Algebraic Expressions and Factorisation Quiz - ICSE Class 8 Mathematics
Free practice quiz: Algebraic Expressions and Factorisation (ICSE Class 8 Mathematics). 10 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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What is the degree of the term $5a^3b^4c^2$?
- 3
- 7
- 9
- 10
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Answer: C. 9
The degree of a term with multiple variables is the sum of the exponents of all the variables in that term. In the term $5a^3b^4c^2$, the variables are $a$, $b$, and $c$, and their respective exponents are 3, 4, and 2. The degree of the term is $3 + 4 + 2 = 9$. The coefficient (5) does not affect the degree of the term.
If $x - \frac{1}{x} = 5$, find the value of $x^2 + \frac{1}{x^2}$.
- 23
- 25
- 27
- 29
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Answer: C. 27
We are given $x - \frac{1}{x} = 5$. To find $x^2 + \frac{1}{x^2}$, we can square both sides of the given equation: $(x - \frac{1}{x})^2 = 5^2$. Expanding the left side using the formula $(a-b)^2 = a^2 - 2ab + b^2$: $x^2 - 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = 25$. Simplifying the middle term: $x^2 - 2 + \frac{1}{x^2} = 25$. Now, isolate $x^2 + \frac{1}{x^2}$ by adding 2 to both sides of the equation: $x^2 + \frac{1}{x^2} = 25 + 2$. Therefore, $x^2 + \frac{1}{x^2} = 27$.
If $a+b=7$ and $ab=10$, find the value of $a^2+b^2$.
- 29
- 39
- 49
- 59
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Answer: A. 29
We know the identity $(a+b)^2 = a^2 + 2ab + b^2$. We are given $a+b=7$ and $ab=10$. Substitute these values into the identity: $(7)^2 = a^2 + 2(10) + b^2$. This simplifies to $49 = a^2 + 20 + b^2$. To find $a^2+b^2$, subtract 20 from both sides: $a^2 + b^2 = 49 - 20$. Therefore, $a^2 + b^2 = 29$.
What is the quotient when $18a^2b^3c$ is divided by $3ab^2$?
- $6ab c$
- $6a^2b c$
- $6abc$
- $6a^2bc$
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Answer: C. $6abc$
To divide $18a^2b^3c$ by $3ab^2$, we divide the coefficients and subtract the exponents of the like variables. Coefficient: $18 \div 3 = 6$. Variable $a$: $a^2 \div a^1 = a^{2-1} = a^1 = a$. Variable $b$: $b^3 \div b^2 = b^{3-2} = b^1 = b$. Variable $c$: $c \div 1 = c$. Combining these, the quotient is $6abc$.
If $x+y=10$ and $x-y=4$, find the value of $x^2-y^2$.
- 14
- 40
- 100
- 16
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Answer: B. 40
We know the difference of squares identity: $x^2 - y^2 = (x+y)(x-y)$. We are given the values of $(x+y)$ and $(x-y)$. Substitute these values directly into the identity: $x^2 - y^2 = (10)(4)$. Therefore, $x^2 - y^2 = 40$.