General chapter practice

The term \(a\) is added three times, so repeated addition becomes multiplication by the number of terms: \(a+a+a=3a\). The option \(a^3\) represents multiplying \(a\) by itself three times, not adding it. In such questions, count the identical terms to find the coefficient.

A variable is a letter or symbol whose value can change. In \(9q\), \(q\) is the variable, while \(9\) is its coefficient and \(9q\) is the complete algebraic term. \(0\) is a constant number. In an exam, identify the letter as the variable and the number multiplying it as the coefficient.

In \(x^3\), the number 3 written above \(x\) is its exponent. It means \(x^3=x\times x\times x\), so the correct answer is 3. The nearby distractor 2 would be the exponent in \(x^2\), not in \(x^3\). Exam tip: identify the small raised number attached to the variable as the exponent.

\(2x\) and \(3x\) are like terms, so their coefficients can be added: \(2x+3x=(2+3)x=5x\). The constant term \(4\) remains separate, giving \(5x+4\). Option D incorrectly combines unlike terms. In an exam, first identify like terms and then add their coefficients.

The area of a rectangle is calculated as length × width. Thus, area = 8 × 5 = 40 cm². Option B, 26 cm², is the perimeter found by 2(8 + 5), not the area. In exams, distinguish carefully between area and perimeter formulas.

Here, \(10r\) and \(3r\) are like terms because both contain the variable \(r\) to the same power. Subtract their coefficients: \(10-3=7\), so \(10r-3r=7r\). Option A incorrectly adds the coefficients, while option C changes the power of \(r\) to \(2\). Exam tip: when combining like terms, add or subtract only their coefficients and keep the variable part unchanged.

A binomial is a polynomial with exactly two unlike terms. In \(p-6\), the two terms are \(p\) and \(-6\), so it is a binomial. \(5z\) and \(8\) are monomials, while \(x^2+x+1\) is a trinomial. Exam tip: count the terms separated by plus or minus signs to identify the type of polynomial.

Using the distributive law, multiply \(5\) by each term inside the bracket: \(5(x-1)=5\times x-5 imes1=5x-5\). Therefore, option B is correct. In option A, \(-1\) has not been multiplied by \(5\). Exam tip: When removing brackets, multiply the outside factor by every term inside.

Like terms must have the same variable raised to the same power; only their coefficients may differ. In \(m^2\), the power of \(m\) is 2, whereas in \(m\) it is 1. Therefore, they are unlike terms. The presence of the same variable alone does not make terms like terms. Exam tip: always compare both the variable and its exponent.

To isolate \(p\) in \(2p=12\), divide both sides by 2: \(p=12\div2=6\). Therefore, the correct answer is 6. Do not multiply 12 by 2; division is required because \(p\) is multiplied by 2. Exam tip: use the inverse operation to undo multiplication.

In \(x^2+5x\), the term containing \(x\) is \(5x\). Therefore, the coefficient of \(x\) is 5, so option D is correct. The term \(x^2\) is separate and is not the term being asked about. Exam tip: the coefficient is the number multiplied by the variable.

Every point on the y-axis has x-coordinate 0. In \(P(0,-6)\), x = 0, so the statement is correct. Points with y = 0 lie on the x-axis. Exam tip: check the x-coordinate first.

The discount is \(\frac{15}{100}\times800=120\) rupees. Hence, selling price \(=800-120=680\) rupees. ₹720 would result after a 10% discount. Exam tip: calculate the discount amount first.

The number \(3\) is common to both terms, \(3x\) and \(6\). Factoring gives \(3x+6=3(x+2)\), so the greatest common factor is \(3\). Option 6 is incorrect because 6 is not a factor of \(3x\) in general. Exam tip: Find the HCF of the numerical coefficients and include only the variables common to every term.

\(\sqrt{49}=7\), and 7 is an integer and hence a rational number. Therefore, a square root sign does not always give an irrational number. \(2,3,5\) are not perfect squares. Exam tip: first check whether the number inside the root is a perfect square.

To divide the monomial \(12t\) by \(3\), divide its coefficient \(12\) by \(3\) while keeping the variable \(t\) unchanged: \(12t\div3=(12\div3)t=4t\). Therefore, \(4t\) is correct. Exam tip: when a monomial is divided by a numerical constant, divide the coefficient and retain the variable.

To isolate \(x\) in \(x-7=2\), add 7 to both sides: \(x-7+7=2+7\), so \(x=9\). Option B is not correct because substituting 7 gives \(7-7=0\), not 2. Exam tip: Always substitute your answer back into the original equation to verify it.

In the expression \(2a+3b\), the term containing \(a\) is \(2a\). The number multiplying \(a\) is 2, so the coefficient of \(a\) is 2. Option 3 is the coefficient of \(b\), not of \(a\). Exam tip: to find a variable’s coefficient, identify the number multiplied by that variable.

Both terms, 4x and 4y, are divisible by 4. Therefore, 4 is their common factor, and the expression can be written as 4(x + y). The variable x appears only in the first term and y only in the second, so neither is a common factor. Exam tip: check which numerical factors and variables occur in every term.

Substituting \\(n=4\\) into the expression gives \\(3n-2=3\\times4-2=12-2=10\\). Therefore, 10 is correct. The value 12 results from stopping at \\(3\\times4\\) without subtracting 2. In such questions, substitute the variable first, then perform multiplication before subtraction.