NIMCET 2020 Previous Year Questions (PYQs) – Page 9 of 12
Study Stuff for MCA Examinations
Solution
Given: \(A = \{x, y, z\}\)
Step 1: Find the number of subsets of \(A\):
If a set has \(n\) elements, its power set has \(2^n\) subsets.
\(|A| = 3 \Rightarrow |P(A)| = 2^3 = 8.\)
Step 2: Now we need the number of subsets of \(P(A)\), i.e. the power set of the power set.
So, \(|P(P(A))| = 2^{|P(A)|} = 2^8 = \boxed{256}.\)
✅ Final Answer: 256
Solution
Solution
Solution
Solution
Let the distance between each pair of cities = d km.
Step 1: Distances & Speeds
- A → B: distance = d, speed = 30 km/h
- B → C: distance = d, speed = 40 km/h
- C → A: distance = d, speed = 50 km/h
Step 2: Total Distance
Total distance = d + d + d = 3d
Step 3: Total Time
Time = (d/30) + (d/40) + (d/50)
= d × (1/30 + 1/40 + 1/50)
LCM of (30,40,50) = 600
⇒ (20 + 15 + 12)/600 = 47/600
Total time = d × (47/600) = (47d / 600)
Step 4: Average Speed
Average speed = Total distance / Total time
= (3d) ÷ (47d/600)
= (3d × 600) / 47d
= 1800 / 47
≈ 38.3 km/h
✅ Final Answer: Average speed ≈ 38.3 km/h
Solution
Solution
Thus, probability of not winning = $1-\frac{1}{25}=\frac{24}{25}$
Solution
Given sets
- A: 2, 3, 7, 1, 3, 2, 3
- B: 7, 5, 9, 12, 5, 3, 8
- C: 4, 4, 11, 7, 2, 3, 4
Set A: sum = 21, n = 7 ⇒ Mean = 21/7 = 3.
Sorted A = (1, 2, 2, 3, 3, 3, 7) ⇒ Median = 3.
Mode(A) = most frequent = 3.
Set B: sorted = (3, 5, 5, 7, 8, 9, 12) ⇒ Median(B) = 7.
Set C: sum = 35, n = 7 ⇒ Mean(C) = 5; Mode(C) = 4.
Check options
- Mean(A) = 3 vs Mode(C) = 4 → False
- Mean(C) = 5 vs Median(B) = 7 → False
- Median(B) = 7 vs Mode(A) = 3 → False
- Mean(A) = Median(A) = Mode(A) = 3 → True
Correct statement: Option 4
| Size of item | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Frequency | 3 | 6 | 9 | 13 | 8 | 5 | 4 |
Solution
Total number of items in the distribution = Σ fi = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.
The Mean (x̅) of the given set = \(\rm \dfrac{\sum f_i x_i}{\sum f_i}\).
⇒ x̅ = \(\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}\) = 9.
Let's calculate the variance using the formula: \(\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2\).
\(\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}\) = 83.58.
∴ σ2 = 83.58 - 92 = 83.58 - 81 = 2.58.
And, Standard Deviation (σ) = \(\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}\) ≈ 1.607.
If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix},$ then for any positive integer $n$, $A^n$ is
Solution
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Commented Feb 06, 2021
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