Surds & Indices – MBA CET questions with solutions

Approach: Reciprocal pair: x + y and xy, then x² + y² = (x + y)² − 2xy.

Solution:
y = 1/x, so xy = 1.
x + y = [(√10+√6)² + (√10−√6)²] / (10 − 6) = 2(10+6)/4 = 32/4 = 8.
x² + y² = 8² − 2 × 1 = 62.

Cetking shortcut: For (√a+√b)/(√a−√b) + reciprocal: x + y = 2(a+b)/(a−b) = 32/4 = 8 directly.

Answer: C

Approach: Square both sides to remove the root.

Solution:
√(1.44/x) = 0.1 → 1.44/x = 0.01.
x = 1.44 / 0.01 = 144.

Cetking shortcut: Dividing by 0.01 = multiplying by 100.

Answer: C

Approach: Reciprocal of a surd by rationalising (conjugate).

Solution:
1/x = 1/(7 − 4√3) × (7 + 4√3)/(7 + 4√3) = (7 + 4√3)/(49 − 48) = 7 + 4√3.
x + 1/x = (7 − 4√3) + (7 + 4√3) = 14.

Why the other options are wrong:

• A, B: surd terms cancel, so no √3 can remain.
• D, E: forget that the conjugate's surd part cancels with x's.

Cetking shortcut: When a² − b = 1 (here 49 − 48), the reciprocal of a − √b is simply a + √b, so x + 1/x = 2a = 14.

Answer: C

Approach: Shift the decimal in pairs: √(13 × 100ⁿ) and √(130 × 100ⁿ) only move the decimal point.

Solution:
√1.3 = √(130/100) = √130 / 10 = 11.40 / 10 = 1.140.
√1300 = √(13 × 100) = 10√13 = 36.05.
√0.013 = √(130/10000) = √130 / 100 = 0.114.
Sum = 1.140 + 36.05 + 0.114 = 37.304.

Cetking shortcut: Count digits in pairs from the decimal point: 1.3 → 130 family, 1300 → 13 family, 0.013 → 130 family. No √10 needed.

Answer: D

Approach: Cube root via powers of 10.

Solution:
0.000216 = 216 × 10⁻⁶.
∛216 = 6 and ∛10⁻⁶ = 10⁻².
∛0.000216 = 6 × 10⁻² = 0.06.
Check: 0.06³ = 0.000216 ✓.

Why the other options are wrong:
A (0.6): cube = 0.216.
C (0.006): cube = 0.000000216.
D, E: 16 is not a cube root of 216.

Answer: B

Approach: Find the largest perfect square not exceeding 36581.

Solution: 190² = 36,100; 191² = 36,481; 192² = 36,864 (> 36,581).
So the largest square uses 191 rows → 36,481 soldiers.
Left over = 36,581 − 36,481 = 100.

Cetking shortcut: 191² = (190 + 1)² = 36,100 + 380 + 1 = 36,481.

Answer: C

Approach: Total = n² paise.

Solution: Rs 59.29 = 5929 paise.
n² = 5929 → n = 77 (since 77² = 5929).

Cetking shortcut: 5929 ends in 9 → root ends in 3 or 7; 70² = 4900 and 80² = 6400 → 77.

Answer: C

Approach: Find the largest square ≤ 549162.

Solution: 740² = 547,600; 741² = 549,081; 742² = 550,564 (too big).
549,162 − 549,081 = 81.

Cetking shortcut: 741² = 740² + 740 + 741 = 547,600 + 1,481 = 549,081.

Answer: D

Approach: Square a difference of square roots, or spot perfect squares of the form (a ± √b)².

Solution:
√(x + 6√2) − √(x − 6√2) = 2√2. Square both sides:
(x + 6√2) + (x − 6√2) − 2√(x² − 72) = 8 → 2x − 8 = 2√(x² − 72) → x − 4 = √(x² − 72).
Square again: x² − 8x + 16 = x² − 72 → 8x = 88 → x = 11.
Check: 11 + 6√2 = (3 + √2)², 11 − 6√2 = (3 − √2)² → difference = 2√2 ✓

Cetking shortcut: 2√2 = (3 + √2) − (3 − √2), and (3 ± √2)² = 11 ± 6√2, so x = 11 directly.

Answer: C

Approach: Group equal bases on each side and factor out the smallest power.

Solution:
9^(x − 1/2) = 3^(2x − 1). Move terms: 3^(2x − 1) + 3^(2x − 3) = 4^x + 2^(2x − 2) = 4^x + 4^(x − 1).
Left = 3^(2x − 3)(9 + 1) = 10 × 3^(2x − 3). Right = 4^(x − 1)(4 + 1) = 5 × 2^(2x − 2).
10 × 3^(2x − 3) = 5 × 2^(2x − 2) → 3^(2x − 3) = 2^(2x − 3) → 2x − 3 = 0 → x = 3/2.
Check: LHS = 9¹ − 2¹ = 7; RHS = 4^(3/2) − 3⁰ = 8 − 1 = 7 ✓

Cetking shortcut: a^k = b^k with a ≠ b forces k = 0; or simply plug in options — x = 3/2 makes every exponent an integer.

Answer: E