CAT Algebra Bible Top 10 Must Solve Algebra Questions covering 100% of Syllabus – Page 2
All 21 Algebra questions from CAT papers. Tap Show answer, or open the detailed solution (free login). Page 2 of 7.
Let \(x, y\), and \(z\) be real numbers satisfying $$\begin{aligned} & 4\left(x^2+y^2+z^2\right)=a \\ & 4(x-y-z)=3+a \end{aligned}$$ Then \(a\) equals
Answer: A. 3
Solution — first steps
Step 1. Substitute a=4(x²+y²+z²) into the second equation.
Step 2. Rearrange to x²−x+y²+y+z²+z+3/4=0.
Step 2. Rearrange to x²−x+y²+y+z²+z+3/4=0.
The full step-by-step working opens with a free login.
For students who have joined a Cetking program.
The roots \( \alpha, \beta \) of the equation \( 3x^2 + \lambda x - 1 = 0 \)
satisfy
\[
\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15.
\]
The value of \( (\alpha^3 + \beta^3)^2 \) is:
satisfy
\[
\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15.
\]
The value of \( (\alpha^3 + \beta^3)^2 \) is:
Answer: 4
Solution — first steps
Try Each Option - Algebra Checking Each Option 🔹 Option 1: \( (\alpha^3 + \beta^3)^2 = 9 \) Let \( \alpha + \beta = s \), \( \alpha\beta = -\frac{1}{3} \) Then \( \alpha^3 + \beta^3 = s^3 - 3\alpha\beta s = s^3 + s \) Try \( s = 2 \Rightarrow \alpha^3 + \beta^3 = 8 + 2 = 10 \Rightarrow (10)^2 = 100 \neq 9 \) Try \( s = 1 \Rightarrow 1 + 1 = 2 \Rightarrow (2)^2 = 4 \neq 9 \) ❌ Option 1 is incorrect 🔹 Option 2: \( (\alpha^3 + \beta^3)^2 = 4 \) Try \( s = 1 \Rightarrow \alpha^3 + \beta^3 = 1 + 1 = 2 \Rightarrow (2)^2 = 4 \) Now check the condition: \[ \alpha^2 + \beta^2 = s^2 - 2\alpha\beta = 1 + \frac{2}{3} = \frac{5}{3}, \quad (\alpha\beta)^2 = \frac{1}{9} \] \[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{5/3}{1/9} = 15 \quad \text{✅ Satisfied} \] ✅ Option 2 is correct 🔹 Option 3: \( (\alpha^3 + \beta^3)^2 = 16 \) Try \( s = 2 \Rightarrow \alpha^3 + \beta^3 = 8 + 2 = 10 \Rightarrow (10)^2 = 100 \) Try \( s = \sqrt{16} = 4 \Rightarrow Not possible, would make cubic sum too large ❌ Option 3 is incorrect 🔹 Option 4: \( (\alpha^3 + \beta^3)^2 = 1 \) Try \( s = 0 \Rightarrow \alpha^3 + \beta^3 = 0 - 0 = 0 \Rightarrow (0)^2 = 0 \) Try \( s = 1 \Rightarrow (1 + 1)^2 = 4 \) ❌ Option 4 is incorrect ✅ Final Answer: Option 2 — \(\boxed{4}\)
The full step-by-step working opens with a free login.
For students who have joined a Cetking program.
If \(a, b\) and \(c\) are positive real numbers such that \(a>10 \geq b \geq c\) and \(\frac{\log _8(a+b)}{\log _2 c}+\frac{\log _{27}(a-b)}{\log _3 c}=\frac{2}{3}\), then the greatest possible integer value of \(a\) is
Answer: 14
Solution — first steps
Step 1. The logarithm denominators require c≠1. Change of base turns the equation into (1/3)log_c(a+b)+(1/3)log_c(a−b)=2/3.
Step 2. Combine logs to get log_c(a²−b²)=2, hence a²=b²+c².
Step 2. Combine logs to get log_c(a²−b²)=2, hence a²=b²+c².
The full step-by-step working opens with a free login.
For students who have joined a Cetking program.
If \((x+6 \sqrt{2})^{\frac{1}{2}}-(x-6 \sqrt{2})^{\frac{1}{2}}=2 \sqrt{2}\), then \(x\) equals
Answer: 11
Solution — first steps
Step 1. Let A=√(x+6√2) and B=√(x-6√2). The equation gives A-B=2√2.
Step 2. Also A²-B²=12√2, so (A-B)(A+B)=12√2 and A+B=6.
Step 2. Also A²-B²=12√2, so (A-B)(A+B)=12√2 and A+B=6.
The full step-by-step working opens with a free login.
For students who have joined a Cetking program.
\( a \) and \( b \) are natural numbers, then the value of \( a + b \) is:
\[
(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}
\]
\[
(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}
\]
Answer: 8
Solution — first steps
Surd Algebra Question \( a \) and \( b \) are natural numbers, then the value of \( a + b \) is: \[ (a + b\sqrt{3})^2 = 52 + 30\sqrt{3} \] ✅ Solution: Expanding the left-hand side: \[ (a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2 \] Match with the right-hand side: \[ a^2 + 3b^2 = 52 \quad \text{(1)} \\ 2ab = 30 \Rightarrow ab = 15 \quad \text{(2)} \] If \( ab = 15 \) and \( a^2 + 3b^2 = 52 \), where \( a \) and \( b \) are natural numbers, find \( a + b \). Solution using Hit and Trial: We try natural number pairs such that \( ab = 15 \).
The full step-by-step working opens with a free login.
For students who have joined a Cetking program.
Last day solving checklist
- Name the unknown. Include its unit and state whether it is a count, amount, rate or parameter.
- Form the relationships. Translate each given condition into an equation or restriction.
- Choose the method. Use the formula, substitution, elimination, a sign chart or a bound appropriate to this concept.
- Check restrictions. Check integer counts, signs, nonzero denominators, domains and all original conditions.
- Verify the answer. Substitute into the original question; for an extreme, show a feasible example reaches it.
All 10 repeating CAT PYQs and how much each adds
| CAT PYQ | Repeating CAT PYQ | Questions | Representative | Cumulative questions | Cumulative share |
|---|---|---|---|---|---|
| 1 | Conjugate surds and difference of squares | 7 | CAT 2021 S1 Q4 | 7 | 21.21% |
| 2 | Symmetric sums products and reciprocal powers | 7 | CAT 2022 S3 Q14 | 14 | 42.42% |
| 3 | Square sum product identities in diagonal geometry | 5 | CAT 2021 S2 Q9 | 19 | 57.58% |
| 4 | Complete squares and use sign or equality conditions | 4 | CAT 2023 S1 Q2 | 23 | 69.70% |
| 5 | Group common expressions and factor product constraints | 4 | CAT 2021 S2 Q3 | 27 | 81.82% |
| 6 | Difference or sum formulas for sequence totals | 2 | CAT 2021 S2 Q20 | 29 | 87.88% |
| 7 | Add and subtract values at opposite polynomial roots | 1 | CAT 2022 S2 Q9 | 30 | 90.91% |
| 8 | Common base exponent identity for all parameter values | 1 | CAT 2022 S3 Q11 | 31 | 93.94% |
| 9 | Square and factor an absolute value inequality | 1 | CAT 2021 S3 Q18 | 32 | 96.97% |
| 10 | Expand a nested GP into simpler series | 1 | CAT 2023 S3 Q20 | 33 | 100% |
Based on CAT Actual Quant Papers.