MAH CET MCA 2023 Previous Year Questions (PYQs) – Page 4 of 10
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$x$ is a G.M., $y$ is a H.M. of two numbers. If $\dfrac{x}{y} = \dfrac{5}{4}$, then the numbers are in ratio
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Solution
Let numbers be $a, b$.
Then $\dfrac{x}{y} = \dfrac{\sqrt{ab}}{\dfrac{2ab}{a+b}} = \dfrac{a+b}{2\sqrt{ab}} = \dfrac{5}{4}$.
Let $\dfrac{a}{b} = r$,
then $\dfrac{r+1}{2\sqrt{r}} = \dfrac{5}{4}$.
On solving, $r = 4$ or $\dfrac{1}{4}$.
Hence ratio = $1:4$.
Sum of squares of odd positive integers $1$ to $100$ belongs to
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Solution
Sum of squares of first $n$ odd numbers $= \dfrac{n(2n-1)(2n+1)}{3}$.
For $n = 50$, sum $= \dfrac{50 \times 99 \times 101}{3} = 166650$.
Hence, it lies between $150000$ and $250000$.
The number of values of $C$ such that the straight line $y = 4x + C$ touches the curve $x^2 + 4y^2 = 4$ is
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Solution
Substitute $y = 4x + C$ in $x^2 + 4y^2 = 4$.
To touch, discriminant $= 0$.
After solving, $C = \pm \dfrac{2}{\sqrt{17}}$.
Hence, there are $2$ possible values.
________ are attempts by individuals to obtain confidential information from you by falsifying their identity.
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Solution
Phishing is a fraudulent attempt to obtain sensitive information by pretending to be a trustworthy entity.
$f(x) = \sqrt{\log_{1/4}\left(\dfrac{5x - x^2}{4}\right)}$, find the domain of $f(x)$.
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Solution
For $\log_{1/4}\left(\dfrac{5x - x^2}{4}\right)$ to be $\ge 0$,
$\dfrac{5x - x^2}{4} \le 1$ since base $< 1$.
Simplifying gives $x = 1, 4$.
Hence, domain $= {1,4}$.
How many times does the number $2$ appear in the range $1$ to $1000$?
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Solution
Between $200$–$300$, digit $2$ appears most frequently (in hundreds, tens, and ones place).
Which of the following is used to move the document?
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Solution
Cut & Paste moves the selected content.
$1 + i,\ 3 + 2i,\ 5 + 3i$ then the points are
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Solution
All points lie on line $y = x - 1$.
Hence, points are collinear.
In the triangle $ABC$, sides $AB$, $BC$, and $CA$ are extended such that $B'$, $C'$, and $A'$ are the new points formed.
If the area of $\triangle ABC = 1$ unit and
$AA' = 2AB,\ BB' = 2BC,\ CC' = 3AC$,
then find the area of $\triangle A'B'C'$.
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Solution
Each side is scaled by 2, 2, and 3 respectively; the area ratio depends on the product of scale factors.
Hence, area $= 18$ times original.
A cone of base radius $1$ and height $15,m$ is melted to form a sphere of equal radius. Find radius nearer to:
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Solution
Volume of cone $= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(1)^2(15)=5\pi$.
Let sphere radius $= R$, then $\dfrac{4}{3}\pi R^3 = 5\pi \Rightarrow R^3 = \dfrac{15}{4} \Rightarrow R \approx 0.5$.
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