If (p=\sqrt{2}+\sqrt{3}) and (q=\sqrt{3}-\sqrt{2}), what is the value of (pq)?
Correct answer: A
Here (pq=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=3-2=1). In exams, use ((a+b)(a-b)=a^{2}-b^{2}).
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
Quiz this set
Up to 25 questions from this page. Select your focus, then start.
25 questions
Choose questions
Hard · Level 5
View options
- (1)
- (\sqrt{6})
- (5)
- (2\sqrt{6})
Hard · Level 5
View options
- (2+\sqrt{3})
- (2-\sqrt{3})
- (\frac{2+\sqrt{3}}{7})
- (\sqrt{3}-2)
Hard · Level 5
View options
- (\frac{9}{8})
- (\frac{8}{9})
- (72)
- (\frac{1}{72})
Hard · Level 5
View options
- \(\frac{y^{8}}{4x^{6}}\)
- \(4x^{6}y^{-8}\)
- \(\frac{x^{6}}{4y^{8}}\)
- \(2x^{3}y^{-4}\)
Hard · Level 5
View options
- (3)
- (4)
- (5)
- (6)
Hard · Level 5
View options
- (6\sqrt{2})
- (5\sqrt{2})
- (4\sqrt{2})
- (3\sqrt{2})
Hard · Level 5
View options
- \(\frac{4}{9}\)
- \(\frac{9}{4}\)
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
Hard · Level 5
View options
- (\sqrt{5}-2)
- (2-\sqrt{5})
- (\frac{\sqrt{5}-2}{9})
- (\sqrt{5}+2)
Hard · Level 5
View options
- (1)
- \(3^{2}\)
- \(3^{4}\)
- \(3^{6}\)
Hard · Level 5
View options
- \(a^{2}b^{-1}\)
- \(a^{4}b^{-1}\)
- \(a^{2}b\)
- \(a^{-2}b^{-1}\)
Hard · Level 5
View options
- (12\sqrt{2})
- (6\sqrt{2})
- (18)
- (4\sqrt{2})
Hard · Level 5
View options
- (2^{5})
- (2^{6})
- (2^{7})
- (2^{8})
Hard · Level 5
View options
- (x)
- (x^{2})
- (1)
- (\sqrt{x})
Hard · Level 5
View options
- \(x^{6}y^{-4}\)
- \(x^{-6}y^{4}\)
- \(x^{2}y^{-2}\)
- \(x^{-2}y^{2}\)
Hard · Level 5
View options
- (14)
- (12)
- (10)
- (8)
Hard · Level 5
View options
- \(2+2\sqrt{5}\)
- \(12+2\sqrt{5}\)
- \(2+\sqrt{5}\)
- \(4+2\sqrt{5}\)
Hard · Level 5
View options
- (6)
- (7)
- (8)
- (9)
Hard · Level 5
View options
- (12)
- (9)
- (6)
- (4)
Hard · Level 5
View options
- \((x-2)(x+2)\)
- \((x-4)(x+1)\)
- \((x-2)^{2}\)
- \((x+2)^{2}\)
Hard · Level 5
View options
- (8+4\sqrt{3})
- (8+2\sqrt{3})
- (4+4\sqrt{3})
- (6+2\sqrt{3})
Hard · Level 5
View options
- (10)
- (100)
- (\frac{1}{10})
- (1)
Hard · Level 5
View options
- \(a^{8}b^{-8}\)
- \(a^{4}b^{-4}\)
- \(a^{6}b^{-4}\)
- \(a^{2}b^{0}\)
Hard · Level 5
View options
- (4)
- (\frac{1}{4})
- (2)
- (\frac{1}{2})
Hard · Level 5
View options
- \(a^{0}=1\)
- \(a^{0}=0\)
- \(a^{-1}=a\)
- \(a^{1}=1\)
Hard · Level 5
View options
- (4)
- (2\sqrt{3})
- (2)
- (1)
What is the rationalized form of (\frac{1}{2-\sqrt{3}})?
Correct answer: A
To rationalize, (\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}). In exams, multiply by the conjugate.
If (x=2^{3}\cdot3^{-2}), then (x^{-1}) is equal to which expression?
Correct answer: A
Here (x=\frac{8}{9}), so (x^{-1}=\frac{9}{8}). In exams, apply (a^{-n}=\frac{1}{a^{n}}) in the correct direction.
What is the simplified form of \(\left(\frac{4x^{2}y^{-3}}{2x^{-1}y}\right)^{-2}\), where \(x\neq0\) and \(y\neq0\)?
Correct answer: A
Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.
If (2^{x+1}+2^{x}=48), what is the value of (x)?
Correct answer: B
Here (2^{x+1}+2^{x}=2\cdot2^{x}+2^{x}=3\cdot2^{x}=48), so (2^{x}=16=2^{4}). In exams, factor the common power (2^{x}).
What is the simplified form of (\sqrt{50}+\sqrt{18}-\sqrt{8})?
Correct answer: A
We get (\sqrt{50}=5\sqrt{2}), (\sqrt{18}=3\sqrt{2}), and (\sqrt{8}=2\sqrt{2}), so the result is (6\sqrt{2}). In exams, combine only like surd terms.
What is the value of \(\left(\frac{27}{8}\right)^{-\frac{2}{3}}\)?
Correct answer: A
Since \(\left(\frac{27}{8}\right)^{\frac{1}{3}}=\frac{3}{2}\), \(\left(\frac{27}{8}\right)^{-\frac{2}{3}}=\left(\frac{3}{2}\right)^{-2}=\frac{4}{9}\). In exams, take the cube root first.
If (x=\sqrt{5}+2), then (\frac{1}{x}) is equal to which expression?
Correct answer: A
Rationalizing gives (\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2). In exams, use the conjugate of the denominator.
What is the value of \(\left(9^{2}\right)^{3}\div 3^{10}\)?
Correct answer: B
Here \(\left(9^{2}\right)^{3}=(3^{2})^{6}=3^{12}\), and \(3^{12}\div3^{10}=3^{2}\). In exams, write (9) as \(3^{2}\).
Which option gives the correct simplified form of \(\left(ab^{-2}\right)^{3}\cdot a^{-1}b^{5}\)?
Correct answer: A
We have \(\left(ab^{-2}\right)^{3}=a^{3}b^{-6}\), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.
If (x=3+\sqrt{2}) and (y=3-\sqrt{2}), what is the value of (x^{2}-y^{2})?
Correct answer: A
Using (x^{2}-y^{2}=(x-y)(x+y)), we get (x-y=2\sqrt{2}) and (x+y=6), so the value is (12\sqrt{2}). In exams, identities reduce calculation.
What is the simplified value of (\frac{(2^{5})^{3}\cdot(4^{-2})}{8^{2}})?
Correct answer: A
Here ((2^{5})^{3}=2^{15}), (4^{-2}=2^{-4}), and (8^{2}=2^{6}), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).
If (\sqrt{x}=x^{\frac{1}{2}}) and (x>0), then (\sqrt{x^{3}}\cdot x^{-\frac{1}{2}}) equals which expression?
Correct answer: A
Since (\sqrt{x^{3}}=x^{\frac{3}{2}}), (x^{\frac{3}{2}}\cdot x^{-\frac{1}{2}}=x^{1}=x). In exams, convert radicals to fractional exponents.
What is the simplified form of \(\left(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}\right)^{-1}\)?
Correct answer: A
Inside, \(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}=x^{-6}y^{4}\), and raising to (-1) gives \(x^{6}y^{-4}\). In exams, subtract exponents during division.
If \(a=2+\sqrt{3}\), what is the value of \(a^{2}+\frac{1}{a^{2}}\)?
Correct answer: A
Here \(\frac{1}{a}=2-\sqrt{3}\), so \(a+\frac{1}{a}=4\) and \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\). In exams, use the identity \(\left(a+\frac{1}{a}\right)^{2}\).
What is the simplified form of \(\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)+\sqrt{20}\)?
Correct answer: A
The first product is \(7-5=2\), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.
If (5^{x}=125) and (2^{y}=32), what is the value of (x+y)?
Correct answer: C
Since (125=5^{3}), (x=3), and since (32=2^{5}), (y=5), so (x+y=8). In exams, write numbers as powers of their prime bases.
What is the value of (\frac{3^{-2}+3^{-1}}{3^{-3}})?
Correct answer: A
Here (3^{-2}+3^{-1}=\frac{1}{9}+\frac{1}{3}=\frac{4}{9}), and (3^{-3}=\frac{1}{27}), so the value is (12). In exams, convert negative powers into fractions.
Which option correctly writes \(\left(x^{2}-4\right)\)?
Correct answer: A
We use \(x^{2}-4=x^{2}-2^{2}=(x-2)(x+2)\). In exams, remember the difference of squares form \((a^{2}-b^{2})\).
If (m=\sqrt{6}+\sqrt{2}), what is the value of (m^{2})?
Correct answer: A
((\sqrt{6}+\sqrt{2})^{2}=6+2+2\sqrt{12}=8+4\sqrt{3}). In exams, do not miss the middle term of ((a+b)^{2}).
If (r=10^{2}\cdot10^{-5}\cdot10^{4}), what is the value of (r)?
Correct answer: A
For the same base (10), the exponent is (2-5+4=1), so (r=10^{1}=10). In exams, add exponents during multiplication.
What is the simplified form of \(\left(\frac{a^{3}b^{-2}}{a^{-1}b^{2}}\right)^{2}\)?
Correct answer: A
Inside, \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), and squaring gives \(a^{8}b^{-8}\). In exams, watch the sign when subtracting negative exponents.
If (u=\frac{\sqrt{3}}{\sqrt{12}}), what is the value of (u^{-2})?
Correct answer: A
(\frac{\sqrt{3}}{\sqrt{12}}=\sqrt{\frac{3}{12}}=\frac{1}{2}), so (u^{-2}=4). In exams, simplify the radical first.
If \(a\) is a nonzero real number, which of the following statements is always true?
Correct answer: A
The zero-exponent law states that \(a^{0}=1\) for every nonzero number, so option A is correct. Option B is false because \(a^{0}\) equals 1, not 0. In option C, \(a^{-1}=\frac{1}{a}\), which is not generally equal to \(a\); and in option D, \(a^{1}=a\). Exam tip: Apply \(a^{0}=1\) only when \(a\ne0\); do not include \(0^{0}\) in this rule.
If (x=2-\sqrt{3}), what is the value of (\frac{1}{x}+x)?
Correct answer: A
Since (\frac{1}{2-\sqrt{3}}=2+\sqrt{3}), (\frac{1}{x}+x=(2+\sqrt{3})+(2-\sqrt{3})=4). In exams, identify conjugate numbers quickly.
QUIZ COMPLETE