CAT Inequalities Bible Top 10 Must Solve Inequalities Questions covering 80% of Syllabus – Page 3
All 38 Inequalities questions from CAT papers. Tap Show answer, or open the detailed solution (free login). Page 3 of 11.
In the XY-plane, the area, in sq. units, of the region defined by the inequalities
\( y \geq x + 4 \) and \( -4 \leq x^2 + y^2 + 4(x - y) \leq 0 \) is:
\( y \geq x + 4 \) and \( -4 \leq x^2 + y^2 + 4(x - y) \leq 0 \) is:
Answer: B. \( 2\pi \)
Solution — first steps
Step 1: Simplify the second inequality
\[
x^2 + y^2 + 4x - 4y = (x + 2)^2 - 4 + (y - 2)^2 - 4 = (x + 2)^2 + (y - 2)^2 - 8
\]
So the inequality becomes:
\[
-4 \leq (x + 2)^2 + (y - 2)^2 - 8 \leq 0
\Rightarrow 4 \leq (x + 2)^2 + (y - 2)^2 \leq 8
\]
This represents an annular region (ring) centered at \( (-2, 2) \), with:
Inner radius: \( \sqrt{4} = 2 \)
Outer radius: \( \sqrt{8} = 2\sqrt{2} \)
Step 2: Apply the line constraint \( y \geq x + 4 \)
The line cuts the annular region approximately in half (passing near the center), so the required area is half of the total area of the ring.
\[
x^2 + y^2 + 4x - 4y = (x + 2)^2 - 4 + (y - 2)^2 - 4 = (x + 2)^2 + (y - 2)^2 - 8
\]
So the inequality becomes:
\[
-4 \leq (x + 2)^2 + (y - 2)^2 - 8 \leq 0
\Rightarrow 4 \leq (x + 2)^2 + (y - 2)^2 \leq 8
\]
This represents an annular region (ring) centered at \( (-2, 2) \), with:
Inner radius: \( \sqrt{4} = 2 \)
Outer radius: \( \sqrt{8} = 2\sqrt{2} \)
Step 2: Apply the line constraint \( y \geq x + 4 \)
The line cuts the annular region approximately in half (passing near the center), so the required area is half of the total area of the ring.
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Let \( x, y, \) and \( z \) be real numbers satisfying:
\( 4(x^2 + y^2 + z^2) = a \),
\( 4(x - y - z) = 3 + a \).
Then \( a \) equals:
\( 4(x^2 + y^2 + z^2) = a \),
\( 4(x - y - z) = 3 + a \).
Then \( a \) equals:
Answer: C. \( 3 \)
Solution — first steps
Hit and Trial Table - MathJax Hit and Trial Table for Solving Using Options Given equations: \[ 4(x^2 + y^2 + z^2) = a \quad \text{and} \quad 4(x - y - z) = 3 + a \] Option \( a \) \( x^2 + y^2 + z^2 = \frac{a}{4} \) \( x - y - z = \frac{3 + a}{4} \) Trial Values for \( x, y, z \) Result 1 \( 1 \) \( \frac{1}{4} \) \( 1 \) \( x = 1, y = 0, z = 0 \) ❌ Too large for (1) 2 \( 4 \) \( 1 \) \( \frac{7}{4} \) \( x = 2, y = \frac{1}{8}, z = \frac{1}{8} \) ❌ Doesn't satisfy (1) 3 ✅ \( 3 \) \( \frac{3}{4} \) \( \frac{3}{2} \) Algebra confirms consistency ✅ Correct 4 \( \frac{4}{3} \) \( \frac{1}{3} \) \( \frac{13}{12} \) No simple values fit both equations ❌ Invalid ✅ Final Answer: \[ \boxed{3} \quad \text{(Option 3)} \]
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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.
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All the values of \( x \) satisfying the inequality \( \frac{1}{x + 5} \leq \frac{1}{2x - 3} \) are:
Answer: D. \( x < -5 \) or \( \frac{3}{2} < x \leq 8 \)
Solution — first steps
Step 1: Try \( x = 0 \)
\[
\frac{1}{0 + 5} \leq \frac{1}{2(0) - 3} \Rightarrow \frac{1}{5} \leq \frac{1}{-3}
\Rightarrow 0.2 \leq -0.333 \quad \text{(False)}
\]
This means \( x = 0 \) does
not
satisfy the inequality.
- Option (1): \( -5 \frac{3}{2} \), so includes 100 ✅
- Option (4): only includes up to \( x \leq 8 \) ❌
So, eliminate Option 3.
✅ Final Answer: Option (4)
\[
x < -5 \quad \text{or} \quad \frac{3}{2} < x \leq 8
\]
\[
\frac{1}{0 + 5} \leq \frac{1}{2(0) - 3} \Rightarrow \frac{1}{5} \leq \frac{1}{-3}
\Rightarrow 0.2 \leq -0.333 \quad \text{(False)}
\]
This means \( x = 0 \) does
not
satisfy the inequality.
- Option (1): \( -5 \frac{3}{2} \), so includes 100 ✅
- Option (4): only includes up to \( x \leq 8 \) ❌
So, eliminate Option 3.
✅ Final Answer: Option (4)
\[
x < -5 \quad \text{or} \quad \frac{3}{2} < x \leq 8
\]
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The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is _____
Answer: 70
Solution — first steps
Sum = 84. Let numbers a Solve → b=25, a+c=59 (c-10)-(a+7)=64 → c-a=81 Add: 2c=140 → c=70 Answer: 70
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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 21 repeating CAT PYQs and how much each adds
| CAT PYQ | Repeating CAT PYQ | Questions | Representative | Cumulative questions | Cumulative share |
|---|---|---|---|---|---|
| 1 | Exponential bounds and smallest integer periods | 7 | CAT 2021 S2 Q15 | 7 | 12.73% |
| 2 | Discriminant bounds for real roots or ranges | 6 | CAT 2022 S1 Q5 | 13 | 23.64% |
| 3 | Linear allocation bounds with nonnegative counts | 6 | CAT 2022 S1 Q8 | 19 | 34.55% |
| 4 | Absolute value intervals and sign cases | 5 | CAT 2021 S1 Q2 | 24 | 43.64% |
| 5 | Ordered values and average bounds | 5 | CAT 2022 S2 Q8 | 29 | 52.73% |
| 6 | Logarithmic bounds and monotonicity | 4 | CAT 2024 S2 Q9 | 33 | 60.00% |
| 7 | Rational sign charts and excluded poles | 3 | CAT 2021 S1 Q1 | 36 | 65.45% |
| 8 | Set overlap bounds and feasible endpoints | 3 | CAT 2022 S1 Q17 | 39 | 70.91% |
| 9 | Finite integer enumeration within bounds | 3 | CAT 2025 S1 Q17 | 42 | 76.36% |
| 10 | Floor intervals and endpoint inclusion | 2 | CAT 2022 S1 Q12 | 44 | 80.00% |
| 11 | Factorial size bounds | 1 | CAT 2022 S2 Q13 | 45 | 81.82% |
| 12 | Bounds on eligible AP terms | 1 | CAT 2022 S3 Q17 | 46 | 83.64% |
| 13 | Remainders with an upper population bound | 1 | CAT 2022 S3 Q19 | 47 | 85.45% |
| 14 | Polygon side inequality | 1 | CAT 2022 S3 Q22 | 48 | 87.27% |
| 15 | Speed bound with a palindrome restriction | 1 | CAT 2023 S1 Q8 | 49 | 89.09% |
| 16 | Translate time bounds to work rate bounds | 1 | CAT 2023 S3 Q8 | 50 | 90.91% |
| 17 | Quadratic percentage growth inequality | 1 | CAT 2023 S3 Q10 | 51 | 92.73% |
| 18 | Geometric regions defined by inequalities | 1 | CAT 2024 S1 Q4 | 52 | 94.55% |
| 19 | Square bound with an equality condition | 1 | CAT 2024 S1 Q8 | 53 | 96.36% |
| 20 | Compare quadratic extrema | 1 | CAT 2025 S1 Q8 | 54 | 98.18% |
| 21 | Range by completing the square and reciprocal bounds | 1 | CAT 2021 S2 Q13 | 55 | 100% |
Based on CAT Actual Quant Papers.