Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.4

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.4

Samacheer Kalvi 8th Maths Book Solutions Exercise 1.4 Question 1.
Fill in the blanks:
(i) (-1 )^{even integer} is…………
(ii) For a ≠ 0, a° is……….
(iii) 4^-3 × 5^-3 =…………
(iv) (-2)⁷ =………….
(v) \((-\frac{1}{3})^{-5}\) =…………
Solution:
(i) 1
(ii) 1
(iii) 20^-3
(iv) \(\frac{-1}{128}\)
(v) -243

Samacheer Kalvi 8th Maths Solutions Term 3 Pdf Question 2.
Say True or False:
(i) If 8^x = \(\frac{1}{64}\), the value of x is -2
(ii) The simplified form of (256)^-1/4 is \(\frac{1}{4}\)
(iii) The standard form of 2 x 10^-4 is 0.0002.
(iv) The scientific form of 123.456 is 1.23456 x 10^-2.
(v) The multiplicative inverse of (3)^-7 is 3⁷.
Solution:
(i) True
(ii) True
(iii) True
(iv) False
(v) True

8th Maths Exercise 1.4 Samacheer Kalvi Question 3.
Evaluate:
(i) \(\left(\frac{1}{2}\right)^{3}\)
Solution:
\(\left(\frac{1}{2}\right)^{3}=\frac{1^{3}}{2^{3}}=\frac{1}{2 \times 2 \times 2}=\frac{1}{8}\)

(ii) \(\left(\frac{1}{2}\right)^{-5}\)
Solution:
\(\left(\frac{1}{2}\right)^{-5}=\frac{1^{-5}}{2^{-5}}=\frac{1}{2^{-5}}=2^{5}\) = 2 × 2 × 2 × 2 × 2 = 32

(iii) (-3)^-3
Solution:
(-3)^-3 = \(\frac{1}{(-3)^{3}}=\frac{1}{-3 \times-3 \times-3}=\frac{1}{-27}=\frac{-1}{27}\)

(iv) (-3)⁴
Solution:
(-3)⁴ = -3 × -3 × -3 × -3 = 81

(v) \(\left(\frac{-5}{6}\right)^{-3}\)
Solution:
\(\left(\frac{-5}{6}\right)^{-3}=\frac{(-5)^{-3}}{6^{-3}}=\frac{6^{3}}{(-5)^{3}}=\frac{6 \times 6 \times 6}{-5 \times-5 \times-5}=-\frac{216}{125}\)

(vi) \(\left( { 2 }^{ -5 }\div { 2 }^{ 7 } \right) \times { 2 }^{ -2 }\)
Solution:
\(\left( { 2 }^{ -5 }\div { 2 }^{ 7 } \right) \times { 2 }^{ -2 }\) = \(\left( { 2 }^{ -5-7 }\right)\times { 2 }^{ -2 }\)
\(2^{-12} \times 2^{-2}=2^{-12+(-2)}=2^{-14}\)

(vii) \(\left( { 2 }^{ -1 }\times { 3 }^{ -1 } \right) \div 6^{ -2 }\)
Solution:
\(\left( { 2 }^{ -1 }\times { 3 }^{ -1 } \right) \div 6^{ -2 }\) = \((2 \times { 3})^{ -1 } \div 6^{ -2 }\)
= \((6)^{-1} \div 6^{-2}=6^{(-1)-(-2)}=6^{1}=6\)

(viii) \(\left(-\frac{1}{3}\right)^{-2}\)
Solution:
\(\left(-\frac{1}{3}\right)^{-2}=\left(-\frac{3}{1}\right)^{2}=\frac{(-3)^{2}}{1^{2}}=\frac{-3 \times-3}{1}\) = 9

Samacheer Kalvi 8th Maths Book Solutions Question 4.
Evaluate:
(i) \(\left(\frac{2}{5}\right)^{4} \times\left(\frac{2}{5}\right)^{2}\)
Solution:
\(\left(\frac{2}{5}\right)^{4} \times\left(\frac{2}{5}\right)^{2}\) = \(\left(\frac{2}{5}\right)^{4+2}=\left(\frac{2}{5}\right)^{6}\)

(ii) \(\left(\frac{4}{5}\right)^{-2} \times\left(\frac{4}{5}\right)^{-3}\)
Solution:
\(\left(\frac{4}{5}\right)^{-2} \times\left(\frac{4}{5}\right)^{-3}\) = \(\left(\frac{4}{5}\right)^{-2+(-3)}=\left(\frac{4}{5}\right)^{-5}\)

(iii) \(\left(\frac{1}{2}\right)^{-3} \times\left(\frac{1}{2}\right)^{7}\)
\(\left(\frac{1}{2}\right)^{-3} \times\left(\frac{1}{2}\right)^{7}\) = \(\left(\frac{1}{2}\right)^{-3+7}=\left(\frac{1}{2}\right)^{4}\)

Samacheer Kalvi Guru 8th Maths Question 5.
Evaluate:
(i) \(\left( { 5 }^{ 0 }+{ 6 }^{ -1 } \right) \times { 3 }^{ 3 }\)
Solution:

(ii) \(\left( 2^{ -1 }\times { 3 }^{ -1 } \right) \div { 6 }^{ -1 }\)
Solution:
\(\left( 2^{ -1 }\times { 3 }^{ -1 } \right) \div { 6 }^{ -1 }\) = \((2 \times 3)^{-1} \div 6^{-1}=6^{-1}+6^{-1}=6^{-1-(-1)}=6^{0}\) = 1

(iii) \(\left( 3^{ -1 }+{ 4 }^{ -2 }+{ 5 }^{ -3 } \right) ^{ 0 }\)
Solution:
\(\left( 3^{ -1 }+{ 4 }^{ -2 }+{ 5 }^{ -3 } \right) ^{ 0 }\) = 1 [ ∴ a° = 1 where a ≠ 0]

Samacheer Kalvi Guru 8th Maths Book Solutions Question 6.
Simplify
(i) \(\left(3^{2}\right)^{3} \times\left(2 \times 3^{5}\right)^{-2} \times(18)^{2}\)
Solution:

Samacheer Kalvi.Guru 8th Maths Question 7.
Solve for x.
(i) \(\frac{10^{x}}{10^{-3}}=10^{9}\)
Solution:
\(\frac{10^{x}}{10^{-3}}=10^{9}\)
\({10^{x+3}}=10^{9}\)
Equating the powers of same base 10
x + 3 = 9
x + 3 – 3 = 9 – 3
x = 6

(ii) \(\frac{2^{2x-1}}{2^{x+2}}\) = 4
Solution:
\(2^{2x-1-1(x+2)}\) = 2²
\(2^{2x-1-x-2}\) = 2²
\(2^{x-3}\) = 2²
Equating the powers of the same base 2.
x – 3 = 2
x – 3 + 3 = 2 + 3
x = 5

(iii) \(\frac{5^{5} \times 5^{-4} \times 5^{x}}{5^{12}}=5^{-5}\)
Solution:
\(\frac{5^{5} \times 5^{-4} \times 5^{x}}{5^{12}}=5^{-5}\) = \(5^{-5} \Rightarrow \frac{5^{5-4+x}}{5^{12}}=5^{-5}\)
\(\Rightarrow \frac{5^{1+x}}{5^{12}}\) = 5^-5
⇒ \(5^{1+x-12}\) = 5^-5
⇒ \(5^{x-11}\) = 5^-5
Equating the powers of same base 5.
x – 11 = -5
x – 11 + 11 = -5 + 11
x = 6

Samacheer Kalvi Guru 8th Maths Guide Question 8.
Expand using exponents:
(i) 6054.321
(ii) 897.14
Solution:

Samacheerkalvi.Guru 8th Maths Question 9.
Find the number is standard form:
(i) 8 x 10⁴ + 7 x 10³ + 6 x 10² + 5 x 10¹+ 2 x 1 + 4 x 10^-2 + 7 x 10^-4
Solution:
8 x 10⁴ + 7 x 10³ + 6 x 10² + 5 x 10¹ + 2 x 1 + 4 x 10^-2 + 7 x 10^-4
= 8 x 10000 + 7 x 1000 + 6 x 100 + 5 x 10 + 2 x 1 + 4 x \(\frac{1}{100}\) + 7 x \(\frac{1}{10000}\)
= 80000 + 7000 + 600 + 50 + 2 + \(\frac{4}{100}\) + \(\frac{7}{10000}\)
= 87652.0407

(ii) 5 x 10³ + 5 x 10¹ + 5 x 10^-1 + 5 x 10^-3
Solution:
5 x 10³ + 5 x 10¹ + 5 x 10^-1 + 5 x 10^-3
= 5 x 1000 + 5 x 10 + 5 x \(\frac{1}{10}\) + 5 x \(\frac{1}{1000}\)
= 5000 + 50 + \(\frac{5}{10}\) + \(\frac{5}{1000}\) = 5050.505

8th Maths Exercise 1.4 Question 10.
The radius of a hydrogen atom is 2.5 x 10 11 m. Express this number in standard notation.
Solution:
Radius of a hydrogen atom = 2.5 x 10¹¹ m
= 2.5 x \(\frac{1}{10^{11}}\) m = \(\frac{2.5}{10^{11}}\) m = 0.000000000025 m

Samacheer Kalvi Guru 8 Maths Question 11.
Write each number in scientific notation:
(i) 467800000000
(ii) 0.000001972
Solution:

Samacheer Kalvi 8th Maths Solutions Question 12.
Write in scientific notation:
(i) Earth’s volume is about 1,083,000,000,000 cubic kilometers.
Solution:

Earth’s volume = 1.083 x 10¹² cubic kilometers.

(ii) If you fill a bucket with dirt, the portion of the whole Earth that is in the bucket will be 0.0000000000000000000000016 kg.
Solution:
Portion of earth in the bucket =  kg = 1.6 x 10^-24 kg.

Objective Type Questions

Samacheer Kalvi 8th Maths Book Solution Question 13.
By what number should (-4)^-1 be multiplied so that the product becomes 10^-1?
(a) \(\frac{2}{3}\)
(b) \(\frac{-2}{5}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{-5}{2}\)
Solution:
(b) \(\frac{-2}{5}\)
Hint:

Question 14.
0.0000000002020 in scientific form is
(a) 2.02 x 10⁹
(b) 2.02 x 10^-9
(c) 2.02 x 10^-8
(d) 2.02 x 10^-10
Solution:
(d) 2.02 x 10^-10
Hint:

Question 15.
(-2)^-3 x (-2)^-2 is
(a) \(\frac{-1}{32}\)
(b) \(\frac{1}{32}\)
(c) 32
(d) -32
Solution:
(a) \(\frac{-1}{32}\)

Question 16.
Which is not correct?
(a) \(\left(\frac{-1}{4}\right)^{2}\) = 4^-2
(b) \(\left(\frac{-1}{4}\right)^{2}\) = \(\left(\frac{1}{2}\right)^{4}\)
(c) \(\left(\frac{-1}{4}\right)^{2}\) = 16^-1
(d) \(-\left(\frac{1}{4}\right)^{2}\) = 16^-1
Solution:
(d) \(-\left(\frac{1}{4}\right)^{2}\) = 16^-1
Hint:
\((-2)-3 x(-2)-2=(-2)-3-2=(-2)-5\left(-\frac{1}{2}\right) 5=-\frac{1}{32}\)

Question 17.
If \(\left(\frac{p}{q}\right)^{1-3 x}=\left(\frac{q}{p}\right)^{\frac{1}{2}}\), then x is
(a) 4^-1
(b) 3^-1
(c) 2^-1
(d) 1^-1
Solution:
(c) 2^-1
Hint: