What is the simplified value of (\frac{6^{5}}{2^{3}\cdot3^{4}})?
Correct answer: A
Since (6^{5}=2^{5}\cdot3^{5}), (\frac{2^{5}3^{5}}{2^{3}3^{4}}=2^{2}\cdot3). In exams, split a composite base into prime bases.
Mathematics
Operations on real numbers and the laws of exponents
वास्तविक संख्याओं पर संक्रियाएँ और घातांक के नियम
In this Class 10 Mathematics topic from the Polynomials chapter, students strengthen their understanding of operations on real numbers and the laws of exponents. They learn to add, subtract, multiply and divide numerical expressions accurately, use exponent rules for products, quotients and powers, and simplify expressions involving positive, zero and negative exponents where appropriate. The topic builds fluency in working with polynomial terms, comparing equivalent forms, and checking calculations through properties such as commutativity, associativity and distributivity. Examples connect numerical rules with algebraic manipulation and related exercises.
TOPIC PRACTICE
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Hard · Level 6
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- (2^{2}\cdot3)
- (2^{3}\cdot3^{2})
- (2\cdot3)
- (2^{5}\cdot3)
Hard · Level 6
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- (2)
- (3)
- (4)
- (6)
Hard · Level 6
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- (4x^{2})
- (8x^{2})
- (4x^{3})
- (2x^{2})
Hard · Level 6
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- \(13-2\sqrt{22}\)
- \(9-2\sqrt{22}\)
- \(13-\sqrt{22}\)
- \(11-2\sqrt{2}\)
Hard · Level 6
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- (27)
- (64)
- (81)
- (12)
Hard · Level 6
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- (x^{2}-1)
- (x^{2})
- (x^{8}-1)
- (x^{2}-x)
Hard · Level 6
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- \(\frac{1}{2}\)
- (2)
- (1)
- \(\frac{1}{4}\)
Hard · Level 6
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- (2+\sqrt{3})
- (3+\sqrt{2})
- (\sqrt{7}+2\sqrt{3})
- (4+\sqrt{3})
Hard · Level 6
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- (x+y)
- (\frac{x+y}{xy})
- (xy(x+y))
- (x^{-1}+y^{-1})
Hard · Level 6
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- (1)
- (25)
- (\sqrt{156})
- (2\sqrt{13})
Hard · Level 6
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- \(x^{-6}y^{5}\)
- \(x^{6}y^{-5}\)
- \(x^{-6}y^{-1}\)
- \(x^{-2}y^{3}\)
Hard · Level 6
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- (4)
- (5)
- (6)
- (10)
Hard · Level 6
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- (1)
- (2)
- (\sqrt{5})
- (5)
Hard · Level 6
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- \((\sqrt{3}-\sqrt{2})^{2}\)
- \((\sqrt{6}-1)^{2}\)
- \((3-\sqrt{2})^{2}\)
- \((2-\sqrt{3})^{2}\)
Hard · Level 6
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- (6)
- \(\frac{3}{2}\)
- \(\frac{27}{8}\)
- \(\frac{2}{3}\)
Hard · Level 6
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- (18)
- (9)
- (6)
- (27)
Hard · Level 6
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- (a^{-2}b)
- (a^{2}b^{-1})
- (a^{-10}b^{5})
- (a^{10}b^{-5})
Hard · Level 6
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- \(9x^{2}\)
- (9x)
- \(81x^{2}\)
- \(3x^{2}\)
Hard · Level 6
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- (2)
- (3)
- (4)
- (5)
Hard · Level 6
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- (2\sqrt{3})
- (2\sqrt{2})
- (\sqrt{3})
- (\sqrt{2})
If \(\left(x^{2}y^{3}\right)^{n}=x^{8}y^{12}\), what is the value of (n)?
Correct answer: C
The left side has exponents (2n) and (3n), so \(2n=8\) and \(3n=12\), giving \(n=4\). In exams, match exponents of both variables.
What is the simplified form of (\sqrt[3]{64x^{6}}), where (x) is real?
Correct answer: A
Since (\sqrt[3]{64}=4) and (\sqrt[3]{x^{6}}=x^{2}), the answer is (4x^{2}). In exams, divide the exponent by (3) for cube roots.
Which option gives the correct expansion of \(\left(\sqrt{11}-\sqrt{2}\right)^{2}\)?
Correct answer: A
\((\sqrt{11}-\sqrt{2})^{2}=11+2-2\sqrt{22}=13-2\sqrt{22}\). In exams, include both \(+b^{2}\) and (-2ab) in \((a-b)^{2}\).
If (2^{a}=8) and (3^{b}=81), what is the value of (a^{b})?
Correct answer: C
From (2^{a}=2^{3}), (a=3), and from (3^{b}=3^{4}), (b=4), so (a^{b}=3^{4}=81). In exams, compare powers using equal bases.
What is the simplified form of (\frac{x^{5}-x^{3}}{x^{3}}), where (x\neq0)?
Correct answer: A
(\frac{x^{5}-x^{3}}{x^{3}}=\frac{x^{3}(x^{2}-1)}{x^{3}}=x^{2}-1). In exams, take out the common factor first.
What is the value of \(\left(16^{-\frac{3}{4}}\right)\cdot8^{\frac{2}{3}}\)?
Correct answer: A
Here \(16^{-\frac{3}{4}}=(2^{4})^{-\frac{3}{4}}=2^{-3}\) and \(8^{\frac{2}{3}}=(2^{3})^{\frac{2}{3}}=2^{2}\), so the value is \(2^{-1}=\frac{1}{2}\). In exams, multiply powers of powers.
If (A=7+4\sqrt{3}), what is the correct simplified form of (\sqrt{A})?
Correct answer: A
Since ((2+\sqrt{3})^{2}=4+3+4\sqrt{3}=7+4\sqrt{3}), (\sqrt{A}=2+\sqrt{3}). In exams, recognize the form ((a+b)^{2}).
Which option gives the simplified form of (\frac{x^{-1}+y^{-1}}{(xy)^{-1}}), where (x\neq0) and (y\neq0)?
Correct answer: A
Here (x^{-1}+y^{-1}=\frac{x+y}{xy}) and ((xy)^{-1}=\frac{1}{xy}), so division gives (x+y). In exams, converting negative exponents to fractions is safer.
If (t=\sqrt{13}+\sqrt{12}), what is the value of (t\cdot(\sqrt{13}-\sqrt{12}))?
Correct answer: A
((\sqrt{13}+\sqrt{12})(\sqrt{13}-\sqrt{12})=13-12=1). In exams, the product of conjugate surds is rational.
What is the simplified form of \(\left(\frac{3x^{-2}}{y^{-1}}\right)^{3}\cdot\frac{y^{2}}{27}\)?
Correct answer: A
\(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), its cube is \(27x^{-6}y^{3}\), and multiplying by \(\frac{y^{2}}{27}\) gives \(x^{-6}y^{5}\). In exams, turn division by a negative power into multiplication.
If (4^{x}=2^{10}), what is the value of (x)?
Correct answer: B
Since (4^{x}=(2^{2})^{x}=2^{2x}), (2x=10) and (x=5). In exams, convert mixed bases into a common base.
What is the value of (\frac{\sqrt{45}-\sqrt{20}}{\sqrt{5}})?
Correct answer: A
(\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the numerator is (\sqrt{5}), and division gives (1). In exams, first make like radicals.
If (z=5-2\sqrt{6}), then (z) can be written as which square?
Correct answer: A
((\sqrt{3}-\sqrt{2})^{2}=3+2-2\sqrt{6}=5-2\sqrt{6}). In exams, identify (a,b) from (a+b) and (2\sqrt{ab}).
What is the value of \(\left(\frac{2}{3}\right)^{-3}\cdot\left(\frac{9}{4}\right)^{-1}\)?
Correct answer: A
\(\left(\frac{2}{3}\right)^{-3}=\left(\frac{3}{2}\right)^{3}=\frac{27}{8}\) and \(\left(\frac{9}{4}\right)^{-1}=\frac{4}{9}\), so the product is (6). In exams, invert the fraction for negative powers.
If (x^{2}=3), what is the value of (x^{6}-x^{4})?
Correct answer: A
Here (x^{6}=(x^{2})^{3}=27) and (x^{4}=(x^{2})^{2}=9), so the difference is (18). In exams, express powers using the given (x^{2}).
What is the simplified form of (\frac{(a^{2}b^{-1})^{-3}}{a^{-4}b^{2}})?
Correct answer: A
((a^{2}b^{-1})^{-3}=a^{-6}b^{3}), then (\frac{a^{-6}b^{3}}{a^{-4}b^{2}}=a^{-2}b). In exams, subtract powers of the same base during division.
What is the simplified form of \(\left(81x^{4}\right)^{\frac{1}{2}}\), where \(x\ge0\)?
Correct answer: A
\(\left(81x^{4}\right)^{\frac{1}{2}}=\sqrt{81x^{4}}=9x^{2}\). In exams, the exponent becomes half under a square root.
If (\frac{2^{x}\cdot2^{x+2}}{2^{3}}=32), what is the value of (x)?
Correct answer: B
The total exponent on the left is (x+x+2-3=2x-1), and (32=2^{5}), so (2x-1=5), giving (x=3). In exams, convert the whole expression into one power.
What is the value of (\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{3}-\sqrt{2}})?
Correct answer: A
The first term becomes (\sqrt{3}-\sqrt{2}), and the second becomes (\sqrt{3}+\sqrt{2}), so the sum is (2\sqrt{3}). In exams, rationalize both denominators separately.
QUIZ COMPLETE