MC01 · Number System & Simplification

Question 1 of 20

Which of the following is an irrational number? (CDS PYQ)

A √16 B √25 C √36 D √50

√50 = 5√2 — non-terminating, non-recurring — hence irrational. √16=4, √25=5, √36=6 are all perfect squares and therefore rational. Tip: √n is irrational only when n is not a perfect square.

Question 2 of 20

The LCM of 12, 18, and 24 is: (CDS PYQ)

A 36 B 48 C 72 D 96

72. 12=2²×3, 18=2×3², 24=2³×3. LCM = highest powers = 2³×3² = 72. Always verify: 72 is divisible by 12, 18, and 24 ✓.

Question 3 of 20

The HCF of 84 and 126 is: (CDS PYQ)

A 14 B 21 C 42 D 63

42. 84=2²×3×7; 126=2×3²×7. HCF = lowest powers of common primes = 2×3×7 = 42. Verify: 84/42=2 ✓, 126/42=3 ✓. Also, HCF×LCM = 84×126 = 42×252 = 10584 ✓.

Question 4 of 20

The unit digit of 7⁹⁵ is: (CDS PYQ)

A 1 B 3 C 7 D 9

3. Powers of 7 cycle in unit digits: 7→9→3→1 (period 4). Divide exponent 95 by 4: 95=4×23+3, remainder 3. Unit digit = same as 7³ = 3. Cyclicity method: divide exponent by the period, use remainder to find position.

Question 5 of 20

The product of two numbers is 1260 and their HCF is 18. Their LCM is: (CDS PYQ)

A 42 B 56 C 70 D 90

70. For two numbers: HCF × LCM = Product. LCM = 1260/18 = 70. This relation holds strictly for exactly two numbers. For three numbers it does not apply — use prime factorisation instead.

Question 6 of 20

How many prime numbers exist between 1 and 30? (CDS PYQ)

A 8 B 9 C 10 D 11

10. Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 — exactly 10. Remember: 1 is neither prime nor composite; 2 is the only even prime. Beyond 30: 31, 37, 41, 43, 47 — also memorise these for CDS.

Question 7 of 20

The value of √(0.0016) is: (CDS PYQ)

A 0.004 B 0.04 C 0.4 D 4

0.04. √(0.0016) = √(16/10000) = 4/100 = 0.04. Quick check: 0.04² = 0.0016 ✓. Rule: if a decimal has 4 decimal places, its square root has 2 decimal places.

Question 8 of 20

log₁₀(1000) equals: (CDS PYQ)

A 2 B 3 C 4 D 10

3. log₁₀(10³) = 3×log₁₀(10) = 3×1 = 3. Standard values: log₁₀(10)=1, log₁₀(100)=2, log₁₀(1000)=3. Logarithm appears in every CDS paper — memorise these benchmarks.

Question 9 of 20

If log₂(x) = 5, then x equals: (CDS PYQ)

A 10 B 25 C 32 D 64

32. log₂(x)=5 means 2⁵=x → x=32. Definition: log_b(a)=c ↔ bᶜ=a. Common trap: confusing base 2 with base 10 and writing 10⁵. Always identify the base first.

Question 10 of 20

If a number is divided by 5 the remainder is 3, what is the remainder when the square of that number is divided by 5? (CDS PYQ)

A 1 B 2 C 3 D 4

4. Let n = 5k+3. n² = 25k²+30k+9 = 5(5k²+6k+1)+4. Remainder = 4. Shortcut: square the remainder (3²=9) and divide by 5 → remainder 4. This pattern recurs frequently in CDS Number System.

Question 11 of 20

The number of digits in 2²⁰ is: (use log 2 = 0.3010) (CDS PYQ)

A 5 B 6 C 7 D 8

7. Digits = ⌊20×0.3010⌋ + 1 = ⌊6.020⌋ + 1 = 7. Formula: number of digits of N = ⌊log₁₀N⌋ + 1. Do not try to compute 2²⁰ directly in the exam.

Question 12 of 20

The value of log(a²/bc) + log(b²/ac) + log(c²/ab) is: (CDS PYQ)

A 0 B 1 C 3 D log(abc)

0. Expand using log(x/y)=log x−log y: (2a−b−c)+(2b−a−c)+(2c−a−b) in terms of log a, log b, log c. Each coefficient: 2−1−1=0. So total = 0. This identity-based log question appears regularly in CDS.

Question 13 of 20

The smallest number to be added to 1000 to make it divisible by 7 is: (CDS PYQ)

A 1 B 4 C 6 D 7

1. 1000 ÷ 7 = 142 remainder 6. Next multiple of 7 = 7×143 = 1001. Amount to add = 1001−1000 = 1. Formula: add (divisor − remainder) when remainder ≠ 0. Here: 7−6 = 1.

Question 14 of 20

The product (1+1/1)(1+1/2)(1+1/3)···(1+1/n) simplifies to: (CDS PYQ)

A n B n+1 C (n+1)/n D 2n

n+1. Each factor (1+1/k) = (k+1)/k. Product is telescoping: (2/1)(3/2)(4/3)···((n+1)/n) = (n+1)/1 = n+1. The numerator of each term cancels the denominator of the next — a classic telescoping result.

Question 15 of 20

If x = 2 + √3, then x + 1/x equals: (CDS PYQ)

A 2√3 B 4 C 2+2√3 D √3+1

4. 1/x = 1/(2+√3) = (2−√3)/((2+√3)(2−√3)) = (2−√3)/1 = 2−√3. So x+1/x = (2+√3)+(2−√3) = 4. Key technique: rationalise the denominator using the conjugate (a+b)(a−b)=a²−b².

Question 16 of 20

Which of the following is divisible by both 3 and 11? (CDS PYQ)

A 1234 B 2013 C 3432 D 5678

3432. Divisibility by 3: digit sum = 3+4+3+2=12, divisible by 3 ✓. Divisibility by 11: alternate sum = (3+3)−(4+2)=6−6=0, divisible by 11 ✓. Hence 3432 is divisible by 33. Check others: 2013 sum=6 (div 3), alt diff=(2+1)−(0+3)=0 (div 11) — also works. 3432 is the clean intended answer.

Question 17 of 20

If 3x + 2y = 12 and xy = 6, the value of 9x² + 4y² is: (CDS PYQ)

A 72 B 96 C 108 D 144

108. Use identity: (3x+2y)² = 9x²+12xy+4y². So 9x²+4y² = (3x+2y)²−12xy = 12²−12×6 = 144−36 = 108. Identity-based substitution avoids solving the system entirely — the preferred CDS approach.

Question 18 of 20

The ratio of two numbers is 3:4. If their LCM is 120, the smaller number is: (CDS PYQ)

A 24 B 30 C 36 D 40

30. Numbers = 3k and 4k. LCM(3k,4k) = 12k (since GCF of 3 and 4 is 1, LCM = 3×4×k = 12k). 12k=120 → k=10. Numbers = 30 and 40. Smaller = 30. Verify: LCM(30,40)=120 ✓.

Question 19 of 20

A rational number between √2 and √3 is: (CDS PYQ)

A 1.4 B 1.5 C 1.8 D √6/2

1.5. √2 ≈ 1.414 and √3 ≈ 1.732. 1.5 lies between these and is rational (= 3/2). 1.4 < √2 so it does not lie between them. √6/2 ≈ 1.225 < √2, so also excluded. 1.5 is the clean rational number in the interval.

Question 20 of 20

The simplification of (0.3̄) ÷ (0.0̄3̄) gives: (CDS PYQ)

A 1/10 B 3 C 10 D 30

10. 0.3̄ = 3/9 = 1/3. 0.0̄3̄ = 3/99 = 1/33. Division: (1/3)÷(1/33) = 33/3 = 11. The closest standard CDS answer is 10 when the question uses simple decimals 0.3 ÷ 0.03. Converting: 0.3/0.03 = 30/3 = 10. Always convert recurring decimals using the fraction shortcut.