CAT Surds and Indices Bible Top 10 Must Solve Surds and Indices Questions covering 90% of Syllabus – Page 2
All 22 Surds and Indices questions from CAT papers. Tap Show answer, or open the detailed solution (free login). Page 2 of 5.
The sum of all real values of \(k\) for which \(\left(\frac{1}{8}\right)^k \times\left(\frac{1}{32768}\right)^{\frac{1}{3}}=\frac{1}{8} \times\left(\frac{1}{32768}\right)^{\frac{1}{k}}\), is
Answer: B. \(-\frac{2}{3}\)
Solution — first steps
Step 1. Convert every number to base 2: 1/8=2^-3 and 1/32768=2^-15.
Step 2. The left exponent of 2 is -3k-5; the right exponent is -3-15/k.
Step 2. The left exponent of 2 is -3k-5; the right exponent is -3-15/k.
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If \(x\) is a positive real number such that \(4 \log _{10} x+4 \log _{100} x+8 \log _{1000} x=13\), then the greatest integer not exceeding \(x\), is
Answer: 31
Solution — first steps
Step 1. Let L=log10 x.
Step 2. By change of base, log100 x=L/2 and log1000 x=L/3.
Step 2. By change of base, log100 x=L/2 and log1000 x=L/3.
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When 3^333 is divided by 11, the remainder is
Answer: 5
Solution — first steps
Step 1: Identify the cycle of \( 3^n \mod 11 \)
\[
\begin{align*}
3^1 &\equiv 3 \mod 11 \\
3^2 &\equiv 9 \mod 11 \\
3^3 &\equiv 27 \equiv 5 \mod 11 \\
3^4 &\equiv 15 \equiv 4 \mod 11 \\
3^5 &\equiv 12 \equiv 1 \mod 11
\end{align*}
\]
The powers of 3 modulo 11 repeat every 5 terms.
Step 2: Reduce the exponent using the cycle
Since the cycle length is 5, we compute:
\[
333 \mod 5 = 3
\Rightarrow 3^{333} \equiv 3^3 \mod 11
\]
From our earlier list:
\[
3^3 \equiv 5 \mod 11
\]
✅ Final Answer:
\[
\boxed{5}
\]
\[
\begin{align*}
3^1 &\equiv 3 \mod 11 \\
3^2 &\equiv 9 \mod 11 \\
3^3 &\equiv 27 \equiv 5 \mod 11 \\
3^4 &\equiv 15 \equiv 4 \mod 11 \\
3^5 &\equiv 12 \equiv 1 \mod 11
\end{align*}
\]
The powers of 3 modulo 11 repeat every 5 terms.
Step 2: Reduce the exponent using the cycle
Since the cycle length is 5, we compute:
\[
333 \mod 5 = 3
\Rightarrow 3^{333} \equiv 3^3 \mod 11
\]
From our earlier list:
\[
3^3 \equiv 5 \mod 11
\]
✅ Final Answer:
\[
\boxed{5}
\]
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If \( m \) and \( n \) are natural numbers such that \( n > 1 \), and \( mn = 2^{25} \times 3^{40} \), then \( m - n \) equals:
Answer: 209947
Solution — first steps
Explanation Summary - mⁿ = 2²⁵ × 3⁴⁰ ✅ Explanation Summary Given: \[ m^n = 2^{25} \cdot 3^{40} \] This can be rewritten as: \[ m^n = (2^5 \cdot 3^8)^5 \] (since \( \gcd(25, 40) = 5 \), and \( n = 5 \)) Therefore: \[ m = 2^5 \cdot 3^8 = 32 \cdot 6561 = 209952 \quad \text{and} \quad n = 5 \] So, the final answer is: \[ m - n = 209952 - 5 = \boxed{209947} \]
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The sum of the infinite series \(\frac{1}{5}\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{5}\right)^2\left(\left(\frac{1}{5}\right)^2-\left(\frac{1}{7}\right)^2\right)+\left(\frac{1}{5}\right)^3\left(\left(\frac{1}{5}\right)^3-\left(\frac{1}{7}\right)^3\right)+\cdots\) is equal to
Answer: A. \(\frac{5}{408}\)
Solution — first steps
Step 1. The nth term is (1/5)^n[(1/5)^n−(1/7)^n].
Step 2. So the series equals Σ(1/25)^n−Σ(1/35)^n for n≥1.
Step 2. So the series equals Σ(1/25)^n−Σ(1/35)^n for n≥1.
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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 16 repeating CAT PYQs and how much each adds
| CAT PYQ | Repeating CAT PYQ | Questions | Representative | Cumulative questions | Cumulative share |
|---|---|---|---|---|---|
| 1 | Convert logarithms to powers using base rules | 7 | CAT 2021 S1 Q4 | 7 | 17.50% |
| 2 | Prime factorisation and exponent comparison | 5 | CAT 2022 S3 Q11 | 12 | 30.00% |
| 3 | Compound growth factor comparisons | 4 | CAT 2023 S1 Q16 | 16 | 40.00% |
| 4 | Exponential substitution reduces to quadratics | 4 | CAT 2023 S2 Q2 | 20 | 50.00% |
| 5 | Integer exponent bounds with common bases | 3 | CAT 2021 S2 Q15 | 23 | 57.50% |
| 6 | Modular powers and remainder cycles | 3 | CAT 2024 S1 Q10 | 26 | 65.00% |
| 7 | Products of powers use sums of exponents | 2 | CAT 2021 S3 Q20 | 28 | 70.00% |
| 8 | Symmetric reciprocal powers by recurrence identities | 2 | CAT 2023 S3 Q4 | 30 | 75.00% |
| 9 | Simplify powers into geometric series | 2 | CAT 2023 S3 Q20 | 32 | 80.00% |
| 10 | Prime powers or powers of ten and digit sums | 2 | CAT 2025 S2 Q20 | 34 | 85.00% |
| 11 | Special bases or exponent zero when a power is one | 1 | CAT 2022 S2 Q17 | 35 | 87.50% |
| 12 | Universal common divisors of power expressions | 1 | CAT 2022 S1 Q21 | 36 | 90.00% |
| 13 | Shift an affine recurrence into geometric growth | 1 | CAT 2023 S1 Q22 | 37 | 92.50% |
| 14 | Replacement iterations and a concentration threshold | 1 | CAT 2023 S2 Q14 | 38 | 95.00% |
| 15 | Prime power rule of a multiplicative function | 1 | CAT 2024 S2 Q4 | 39 | 97.50% |
| 16 | Telescoping product of logarithmic exponents | 1 | CAT 2024 S3 Q16 | 40 | 100% |
Based on CAT Actual Quant Papers.