CAT Surds and Indices Bible Top 10 Must Solve Surds and Indices Questions covering 90% of Syllabus – Page 3
All 22 Surds and Indices questions from CAT papers. Tap Show answer, or open the detailed solution (free login). Page 3 of 5.
Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is
Answer: B. 20808
Solution — first steps
Step 1. At 4% per annum compounded half-yearly, each half-year multiplier is1.02.
Step 2. Anil receives 22000(1.02)^12 after six years.
Step 2. Anil receives 22000(1.02)^12 after six years.
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If 10^68 is divided by 13, the remainder is
Answer: B. 9
Solution — first steps
Step 1: Find powers of 10 modulo 13
10¹ mod 13 = 10
10² mod 13 = (10 × 10) mod 13 = 100 mod 13
100 ÷ 13 = 7 remainder 9 → 10² mod 13 = 9
10³ mod 13 = (10² × 10) mod 13 = (9 × 10) mod 13 = 90 mod 13
90 ÷ 13 = 6 remainder 12 → 10³ mod 13 = 12
10⁴ mod 13 = (12 × 10) mod 13 = 120 mod 13
120 ÷ 13 = 9 remainder 3 → 10⁴ mod 13 = 3
10⁵ mod 13 = (3 × 10) mod 13 = 30 mod 13
30 ÷ 13 = 2 remainder 4 → 10⁵ mod 13 = 4
10⁶ mod 13 = (4 × 10) mod 13 = 40 mod 13
40 ÷ 13 = 3 remainder 1 → 10⁶ mod 13 = 1
✅ The cycle length is 6:
10^6 mod 13 = 1
Step 2: Find position in cycle
10^68 mod 13 = 10^(68 mod 6) mod 13
68 ÷ 6 = 11 remainder 2
So 10^68 mod 13 = 10² mod 13 = 9
✅ Final answer:
👉 The remainder is 9
Correct option: 2) 9
10¹ mod 13 = 10
10² mod 13 = (10 × 10) mod 13 = 100 mod 13
100 ÷ 13 = 7 remainder 9 → 10² mod 13 = 9
10³ mod 13 = (10² × 10) mod 13 = (9 × 10) mod 13 = 90 mod 13
90 ÷ 13 = 6 remainder 12 → 10³ mod 13 = 12
10⁴ mod 13 = (12 × 10) mod 13 = 120 mod 13
120 ÷ 13 = 9 remainder 3 → 10⁴ mod 13 = 3
10⁵ mod 13 = (3 × 10) mod 13 = 30 mod 13
30 ÷ 13 = 2 remainder 4 → 10⁵ mod 13 = 4
10⁶ mod 13 = (4 × 10) mod 13 = 40 mod 13
40 ÷ 13 = 3 remainder 1 → 10⁶ mod 13 = 1
✅ The cycle length is 6:
10^6 mod 13 = 1
Step 2: Find position in cycle
10^68 mod 13 = 10^(68 mod 6) mod 13
68 ÷ 6 = 11 remainder 2
So 10^68 mod 13 = 10² mod 13 = 9
✅ Final answer:
👉 The remainder is 9
Correct option: 2) 9
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Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value after 5 years is 25:36, then the minimum number of years required for the value to exceed Rs 20000 is _____
Answer: 9
Solution — first steps
16 → n=9
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If \(3^a=4,4^b=5,5^c=6,6^d=7,7^e=8\) and \(8^f=9\), then the value of the product abcdef is
Answer: 2
Solution — first steps
Step 1. Rewrite each equation logarithmically: a=log_3 4, b=log_4 5, …, f=log_8 9.
Step 2. Multiply the six expressions.
Step 2. Multiply the six expressions.
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The sum of all distinct real values of \(x\) that satisfy the equation \(10^x+\frac{4}{10^x}=\frac{17}{2}\), is
Answer: A. \(2 \log _{10} 2\)
Solution — first steps
Step 1. Let t=10^x>0. The equation becomes t+4/t=17/2.
Step 2. Multiply by 2t: 2t²−17t+8=0.
Step 2. Multiply by 2t: 2t²−17t+8=0.
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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.