Jamia Millia Islamia MCA 2024 Previous Year Questions (PYQs) – Page 7 of 10
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An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement.
What is the probability that both drawn balls are black?
What is the probability that both drawn balls are black?
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Solution
Total balls $= 15$
Probability of drawing two black balls:
$P = \dfrac{10}{15} \times \dfrac{9}{14} = \dfrac{90}{210} = \dfrac{3}{7}$
The absolute maximum value of $y = x^3 - 3x + 2$ in $0 \le x \le 2$ is
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Solution
$y' = 3x^2 - 3 = 0 \Rightarrow x = 1$
Now,
$y(0) = 2$, $y(1) = 0$, $y(2) = 8 - 6 + 2 = 4$
Hence, maximum value = 4.
$\displaystyle \int_{0}^{\pi} \sin^2 x,dx =$
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Solution
$\sin^2 x = \dfrac{1 - \cos 2x}{2} \Rightarrow \int_{0}^{\pi} \sin^2 x,dx = \dfrac{1}{2} \left[ x - \dfrac{\sin 2x}{2} \right]_{0}^{\pi} = \dfrac{1}{2}(\pi - 0) = \dfrac{\pi}{2}$
If$P(A) = 0.4,\ P(B) = 0.7,\ P(B|A) = 0.6$, then $P(A \cup B)$ is
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Solution
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cap B) = P(A) \times P(B|A) = 0.4 \times 0.6 = 0.24$
$\Rightarrow P(A \cup B) = 0.4 + 0.7 - 0.24 = 0.86$
The area bounded by the curves $y^2 = 4x$ and $y = x$ is equal to
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Solution
For intersection: $y^2 = 4x$ and $y = x \Rightarrow x^2 = 4x \Rightarrow x = 0, 4$
Area $= \int_{0}^{4} (\sqrt{4x} - x),dx = \int_{0}^{4} (2\sqrt{x} - x),dx = \left[\dfrac{4}{3}x^{3/2} - \dfrac{x^2}{2}\right]_{0}^{4} = \dfrac{32}{3} - 8 = \dfrac{8}{3}$
What is the order of the differential equation $y'' + 5y' + 6 = 0$?
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Solution
The order of a differential equation is the highest derivative present.
Here, the highest derivative is $y''$ $\Rightarrow$ Order = 2.
The side of an equilateral triangle is increasing at the rate of $2\ \text{cm/s}$.
The rate at which area increases when the side is $10\ \text{cm}$ will be —
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Solution
For an equilateral triangle, area $A = \dfrac{\sqrt{3}}{4}s^2$
Differentiate with respect to time $t$:
$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2}s\dfrac{ds}{dt}$
Given $\dfrac{ds}{dt} = 2$ and $s = 10$,
$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times 10 \times 2 = 10\sqrt{3}$
Hence, rate = $10\sqrt{3}$ cm$^2$/s
The scalar product of $(5i + j - 3k)$ and $(3i - 4j + 7k)$ is —
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Solution
Dot product $= (5)(3) + (1)(-4) + (-3)(7)$
$= 15 - 4 - 21 = -10$
Wait — recheck:
$15 - 4 - 21 = -10$ (not -15).
Let’s verify carefully — if given answer key shows B (-15), check question once more:
If second vector was $(3i - 4j + 7k)$ indeed,
then $5×3 = 15$, $1×(-4) = -4$, $(-3)×7 = -21$, total = $-10$.
So true answer is $-10$, though the paper marks B ($-15$).
Maybe a misprint.
If $a_n = 4n + 6$, find the $15^{th}$ term of the sequence.
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Solution
$a_{15} = 4(15) + 6 = 60 + 6 = 66$
If A.M. of two numbers is $15/2$ and their G.M. is $6$, then find the two numbers.
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Solution
Let the numbers be $a$ and $b$.
$\dfrac{a + b}{2} = \dfrac{15}{2} \Rightarrow a + b = 15$
and $\sqrt{ab} = 6 \Rightarrow ab = 36$
Now, $a$ and $b$ are roots of $x^2 - 15x + 36 = 0$
$\Rightarrow x = 12, 3$
Hence, the numbers are 12 and 3.
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