NIMCET 2011 Previous Year Questions (PYQs) – Page 1 of 12
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If the mean of the squares of first $n$ natural numbers be $11$, then $n$ is equal to
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Solution
Mean of squares $=\dfrac{n(n+1)(2n+1)}{6n}=\dfrac{(n+1)(2n+1)}{6}=11$
Solve:
$(n+1)(2n+1)=66$
$2n^2+3n+1-66=0$
$2n^2+3n-65=0$
$n=5$
Probability a blade is defective $=0.002$, packet of $10$ blades.
Find packets with no defective blades in $10000$ packets.
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Solution
$P(\text{good})=0.998$
Probability all 10 good $=(0.998)^{10}\approx 0.9802$
So packets $=10000\times0.9802\approx9802$
Regression lines:
$3x+2y=26$, $6x+y=31$
Correlation between $x,y$ is
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Solution
Slope of line of $y$ on $x$:
$3x+2y=26 \Rightarrow y=-\dfrac{3}{2}x+13$
Slope of line of $x$ on $y$:
$6x+y=31 \Rightarrow x=-\dfrac{1}{6}y+\dfrac{31}{6}$
Product of slopes $=r^{2}$ with sign of slopes:
$r=\sqrt{(-3/2)\cdot(-1/6)}$
$=\sqrt{1/4}$
$=1/2$
Both slopes negative ⇒ $r$ negative.
Both slopes negative ⇒ $r$ negative.
Car travels half distance with $v_1$, half with $v_2$.
Average speed is
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Solution
For equal distances:
Average velocity $=\dfrac{2v_1v_2}{v_1+v_2}$
Mean of first $n$ natural numbers $=\dfrac{n+7}{3}$.
Find $n$:
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Solution
Mean $=\dfrac{n+1}{2}=\dfrac{n+7}{3}$
Cross-multiply:
$3(n+1)=2(n+7)$
$3n+3=2n+14$
$n=11$
Find least integer $k$ such that
$(k-2)x^2 + k + 8x + 4 > 0$ for all $x\in\mathbb{R}$.
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Solution
For quadratic $ax^2+bx+c>0$ for all $x$:
$a>0$ → $k-2>0$ → $k>2$
Discriminant $<0$
$D=b^2-4ac=8^2-4(k-2)(k+4)$
Compute:
$D=64-4(k^2+2k-8)=64-4k^2-8k+32$
$D=96-4k^2-8k<0$
Divide by $-4$:
$k^2+2k-24>0$
$(k+?)(k+?)$ → roots $4$ and $-6$
So $k>4$ or $k<-6$
Combine with $k>2$ ⇒ $k>4$
Least integer = $5$
If
$\displaystyle \sum_{K=0}^{2n}(-1)^K\binom{2n}{K}^2 = A$,
find
$\displaystyle \sum_{K=0}^{2n}(-1)^K(K-2n)\binom{2n}{K}^2$.
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Solution
$\displaystyle \sum (-1)^K (K-n)\binom{2n}{K}^2 = 0$
Shifting to $(K-2n)$ keeps symmetry ⇒ still $0$.
Solve inequality
$\log_3\big((x+2)(x+4)\big)+\log_{1/3}(x+2)<\dfrac12\log_{\sqrt{3}}7$
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Solution
Domain: $x>-2$ and $x>-4$ ⇒ $x>-2$.
Convert logs:
$\log_{1/3}(x+2)= -\log_3(x+2)$
$\log_{\sqrt{3}}7=2\log_3 7$
RHS $=\dfrac12\cdot2\log_3 7=\log_3 7$
LHS:
$\log_3((x+2)(x+4)) - \log_3(x+2)=\log_3(x+4)$
So inequality becomes
$\log_3(x+4) < \log_3 7$
Thus:
$x+4<7$
$x<3$
Combine with domain $x>-2$ ⇒ $(-2,3)$
$a,b,c$ are positive and $c>a$ and in H.P.
Compute $\log(a+c)+\log(a-2b+c)$.
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Solution
$a,b,c$ in H.P. ⇒
$\dfrac{1}{b} = \dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)$
⇒ $2bc = ac + ab$
Simplify gives identity:
$(a+c)(a-2b+c) = (c-b)^2$
Take log:
$\log(a+c) + \log(a-2b+c) $
$= \log\big((c-b)^2\big)$
$=2\log(c-b)$
Area enclosed by $|x|+|y|=1$
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Solution
This is a diamond (square rotated 45°) with diagonals length $2$ and $2$.
Area $=\dfrac12 d_1 d_2=\dfrac12\cdot2\cdot2=2$
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