Showing 1–9 of 9 questions
in Mathematics (NDA)
Q.1
Hard
Mathematics (NDA)
In a frequency distribution, if mean = 25, mode = 23, find median using the empirical relationship.
A
24
B
24.33
C
24.67
D
25.33
Explanation:
Mode = 3×Median - 2×Mean, 23 = 3×Median - 50, Median = 24.33. Actually: Median = (Mode + 2×Mean)/3 = (23 + 50)/3 = 24.33, checking: mode≈3med-2mean gives med≈24.67
Q.2
Hard
Mathematics (NDA)
If a, b, c are in geometric progression with common ratio r, and a = 2, find the sum of first 5 terms when r = 2.
Explanation:
S₅ = a(r⁵ - 1)/(r - 1) = 2(32 - 1)/(2 - 1) = 2×31 = 62
Q.3
Hard
Mathematics (NDA)
If 2sin²θ - sin θ - 1 = 0, find sin θ (θ is acute).
Explanation:
2sin²θ - sin θ - 1 = 0. Factoring: (2sinθ + 1)(sinθ - 1) = 0. Since θ is acute, sin θ = 1.
Q.4
Hard
Mathematics (NDA)
If tan θ + cot θ = 4, find tan² θ + cot² θ.
Explanation:
(tan θ + cot θ)² = tan²θ + 2 + cot²θ = 16. So tan²θ + cot²θ = 14.
Q.5
Hard
Mathematics (NDA)
If sin 3θ = cos θ, where 0 < θ < 90°, find θ.
A
22.5°
B
30°
C
45°
D
60°
Explanation:
sin 3θ = cos θ = sin(90° - θ). So 3θ = 90° - θ → 4θ = 90° → θ = 22.5°.
Q.6
Hard
Mathematics (NDA)
If sec θ - tan θ = 1/5, find sec θ + tan θ.
Explanation:
We know sec²θ - tan²θ = 1. So (sec θ - tan θ)(sec θ + tan θ) = 1. Given sec θ - tan θ = 1/5, we get (1/5)(sec θ + tan θ) = 1, so sec θ + tan θ = 5
Q.7
Hard
Mathematics (NDA)
If nC2 = 21, find n.
Explanation:
nC2 = n(n-1)/2 = 21 means n(n-1) = 42. For n=7: 7×6 = 42. So n = 7
Q.8
Hard
Mathematics (NDA)
Solve: 5^(2x) - 6×5^x + 5 = 0
A
x = 0 or x = 1
B
x = 1 or x = 2
C
x = 0 or x = -1
D
x = 2 or x = 3
Correct Answer:
A. x = 0 or x = 1
Explanation:
Let y = 5^x. Then y² - 6y + 5 = 0, so (y-5)(y-1) = 0. Thus y = 5 or y = 1, giving x = 1 or x = 0
Q.9
Hard
Mathematics (NDA)
Find the remainder when 2⁵⁰ is divided by 7.
Explanation:
By Fermat's Little Theorem, 2⁶ ≡ 1 (mod 7). So 2⁵⁰ = 2⁴⁸ × 2² = (2⁶)⁸ × 4 ≡ 1⁸ × 4 ≡ 4 (mod 7).
