :
$\begin{array}{l|l}2 & 3364 \\
\hline 2 & 1682 \\
\hline 29 & 841 \\
\hline 29 & 29 \\
\hline & 1 \\
\end{array}$
√3364=√2×2×29×29=2×29=58
(iv) 5625
Sol :
$\begin{array}{l|l}3 & 5625 \\
\hline 3 & 1875 \\
\hline 5 & 625 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1 \\
\end{array}$
√5625=√3×3×5×5×5×5=3×5×5=75
(v) 7744
Sol :
$\begin{array}{l|l}2 & 7744 \\
\hline 2 & 3872 \\
\hline 2 & 1936 \\
\hline 2 & 968 \\
\hline 2 & 484 \\
\hline 2 & 242 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1 \\
\end{array}$
√7744=√2×2×2×2×2×2×11×11=2×2×2×11=88
(vi) 12544
Sol :
$\begin{array}{l|l}2 & 12544 \\
\hline 2 & 6272 \\
\hline 2 & 3136 \\
\hline 2 & 1568 \\
\hline 2 & 784 \\
\hline 2 & 392 \\
\hline 2 & 196 \\
\hline 2 & 98 \\
\hline 7 & 49 \\ \hline 7&7\\ \hline &1
\end{array}$
√12544=√2×2×2×2×2×2×2×2×7×7=2×2×2×2×7=112
Question 6
Find the square root of the following fractions.
(i) \( \dfrac{529}{1225} \)
(ii) \( \dfrac{1444}{3249} \)
(iii) \( 1\dfrac{396}{9604} \)
(iv) 0.01
(v) 0.1764
(vi) 0.00003136
Sol :
(i)
$=\sqrt{\frac{23 \times 23}{5^{2} \times 7^{2}}}$
$=\frac{23}{5 \times 7}$
$=\frac{23}{35}$
(ii)
$=\sqrt{\frac{2^{2} \times 19^{2}}{3^{2} \times 19^{2}}}$
$=\frac{2 \times 19}{3 \times 19}$
$=\frac{2}{3}$
(iii)
$=\sqrt{\frac{9604 \times 1+396}{9604}}$
$=\sqrt{\frac{10000}{9604}}$
$=\sqrt{\frac{2^{4} \times 5^{4}}{2^{2} \times 7^{4}}}$
$=\frac{2 \times 2 \times 5 \times 5}{2 \times 7 \times 1}$
(iv)
$=\sqrt{\frac{0.01}{1 \times 100}}$
$=\frac{1}{100}$
$\frac{1}{\sqrt{2^{2} \times 5^{2}}}$
$=\frac{1}{2 \times 5}$
$=\frac{1}{10}$
(v)
$=\sqrt{\frac{0.1764 \times 10000}{10000}}$
$=\sqrt{\frac{1764}{10000}}$
$=\sqrt{ \frac{2^{2} \times 3^{2} \times 7^{2}}{2^{4} \times 5^{4}}}$
$=\frac{2 \times 3 \times 7}{2 \times 2 \times 5 \times 5}$
$=\frac{21}{50}$
=0.42
(vi)
$=\sqrt{\frac{0.00003136 \times 100000000}{100000000}}$
$=\sqrt{\frac{3136}{100000000}}$
$=\sqrt{\frac{2^{6} \times 7^{2}}{2^{8} \times 5^{8}}}$
$=\frac{2^{3} \times 7}{2^{4} \times 5^{4}}$
$=\frac{7}{2 \times 5^{4}}$
$=\frac{7}{1250}$
=0.0056
Question 7
Simplify:
(i) \( \sqrt{10^2-{6^2}} \)
(ii) \( \Bigg(-\sqrt{\dfrac{9}{16}}\Bigg)\Bigg(\sqrt{\dfrac{64}{81}}\Bigg) \)
(iii) \( \sqrt{\dfrac{16}{36}-\dfrac{1}{4}} \)
Sol :
(i)
=√100-36
=√64
=√8×8
=8
(ii)
$=\left(-\sqrt{\frac{3 \times 3}{4 \times 4}}\right)\left(\sqrt{\frac{8 \times 8}{9 \times 9}}\right)$
$=-\frac{3}{4} \times \frac{8}{9}$
$=-\frac{2}{3}$
(iii)
L.C.M of 36 and 4 is 36
$=\sqrt{\frac{16 \times 1+1 \times 9}{36}}$
$=\sqrt{\frac{16+9}{36}}=\sqrt{\frac{25}{36}}$
$=\sqrt{\frac{5 \times 5}{6 \times 6}}=\frac{5}{6}$
Question 8
Find the square root of 1024. Hence find the value of \( \sqrt{10.24} + \sqrt{0.1024}+\sqrt{10240000} \).
Sol :
$\begin{array}{l|l}2&1024\\ \hline 2&512 \\ \hline 2& 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2& 32 \\ \hline 2& 16 \\ \hline 2& 8 \\ \hline 2 & 4\\ \hline 2& 2 \\ \hline &1 \end{array}$
=√210
=25
=32
=√10.24+=√0.1024+=√10240000
$=\sqrt{\frac{1024}{100}}+\sqrt{\frac{1024}{10000}}+\sqrt{1024 \times 10000}$
$=\frac{32}{10}+\frac{32}{100}+32 \times 100$
=3.2+0.32+3200
=3203.52
Question 9
Find the smallest number by which each of the given number s ,ust be multiplied so that the product is a perfect square. Also find the square root of this new number.
(i) 175
Sol :
$\begin{array}{l|l}5&175 \\ \hline 5& 35 \\ \hline 7& 7 \\ \hline & 1\end{array}$
Resolving prime factors
175=5×5×7
In the above product 5 occur in pairs whereas 7 does not exist in pair.So we have to multiply 7 to 175 to make it a perfect square
Perfect square number=1225
Square root=√1225
=√5×5×7×7
=5×7
=35
(ii) 325
Sol :
$\begin{array}{l|l}5&325\\ \hline 5&65 \\ \hline 13& 13\\ \hline &1\end{array}$
Resolving prime factors
325=5×5×13
In the above product 5 occur in pairs whereas 13 does not exist in pair.So we have to multiply 325 by 13 to make it a perfect square
Perfect square number=4225
Square root=√4225
=√5×5×13×13
=5×13
=65
(iii) 720
Sol :
$\begin{array}{l|l}2 & 720 \\
\hline 2 & 360 \\
\hline 2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline \\
\end{array}$
Resolving Prime factors
720= (2×2)×(2×2)×(3×3)×5
Above ,product of 2 and 3 occur in pairs whereas 5 does not exist in pair so we have to multiply 720 by 5 to make it a perfect square
Perfect square number=3600
Square root=√(2×2)×(2×2)×(3×3)×(5×5)
=2×2×3×5 =60
(iv) 3150
Sol :
$\begin{array}{l|l}
2 & 3150 \\
\hline 3 & 1575 \\
\hline 3 & 525 \\
\hline 5 & 175 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1 \\
\end{array}$
Resolving Prime factors
3150= 2×(3×3)×(5×5)×7
Above ,product of 3 and 5 occur in pairs whereas 2 and 7 does not exist in pair so we have to multiply 3150 by 2 and 7 (i.e. 14) to make it a perfect square
Perfect square number=44100
Square root=√(2×2)×(3×3)×(5×5)×(7×7)
=2×3×5×7 =210
Question 10
Find the smallest number by which each of the given numbers must be divided so that the quotient is a perfect square.Also find the square root of this quotient.
(i) 2700
Sol :
$\begin{array}{l|l}
2 & 2700 \\
\hline 2 & 1350 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline \\
\end{array}$
Resolving Prime factors
2700= (2×2)×(3×3)×3×(5×5)
Above ,product of 2 , 3 and 5 occur in pairs but their is a single 3 extra so we have to divide 2700 by 3 to make it a perfect square
Perfect square number=900
Square root=√(2×2)×(3×3)×(5×5)
=2×3×5 =30
(ii) 5488
Sol :
$\begin{array}{l|l}
2 & 5488 \\
\hline 2 & 2744 \\
\hline 2 & 1372 \\
\hline 2 & 686 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1 \\
\end{array}$
Resolving Prime factors
5488= (2×2)×(2×2)×(7×7)×7
Above ,product of 2 and 7 occur in pairs but their is a single 7 extra so we have to divide 5488 by 7 to make it a perfect square
Perfect square number=784
Square root=√(2×2)×(2×2)×(7×7)
=2×2×7 =28
(iii) 7203
Sol :
$\begin{array}{l|l}
3 & 7203 \\
\hline 7 & 2401 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\ \hline & 1
\end{array}$
Resolving Prime factors
7203= 3×(7×7)×(7×7)
Above ,product of 7 occur in pairs but their is a single 3 extra so we have to divide 7203 by 3 to make it a perfect square
Perfect square number=2401
Square root=√(7×7)×(7×7)
=7×7 =49
(iv) 20886
Sol :
$\begin{array}{l|l}
2 & 20886 \\
\hline 3 & 10443 \\
\hline 59 & 3481 \\
\hline 59 & 59 \\
\hline & 1 \\
\end{array}$
Resolving Prime factors
7203= 3×(7×7)×(7×7)
Above ,product of 7 occur in pairs but their is a single 3 extra so we have to divide 7203 by 3 to make it a perfect square
Perfect square number=2401
Square root=√(7×7)×(7×7)
=7×7 =49
Question11
Find the smallest number which is completely divisible by each of the numbers 10 , 16 and 24.
Sol :
Here, we need to find the perfect number divisible by square of L.C.M of 10 , 16 and 24
L.C.M of 10 ,16 , 24 is =2×2×2×2×3×5 =240
$\begin{array}{l|l}2 & 10,16,24 \\
\hline 2 & 5,8,12 \\
\hline 2 & 5,4,6 \\
\hline 2 & 5,2,3 \\
\hline 3 & 5,1,3 \\
\hline 5 & 5,1,1 \\
\hline &1,1,1
\end{array}$
Resolving Prime factors
240= (2×2)×(2×2)×3×5
To make it into a perfect square is should be multiplied by 3×5=15
∴Required square number =240×15=3600
Question12
The children of class VIII of a school contributed ₹ 3025 for the Prime Minister's National Relief Fund, contributing as much money as there are children in the class.How many children are there in the class ?
Sol :
Let the number of children be x
Then, according to question,
Money donated by individual = number of
children = x
Now,
Total money donated = money donated by individual × number of
children
⇒x×x=3025
⇒x2=3025
⇒x=√3025
⇒x=√5×5×11×11
⇒x=5×11=55
Hence there are 55 children in the class and all of them donated 55 ₹ each
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