In Text Questions (Text Book Page No. 27,30,32,33,34,37,41,42,44) - Chapter 1 Numbers 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Think (Text Book Page No. 27)
Consider the claim: "Between the squares of the consecutive numbers $n$ and $(n+1)$, there are $2 n$ non-square numbers' Can it be true? FInd how many non-square numbers are there
(i) between 4 and 9 ?
(ii) between 49 and 64 ? and Verify the claim.
Answer:

Therefore we conclude that there are $2 \mathrm{n}$ non-square numbers between two consecutive square numbers.
 

Think (Text Book Page No. 30)
In this case, if we want to find the smallest factor with which we can multiply or divide 108 to get a square number, what should we do?
Answer:
$108=2 \times 2 \times 3 \times 3=2^{2} \times 3^{2} \times 3$
If we multiply the factors by 2 , then we get
$2^{2} \times 3^{2} \times 3 \times 3=2^{2} \times 3^{2} \times 3^{2}=(2 \times 3 \times 3)^{2}$
Which is perfect square.
$\therefore$ Again if we divide by 3 then we get $2^{2} \times 3^{2} \Rightarrow(2 \times 3)^{2}$, a perfect square.
$\therefore$ We have to multiply or divide 108 by 3 to get a perfect square.

Try These (Text Book Page No. 32)
Find the square root by long division method.
Question 1.
400
Answer:

$\sqrt{400}=20$

Question 2.
$1764$
Answer:

Question $3 .$
9801
Answer:

Try These (Text Book Page No. 32)
without calculating the square root, guess the number of digits in the square root of the following numbers: Question 1.
14400
Answer:
$\begin{aligned}
&\sqrt{14400}=\sqrt{144 \times 100} \\
&=\sqrt{144} \times \sqrt{100} \\
&=12 \times 10 \\
&=120
\end{aligned}$

Question $2 .$
390625
Answer:
$\begin{aligned}
&\sqrt{390625}=\sqrt{25 \times 25 \times 25 \times 25} \\
&=\sqrt{25 \times 25} \times \sqrt{25 \times 25} \\
&=25 \times 25 \\
&=625
\end{aligned}$

Question $3 .$
100000000
Answer:
$\begin{aligned}
&\sqrt{100000000}=\sqrt{10000 \times 10000} \\
&=\sqrt{10000} \times \sqrt{10000} \\
&=100 \times 100 \\
&=10,000
\end{aligned}$

Try These (Text Book Page No. 33)
Find the square root of
Question $1 .$

$5.4756$
Answer:

Question $2 .$
$19.36$
Answer:

Question $3 .$
$116.64$
Answer:

Think (Text Book Page No. 33)
Try to fill in the blanks using $\sqrt{a b}=\sqrt{a} \times \sqrt{b}$

Answer:

Try These (Text Book Page No. 34)
Using this method, find the square root of the numbers $1.2321$ and $11.9025$.
Answer:
(i) $1.2321$
$\sqrt{1.2321}=\sqrt{\frac{12321}{10000}}$
$=\frac{111}{100}=1.11$
(ii) $11.9025$
$\sqrt{11.9025}=\frac{\sqrt{119025}}{\sqrt{10000}}$
$=\frac{345}{100}=3.45$
 

Try These (Text Book Page No. 34)
Write the numbers in ascending order.
Question $1 .$
$4, \sqrt{14}, 5$
Answer:
Squaring all the numbers we get $4^{2},(\sqrt{14})^{2}, 5^{2} \Rightarrow 16,14,25$
$\therefore$ Ascending order: $14,16,25$
Ascending order: $\sqrt{14}, 4,5$
 

Question $2 .$
$7, \sqrt{65}, 8$
Answer:
Squaring $7, \sqrt{65}$ and 8 we get $7^{2},(\sqrt{65})^{2}, 8^{2} \Rightarrow 49,65,64$
Ascending order : $49,64,65$
Ascending order: $7,8, \sqrt{65}$

Try These (Text Book Page No. 37)
Find the ones digit in the cubes of each of the following numbers.
(i) 12
(ii) 27
(iii) 38
(iv) 53
(v) 71
(vi) 84
Answer:
(i) 12
12 ends with 2 , so its cube ends with 8 i.e. ones digit in $12^{3}$ is 8 .
(ii) 27
27 ends with 7 , so its cube end with 3. i.e., ones digit in $27^{3}$ is 3 .
(iii) 38
38 ends with 8 , so its cube ends with 2 i.e. ones digit in $38^{3}$ is 2 .
(iv) 53
53 ends with 3 , so its cube ends with 7. i.e, ones digit in $53^{3}$ is 7 .
(v) 71
71 ends with 1 , so its cube ends with 1 . i.e. ones digit in $71^{3}$ is 1

(vi) 84
84 ends with 4 , so its cube ends with 4 . i.e, ones digit in $84^{3}$ is 4 .
 

Try These (Text Book Page No. 41)
Expand the following numbers using exponents:
(i) 8120
(ii) 20,305
(iii) $3652.01$
(iv) $9426.521$
Answer:
(i) 8120
$8120=(8 \times 1000)+(1 \times 100)+(2 \times \times 10)+0 \times 1$
$=\left(8 \times 10^{3}\right)+\left(1 \times 10^{2}\right)+\left(2 \times 10^{1}\right)$
(ii) 20,305
$20305=(2 \times 10000)+(0 \times 1000)+(3 \times 100)+(0 \times 10)+(5 \times 1)$
$=\left(2 \times 10^{4}\right)+\left(3 \times 10^{2}\right)+5$
(iii) $3652.01$
$3652.01=3000+600+50+2+\frac{0}{10}+\frac{1}{100}$
$=(3 \times 1000)+(6 \times 100)+(5 \times 10)+(2 \times 1)+\left(1 \times \frac{1}{100}\right)$
$=\left(3 \times 10^{3}\right)+\left(6 \times 10^{2}\right)+\left(5 \times 10^{1}\right)+2+\left(1 \times 10^{-2}\right)$
(iv) $9426.521$
$=(9 \times 1000)+(4 \times 100)+(2 \times 10)+(6 \times 1)+\left(\frac{5}{10}\right)+\left(\frac{2}{100}\right)+\left(\frac{1}{1000}\right)$
$=\left(9 \times 10^{3}\right)+\left(4 \times 10^{2}\right)+\left(2 \times 10^{1}\right)+6+\left(5 \times 10^{-1}\right)+\left(2 \times 10^{-2}\right)+\left(1 \times 10^{-3}\right)$

Try These (Text Book Page No. 42)
Verify the following rules (as we did above). Here, a,b are non-zero integers and are any integers.
1. Product of same powers to power of product rule: $a^{m} \times b^{m}=(a b)^{\mathrm{m}}$
2. Quotient of same powers to power of quotient rule: $\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}$
3. Zero exponent rule: $a^{0}=1$.
Answer:
let $\mathrm{a}=2 ; \mathrm{b}=3 ; \mathrm{m}=2$
$\begin{aligned}
&\text { 1. } a^{\mathrm{m}} \times b^{\mathrm{m}}=2^{2} \times 3^{2} \\
&=4 \times 9=36=(2 \times 3)^{2}
\end{aligned}$
2. $\frac{a^{m}}{b^{m}}=\frac{2^{2}}{3^{2}}=\frac{4}{9}=\left(\frac{2}{3}\right)^{2}$
3. $a^{0}=2^{0}=1$.

Try These (Text Book Page No. 44)
Question $1 .$
Write in standard form: Mass of planet Uranus is $8.68 \times 10^{25} \mathrm{~kg}$.
Answer:
Mass of Planet Uranus $=86800000000000000000000000 \mathrm{~kg}$
[23 zeros after 88 ]
 

Question 2.
Write in scientific notation:
(i) $0.000012005$
Answer:
$0.000012005=1.2005 \times 10^{-5}$
(ii) $4312.345$
Answer:
$4312.345=4.312345 \times 10^{3}$
(iii) $0.10524$
Answer:
$0.10524=1.0524 \times 10^{-1}$
(iv) The distance between the Sun and the planet Saturn $1.4335 \times 10^{12}$ miles.