1
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=2 x^3+\mathrm{m} x^2-13 x+\mathrm{n}$ and 2,3 are the roots of the equation $\mathrm{f}(x)=0$ then the value of $4 m+5 n$ is

A
30
B
100
C
130
D
150
2
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

4 red balls and 5 green balls are selected from $n$ balls. If the sum of both the selections is greater than ${ }^{n+1} C_4$ then the value of $n$ is equal to

A
$\mathrm{n}>8$
B
$\mathrm{n}<8$
C
$n>10$
D
$\mathrm{n}>12$
3
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Two numbers are selected at random, without replacement from the first 6 positive integers. Let $X$ denote the larger of the two numbers. Then $\mathrm{E}(\mathrm{X})=$

A
$\frac{14}{3}$
B
$\frac{3}{14}$
C
$\frac{14}{5}$
D
$\frac{15}{41}$
4
MHT CET 2025 20th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

For $\mathrm{k}=1,2,3$ the box $\mathrm{B}_{\mathrm{k}}$ contains k red balls and $(k+1)$ white balls. Let $P\left(B_1\right)=\frac{1}{2}, P\left(B_2\right)=\frac{1}{3}$ and $\mathrm{P}\left(\mathrm{B}_3\right)=\frac{1}{6} . \mathrm{A}$ box is selected at random and a ball is drawn from it. If a red ball is drawn from it, then the probability that it comes from box $\mathrm{B}_2$ is

A
$\frac{35}{78}$
B
$\frac{14}{39}$
C
$\frac{10}{13}$
D
$\frac{12}{13}$
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