# 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

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

Question 2.
Say True or False:
(i) If 8x = $$\frac{1}{64}$$, 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

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

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

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]

Question 6.
Simplify
(i) $$\left(3^{2}\right)^{3} \times\left(2 \times 3^{5}\right)^{-2} \times(18)^{2}$$
Solution:

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

Question 8.
Expand using exponents:
(i) 6054.321
(ii) 897.14
Solution:

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

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

Question 11.
Write each number in scientific notation:
(i) 467800000000
(ii) 0.000001972
Solution:

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

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:

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