Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
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 20 questions from this page. Select your focus, then start.
20 questions
Choose questions
Medium · Level 43 · polynomials,algebraic identities,difference of squares,mental calculationView options
2491
2509
2391
2809
Easy · Level 43 · polynomials,difference-of-squares,decimal-arithmetic,Operations on real numbers and the laws of exponents,Mathematics,Class 10 MCQView options
1
2
2.5
3
Medium · Level 43 · laws of exponents,negative exponents,real numbers,fractionsView options
\(5\)
\(13\)
\(\frac{13}{36}\)
\(36\)
Medium · Level 43 · polynomials,exponents,real numbers,exponent laws,simplificationView options
2
4
8
16
Medium · Level 43 · polynomials,common factor,factorisation,algebraic identities,exponentsView options
\(2x^2(x+3)\)
\(2x(x^2+3)\)
\(x^2(2x+6x)\)
\(6x^2(x+1)\)
Medium · Level 43 · polynomials,monomial division,exponent laws,algebraic simplificationView options
\(3a^2b\)
\(3a^2b^2\)
\(10a^2b\)
\(3a^6b^3\)
Medium · Level 43 · polynomials,monomial division,laws of exponents,algebraic expressionsView options
\(3a^3b^2\)
\(3a^7b^4\)
\(12a^3b^2\)
\(3a^2b^3\)
Medium · Level 43 · polynomials,substitution,value of expression,exponents,order of operationsView options
11
13
15
17
Medium · Level 43 · polynomials,exponents,substitution,negative numbers,algebraic expressionsView options
0
16
32
48
Medium · Level 43 · polynomials,negative exponents,laws of exponents,real numbers,algebraic simplificationView options
\(x^3y^2\)
\(x^7y^{-6}\)
\(x^3y^{-6}\)
\(x^7y^2\)
Medium · Level 43 · polynomials,negative exponents,laws of exponents,real numbers,algebraic simplificationView options
\(\frac{x^{-2}}{y^3}\)
\(x^2y^3\)
\(\frac{y^3}{x^2}\)
\(\frac{x^2}{y^3}\)
Medium · Level 43 · polynomials,zero exponent,negative exponent,real numbers,laws of exponentsView options
\(\frac{1}{5}\)
\(\frac{6}{5}\)
\(8\)
\(\frac{9}{5}\)
Medium · Level 43 · polynomials,negative exponents,real numbersView options
\(\frac{1}{2}\)
\(\frac{1}{4}\)
\(\frac{3}{4}\)
\(6\)
Medium · Level 43 · polynomials,real numbers,exponents,negative exponents,powers of twoView options
4
8
16
32
Medium · Level 43 · polynomials,algebraic identities,binomial expansion,laws of exponentsView options
\(4x^2-12x+9\)
\(4x^2-6x+9\)
\(2x^2-12x+9\)
\(4x^2-9\)
Medium · Level 43 · polynomials,algebraic identities,exponents,binomial squareView options
\(9x^2+12x+4\)
\(9x^2+6x+4\)
\(3x^2+12x+4\)
\(9x^2+4\)
Medium · Level 43 · laws of exponents,real numbers,algebraic simplification,indices,grade 10 mathematicsView options
x⁴
x⁸
x¹²
x²
Medium · Level 43 · polynomials,laws of exponents,real numbers,exponent simplificationView options
27
81
243
729
Medium · Level 43 · polynomials,exponents,exponential equations,real numbersView options
2
3
4
5
Medium · Level 43 · laws of exponents,negative exponents,real numbers,reciprocals,polynomials,class 10 mathematicsView options
\((ab)^{-1}=a^{-1}b^{-1}\)
\((a+b)^{-1}=a^{-1}+b^{-1}\)
\(a^{-1}+b^{-1}=(a+b)^{-1}\)
\((a-b)^{-1}=a^{-1}-b^{-1}\)
Question 1MediumLevel 43
Using an algebraic identity, find the value of \(53\cdot47\).
Correct answer: A
\(53\cdot47=(50+3)(50-3)\). Using the difference-of-squares identity \((a+b)(a-b)=a^2-b^2\), we get \(50^2-3^2=2500-9=2491\). Therefore, option A is correct. Option B, 2509, may result from adding 9 instead of subtracting it. Exam tip: When two numbers are equally above and below a common number, use the difference-of-squares identity.
The governing concept is the difference-of-squares identity, a^2-b^2=(a-b)(a+b). Apply it with a=1.5 and b=0.5: (1.5)^2-(0.5)^2=(1.5-0.5)(1.5+0.5). The first factor is 1 and the second factor is 2, so their product is 1×2=2. Therefore option B is correct. Direct checking gives 1.5^2=2.25 and 0.5^2=0.25, whose difference is also 2. Option A may result from calculating only the difference of the numbers; option C can come from adding incorrectly, and option D reflects an arithmetic error. The identity works equally well with decimal numbers, provided the signs and parentheses are handled carefully.
What is the value of \(\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{2}\right)^{-2}\)?
Correct answer: B
Using the negative-exponent rule \(a^{-n}=\frac{1}{a^n}\), we get \(\left(\frac{1}{3}\right)^{-2}=3^2=9\) and \(\left(\frac{1}{2}\right)^{-2}=2^2=4\). Therefore, the sum is \(9+4=13\). The value \(\frac{13}{36}\) results from mishandling the negative powers of the fractions. Exam tip: for a negative exponent, take the reciprocal of the base before applying the exponent.
What is the value of \(\dfrac{2^3\times 2^4}{2^5}\) after simplification?
Correct answer: B
For powers with the same base, exponents are added during multiplication and subtracted during division. Thus, \(\dfrac{2^3\times2^4}{2^5}=2^{3+4-5}=2^2=4\). Therefore, the correct answer is 4. The value 8 may result from forgetting to subtract the exponent in the denominator. Exam tip: When the bases are the same, simplify the exponents before calculating the numerical value.
What is the factorised form obtained by taking the common factor out of the polynomial \(2x^3+6x^2\)?
Correct answer: A
Both terms, \(2x^3\) and \(6x^2\), contain 2 and \(x^2\), so the common factor is \(2x^2\). Dividing each term by it gives \(2x^3+6x^2=2x^2(x+3)\). Option B does not reproduce the second term correctly. Exam tip: take the HCF of the numerical coefficients and the lowest power of the common variable.
If \(a\neq0\) and \(b\neq0\), what is the simplified form of \(\frac{15a^4b^2}{5a^2b}\)?
Correct answer: A
Divide the numerical coefficients to get \(15\div5=3\). For like bases, subtract the exponents: \(a^{4-2}=a^2\) and \(b^{2-1}=b\). Therefore, the simplified form is \(3a^2b\). Option B is incorrect because it does not reduce the exponent of \(b\). Exam tip: when dividing monomials, divide the coefficients and subtract the exponents of like variables.
Dividing the coefficients gives \(18\div6=3\). When powers with the same base are divided, their exponents are subtracted: \(a^5\div a^2=a^{5-2}=a^3\) and \(b^3\div b=b^{3-1}=b^2\). Hence, the simplified form is \(3a^3b^2\). Option B adds the exponents instead of subtracting them, while option C uses an incorrect coefficient. Exam tip: for division of like bases, subtract the denominator exponent from the numerator exponent.
Substituting x = 2 gives 3(2³) − 4(2²) + 5 = 3(8) − 4(4) + 5 = 24 − 16 + 5 = 13. Therefore, 13 is correct. Values such as 15 or 17 may result from an error in evaluating the powers or following the order of operations. Exam tip: evaluate powers first, then multiplication, and finally addition or subtraction.
Substituting \(x=-2\) gives \((-2)^4-2(-2)^3=16-2(-8)=16+16=32\). Therefore, the correct answer is 32. Remember that a negative number has a positive value when raised to an even power, but remains negative when raised to an odd power.
What is the simplified form of \(\frac{x^5y^{-2}}{x^2y^{-4}}\), if \(x\neq0\) and \(y\neq0\)?
Correct answer: A
When dividing powers with the same base, subtract the exponents: \(x^{5-2}=x^3\) and \(y^{-2-(-4)}=y^2\). Therefore, the simplified form is \(x^3y^2\). Option B results from adding the exponents, but division requires their subtraction. Exam tip: Use parentheses carefully when subtracting a negative exponent, as in \(-2-(-4)=2\).
If \(x\neq0\) and \(y\neq0\), what is the simplified form of \(\left(\frac{x^{-2}}{y^{-3}}\right)^{-1}\)?
Correct answer: D
First, \(\frac{x^{-2}}{y^{-3}}=x^{-2}\times y^3=\frac{y^3}{x^2}\). Raising this expression to the power \(-1\) gives its reciprocal: \(\left(\frac{y^3}{x^2}\right)^{-1}=\frac{x^2}{y^3}\). Option C is only the form inside the outer power, not the final answer. Exam tip: \(a^{-1}\) means \(\frac{1}{a}\).
By the zero-exponent law, the zeroth power of any non-zero number is 1, so \((2^3)^0=1\). By the negative-exponent law, \(5^{-1}=\frac{1}{5}\). Therefore, \(1+\frac{1}{5}=\frac{6}{5}\), so option B is correct. Exam tip: remember that \(a^{-n}=\frac{1}{a^n}\) and \(a^0=1\) for \(a\ne0\).
Use the law of negative exponents: \(a^{-n}=\frac{1}{a^n}\). Thus, \(4^{-1}=\frac{1}{4}\) and \(2^{-2}=\frac{1}{2^2}=\frac{1}{4}\). Therefore, \(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\), so option A is correct. Exam tip: Convert negative powers into reciprocals before performing the addition.
Using the law of negative exponents, \(4^{-1}=(2^2)^{-1}=2^{-2}\). Therefore, \(2^5\cdot4^{-1}=2^5\cdot2^{-2}=2^{5-2}=2^3=8\), so option B is correct. Exam tip: when multiplying powers with the same base, add their exponents; when dividing, subtract them.
Apply the identity \((a-b)^2=a^2-2ab+b^2\). Here, \(a=2x\) and \(b=3\), so \((2x-3)^2=(2x)^2-2(2x)(3)+3^2=4x^2-12x+9\). Option B has an incorrect middle term because it does not evaluate \(-2ab\) correctly. Exam tip: in \((a-b)^2\), the middle term is always \(-2ab\).
Use the identity \((a+b)^2=a^2+2ab+b^2\). With \(a=3x\) and \(b=2\), we get \((3x)^2+2(3x)(2)+2^2=9x^2+12x+4\). Therefore, option A is correct. Option B incorrectly uses only \(ab\) instead of the middle term \(2ab\). Exam tip: in \((a+b)^2\), the middle term is always \(2ab\).
If x ≠ 0, what is the simplest form of (x³ × x⁵) / x⁴?
Correct answer: A
By the laws of exponents, powers with the same base are multiplied by adding their exponents and divided by subtracting the exponent in the denominator. Thus, (x³ × x⁵) / x⁴ = x^(3+5−4) = x⁴. Option B results from ignoring the denominator x⁴. Exam tip: Add exponents for multiplication of like bases and subtract them for division.
For powers with the same base, exponents are added during multiplication and subtracted during division. Thus, \(\frac{3^4 \times 3^2}{3^3}=3^{4+2-3}=3^3=27\). Exam tip: apply the exponent laws before evaluating the final power.
Since \(81=3^4\), the equation becomes \(3^{x+1}=3^4\). Because the bases are equal and positive, their exponents must be equal: \(x+1=4\), so \(x=3\). Option C is a common error because substituting \(x=4\) gives an exponent of \(5\), not \(4\). Exam tip: Express both sides with the same base before equating the exponents.
Which of the following statements is true for all non-zero real numbers \(a\) and \(b\), according to the laws of exponents?
Correct answer: A
A negative exponent represents a reciprocal. Thus, \((ab)^{-1}=\frac{1}{ab}=\frac{1}{a}\cdot\frac{1}{b}=a^{-1}b^{-1}\). No similar rule applies to a sum or difference. Exam tip: distinguish a product from a sum before using exponent laws.
Google Analytics helps us understand site usage. Google may send limited cookie-free signals before your choice. The Live Visitors widget operates independently of this analytics choice; see the privacy policy for its provider and fallback details. Essential site features work without analytics cookies. You can change your choice later in Privacy choices. Privacy policy