What is the simplified form of (\frac{(3x^{2})^{3}(2x^{-1})^{2}}{6x^{4}})?
Correct answer: A
The numerator is ((3x^{2})^{3}(2x^{-1})^{2}=27x^{6}\cdot4x^{-2}=108x^{4}). Then (\frac{108x^{4}}{6x^{4}}=18), so check cancellation of powers.
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.
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Expert · Level 2
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- (18)
- (12x^{2})
- (18x^{0})
- (36)
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- (9)
- (7)
- (5)
- (3)
Expert · Level 2
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- (6)
- (4\sqrt{2})
- (3)
- (2\sqrt{2})
Expert · Level 2
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- \(\frac{y}{xz}\)
- \(\frac{x}{yz}\)
- \(\frac{yz}{x}\)
- \(\frac{1}{xyz}\)
Expert · Level 2
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- \(\frac{1}{2}\)
- (2)
- (4)
- \(\frac{1}{4}\)
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- 3
- 4
- 5
- 6
Expert · Level 2
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- (5a^{3}b^{2})
- (25a^{3}b^{2})
- (5a^{2}b^{3})
- (5a^{6}b^{3})
Expert · Level 2
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- (x^{3}+1)
- (x^{3}-1)
- (x^{2}+1)
- (x^{6}+1)
Expert · Level 2
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- (2\sqrt{15})
- (8)
- (\sqrt{15})
- (4\sqrt{15})
Expert · Level 2
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- (6)
- (8)
- (10)
- (12)
Expert · Level 2
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- (7)
- (3)
- (5)
- (2)
Expert · Level 2
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- \(5m^{6}n^{-4}\)
- \(\frac{1}{5}m^{-6}n^{4}\)
- \(5m^{-6}n^{4}\)
- \(\frac{1}{5}m^{6}n^{-4}\)
Expert · Level 2
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- (\frac{9}{2})
- (\frac{7}{2})
- (4)
- (5)
Expert · Level 2
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- (0)
- (2)
- (4)
- (6)
Expert · Level 2
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- (2^{3}\cdot3)
- (2^{4}\cdot3)
- (2^{3}\cdot3^{2})
- (2^{2}\cdot3)
Expert · Level 2
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- (8\sqrt{7})
- (4\sqrt{7})
- (16\sqrt{7})
- (14)
Expert · Level 2
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- \(\frac{2xy^{2}}{3}\)
- \(\frac{3}{2xy^{2}}\)
- \(\frac{3x}{2y^{2}}\)
- \(\frac{2y^{2}}{3x}\)
Expert · Level 2
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- (6)
- (4)
- (8)
- (2)
Expert · Level 2
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- \(a\)
- \(a^{-1}\)
- \(a^2\)
- \(1\)
Expert · Level 2
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- (54\sqrt{2})
- (18\sqrt{2})
- (27\sqrt{2})
- (36\sqrt{2})
Expert · Level 2
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- \(\frac{706}{225}\)
- \(\frac{34}{15}\)
- \(\frac{225}{706}\)
- \(\frac{106}{45}\)
Expert · Level 2
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- \(\frac{7}{3}\)
- (2)
- \(\frac{5}{2}\)
- (3)
Expert · Level 2
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- (16)
- (14)
- (12)
- (10)
Expert · Level 2
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- (1)
- (2)
- (\sqrt{a+b})
- (a+b)
Expert · Level 2
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- \(\frac{x^{6}}{4}\)
- \(\frac{x^{10}}{4}\)
- \(4x^{-6}\)
- \(4x^{6}\)
What is the value of (\frac{\sqrt{75}+\sqrt{48}}{\sqrt{3}})?
Correct answer: A
Here (\sqrt{75}=5\sqrt{3}) and (\sqrt{48}=4\sqrt{3}), so the numerator is (9\sqrt{3}). Dividing by (\sqrt{3}) gives (9).
If (y=3+2\sqrt{2}), what is the value of (y+\frac{1}{y})?
Correct answer: A
Since (\frac{1}{3+2\sqrt{2}}=3-2\sqrt{2}), the sum is (6). In exams, use the conjugate quickly for such reciprocals.
What is the simplified form of \(\left(\frac{x^{3}y^{-2}}{z^{-1}}\right)^{-1}\cdot\frac{x^{2}}{yz^{2}}\)?
Correct answer: A
Inside, \(\frac{x^{3}y^{-2}}{z^{-1}}=x^{3}y^{-2}z\), so its reciprocal is \(x^{-3}y^{2}z^{-1}\). Multiplying by \(\frac{x^{2}}{yz^{2}}\) gives \(\frac{y}{xz^{3}}\), so the (z)-power must be checked carefully.
What is the value of \(\left(32^{\frac{2}{5}}\right)\cdot\left(4^{-\frac{3}{2}}\right)\)?
Correct answer: A
Here \(32^{\frac{2}{5}}=(2^{5})^{\frac{2}{5}}=2^{2}=4\), and \(4^{-\frac{3}{2}}=(2^{2})^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}\). The product is \(\frac{1}{2}\).
If \((x^{2}y^{-1})^{k}=x^{10}y^{-5}\), where \(x,y\neq 0\), what is the value of \(k\)?
Correct answer: C
Using the exponent rule \((a^m)^n=a^{mn}\), we get \((x^{2}y^{-1})^k=x^{2k}y^{-k}\). Comparing the exponents of the same bases gives \(2k=10\) and \(-k=-5\), so both equations yield \(k=5\). Option 4 is incorrect because it would produce \(x^8y^{-4}\). Exam tip: multiply every exponent inside the bracket by the outside exponent before comparing like bases.
What is the simplified form of (\sqrt[3]{125a^{9}b^{6}})?
Correct answer: A
We have (\sqrt[3]{125}=5), (\sqrt[3]{a^{9}}=a^{3}), and (\sqrt[3]{b^{6}}=b^{2}). In exams, divide exponents by (3) under a cube root.
Which option is the simplified form of (\frac{x^{6}-1}{x^{3}-1}), where (x^{3}\neq1)?
Correct answer: A
Since (x^{6}-1=(x^{3}-1)(x^{3}+1)), cancelling the common factor gives (x^{3}+1). In exams, recognize the (A^{2}-B^{2}) form.
If (p=4-\sqrt{15}), what is the value of (\frac{1}{p}-p)?
Correct answer: A
Since (\frac{1}{4-\sqrt{15}}=4+\sqrt{15}), (\frac{1}{p}-p=(4+\sqrt{15})-(4-\sqrt{15})=2\sqrt{15}). In exams, the conjugate gives the reciprocal directly when the denominator product is (1).
What is the value of (\frac{(2^{-3}+2^{-4})}{2^{-5}})?
Correct answer: A
Here (2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}), and (2^{-5}=\frac{1}{32}). Therefore, the value is (\frac{3}{16}\div\frac{1}{32}=6).
If (x=\sqrt{5}-\sqrt{2}), what is the value of (x^{2}+2\sqrt{10})?
Correct answer: A
Since (x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}), (x^{2}+2\sqrt{10}=7). In exams, write the middle term of ((a-b)^{2}) carefully.
What is the simplified form of \(\left(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}\right)^{-1}\)?
Correct answer: A
Inside, \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), so raising to (-1) gives \(5m^{6}n^{-4}\). In exams, do not forget to invert the coefficient too.
If (6^{x}=216) and (36^{y}=216), what is the value of (x+y)?
Correct answer: A
Since (216=6^{3}), (x=3). Also (36^{y}=(6^{2})^{y}=6^{2y}=6^{3}), so (y=\frac{3}{2}) and the sum is (\frac{9}{2}).
What is the value of \(\left(\sqrt{13}+\sqrt{3}\right)\left(\sqrt{13}-\sqrt{3}\right)-\sqrt{100}\)?
Correct answer: A
The conjugate product is \(13-3=10\), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
What is the simplified form of (\frac{12^{4}}{2^{5}\cdot3^{3}})?
Correct answer: A
Since (12^{4}=(2^{2}\cdot3)^{4}=2^{8}\cdot3^{4}), division leaves (2^{3}\cdot3). In exams, prime-factorize first.
If (s=2+\sqrt{7}), what is the value of (s^{2}-\frac{1}{s^{2}})?
Correct answer: A
Here (\frac{1}{s}=\sqrt{7}-2), so (s-\frac{1}{s}=4) and (s+\frac{1}{s}=2\sqrt{7}). Thus (s^{2}-\frac{1}{s^{2}}=8\sqrt{7}).
What is the simplified form of \(\left(\frac{27x^{-3}}{8y^{6}}\right)^{-\frac{1}{3}}\)?
Correct answer: A
We get \(\left(\frac{27x^{-3}}{8y^{6}}\right)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}\), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.
If \(x^{2}-\frac{1}{x^{2}}=24\) and \(x-\frac{1}{x}=4\), what is the value of \(x+\frac{1}{x}\)?
Correct answer: A
Since \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\), \(24=4\left(x+\frac{1}{x}\right)\). In exams, use the difference of squares identity.
What is the simplified form of \(\frac{2a^{-1}+3a^{-1}}{5a^{-2}}\), where \(a\neq 0\)?
Correct answer: A
Combining like terms gives \(2a^{-1}+3a^{-1}=5a^{-1}\). Therefore, \(\frac{5a^{-1}}{5a^{-2}}=a^{-1-(-2)}=a^1=a\), so option A is correct. Exam tip: when dividing powers with the same non-zero base, subtract the exponents: \(a^m/a^n=a^{m-n}\).
If (\sqrt{x}=3\sqrt{2}), what is the value of (x^{\frac{3}{2}})?
Correct answer: A
From (\sqrt{x}=3\sqrt{2}), (x=18), and (x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}). In exams, write (x^{\frac{3}{2}}) as (x\sqrt{x}).
What is the value of \(\left(\frac{3}{5}\right)^{-2}+\left(\frac{5}{3}\right)^{-2}\)?
Correct answer: A
Here \(\left(\frac{3}{5}\right)^{-2}=\frac{25}{9}\) and \(\left(\frac{5}{3}\right)^{-2}=\frac{9}{25}\), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.
If \(\left(3^{x}\right)^{2}\cdot3^{x-1}=729\), what is the value of (x)?
Correct answer: A
The left side is \(3^{2x}\cdot3^{x-1}=3^{3x-1}\), and \(729=3^{6}\). Hence \(3x-1=6\) and \(x=\frac{7}{3}\).
What is the value of (\frac{\sqrt{147}-2\sqrt{12}+3\sqrt{27}}{\sqrt{3}})?
Correct answer: A
Here (\sqrt{147}=7\sqrt{3}), (2\sqrt{12}=4\sqrt{3}), and (3\sqrt{27}=9\sqrt{3}), so the numerator is (12\sqrt{3}). Therefore, the value should be (12).
If (\frac{1}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}) and (a>b>0), what is the value of (a-b)?
Correct answer: A
Multiplying both sides by (\sqrt{a}+\sqrt{b}), we get (1=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b). In exams, apply the conjugate product directly.
If \(x\neq0\), what is the simplified form of \(\left(\frac{2x^{-3}}{x^{2}}\right)^{-2}\cdot x^{-4}\)?
Correct answer: A
Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so \(\left(2x^{-5}\right)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}\). In exams, subtract the inner exponents first.
QUIZ COMPLETE