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Solution

Case I: Exactly one digit repeated thrice $\rightarrow{ }^{5} \mathrm{C}_{1} \times \frac{7!}{3!}=4200$
Case II: Exactly two digits repeated twice $\rightarrow{ }^{5} \mathrm{C}_{2} \times \frac{7!}{2!2!}=12600$
Total such numbers = 16800

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Answer

(3)

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Solution

$\mathrm{nx}^{2}+\mathrm{x}(2+4+6+\ldots . .+2 \mathrm{n})+(1.3+2.4 \ldots . .+(\mathrm{n}-1)(\mathrm{n}+1))=4 \mathrm{n}$

\[ \begin{aligned} & \therefore 1.3+2.4+\cdots .+(n-1)(n+1) \\ &=\sum_{r=1}^{n-1} r(r+2)=\sum_{r=1}^{n-1} r^{2}+\sum_{r=1}^{n-1} 2 r=\frac{(n-1)(n)(2 n-1)}{6}+\frac{2(n-1) n}{2}=\frac{n(n-1)(2 n+5)}{6} \\ & \Rightarrow n x^{2}+n(n+1) x+\frac{n(n-1)(2 n+5)}{6}=4 n \Rightarrow x^{2}+(n+1) x+\frac{(n-1)(2 n+5)}{6}=4 \end{aligned} \]

As roots are integers
D must be a perfect square $\mathrm{D}=(n+1)^{2}-4\left[\frac{(\mathrm{n}-1)(2 \mathrm{n}+5)}{6}-4\right]=16+\frac{13-\mathrm{n}^{2}}{3}$

\[ \begin{aligned} & =20-\left(\frac{\mathrm{n}^{2}-1}{3}\right)=\mathrm{P}^{2} ; \mathrm{P} \in \mathrm{I} ; \mathrm{P} \in\{1,2,3,4\} \\ & \Rightarrow \mathrm{n}^{2}=61-3 \mathrm{P}^{2} \text { For integer, } \mathrm{P}=2, \mathrm{n}=7 \\ & \mathrm{x}^{2}+8 \mathrm{x}+15=0 \Rightarrow \mathrm{x}=-3,-5 \Rightarrow \alpha=-5, \alpha+2=-3 \therefore \alpha+\mathrm{n}=7-5=2 \end{aligned} \]

[CHAPTER M06 COMPLEX NUMBER TOPIC T1 ]

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Answer

(3)

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Solution

$\quad 50\left(\frac{2 \mathrm{x}(1-3 \mathrm{i})}{10}-\frac{\mathrm{y}(1+2 \mathrm{i})}{5}\right)=31+17 \mathrm{i} \Rightarrow 10 \mathrm{x}-(30 \mathrm{x}) \mathrm{i}-10 \mathrm{y}-20 \mathrm{yi}=31+17 \mathrm{i}$

\[ \begin{aligned} & 10(x-y)=31 \quad \ldots(1) \text { and }-30 x-20 y=17 \\ & \text { Eq.(1) } x 3+\text { Eq.(2) } \\ & \Rightarrow 30 x-30 y-30 x-20 y=93+17 \\ & \Rightarrow y=-\frac{110}{50}=-2.2, x=0.9 \\ & \therefore 10(x-3 y)=75 \end{aligned} \]

[CHAPTER M25 DETERMINANT TOPIC T2 ]

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Answer

(4)

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Solution

for no solution $\Delta=\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta\end{array}\right|=4 \alpha-\beta=0 \Rightarrow \beta=4 \alpha$
Second eq. $2 \mathrm{x}+\mathrm{y}+\alpha \mathrm{z}=5$
third eq. $8 x+4 y+4 \alpha z=18 \Rightarrow 2 x+y+\alpha z=\frac{9}{2}$ both are parallel planes $\therefore$ no solution
[CHAPTER M26 MATRICES TOPIC T2 ]

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Answer

(2)

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Solution

For 2 × 2 order matrix M

\[ \begin{aligned} & M^{2}-\operatorname{tr}(M) M+\operatorname{det}(M) I=0 \\ & A=\left[\begin{array}{ll} 1 & 2 \\ 1 & \alpha \end{array}\right] \Rightarrow A^{2}-4 A+I=0 ; \text { trace of matrix } \alpha+1=4 \Rightarrow \alpha=3 \\ & B=\left[\begin{array}{ll} 3 & 3 \\ \beta & 2 \end{array}\right] \Rightarrow B^{2}-5 B-6 I=0 ; \operatorname{det} .(B)=6-3 \beta=-6 \Rightarrow 3 \beta=12 \Rightarrow \beta=4 \\ & B-A=\left[\begin{array}{ll} 2 & 1 \\ 3 & -1 \end{array}\right] ; B+A=\left[\begin{array}{ll} 4 & 5 \\ 5 & 5 \end{array}\right] \Rightarrow \operatorname{adj}(A+B)=\left[\begin{array}{cc} 5 & -5 \\ -5 & 4 \end{array}\right] \\ & (B-A)(B+A)=\left[\begin{array}{cc} 13 & 15 \\ 7 & 10 \end{array}\right] \end{aligned} \]

[CHAPTER M05 PROGRESSION TOPIC T1 ]

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Answer

(2)

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Solution

\[ \mathrm{AP}_{1} \text { Set } \mathrm{A}=\{1,6,11,16 \ldots \ldots \ldots \ldots\}, \mathrm{n}(\mathrm{~A})=101 \]

\[ \begin{aligned} & \mathrm{T}_{101}^{\mathrm{A}}=1+100 \times 5=501 \\ & \mathrm{AP}_{2} \text { Set } \mathrm{B}=\{9,16 \ldots \ldots \ldots\} \mathrm{N}(\mathrm{~B})=71 \\ & \mathrm{~T}_{71}^{\mathrm{B}}=9+70 \times 7=499 \end{aligned} \]

Common A.P., common difference $\mathrm{D}=\mathrm{L} . \mathrm{C} . \mathrm{M}\left\{\mathrm{d}_{1}, \mathrm{~d}_{2}\right\}=35$;
$1^{\text {st }}$ Common term is 16

\[ \begin{aligned} & 16+(n-1) 35 \leq 499 \Rightarrow n \leq 14.8 \Rightarrow n=14 \\ & T_{n}^{A \cap B}=16+(n-1) 35=35 n-19 \text { is divisible by } 3 \\ & =3(11 n-6)+2 n-1 \text { is divisible by } 3 \\ & 2 n-1 \text { must be divisible by } 3 \\ & n \in\{2,5,8,11,14\} \therefore 5 \text { such terms } \end{aligned} \]

[CHAPTER M08 PERMUTATION \& COMBINATION TOPIC T3 ]

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Answer

(2)

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Solution

\[ \begin{aligned} & X=\{1,2,3,4, \ldots, 19\} \\ & \operatorname{Mean}(\mu)=\frac{1+2+3+\ldots+19}{19}=\frac{19(1+19)}{2 \times 19}=10 \\ & \text { Variance }=\frac{\sum x_{1}^{2}}{\mathrm{n}}-(\mu)^{2}=\frac{1^{2}+2^{2}+3^{2}+\ldots+19^{2}}{19}-10^{2} \\ & \Rightarrow \sigma^{2}=\frac{19(19+1)(38+1)}{6 \times 19}-100 \Rightarrow \sigma^{2}=30 \end{aligned} \]

Replace $\mathrm{x}_{\mathrm{i}} \longrightarrow \mathrm{ax}_{\mathrm{i}}+\mathrm{b}$
New Mean $=\mathrm{a}($ Old Mean $)+\mathrm{b}$

\[ 30=a(10)+b \Rightarrow 10 a+b=30 \]

NewVariance $=\mathrm{a}^{2}$ (old Vanance)

\[ \begin{aligned} & \Rightarrow 750=a^{2}(30) \Rightarrow a^{2}=25 \Rightarrow a \pm 5 \\ & a=5,10(5)+b=30 \Rightarrow b=-20 \\ & a=-5,10(-5)+b=30 \Rightarrow b=80 \end{aligned} \]

sum of all possible values of $b=-20+80=60$ Answer(4)

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Solution

\[ \begin{aligned} & 2,3,5,10,11,13,15,21, a \& b \\ & \text { Mean }=\frac{\sum x_{i}}{10}=\frac{2+3+5+10+11+13+15+21+a+b}{10}=9 \end{aligned} \]

\[ \begin{aligned} & \Rightarrow 80+a+b=90 \Rightarrow a+b=10 \\ & \text { Variance }=\frac{1}{n} \sum x_{i}^{2}-(\text { Meau })^{2}=34.2 \\ & \Rightarrow \frac{1}{10}\left(4+9+25+100+121+169+225+441+a^{2}+b^{2}\right)-81=34.2 \\ & \Rightarrow a^{2}+b^{2}=58 \Rightarrow a^{2}+(10-a)^{2}=58 \\ & \Rightarrow 2 a^{2}-20 a+100=58 \Rightarrow a^{2}-10 a+21=0 \\ & \Rightarrow(a-3)(a-7)=0 \\ & a=3 \quad b=7 \\ & b=7 \quad b=3 \end{aligned} \]

All number 2,3,3,5,7,10,11,13,15,21

\[ \text { Median }=\frac{\mathrm{T}_{5}+\mathrm{T}_{6}}{2}=\frac{7+10}{2}=8.5 \]

Mean derivation about median $=\frac{\sum\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{n}}$

\[ \begin{aligned} & =\frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10} \\ & =5 \text { Answer }(1) \end{aligned} \]

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Solution

\[ \begin{aligned} & \operatorname{Mean}(\mu)=\frac{2+4+10+x+12+14+y}{7}=8 \\ & \Rightarrow x+y=14 \quad \ldots(1) \end{aligned} \]

\[ \begin{aligned} & \text { Variance }=\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-\mu^{2} \\ & \Rightarrow \frac{1}{7}\left[4+16+100+x^{2}+144+196+y^{2}\right]-8^{2}=16 \\ & \Rightarrow x^{2}+y^{2}+264+196=7(64+16)=560 \\ & \Rightarrow x^{2}+y^{2}=100 \end{aligned} \]

Solving (1) \& (2)

\[ \begin{aligned} & x^{2}+(14-x)^{2}=100 \Rightarrow x^{2}-14 x+48=0 \\ & \Rightarrow(x-8)(x-6)=0 \\ & \Rightarrow x=8, y=6(\text { accepted })\{1,2,3,4,6,5\} \\ & \Rightarrow x=6, y=8(\text { rejected because } x>y) \end{aligned} \]

Required probability $=\mathrm{P}$ (both less than 4) $+\mathrm{P}($ one number less than 4 \& other number $\geq 4)$

\[ =\left(\frac{3}{6}\right)\left(\frac{2}{5}\right)+\left(\frac{3}{6}\right)\left(\frac{3}{5}\right) \times 2=\frac{6}{30}+\frac{18}{30}=\frac{4}{5} \text { Answer(2) } \]

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