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Chapter 5

Exponents

Class - 7 Concise Mathematics Selina



Exercise 5(A)

Question 1

Find the value of :

(i) 62

(ii) 73

(iii) 44

(iv) 55

(v) 83

(vi) 75

Answer

(i) Solving,

⇒ 62 = 6 × 6 = 36.

Hence, 62 = 36.

(ii) Solving,

⇒ 73 = 7 × 7 × 7 = 343.

Hence, 73 = 343.

(iii) Solving,

⇒ 44 = 4 × 4 × 4 × 4 = 256.

Hence, 44 = 256.

(iv) Solving,

⇒ 55 = 5 × 5 × 5 × 5 × 5 = 3125.

Hence, 55 = 3125.

(v) Solving,

⇒ 83 = 8 × 8 × 8 = 512.

Hence, 83 = 512.

(vi) Solving,

⇒ 75 = 7 × 7 × 7 × 7 × 7 = 16807.

Hence, 75 = 16807.

Question 2

Evaluate :

(i) 23 × 42

(ii) 23 × 52

(iii) 33 × 52

(iv) 22 × 33

(v) 32 × 53

(vi) 53 × 24

(vii) 32 × 42

(viii) (4 × 3)3

(ix) (5 × 4)2

Answer

(i) Solving,

⇒ 23 × 42 = (2 × 2 × 2) × (4 × 4) = 8 × 16 = 128.

Hence, 23 × 42 = 128.

(ii) Solving,

⇒ 23 × 52 = (2 × 2 × 2) × (5 × 5) = 8 × 25 = 200.

Hence, 23 × 52 = 200.

(iii) Solving,

⇒ 33 × 52 = (3 × 3 × 3) × (5 × 5) = 27 × 25 = 675.

Hence, 33 × 52 = 675.

(iv) Solving,

⇒ 22 × 33 = (2 × 2) × (3 × 3 × 3) = 4 × 27 = 108.

Hence, 22 × 33 = 108.

(v) Solving,

⇒ 32 × 53 = (3 × 3) × (5 × 5 × 5) = 9 × 125 = 1125.

Hence, 32 × 53 = 1125.

(vi) Solving,

⇒ 53 × 24 = (5 × 5 × 5) × (2 × 2 × 2 × 2) = 125 × 16 = 2000.

Hence, 53 × 24 = 2000.

(vii) Solving,

⇒ 32 × 42 = (3 × 3) × (4 × 4) = 9 × 16 = 144.

Hence, 32 × 42 = 144.

(viii) Solving,

⇒ (4 × 3)3 = 123 = 12 × 12 × 12 = 1728.

Hence, (4 × 3)3 = 1728.

(ix) Solving,

⇒ (5 × 4)2 = 202 = 20 × 20 = 400.

Hence, (5 × 4)2 = 400.

Question 3(i)

Evaluate :

(34)4\left(\dfrac{3}{4}\right)^4

Answer

Solving,

(34)4=3×3×3×34×4×4×4=81256\Rightarrow \left(\dfrac{3}{4}\right)^4 \\[1em] = \dfrac{3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4} \\[1em] = \dfrac{81}{256}

Hence, (34)4=81256\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256}.

Question 3(ii)

Evaluate :

(56)5\left(-\dfrac{5}{6}\right)^5

Answer

Solving,

(56)5=(5)×(5)×(5)×(5)×(5)6×6×6×6×6=31257776\Rightarrow \left(-\dfrac{5}{6}\right)^5 \\[1em] = \dfrac{(-5) \times (-5) \times (-5) \times (-5) \times (-5)}{6 \times 6 \times 6 \times 6 \times 6} \\[1em] = -\dfrac{3125}{7776}

Hence, (56)5=31257776\left(-\dfrac{5}{6}\right)^5 = -\dfrac{3125}{7776}.

Question 3(iii)

Evaluate :

(35)3\left(\dfrac{-3}{-5}\right)^3

Answer

Solving,

(35)3=(35)3=3×3×35×5×5=27125\Rightarrow \left(\dfrac{-3}{-5}\right)^3\\[1em] = \left(\dfrac{3}{5}\right)^3 \\[1em] = \dfrac{3 \times 3 \times 3}{5 \times 5 \times 5} \\[1em] = \dfrac{27}{125}

Hence, (35)3=27125\left(\dfrac{-3}{-5}\right)^3 = \dfrac{27}{125}.

Question 4(i)

Evaluate :

(23)3×(34)2\left(\dfrac{2}{3}\right)^3 \times \left(\dfrac{3}{4}\right)^2

Answer

Solving,

(23)3×(34)2=2×2×23×3×3×3×34×4=2×2×2×3×33×3×3×4×4=83×16=848=16\Rightarrow \left(\dfrac{2}{3}\right)^3 \times \left(\dfrac{3}{4}\right)^2\\[1em] = \dfrac{2 \times 2 \times 2}{3 \times 3 \times 3} \times \dfrac{3 \times 3}{4 \times 4}\\[1em] = \dfrac{2 \times 2 \times 2 \times 3 \times 3}{3 \times 3 \times 3 \times 4 \times 4}\\[1em] = \dfrac{8}{3 \times 16}\\[1em] = \dfrac{8}{48}\\[1em] = \dfrac{1}{6}

Hence, (23)3×(34)2=16\left(\dfrac{2}{3}\right)^3 \times \left(\dfrac{3}{4}\right)^2 = \dfrac{1}{6}.

Question 4(ii)

Evaluate :

(34)3×(23)4\left(-\dfrac{3}{4}\right)^3 \times \left(\dfrac{2}{3}\right)^4

Answer

Solving,

(34)3×(23)4=(3)×(3)×(3)4×4×4×2×2×2×23×3×3×3=2764×1681=27×1664×81=14×3=112\Rightarrow \left(-\dfrac{3}{4}\right)^3 \times \left(\dfrac{2}{3}\right)^4\\[1em] = \dfrac{(-3) \times (-3) \times (-3)}{4 \times 4 \times 4} \times \dfrac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3}\\[1em] = -\dfrac{27}{64} \times \dfrac{16}{81}\\[1em] = -\dfrac{27 \times 16}{64 \times 81}\\[1em] = -\dfrac{1}{4 \times 3}\\[1em] = -\dfrac{1}{12}

Hence, (34)3×(23)4=112\left(-\dfrac{3}{4}\right)^3 \times \left(\dfrac{2}{3}\right)^4 = -\dfrac{1}{12}.

Question 4(iii)

Evaluate :

(35)2×(23)3\left(\dfrac{3}{5}\right)^2 \times \left(-\dfrac{2}{3}\right)^3

Answer

Solving,

(35)2×(23)3=3×35×5×(2)×(2)×(2)3×3×3=925×(827)=9×825×27=825×3=875\Rightarrow \left(\dfrac{3}{5}\right)^2 \times \left(-\dfrac{2}{3}\right)^3\\[1em] = \dfrac{3 \times 3}{5 \times 5} \times \dfrac{(-2) \times (-2) \times (-2)}{3 \times 3 \times 3}\\[1em] = \dfrac{9}{25} \times \left(-\dfrac{8}{27}\right)\\[1em] = -\dfrac{9 \times 8}{25 \times 27}\\[1em] = -\dfrac{8}{25 \times 3}\\[1em] = -\dfrac{8}{75}

Hence, (35)2×(23)3=875\left(\dfrac{3}{5}\right)^2 \times \left(-\dfrac{2}{3}\right)^3 = -\dfrac{8}{75}.

Question 5

Which is greater :

(i) 23 or 32

(ii) 25 or 52

(iii) 43 or 34

(iv) 54 or 45

Answer

(i) Solving,

⇒ 23 = 2 × 2 × 2 = 8

⇒ 32 = 3 × 3 = 9.

Since 9 > 8, therefore 32 > 23.

Hence, the greater number is 32.

(ii) Solving,

⇒ 25 = 2 × 2 × 2 × 2 × 2 = 32

⇒ 52 = 5 × 5 = 25.

Since 32 > 25, therefore 25 > 52.

Hence, the greater number is 25.

(iii) Solving,

⇒ 43 = 4 × 4 × 4 = 64

⇒ 34 = 3 × 3 × 3 × 3 = 81.

Since 81 > 64, therefore 34 > 43.

Hence, the greater number is 34.

(iv) Solving,

⇒ 54 = 5 × 5 × 5 × 5 = 625

⇒ 45 = 4 × 4 × 4 × 4 × 4 = 1024.

Since 1024 > 625, therefore 45 > 54.

Hence, the greater number is 45.

Question 6(i)

Express 512 in exponential form.

Answer

By prime factorisation of 512:

2512225621282642322162824221\begin{array}{l|r} 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}

⇒ 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29.

Hence, 512 = 29.

Question 6(ii)

Express 1250 in exponential form.

Answer

By prime factorisation of 1250:

2125056255125525551\begin{array}{l|r} 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}

⇒ 1250 = 2 × 5 × 5 × 5 × 5 = 21 × 54.

Hence, 1250 = 2 × 54.

Question 6(iii)

Express 1458 in exponential form.

Answer

By prime factorisation of 1458:

214583729324338132739331\begin{array}{l|r} 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}

⇒ 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3 = 21 × 36.

Hence, 1458 = 2 × 36.

Question 6(iv)

Express 3600 in exponential form.

Answer

By prime factorisation of 3600:

2360021800290024503225375525551\begin{array}{l|r} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}

⇒ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52.

Hence, 3600 = 24 × 32 × 52.

Question 6(v)

Express 1350 in exponential form.

Answer

By prime factorisation of 1350:

2135036753225375525551\begin{array}{l|r} 2 & 1350 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}

⇒ 1350 = 2 × 3 × 3 × 3 × 5 × 5 = 21 × 33 × 52.

Hence, 1350 = 2 × 33 × 52.

Question 6(vi)

Express 1176 in exponential form.

Answer

By prime factorisation of 1176:

21176258822943147749771\begin{array}{l|r} 2 & 1176 \\ \hline 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}

⇒ 1176 = 2 × 2 × 2 × 3 × 7 × 7 = 23 × 31 × 72.

Hence, 1176 = 23 × 3 × 72.

Question 7

If a = 2 and b = 3, find the value of :

(i) (a + b)2

(ii) (b - a)3

(iii) (a × b)a

(iv) (a × b)b

Answer

(i) Solving,

⇒ (a + b)2 = (2 + 3)2 = 52 = 5 × 5 = 25.

Hence, (a + b)2 = 25.

(ii) Solving,

⇒ (b - a)3 = (3 - 2)3 = 13 = 1 × 1 × 1 = 1.

Hence, (b - a)3 = 1.

(iii) Solving,

⇒ (a × b)a = (2 × 3)2 = 62 = 6 × 6 = 36.

Hence, (a × b)a = 36.

(iv) Solving,

⇒ (a × b)b = (2 × 3)3 = 63 = 6 × 6 × 6 = 216.

Hence, (a × b)b = 216.

Question 8

Express :

(i) 1024 as a power of 2.

(ii) 343 as a power of 7.

(iii) 729 as a power of 3.

Answer

(i) By prime factorisation of 1024:

210242512225621282642322162824221\begin{array}{l|r} 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}

⇒ 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210.

Hence, 1024 = 210.

(ii) By prime factorisation of 343:

7343749771\begin{array}{l|r} 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}

⇒ 343 = 7 × 7 × 7 = 73.

Hence, 343 = 73.

(iii) By prime factorisation of 729:

3729324338132739331\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}

⇒ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36.

Hence, 729 = 36.

Question 9

If 27 × 32 = 3x × 2y; find the values of x and y.

Answer

By prime factorisation of 27:

32739331\begin{array}{l|r} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}

⇒ 27 = 3 × 3 × 3 = 33.

By prime factorisation of 32:

2322162824221\begin{array}{l|r} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}

⇒ 32 = 2 × 2 × 2 × 2 × 2 = 25.

So,

⇒ 27 × 32 = 3x × 2y

⇒ 33 × 25 = 3x × 2y.

Comparing the powers of the same base on both sides, we get :

⇒ x = 3 and y = 5.

Hence, x = 3 and y = 5.

Question 10

If 64 × 625 = 2a × 5b; find :

(i) the values of a and b.

(ii) 2b × 5a

Answer

(i) Solving,

⇒ 64 × 625 = 2a × 5b

⇒ (2 × 2 × 2 × 2 × 2 × 2) × (5 × 5 × 5 × 5) = 2a × 5b

⇒ 26 × 54 = 2a × 5b.

Comparing the powers of the same base on both sides, we get :

⇒ a = 6 and b = 4.

Hence, a = 6 and b = 4.

(ii) Solving,

⇒ 2b × 5a = 24 × 56

⇒ (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5 × 5 × 5)

⇒ 16 × 15625

⇒ 250000.

Hence, 2b × 5a = 250000.

Exercise 5(B)

Question 1

Fill in the blanks :

(i) In 52 = 25, base = .................... and index = ........................

(ii) If index = 3x and base = 2y, the number = ........................

Answer

As we know, in the expression an, a is called the base and n is called the index (or exponent).

(i) In 52 = 25, base = 5 and index = 2

(ii) If index = 3x and base = 2y, the number = (2y)3x

Question 2(i)

Evaluate :

28 ÷ 23

Answer

Solving,

28 ÷ 23

= 2823\dfrac{2^8}{2^3}

= 2(8 - 3)

= 25

= 32

Hence, 28 ÷ 23 = 25 = 32.

Question 2(ii)

Evaluate :

23 ÷ 28

Answer

Solving,

23÷28=2328=1283=125=132\Rightarrow 2^3 ÷ 2^8 \\[1em] = \dfrac{2^3}{2^8} \\[1em] = \dfrac{1}{2^{8-3}} \\[1em] = \dfrac{1}{2^5} \\[1em] = \dfrac{1}{32}

Hence, 23÷28=125=1322^3 ÷ 2^8 = \dfrac{1}{2^5} = \dfrac{1}{32}.

Question 2(iii)

Evaluate :

(26)0

Answer

Solving,

As any non-zero base raised to the power zero is equal to 1,

(26)0=1\Rightarrow (2^6)^0 \\[1em] = 1

Hence, (26)0 = 1.

Question 2(iv)

Evaluate :

(30)6

Answer

Solving,

(30)6=16=1\Rightarrow (3^0)^6 \\[1em] = 1^6 \\[1em] = 1

Hence, (30)6 = 1.

Question 2(v)

Evaluate :

83 × 8-5 × 84

Answer

Solving,

83×85×84=83+(5)+4=82=64\Rightarrow 8^3 \times 8^{-5} \times 8^4 \\[1em] = 8^{3 + (-5) + 4} \\[1em] = 8^2 \\[1em] = 64

Hence, 83 × 8-5 × 84 = 82 = 64.

Question 2(vi)

Evaluate :

54 × 53 ÷ 55

Answer

Solving,

54×53÷55=54×5355=5755=575=52=25\Rightarrow 5^4 \times 5^3 \div 5^5 \\[1em] = \dfrac{5^4 \times 5^3}{5^5} \\[1em] = \dfrac{5^7}{5^5} \\[1em] = 5^{7 - 5} \\[1em] = 5^2 \\[1em] = 25

Hence, 54 × 53 ÷ 55 = 52 = 25.

Question 2(vii)

Evaluate :

54 ÷ 53 × 55

Answer

Solving,

54÷53×55=54×5553=5953=593=56=15625\Rightarrow 5^4 \div 5^3 \times 5^5 \\[1em] = \dfrac{5^4 \times 5^5}{5^3} \\[1em] = \dfrac{5^9}{5^3} \\[1em] = 5^{9 - 3} \\[1em] = 5^6 \\[1em] = 15625

Hence, 54 ÷ 53 × 55 = 56 = 15625.

Question 2(viii)

Evaluate :

44 ÷ 43 × 40

Answer

Solving,

44÷43×40=44×4043=4443=443=41=4\Rightarrow 4^4 \div 4^3 \times 4^0 \\[1em] = \dfrac{4^4 \times 4^0}{4^3} \\[1em] = \dfrac{4^4}{4^3} \\[1em] = 4^{4 - 3} \\[1em] = 4^1 \\[1em] = 4

Hence, 44 ÷ 43 × 40 = 4.

Question 2(ix)

Evaluate :

(35 × 47 × 58)0

Answer

Solving,

As any non-zero base raised to the power zero is equal to 1,

(35×47×58)0=1\Rightarrow (3^5 \times 4^7 \times 5^8)^0 \\[1em] = 1

Hence, (35 × 47 × 58)0 = 1.

Question 3(i)

Simplify, giving answers with positive index :

2b6 · b3 · 5b4

Answer

Solving,

⇒ 2b6 · b3 · 5b4

⇒ (2 × 5) × b6 + 3 + 4

⇒ 10b13.

Hence, 2b6 · b3 · 5b4 = 10b13.

Question 3(ii)

Simplify, giving answers with positive index :

x2y3 · 6x5y · 9x3y4

Answer

Solving,

⇒ x2y3 · 6x5y · 9x3y4

⇒ (1 × 6 × 9) × x2 + 5 + 3 × y3 + 1 + 4

⇒ 54x10y8.

Hence, x2y3 · 6x5y · 9x3y4 = 54x10y8.

Question 3(iii)

Simplify, giving answers with positive index :

(-a5) (a2)

Answer

Solving,

⇒ (-a5) (a2)

⇒ (-1) (a5) (a2)

⇒ -a5 + 2

⇒ -a7.

Hence, (-a5) (a2) = -a7.

Question 3(iv)

Simplify, giving answers with positive index :

(-y2) (-y3)

Answer

Solving,

⇒ (-y2) (-y3)

⇒ (-1) × (-1) × y2 + 3

⇒ y5.

Hence, (-y2) (-y3) = y5.

Question 3(v)

Simplify, giving answers with positive index :

(-3)2 (3)3

Answer

Solving,

⇒ (-3)2 (3)3

⇒ (-1)2 (3)2 (3)3

⇒ 32 × 33

⇒ 32 + 3

⇒ 35.

Hence, (-3)2 (3)3 = 35.

Question 3(vi)

Simplify, giving answers with positive index :

(-4x) (-5x2)

Answer

Solving,

⇒ (-4x) (-5x2)

⇒ (-4) × (-5) × x1 + 2

⇒ 20x3.

Hence, (-4x) (-5x2) = 20x3.

Question 3(vii)

Simplify, giving answers with positive index :

(5a2b) (2ab2) (a3b)

Answer

Solving,

⇒ (5a2b) (2ab2) (a3b)

⇒ (5 × 2 × 1) × a2 + 1 + 3 × b1 + 2 + 1

⇒ 10a6b4.

Hence, (5a2b) (2ab2) (a3b) = 10a6b4.

Question 3(viii)

Simplify, giving answers with positive index :

x2a + 7 · x2a - 8

Answer

Solving,

⇒ x2a + 7 · x2a - 8

⇒ x(2a + 7) + (2a - 8)

⇒ x4a - 1.

Hence, x2a + 7 · x2a - 8 = x4a - 1.

Question 3(ix)

Simplify, giving answers with positive index :

3y · 32 · 3-4

Answer

Solving,

⇒ 3y · 32 · 3-4

⇒ 3y + 2 + (-4)

⇒ 3y - 2.

Hence, 3y · 32 · 3-4 = 3y - 2.

Question 3(x)

Simplify, giving answers with positive index :

24a · 23a · 2-a

Answer

Solving,

⇒ 24a · 23a · 2-a

⇒ 24a + 3a + (-a)

⇒ 26a.

Hence, 24a · 23a · 2-a = 26a.

Question 3(xi)

Simplify, giving answers with positive index :

4x2y2 ÷ 9x3y3

Answer

Solving,

4x2y2÷9x3y3=4x2y29x3y3=49×x23×y23=49×x1×y1=49xy\Rightarrow 4x^2y^2 ÷ 9x^3y^3\\[1em] = \dfrac{4x^2y^2}{9x^3y^3}\\[1em] = \dfrac{4}{9} \times x^{2-3} \times y^{2-3}\\[1em] = \dfrac{4}{9} \times x^{-1} \times y^{-1}\\[1em] = \dfrac{4}{9xy}

Hence, 4x2y2÷9x3y3=49xy4x^2y^2 ÷ 9x^3y^3 = \dfrac{4}{9xy}.

Question 3(xii)

Simplify, giving answers with positive index :

(102)3 (x8)12

Answer

Solving,

⇒ (102)3 (x8)12

⇒ 102 × 3 × x8 × 12

⇒ 106x96.

Hence, (102)3 (x8)12 = 106x96.

Question 3(xiii)

Simplify, giving answers with positive index :

(a10)10 (16)10

Answer

Solving,

⇒ (a10)10 (16)10

⇒ a10 × 10 × 16 × 10

⇒ a100 × 1

⇒ a100.

Hence, (a10)10 (16)10 = a100.

Question 3(xiv)

Simplify, giving answers with positive index :

(n2)2 (-n2)3

Answer

Solving,

⇒ (n2)2 (-n2)3

⇒ n2 × 2 × (-1)3 × n2 × 3

⇒ n4 × (-1) × n6

⇒ -n4 + 6

⇒ -n10.

Hence, (n2)2 (-n2)3 = -n10.

Question 3(xv)

Simplify, giving answers with positive index :

  • (3ab)2 (-5a2bc4)2

Answer

Solving,

⇒ - (3ab)2 (-5a2bc4)2

⇒ - [32a2b2] × [(-5)2a2 × 2b2c4 × 2]

⇒ - [9a2b2] × [25a4b2c8]

⇒ - (9 × 25) × a2 + 4 × b2 + 2 × c8

⇒ -225a6b4c8.

Hence, - (3ab)2 (-5a2bc4)2 = -225a6b4c8.

Question 3(xvi)

Simplify, giving answers with positive index :

(-2)2 × (0)3 × (3)3

Answer

Solving,

As 0 multiplied with any number gives 0,

⇒ (-2)2 × (0)3 × (3)3

⇒ 4 × 0 × 27

⇒ 0.

Hence, (-2)2 × (0)3 × (3)3 = 0.

Question 3(xvii)

Simplify, giving answers with positive index :

(2a3)4 (4a2)2

Answer

Solving,

⇒ (2a3)4 (4a2)2

⇒ [24a3 × 4] × [42a2 × 2]

⇒ [16a12] × [16a4]

⇒ (16 × 16) × a12 + 4

⇒ 256a16.

Hence, (2a3)4 (4a2)2 = 256a16.

Question 3(xviii)

Simplify, giving answers with positive index :

(4x2y3)3 ÷ (3x2y3)3

Answer

Solving,

(4x2y3)3÷(3x2y3)3=(4x2y3)3(3x2y3)3=(4x2y33x2y3)3=(43)3=6427\Rightarrow (4x^2y^3)^3 ÷ (3x^2y^3)^3\\[1em] = \dfrac{(4x^2y^3)^3}{(3x^2y^3)^3}\\[1em] = \left(\dfrac{4x^2y^3}{3x^2y^3}\right)^3\\[1em] = \left(\dfrac{4}{3}\right)^3\\[1em] = \dfrac{64}{27}

Hence, (4x2y3)3÷(3x2y3)3=6427(4x^2y^3)^3 ÷ (3x^2y^3)^3 = \dfrac{64}{27}.

Question 3(xix)

Simplify, giving answers with positive index :

(12x)3×(6x)2\left(\dfrac{1}{2x}\right)^3 \times (6x)^2

Answer

Solving,

(12x)3×(6x)2=123x3×62x2=18x3×36x2=36x28x3=368×x23=92×x1=92x\Rightarrow \left(\dfrac{1}{2x}\right)^3 \times (6x)^2\\[1em] = \dfrac{1}{2^3x^3} \times 6^2x^2\\[1em] = \dfrac{1}{8x^3} \times 36x^2\\[1em] = \dfrac{36x^2}{8x^3}\\[1em] = \dfrac{36}{8} \times x^{2-3}\\[1em] = \dfrac{9}{2} \times x^{-1}\\[1em] = \dfrac{9}{2x}

Hence, (12x)3×(6x)2=92x\left(\dfrac{1}{2x}\right)^3 \times (6x)^2 = \dfrac{9}{2x}.

Question 3(xx)

Simplify, giving answers with positive index :

(14ab2c)2÷(32a2bc2)4\left(\dfrac{1}{4ab^2c}\right)^2 \div \left(\dfrac{3}{2a^2bc^2}\right)^4

Answer

Solving,

(14ab2c)2÷(32a2bc2)4=142a2b4c2÷3424a8b4c8=116a2b4c2÷8116a8b4c8=116a2b4c2×16a8b4c881=16a8b4c816×81×a2b4c2=181×a82×b44×c82=181×a6×b0×c6=a6c681\Rightarrow \left(\dfrac{1}{4ab^2c}\right)^2 ÷ \left(\dfrac{3}{2a^2bc^2}\right)^4\\[1em] = \dfrac{1}{4^2a^2b^4c^2} ÷ \dfrac{3^4}{2^4a^8b^4c^8}\\[1em] = \dfrac{1}{16a^2b^4c^2} ÷ \dfrac{81}{16a^8b^4c^8}\\[1em] = \dfrac{1}{16a^2b^4c^2} \times \dfrac{16a^8b^4c^8}{81}\\[1em] = \dfrac{16a^8b^4c^8}{16 \times 81 \times a^2b^4c^2}\\[1em] = \dfrac{1}{81} \times a^{8-2} \times b^{4-4} \times c^{8-2}\\[1em] = \dfrac{1}{81} \times a^6 \times b^0 \times c^6\\[1em] = \dfrac{a^6c^6}{81}

Hence, (14ab2c)2÷(32a2bc2)4=a6c681\left(\dfrac{1}{4ab^2c}\right)^2 ÷ \left(\dfrac{3}{2a^2bc^2}\right)^4 = \dfrac{a^6c^6}{81}.

Question 3(xxi)

Simplify, giving answers with positive index :

(5x7)3(10x2)2(2x6)7\dfrac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7}

Answer

Solving,

(5x7)3(10x2)2(2x6)7=53x21102x427x42=125x21100x4128x42=125×100128×x21+4x42=12500128×x2542=312532×x17=312532x17\Rightarrow \dfrac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7}\\[1em] = \dfrac{5^3x^{21} \cdot 10^2x^4}{2^7x^{42}}\\[1em] = \dfrac{125x^{21} \cdot 100x^4}{128x^{42}}\\[1em] = \dfrac{125 \times 100}{128} \times \dfrac{x^{21 + 4}}{x^{42}}\\[1em] = \dfrac{12500}{128} \times x^{25 - 42}\\[1em] = \dfrac{3125}{32} \times x^{-17}\\[1em] = \dfrac{3125}{32x^{17}}

Hence, (5x7)3(10x2)2(2x6)7=312532x17\dfrac{(5x^7)^3 \cdot (10x^2)^2}{(2x^6)^7} = \dfrac{3125}{32x^{17}}.

Question 3(xxii)

Simplify, giving answers with positive index :

(7p2q9r5)2(4pqr)3(14p6q10r4)2\dfrac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2}

Answer

Solving,

(7p2q9r5)2(4pqr)3(14p6q10r4)2=72p4q18r10×43p3q3r3142p12q20r8=49p4q18r10×64p3q3r3196p12q20r8=49×64196×p4+3q18+3r10+3p12q20r8=3136196×p712×q2120×r138=16×p5×q1×r5=16qr5p5\Rightarrow \dfrac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2}\\[1em] = \dfrac{7^2p^4q^{18}r^{10} \times 4^3p^3q^3r^3}{14^2p^{12}q^{20}r^8}\\[1em] = \dfrac{49p^4q^{18}r^{10} \times 64p^3q^3r^3}{196p^{12}q^{20}r^8}\\[1em] = \dfrac{49 \times 64}{196} \times \dfrac{p^{4 + 3}q^{18 + 3}r^{10 + 3}}{p^{12}q^{20}r^8}\\[1em] = \dfrac{3136}{196} \times p^{7 - 12} \times q^{21 - 20} \times r^{13 - 8}\\[1em] = 16 \times p^{-5} \times q^1 \times r^5\\[1em] = \dfrac{16qr^5}{p^5}

Hence, (7p2q9r5)2(4pqr)3(14p6q10r4)2=16qr5p5\dfrac{(7p^2q^9r^5)^2 (4pqr)^3}{(14p^6q^{10}r^4)^2} = \dfrac{16qr^5}{p^5}.

Question 4(i)

Simplify and express the answer in the positive exponent form :

(3)3×266×23\dfrac{(-3)^3 \times 2^6}{6 \times 2^3}

Answer

Solving,

(3)3×266×23=(1)3×33×262×3×23=33×263×24=331×264=32×22=(22×32)\Rightarrow \dfrac{(-3)^3 \times 2^6}{6 \times 2^3}\\[1em] = \dfrac{(-1)^3 \times 3^3 \times 2^6}{2 \times 3 \times 2^3}\\[1em] = \dfrac{-3^3 \times 2^6}{3 \times 2^4}\\[1em] = -3^{3-1} \times 2^{6-4}\\[1em] = -3^2 \times 2^2\\[1em] = -(2^2 \times 3^2)

Hence, (3)3×266×23=(22×32)\dfrac{(-3)^3 \times 2^6}{6 \times 2^3} = -(2^2 \times 3^2).

Question 4(ii)

Simplify and express the answer in the positive exponent form :

(23)5×5443×52\dfrac{(2^3)^5 \times 5^4}{4^3 \times 5^2}

Answer

Solving,

(23)5×5443×52=23×5×54(22)3×52=215×5426×52=2156×542=29×52\Rightarrow \dfrac{(2^3)^5 \times 5^4}{4^3 \times 5^2}\\[1em] = \dfrac{2^{3 \times 5} \times 5^4}{(2^2)^3 \times 5^2}\\[1em] = \dfrac{2^{15} \times 5^4}{2^6 \times 5^2}\\[1em] = 2^{15-6} \times 5^{4-2}\\[1em] = 2^9 \times 5^2

Hence, (23)5×5443×52=29×52\dfrac{(2^3)^5 \times 5^4}{4^3 \times 5^2} = 2^9 \times 5^2.

Question 4(iii)

Simplify and express the answer in the positive exponent form :

36×(6)2×36123×35\dfrac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5}

Answer

Solving,

36×(6)2×36123×35=(22×32)×(22×32)×36(22×3)3×35=24×31026×33×35=24×31026×38=246×3108=22×32=3222=(32)2\Rightarrow \dfrac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5}\\[1em] = \dfrac{(2^2 \times 3^2) \times (2^2 \times 3^2) \times 3^6}{(2^2 \times 3)^3 \times 3^5}\\[1em] = \dfrac{2^4 \times 3^{10}}{2^6 \times 3^3 \times 3^5}\\[1em] = \dfrac{2^4 \times 3^{10}}{2^6 \times 3^8}\\[1em] = 2^{4-6} \times 3^{10-8}\\[1em] = 2^{-2} \times 3^2\\[1em] = \dfrac{3^2}{2^2}\\[1em] = \left(\dfrac{3}{2}\right)^2

Hence, 36×(6)2×36123×35=(32)2\dfrac{36 \times (-6)^2 \times 3^6}{12^3 \times 3^5} = \left(\dfrac{3}{2}\right)^2.

Question 4(iv)

Simplify and express the answer in the positive exponent form :

1282187-\dfrac{128}{2187}

Answer

Solving,

1282187=2×2×2×2×2×2×23×3×3×3×3×3×3=2737=(23)7=(23)7\Rightarrow -\dfrac{128}{2187}\\[1em] = -\dfrac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}\\[1em] = -\dfrac{2^7}{3^7}\\[1em] = -\left(\dfrac{2}{3}\right)^7\\[1em] = \left(-\dfrac{2}{3}\right)^7

Hence, 1282187=(23)7-\dfrac{128}{2187} = \left(-\dfrac{2}{3}\right)^7.

Question 4(v)

Simplify and express the answer in the positive exponent form :

a7×b7×c5×d4a3×b5×c3×d8\dfrac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8}

Answer

Solving,

a7×b7×c5×d4a3×b5×c3×d8=a73×b7(5)×c5(3)×d48=a10×b2×c8×d4=c8a10×b2×d4\Rightarrow \dfrac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8}\\[1em] = a^{-7-3} \times b^{-7-(-5)} \times c^{5-(-3)} \times d^{4-8}\\[1em] = a^{-10} \times b^{-2} \times c^8 \times d^{-4}\\[1em] = \dfrac{c^8}{a^{10} \times b^2 \times d^4}

Hence, a7×b7×c5×d4a3×b5×c3×d8=c8a10b2d4\dfrac{a^{-7} \times b^{-7} \times c^5 \times d^4}{a^3 \times b^{-5} \times c^{-3} \times d^8} = \dfrac{c^8}{a^{10}b^2d^4}.

Question 4(vi)

Simplify and express the answer in the positive exponent form :

(a3b-5)-2

Answer

Solving,

(a3b5)2=a3×(2)×b5×(2)=a6×b10=b10a6\Rightarrow (a^3b^{-5})^{-2} \\[1em] = a^{3 \times (-2)} \times b^{-5 \times (-2)} \\[1em] = a^{-6} \times b^{10} \\[1em] = \dfrac{b^{10}}{a^6}

Hence, (a3b5)2=b10a6(a^3b^{-5})^{-2} = \dfrac{b^{10}}{a^6}.

Question 5(i)

Evaluate :

6-2 ÷ (4-2 × 3-2)

Answer

Solving,

62÷(42×32)=162÷(142×132)=136÷(116×19)=136÷1144=136×144=4\Rightarrow 6^{-2} ÷ (4^{-2} \times 3^{-2})\\[1em] = \dfrac{1}{6^2} ÷ \left(\dfrac{1}{4^2} \times \dfrac{1}{3^2}\right)\\[1em] = \dfrac{1}{36} ÷ \left(\dfrac{1}{16} \times \dfrac{1}{9}\right)\\[1em] = \dfrac{1}{36} ÷ \dfrac{1}{144}\\[1em] = \dfrac{1}{36} \times 144\\[1em] = 4

Hence, 6-2 ÷ (4-2 × 3-2) = 4.

Question 5(ii)

Evaluate :

[(56)2×94]÷[(32)2×125216]\left[\left(\dfrac{5}{6}\right)^2 \times \dfrac{9}{4}\right] \div \left[\left(-\dfrac{3}{2}\right)^2 \times \dfrac{125}{216}\right]

Answer

Solving,

[(56)2×94]÷[(32)2×125216]=[2536×94]÷[94×125216]=2516÷12596=2516×96125=25×9616×125=65=115\Rightarrow \left[\left(\dfrac{5}{6}\right)^2 \times \dfrac{9}{4}\right] \div \left[\left(-\dfrac{3}{2}\right)^2 \times \dfrac{125}{216}\right]\\[1em] = \left[\dfrac{25}{36} \times \dfrac{9}{4}\right] \div \left[\dfrac{9}{4} \times \dfrac{125}{216}\right]\\[1em] = \dfrac{25}{16} \div \dfrac{125}{96}\\[1em] = \dfrac{25}{16} \times \dfrac{96}{125}\\[1em] = \dfrac{25 \times 96}{16 \times 125}\\[1em] = \dfrac{6}{5}\\[1em] = 1\dfrac{1}{5}

Hence, the value is 1151\dfrac{1}{5}.

Question 5(iii)

Evaluate :

53 × 32 + (17)0 × 73

Answer

Solving,

53×32+(17)0×73=(125×9)+(1×343)=1125+343=1468\Rightarrow 5^3 \times 3^2 + (17)^0 \times 7^3 \\[1em] = (125 \times 9) + (1 \times 343) \\[1em] = 1125 + 343 \\[1em] = 1468

Hence, 53 × 32 + (17)0 × 73 = 1468.

Question 5(iv)

Evaluate :

25 × 150 + (-3)3 - (27)2\left(\dfrac{2}{7}\right)^{-2}

Answer

Solving,

25×150+(3)3(27)2=(32×1)+(27)(72)2=3227494=5494=204494=20494=294\Rightarrow 2^5 \times 15^0 + (-3)^3 - \left(\dfrac{2}{7}\right)^{-2}\\[1em] = (32 \times 1) + (-27) - \left(\dfrac{7}{2}\right)^2\\[1em] = 32 - 27 - \dfrac{49}{4}\\[1em] = 5 - \dfrac{49}{4}\\[1em] = \dfrac{20}{4} - \dfrac{49}{4}\\[1em] = \dfrac{20 - 49}{4}\\[1em] = -\dfrac{29}{4}

Hence, the value is 294-\dfrac{29}{4}.

Question 5(v)

Evaluate :

(22)0 + 2-4 ÷ 2-6 + (12)3\left(\dfrac{1}{2}\right)^{-3}

Answer

Solving,

(22)0+24÷26+(12)3=1+24(6)+23=1+22+8=1+4+8=13\Rightarrow (2^2)^0 + 2^{-4} \div 2^{-6} + \left(\dfrac{1}{2}\right)^{-3} \\[1em] = 1 + 2^{-4 - (-6)} + 2^3 \\[1em] = 1 + 2^2 + 8 \\[1em] = 1 + 4 + 8 \\[1em] = 13

Hence, the value is 13.

Question 5(vi)

Evaluate :

5n × 25n-1 ÷ (5n-1 × 25n-1)

Answer

Solving,

5n×25n1÷(5n1×25n1)=5n×25n15n1×25n1=5n5n1=5n(n1)=51=5\Rightarrow 5^n \times 25^{n-1} ÷ (5^{n-1} \times 25^{n-1})\\[1em] = \dfrac{5^n \times 25^{n-1}}{5^{n-1} \times 25^{n-1}}\\[1em] = \dfrac{5^n}{5^{n-1}}\\[1em] = 5^{n-(n-1)}\\[1em] = 5^1\\[1em] = 5

Hence, 5n × 25n-1 ÷ (5n-1 × 25n-1) = 5.

Question 6(i)

If m = - 2 and n = 2; find the value of :

m2 + n2 - 2mn

Answer

Solving,

m2+n22mn=(2)2+(2)22×(2)×(2)=4+4(8)=8+8=16\Rightarrow m^2 + n^2 - 2mn \\[1em] = (-2)^2 + (2)^2 - 2 \times (-2) \times (2) \\[1em] = 4 + 4 - (-8) \\[1em] = 8 + 8 \\[1em] = 16

Hence, m2 + n2 - 2mn = 16.

Question 6(ii)

If m = - 2 and n = 2; find the value of :

mn + nm

Answer

Solving,

mn+nm=(2)2+(2)2=4+122=4+14=164+14=174=414\Rightarrow m^n + n^m\\[1em] = (-2)^2 + (2)^{-2}\\[1em] = 4 + \dfrac{1}{2^2}\\[1em] = 4 + \dfrac{1}{4}\\[1em] = \dfrac{16}{4} + \dfrac{1}{4}\\[1em] = \dfrac{17}{4}\\[1em] = 4\dfrac{1}{4}

Hence, mn+nm=414m^n + n^m = 4\dfrac{1}{4}.

Question 6(iii)

If m = - 2 and n = 2; find the value of :

6m-3 + 4n2

Answer

Solving,

6m3+4n2=6×(2)3+4×(2)2=6(2)3+4×4=68+16=34+16=34+644=614=1514\Rightarrow 6m^{-3} + 4n^2\\[1em] = 6 \times (-2)^{-3} + 4 \times (2)^2\\[1em] = \dfrac{6}{(-2)^3} + 4 \times 4\\[1em] = \dfrac{6}{-8} + 16\\[1em] = -\dfrac{3}{4} + 16\\[1em] = -\dfrac{3}{4} + \dfrac{64}{4}\\[1em] = \dfrac{61}{4}\\[1em] = 15\dfrac{1}{4}

Hence, 6m3+4n2=15146m^{-3} + 4n^2 = 15\dfrac{1}{4}.

Question 6(iv)

If m = - 2 and n = 2; find the value of :

2n3 - 3m

Answer

Solving,

2n33m=2×(2)33×(2)=2×8(6)=16+6=22\Rightarrow 2n^3 - 3m \\[1em] = 2 \times (2)^3 - 3 \times (-2) \\[1em] = 2 \times 8 - (-6) \\[1em] = 16 + 6 \\[1em] = 22

Hence, 2n3 - 3m = 22.

Question 7

State true or false :

(i) 8 × 815 = 6416

(ii) 168 ÷ 42 = 46

(iii) 270 = 549030

(iv) (-1)n = 1, if n is an even whole number

(v) (-1)n = -1, if n is an odd or even whole number

(vi) (-3)-3 = +9

(vii) 4-4 = -16

Answer

(i) False. 8 × 815 = 81 + 15 = 816, whereas 6416 = (82)16 = 832. Since 816 ≠ 832, the statement is false.

(ii) False. 168 ÷ 42 = (42)8 ÷ 42 = 416 ÷ 42 = 414, which is not equal to 46.

(iii) True. Any non-zero base raised to the power zero is equal to 1, so 270 = 1 = 549030.

(iv) True. When n is an even whole number, (-1)n = 1.

(v) False. Here the expression is read as (-1)n. For an even whole number n, (-1)n = 1 and not -1; it equals -1 only when n is odd. So the value is not -1 for every odd or even whole number n.

(vi) False. (3)3=1(3)3=127=127(-3)^{-3} = \dfrac{1}{(-3)^3} = \dfrac{1}{-27} = -\dfrac{1}{27}, which is not +9.

(vii) False. 44=144=12564^{-4} = \dfrac{1}{4^4} = \dfrac{1}{256}, which is not -16.

Multiple Choice Questions

Question 1

(23)3\left(-\dfrac{2}{3}\right)^{-3} is equal to :

  1. 278\dfrac{27}{8}

  2. 827\dfrac{-8}{27}

  3. 278\dfrac{-27}{8}

  4. 827\dfrac{8}{27}

Answer

Solving,

(23)3=(32)3=(3)×(3)×(3)2×2×2=278\Rightarrow \left(-\dfrac{2}{3}\right)^{-3}\\[1em] = \left(-\dfrac{3}{2}\right)^{3}\\[1em] = \dfrac{(-3) \times (-3) \times (-3)}{2 \times 2 \times 2}\\[1em] = \dfrac{-27}{8}

Hence, Option 3 is the correct option.

Question 2

(-3)2 ÷ (12)3\left(-\dfrac{1}{2}\right)^3 is equal to :

  1. 98-\dfrac{9}{8}

  2. 98\dfrac{9}{8}

  3. -72

  4. 72

Answer

Solving,

(3)2÷(12)3=9÷(18)=9×(8)=72\Rightarrow (-3)^2 ÷ \left(-\dfrac{1}{2}\right)^3\\[1em] = 9 ÷ \left(-\dfrac{1}{8}\right)\\[1em] = 9 \times (-8)\\[1em] = -72

Hence, Option 3 is the correct option.

Question 3

The reciprocal of (-2)5 is :

  1. -32

  2. 132-\dfrac{1}{32}

  3. 32

  4. 132\dfrac{1}{32}

Answer

Solving,

⇒ (-2)5 = (-2) × (-2) × (-2) × (-2) × (-2) = -32.

Reciprocal of -32 = 132=132\dfrac{1}{-32} = -\dfrac{1}{32}

Hence, Option 2 is the correct option.

Question 4

If (15)3×(15)x+3\left(\dfrac{1}{5}\right)^3 \times \left(\dfrac{1}{5}\right)^{x+3} = 5-2, then x is equal to :

  1. 4

  2. 14\dfrac{1}{4}

  3. 14-\dfrac{1}{4}

  4. -4

Answer

Solving,

(15)3×(15)x+3=52(15)3+x+3=52(15)x+6=52(51)x+6=525(x+6)=52\Rightarrow \left(\dfrac{1}{5}\right)^3 \times \left(\dfrac{1}{5}\right)^{x+3} = 5^{-2}\\[1em] \Rightarrow \left(\dfrac{1}{5}\right)^{3 + x + 3} = 5^{-2}\\[1em] \Rightarrow \left(\dfrac{1}{5}\right)^{x + 6} = 5^{-2}\\[1em] \Rightarrow (5^{-1})^{x + 6} = 5^{-2}\\[1em] \Rightarrow 5^{-(x + 6)} = 5^{-2}

Comparing the powers of the same base,

⇒ -(x + 6) = -2

⇒ x + 6 = 2

⇒ x = -4

Hence, Option 4 is the correct option.

Question 5

40 + 60 - 80 is equal to :

  1. 0

  2. 1

  3. 2

  4. 12-\dfrac{1}{2}

Answer

As any non-zero base raised to the power zero is equal to 1,

⇒ 40 + 60 - 80

⇒ 1 + 1 - 1

⇒ 1.

Hence, Option 2 is the correct option.

Question 6

(23)2 ÷ (22)4 is equal to :

  1. 4

  2. -4

  3. 14-\dfrac{1}{4}

  4. 14\dfrac{1}{4}

Answer

Solving,

⇒ (23)2 ÷ (22)4

= 23 × 2 ÷ 22 × 4

= 26 ÷ 28

= 26-8

= 2-2

= 122=14\dfrac{1}{2^2} = \dfrac{1}{4}

Hence, Option 4 is the correct option.

Question 7

If (23)x=24332\left(-\dfrac{2}{3}\right)^x = -\dfrac{243}{32}, then x is equal to :

  1. -5

  2. 5

  3. 15\dfrac{1}{5}

  4. 15-\dfrac{1}{5}

Answer

Solving,

(23)x=24332(23)x=3525(23)x=(32)5(23)x=(23)5\Rightarrow \left(-\dfrac{2}{3}\right)^x = -\dfrac{243}{32}\\[1em] \Rightarrow \left(-\dfrac{2}{3}\right)^x = -\dfrac{3^5}{2^5}\\[1em] \Rightarrow \left(-\dfrac{2}{3}\right)^x = \left(-\dfrac{3}{2}\right)^5\\[1em] \Rightarrow \left(-\dfrac{2}{3}\right)^x = \left(-\dfrac{2}{3}\right)^{-5}

Comparing the powers of the same base,

⇒ x = -5.

Hence, Option 1 is the correct option.

Question 8

(-4)3 ÷ (4)4 is equal to :

  1. 4

  2. -4

  3. 14-\dfrac{1}{4}

  4. 14\dfrac{1}{4}

Answer

Solving,

(4)3÷(4)4=(4)344=64256=14\Rightarrow (-4)^3 ÷ (4)^4\\[1em] = \dfrac{(-4)^3}{4^4}\\[1em] = \dfrac{-64}{256}\\[1em] = -\dfrac{1}{4}

Hence, Option 3 is the correct option.

Question 9

70 × 5 - (-2)3 - 80 is equal to :

  1. 12

  2. -12

  3. 112\dfrac{1}{12}

  4. 112-\dfrac{1}{12}

Answer

Solving,

⇒ 70 × 5 - (-2)3 - 80

⇒ (1 × 5) - (-8) - 1

⇒ 5 + 8 - 1

⇒ 12.

Hence, Option 1 is the correct option.

Question 10

If (910)2×(109)5=(910)1m\left(\dfrac{9}{10}\right)^2 \times \left(\dfrac{10}{9}\right)^5 = \left(\dfrac{9}{10}\right)^{1-m}, then m is equal to :

  1. 4

  2. -4

  3. 14\dfrac{1}{4}

  4. 14-\dfrac{1}{4}

Answer

Solving,

(910)2×(109)5=(910)1m(910)2×(910)5=(910)1m(910)25=(910)1m(910)3=(910)1m\Rightarrow \left(\dfrac{9}{10}\right)^2 \times \left(\dfrac{10}{9}\right)^5 = \left(\dfrac{9}{10}\right)^{1-m}\\[1em] \Rightarrow \left(\dfrac{9}{10}\right)^2 \times \left(\dfrac{9}{10}\right)^{-5} = \left(\dfrac{9}{10}\right)^{1-m}\\[1em] \Rightarrow \left(\dfrac{9}{10}\right)^{2 - 5} = \left(\dfrac{9}{10}\right)^{1-m}\\[1em] \Rightarrow \left(\dfrac{9}{10}\right)^{-3} = \left(\dfrac{9}{10}\right)^{1-m}

Comparing the powers of the same base,

⇒ 1 - m = -3

⇒ m = 4

Hence, Option 1 is the correct option.

Question 11

(35)1÷(52)1\left(\dfrac{3}{5}\right)^{-1} \div \left(\dfrac{-5}{2}\right)^{-1} is equal to :

  1. 256\dfrac{25}{6}

  2. 256-\dfrac{25}{6}

  3. 625\dfrac{6}{25}

  4. 625-\dfrac{6}{25}

Answer

Solving,

(35)1÷(52)1=53÷(25)=53×(52)=256\Rightarrow \left(\dfrac{3}{5}\right)^{-1} \div \left(\dfrac{-5}{2}\right)^{-1}\\[1em] = \dfrac{5}{3} \div \left(-\dfrac{2}{5}\right)\\[1em] = \dfrac{5}{3} \times \left(-\dfrac{5}{2}\right)\\[1em] = -\dfrac{25}{6}

Hence, Option 2 is the correct option.

Question 12

If (-3)x - 1 = - 243, then (243)x-6 is equal to :

  1. (243)-12

  2. 243

  3. 0

  4. 1

Answer

Solving,

⇒ (-3)x - 1 = -243

⇒ (-3)x - 1 = (-3)5

Comparing the powers of the same base,

⇒ x - 1 = 5

⇒ x = 6

Now,

⇒ (243)x-6 = (243)6 - 6 = (243)0 = 1.

Hence, Option 4 is the correct option.

Question 13

{(52)3 × 54} ÷ 53 is equal to :

  1. 513

  2. 57

  3. 521

  4. none of these

Answer

⇒ {(52)3 × 54} ÷ 53

⇒ {52 × 3 × 54} ÷ 53

⇒ {56 × 54} ÷ 53

⇒ 56 + 4 ÷ 53

⇒ 510 ÷ 53

⇒ 510 - 3

⇒ 57.

Hence, Option 2 is the correct option.

Question 14

If (3-1) × x = (6)-1, then the value of x is :

  1. 12\dfrac{1}{2}

  2. 12-\dfrac{1}{2}

  3. 2

  4. -2

Answer

Solving,

(31)×x=(6)113×x=16x=16×3x=36x=12\Rightarrow (3^{-1}) \times x = (6)^{-1}\\[1em] \Rightarrow \dfrac{1}{3} \times x = \dfrac{1}{6}\\[1em] \Rightarrow x = \dfrac{1}{6} \times 3\\[1em] \Rightarrow x = \dfrac{3}{6}\\[1em] \Rightarrow x = \dfrac{1}{2}

Hence, Option 1 is the correct option.

Question 15

If (-9)-1 ÷ x = (18)-1, then the value of x is :

  1. 2

  2. -2

  3. 12\dfrac{1}{2}

  4. 12-\dfrac{1}{2}

Answer

Solving,

(9)1÷x=(18)119÷x=11819×1x=1181x=118×(9)1x=9181x=12x=2\Rightarrow (-9)^{-1} ÷ x = (18)^{-1}\\[1em] \Rightarrow -\dfrac{1}{9} ÷ x = \dfrac{1}{18}\\[1em] \Rightarrow -\dfrac{1}{9} \times \dfrac{1}{x} = \dfrac{1}{18}\\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{18} \times (-9)\\[1em] \Rightarrow \dfrac{1}{x} = -\dfrac{9}{18}\\[1em] \Rightarrow \dfrac{1}{x} = -\dfrac{1}{2}\\[1em] \Rightarrow x = -2

Hence, Option 2 is the correct option.

Statement I-II Type Questions

Question 16

Statement 1 : Exponential form of 2048 is 212.

Statement 2 : For any number 'a' and a positive integer 'n', we define an = a × a × a × .... a(n times).

Which of the following options is correct ?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

By prime factorisation,

22048210242512225621282642322162824221\begin{array}{l|r} 2 & 2048 \\ \hline 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}

2048 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 211, not 212.

Thus, Statement 1 is false.

For any number 'a' and a positive integer 'n', an = a × a × a × .... a (n times) is the correct definition.

Thus, Statement 2 is true.

∴ Statement 1 is false, and statement 2 is true.

Hence, Option 4 is the correct option.

Question 17

Statement 1 : (a2)14÷(a3)16=1\left(a^2\right)^{\dfrac{1}{4}} \div \left(a^3\right)^{\dfrac{1}{6}} = 1

Statement 2 : If exponent of the exponent is given, then we multiply the exponents, i.e. (an)m = amn.

Which of the following options is correct ?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Solving,

(a2)14÷(a3)16=a2×14÷a3×16=a12÷a12=a1212=a0=1\left(a^2\right)^{\frac{1}{4}} \div \left(a^3\right)^{\frac{1}{6}}\\[1em] = a^{2 \times \frac{1}{4}} \div a^{3 \times \frac{1}{6}}\\[1em] = a^{\frac{1}{2}} \div a^{\frac{1}{2}}\\[1em] = a^{\frac{1}{2} - \frac{1}{2}}\\[1em] = a^0\\[1em] = 1

Thus, Statement 1 is true.

The power law states that (an)m = amn, which is correct.

Thus, Statement 2 is true.

∴ Both the statements are true.

Hence, Option 1 is the correct option.

Assertion-Reason Type Questions

Question 18

Assertion (A) : (90 + 80) ÷ (72)0 = 1

Reason (R) : Any non-zero base raised to the power zero is equal to unity (i.e. 1).

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Solving,

⇒ (90 + 80) ÷ (72)0

⇒ (1 + 1) ÷ 1

⇒ 2 ÷ 1

⇒ 2.

Since the value is 2 and not 1,

Thus, Assertion (A) is false.

Any non-zero base raised to the power zero is equal to 1, which is a correct statement.

Thus, Reason (R) is true.

∴ A is false, R is true.

Hence, Option 2 is the correct option.

Question 19

Assertion (A) : -1n is always equal to -1, n is even or odd whole number.

Reason (R) : (-1)n = -1n for all n ∈ W.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Here -1n means -(1n). Since 1n = 1 for every whole number n, -1n = -1 for all whole numbers n.

Thus, Assertion (A) is true.

(-1)n is +1 when n is even and -1 when n is odd. So (-1)n is not equal to -1n for all n ∈ W.

Thus, Reason (R) is false.

∴ A is true, R is false.

Hence, Option 1 is the correct option.

Question 20

Assertion (A) : (34)×(34)×(34)×\left(-\dfrac{3}{4}\right) \times \left(-\dfrac{3}{4}\right) \times \left(-\dfrac{3}{4}\right) \times .... up to 10 terms = 3a2b\dfrac{3^a}{2^b}, then a - b = -10.

Reason (R) : If an expression is in the form of fraction with the same power, then that power will be taken for numerator and denominator both, i.e. (pq)n=pnqn\left(\dfrac{p}{q}\right)^n = \dfrac{p^n}{q^n}.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Solving,

(34)×(34)×(34)× up to 10 terms=(34)10=(3)10410=310(22)10=310220\left(-\dfrac{3}{4}\right) \times \left(-\dfrac{3}{4}\right) \times \left(-\dfrac{3}{4}\right) \times \dots \text{ up to 10 terms}\\[1em] = \left(-\dfrac{3}{4}\right)^{10}\\[1em] = \dfrac{(-3)^{10}}{4^{10}}\\[1em] = \dfrac{3^{10}}{(2^2)^{10}}\\[1em] = \dfrac{3^{10}}{2^{20}}

Comparing with 3a2b\dfrac{3^a}{2^b}, we get a = 10 and b = 20.

⇒ a - b = 10 - 20 = -10.

Thus, Assertion (A) is true.

For an expression in the form of a fraction with the same power, (pq)n=pnqn\left(\dfrac{p}{q}\right)^n = \dfrac{p^n}{q^n}, which is correct.

Thus, Reason (R) is true.

∴ Both A and R are true.

Hence, Option 3 is the correct option.

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