MAH CET MCA 2025 Previous Year Questions (PYQs) – Page 1 of 10
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The probability distribution function of a random variable $X$ is given by
$f(x) = \dfrac{x}{18}, ; 0 \le x \le 6$
$= 0, \text{ otherwise}$
Then the value of $P(X > 2)$ is
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Solution
$P(X > 2) = \int_{2}^{6} \dfrac{x}{18} , dx = \dfrac{1}{18} \left[\dfrac{x^2}{2}\right]_2^6 = \dfrac{1}{36}(36 - 4) = \dfrac{32}{36} = \dfrac{8}{9}$
Find the probability of guessing correctly at least six of 10 answers in a True or False objective test.
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Solution
Let $X \sim \text{Binomial}(n = 10, p = 0.5)$
$P(X \ge 6) = \sum_{r=6}^{10} \binom{10}{r} (0.5)^{10} = 0.377$
If, for a binomial distribution, the number of trials is $9$, the variance is $2$ and the probability of success is greater than that of failure, find the probability of both success.
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Solution
For a binomial distribution, variance $= npq = 2$ and $n = 9$.
So, $9p(1 - p) = 2 \Rightarrow 9p - 9p^2 = 2 \Rightarrow 9p^2 - 9p + 2 = 0$.
$\Rightarrow p = \dfrac{9 \pm \sqrt{81 - 72}}{18} = \dfrac{9 \pm 3}{18}$.
Hence, $p = \dfrac{2}{3}$ or $\dfrac{1}{3}$.
Since probability of success is greater than failure, $p = \dfrac{2}{3}$.
Two customers, Rachana and Bhakti, are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on consecutive days?
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Solution
Total days = 5 (Tuesday to Saturday).
Total possible pairs $= 5 \times 5 = 25$.
Consecutive day pairs = $(T,W), (W,T), (W,Th), (Th,W), (Th,F), (F,Th), (F,S), (S,F)$ → total 8.
So, $P = \dfrac{8}{25}$.
A pendulum swings through an angle of $30^\circ$ and describes an arc of $17.6\ \text{cm}$ in length. Find the length of the pendulum.
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Solution
Arc length $l = r\theta$.
Here $\theta = 30^\circ = \dfrac{\pi}{6}\ \text{radians}$ and $l = 17.6$.
So, $r = \dfrac{l}{\theta} = \dfrac{17.6}{\pi/6} = \dfrac{17.6 \times 6}{3.14} \approx 33.6\ \text{cm}$.
Two cubes each of volume $64\ \text{cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.
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Solution
Each cube side $a = \sqrt[3]{64} = 4\ \text{cm}$.
When joined end to end, dimensions of cuboid $= 8 \times 4 \times 4$.
Surface area $= 2(lb + bh + hl) = 2(8\times4 + 4\times4 + 4\times8) = 2(32 + 16 + 32) = 160\ \text{cm}^2$.
The incomes of $A$, $B$ and $C$ are in the ratio $7 : 9 : 12$ and their spending are in the ratio $8 : 9 : 15$. If $A$ saves $\dfrac{1}{4}$ of his income, then the savings of $A$, $B$ and $C$ are in the ratio of
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Solution
Let incomes be $7x, 9x, 12x$ and expenditures be $8y, 9y, 15y$.
Saving of $A = 7x - 8y = \dfrac{1}{4}(7x) \Rightarrow 7x - 8y = \dfrac{7x}{4} \Rightarrow 8y = \dfrac{21x}{4} \Rightarrow y = \dfrac{21x}{32}$.
Savings:
$A = \dfrac{7x}{4}$,
$B = 9x - 9y = 9x - 9\left(\dfrac{21x}{32}\right) = \dfrac{99x}{32}$,
$C = 12x - 15y = 12x - 15\left(\dfrac{21x}{32}\right) = \dfrac{69x}{32}$.
Hence ratio $= 56 : 99 : 69$.
Two trains, A and B, start simultaneously from two stations $300\ \text{km}$ apart. Train A travels at $60\ \text{km/h}$, and Train B travels at $90\ \text{km/h}$. If a bird starts flying from Train A towards Train B at $120\ \text{km/h}$ and immediately turns back upon reaching Train B, continuing this until the trains meet, what total distance does the bird cover?
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Solution
Time for trains to meet $= \dfrac{300}{60+90} = 2\ \text{h}$.
Bird flies the whole time at $120\ \text{km/h}$, so distance $= 120 \times 2 = 240\ \text{km}$.
Find the distance from the eye at which a coin of $2\ \text{cm}$ diameter should be held so as to conceal the full moon whose angular diameter is $31'$.
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Solution
Small–angle $\theta\ (\text{radians}) \approx \dfrac{\text{diameter}}{\text{distance}}$.
$\theta = 31' = \dfrac{31\pi}{180\times60}$.
$d = \dfrac{0.02}{\theta} = \dfrac{0.02\times10800}{31\pi} \approx 2.217\ \text{m}$.
A circular wire of radius $7.5\ \text{cm}$ is cut and bent to lie along the circumference of a hoop of radius $120\ \text{cm}$. Find (in degrees) the angle subtended at the hoop’s centre.
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Solution
Arc length $s = 2\pi(7.5)=15\pi$.
On hoop of radius $120$, $\theta = \dfrac{s}{r} = \dfrac{15\pi}{120}=\dfrac{\pi}{8}=22^\circ 30'$.
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