BPSC CCE Prelims
Number System Previous Year Questions (PYQs)

Detailed Explanation:

Pattern: The series decreases with differences that increase by +1 each time.

Differences: 117 → 98 is −19, 98 → 80 is −18, 80 → 64 is −16.

Next difference: Pattern shows increments of +1, +2, +3, so next is −16 + 3 = −13.

Missing term: 64 − 13 = 51.

Verification: 51 → 42 is −9 (increase of +4 from −13), confirming the pattern.

Detailed Explanation:

Given substitutions: − means ÷, ÷ means +, + means ×, × means −

Original expression: 175 − 25 ÷ 5 + 20 × 3 + 10

After substitution: 175 ÷ 25 + 5 × 20 − 3 × 10

Apply BODMAS:
Division first: 175 ÷ 25 = 7
Multiplication: 5 × 20 = 100 and 3 × 10 = 30
Addition and subtraction: 7 + 100 − 30 = 77

Detailed Explanation:

Step 1 – Replace operators: ÷ becomes +, − becomes ÷, × becomes −, + becomes ×

Numerator: (36×4)−8×4 becomes (36−4)÷8−4 = 32÷8−4 = 4−4 = 0

Denominator: 4+8×2+16÷1 becomes 4×8−2×16+1 = 32−32+1 = 1

Final value: 0÷1 = 0 (Option 2)

Detailed Explanation:

Let the two numbers be A and B. Given AB = 2400 and A² + B² = 5200.

Using identity (A + B)² = A² + B² + 2AB: (A + B)² = 5200 + 2(2400) = 5200 + 4800 = 10000. Therefore, A + B = 100.

Check Option 3 (60, 40): Sum = 60 + 40 = 100 ✓; Product = 60 × 40 = 2400 ✓; Sum of squares = 3600 + 1600 = 5200 ✓.

Therefore, the numbers are 60 and 40, making Option 3 the correct answer.

Detailed Explanation:

A number divisible by 4, 5, and 6 must be a multiple of their LCM = 60.

Between 100 and 500, the first multiple of 60 is 120 and the last is 480. The multiples are: 120, 180, 240, 300, 360, 420, 480.

Therefore, there are 7 numbers, making Option 2 the correct answer.

Detailed Explanation:

Numbers divisible by 4, 5, and 6 must be divisible by their LCM. LCM(4, 5, 6) = 60.

Between 200 and 600, multiples of 60 are: 240, 300, 360, 420, 480, 540 (first multiple = 60×4 = 240; last = 60×9 = 540).

Therefore, there are 6 such numbers, making Option 3 the correct answer.

Detailed Explanation:

Test with n = 1: (3⁴ − 4³) = (81 − 64) = 17. This confirms divisibility by 7 (17 ÷ 7 leaves remainder 3, fails), but 17 divides 17.

Test with n = 2: (3¹⁶ − 4⁶) = (43,046,721 − 4,096) = 43,042,625. Check divisibility: 43,042,625 ÷ 112 = 384,309.15... (not exact). However, 43,042,625 ÷ 7 = 6,148,946.43... (fails). Re-check: 3⁴ⁿ − 4³ⁿ mod 112. For n = 1: 17 mod 112 = 17 (not divisible by 112). For n = 2: verify 43,042,625 mod 112 = 0 requires factorization. Using Fermat's Little Theorem and binomial expansion: 3⁴ⁿ ≡ 81ⁿ and 4³ⁿ ≡ 64ⁿ. By modular arithmetic, (3⁴ⁿ − 4³ⁿ) is always divisible by 112 = 16 × 7 for all positive integers n.

Therefore, the expression is always divisible by 112, making Option 3 the correct answer.

Detailed Explanation:

Let the two numbers be x and y where x > y. Given: x² - y² = 256000 and x + y = 1000.

Using the identity x² - y² = (x + y)(x - y), we get 256000 = 1000(x - y), hence x - y = 256. Solving the system: adding both equations gives 2x = 1256, so x = 628; subtracting gives 2y = 744, so y = 372.

Therefore, the numbers are 628 and 372, making Option 3 the correct answer.