Hi IOQM aspirant, Welcome to Amans Maths Blogs (AMBIPie). In this post, you will get IOQM Algebra Practice Questions with Solutions. First, you will need to know what IOQM Algebra topics, syllabus are there in this exam.
IOQM Algebra Syllabus
Since the topics of IOQM Algebra syllabus include all those concepts which are useful for for improving your problem solving skills, you must learn all of these concepts before solving IOQM Algebra Questions.
There are following topics included in IOQM Algebra syllabus.
| Algebraic Identities | Linear Equations in One, Two, Three Variables |
| Factorization | Telescopic Series |
| Wavy Curve Method | Modulus Equation and Inequalities |
| Exponents or Powers | Logarithms |
| Binomial Theorems (Terms Related Properties Only) | Polynomials |
| Quadratic Polynomial & Equations | Cubic and Higher Degree Polynomials & Equations |
| Sequence and Series | Arithmetic |
| Diophantine Equations | Complex Numbers (Iota and Cube Properties Only) |
| Functional Equations | Recurrence Relations |
IOQM Algebra Practice Questions with Solutions
The answers of all IOQM Algebra questions are of two-digits numbers. If some questions have only single digits, then add zero before that.
IOQM Algebra Questions based on Algebraic Identities
Question No 1:
If \(a + b + c = 0\), then find the value of \(a^4 + b^4 + c^4 – 2a^2b^2 – 2b^2c^2 – 2c^2a^2 \).
Answer 1: 00
Solution 1:
Using the identities \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\), we get
$$ a^2 + b^2 + c^2 = – 2(ab + bc + ca) $$
$$ \Rightarrow a^4 + b^4 + c^4 – 2a^2b^2 – 2b^2c^2 – 2c^2a^2 = 00 $$
Question No 2:
If \( a \) be a positive real numbers such that \( \frac{a^2}{a^4 – a^2 + 1} = \frac{4}{37} \) and if \( \frac{a^3}{a^6 – a^3 + 1} = \frac{p}{q} \), where p and q are coprime to each other, then find the sum of the digits of \( (p + q) \).
Answer 2: 16
Solution 2:
On simplifying \( \frac{a^2}{a^4 – a^2 + 1} = \frac{4}{37} \), we get
$$ a + \frac{1}{a} = \frac{7}{2} $$
On cubing both sides, we get
$$ a^3 + \frac{1}{a^3} = \frac{259}{8} $$
Thus, we get
$$ \frac{a^3}{a^6 – a^3 + 1} = \frac{8}{251} = \frac{p}{q} \Rightarrow p + q = 259 $$
Question No 3:
If \( 169(157 – 77x)^2 + 100(201 – 100x)^2 = 26(77x – 157)(1000x – 2010) \), then find the value of \( x \).
Answer 3: 31
Solution 3:
On simplifying \( 169(157 – 77x)^2 + 100(201 – 100x)^2 = 26(77x – 157)(1000x – 2010) \) , we get
$$ [13(77x – 157) – 10(100x – 201)]^2 = 0 $$
On solving for x, we get
$$ x = 31 $$
