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Practice each question, then open the answer or worked solution.
For a non-zero integer x, x7 ÷ x12 is equal to
(a) x5
(b) x19
(c) x–5
(d) x–19
Ans: Option (c) is correct.
Use law of exponent $\dfrac{{{a^m}}}{{{a^n}}} = {\left( a \right)^{m - n}}$, where a is a non zero integer.
Thus, $\dfrac{{{x^7}}}{{{x^{12}}}} = {x^{7 - 12}} = {x^{ - 5}}$.
For a non-zero integer x, (x4)–3 is equal to
(a) x12
(b) x–12
(c) x64
(d) x–64
Ans: Option (b) is correct.
Use law of exponent ${\left( {{x^m}} \right)^n} = {\left( x \right)^{m \times n}} = {\left( x \right)^{mn}}$, where a is a non zero integer.
Thus, ${\left( {{x^4}} \right)^{ - 3}} = {x^{4 \times \left( { - 3} \right)}} = {x^{ - 12}}$
The value of (7–1 – 8–1)–1 – (3–1 – 4–1)–1 is
(a) 44
(b) 56
(c) 68
(d) 12
Ans: Option (a) is correct.
Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.
Thus,
${\left( {\dfrac{1}{7} - \dfrac{1}{8}} \right)^{ - 1}} - {\left( {\dfrac{1}{3} - \dfrac{1}{4}} \right)^{ - 1}} = {\left( {\dfrac{{8 - 7}}{{56}}} \right)^{ - 1}} - {\left( {\dfrac{{4 - 3}}{{12}}} \right)^{ - 1}}$
= ${\left( {\dfrac{1}{{56}}} \right)^{ - 1}} - {\left( {\dfrac{1}{{12}}} \right)^{ - 1}} = 56 - 12 = 44$
The standard form for 0.000064 is
(a) 64 × 104
(b) 64 × 10–4
(c) 6.4 × 105
(d) 6.4 × 10–5
Ans: Option (d) is correct.
Given, 0.000064 = 0.64 × 10-4 = 6.4 × 10-5
The standard form for 234000000 is
(a) 2.34 × 108
(b) 0.234 × 109
(c) 2.34 × 10–8
(d) 0.234 × 10–9
Ans: Option (a) is correct.
234000000 = 234 x 106 = 2.34 x 102+6
= 2.34 x 108
The usual form for 2.03 × 10–5
(a) 0.203
(b) 0.00203
(c) 203000
(d) 0.0000203
Ans: Option (d) is correct.
Use the law of exponent ${a^{ - m}} = \dfrac{1}{{{a^m}}}$, where a is a non zero integer.
Thus,
$2.03 \times {10^{ - 5}} = \dfrac{{2.03}}{{{{10}^5}}} = 0.0000203$
${\left( {\dfrac{1}{{10}}} \right)^0}$is equal to
(a) 0
(b) $\dfrac{1}{{10}}$
(c) 1
(d) 10
Ans: Option (c) is correct.
Use the law of exponent ${a^0} = 1$, where a is a non zero integer.
Thus,
${\left( {\dfrac{1}{{10}}} \right)^0} = 1$
${\left( {\dfrac{3}{4}} \right)^5} \div {\left( {\dfrac{5}{3}} \right)^5}$is equal to
(a) ${\left( {\dfrac{3}{4} \div \dfrac{5}{3}} \right)^5}$
(b) ${\left( {\dfrac{3}{4} \div \dfrac{5}{3}} \right)^1}$
(c) ${\left( {\dfrac{3}{4} \div \dfrac{5}{3}} \right)^0}$
(d) ${\left( {\dfrac{3}{4} \div \dfrac{5}{3}} \right)^{10}}$
Ans: Option (a) is correct.
Use the law of exponent ${a^m} \div {b^m} = {\left( {a \div b} \right)^m}$, where a is a non zero integer.
Thus,
${\left( {\dfrac{3}{4}} \right)^5} \div {\left( {\dfrac{5}{3}} \right)^5} = {\left( {\dfrac{3}{4} \div \dfrac{5}{3}} \right)^5}$
For any two non-zero rational numbers x and y, x4 ÷ y4 is equal to
(a) (x ÷ y)0
(b) (x ÷ y)1
(c) (x ÷ y)4
(d) (x ÷ y)8
Ans: Option (c) is correct.
Use the law of exponent ${a^m} \div {b^m} = {\left( {a \div b} \right)^m}$, where a is a non zero integer.
Thus,
${x^4} \div {y^4} = {\left( {\dfrac{x}{y}} \right)^4} = {\left( {x \div y} \right)^4}$
For a non-zero rational number p, p13 ÷ p8 is equal to
(a) p5
(b) p21
(c) p–5
(d) p–19
Ans: Option (a) is correct.
Use the law of exponent ${a^m} \div {a^n} = {\left( a \right)^{m - n}}$, where a is a non zero integer.
Thus,
${p^{13}} \div {p^8} = {\left( p \right)^{13 - 8}} = {p^5}$