Algebraic Expressions
ICSE · Class 8 · Mathematics
Most important questions from Algebraic Expressions for ICSE Class 8 Mathematics board exam 2026. MCQs, short answer, and long answer questions with marks.
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Sample Questions
Simplify 84 - 7[-11x - 4{-17x + 3(8 - (9 - 5x))}].
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21x
First remove the innermost bracket: 8 - (9 - 5x) = 8 - 9 + 5x = 5x - 1. Then simplify inside the curly brackets: -17x + 3(5x - 1) = -17x + 15x - 3 = -2x - 3. Then the square bracket: -11x - 4(-2x - 3) = -11x + 8x + 12 = -3x + 12. Finally: 84 - 7(-3x + 12) = 84 + 21x - 84 = 21x.
Find the result of (7x + 4y - 5z + 5) + (6x - 7z - 8) - (12x + 3y - z).
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x + y - 11z - 3
First add the first two expressions: (7x + 4y - 5z + 5) + (6x - 7z - 8) = 13x + 4y - 12z - 3. Now subtract the third expression by changing its signs: 13x + 4y - 12z - 3 - 12x - 3y + z. Combine like terms: x + y - 11z - 3.
Divide -50a^2b^3 by -15a^4b^2 and simplify the result.
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10b / 3a^2
Use division of a monomial by another monomial. Divide the coefficients: -50 ÷ -15 = 10/3. Divide the literals: a^2 ÷ a^4 = 1/a^2 and b^3 ÷ b^2 = b. So the result is 10b / 3a^2.
Multiply -5xy^2, -3x^2yz and 8yz^2. What is the product?
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120x^3y^4z^3
Multiply the numerical coefficients first: -5 × -3 × 8 = 120. Now multiply the literal parts: x × x^2 = x^3, y^2 × y × y = y^4, z × z^2 = z^3. So the product is 120x^3y^4z^3.
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