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This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Here,wehaveC{V}_1=50,C{V}_2=60\\ \overline{x_1}=30and\overline{x_2}=25\\ \therefore C{V}_1=\frac{{\sigma }_1}{\overline{x_1}}\times 100\Rightarrow 50=\frac{{\sigma }_1}{30}\times 100\Rightarrow {\sigma }_1=\frac{50\times 30}{100}=15\\ andC{V}_1=\frac{{\sigma }_2}{\overline{x_2}}\times 100\Rightarrow 60=\frac{{\sigma }_2}{25}\times 100\Rightarrow {\sigma }_2=\frac{60\times 25}{100}=15\\ \therefore Difference{\sigma }_1-{\sigma }_2=15-15=0\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

 $[x{]}^2+2[x+2]-7=0$

$\begin{array}{rr}& \Rightarrow [x{]}^2+2[x]+4-7=0\\ & \Rightarrow \left[x\right]=1,-3\\ & \Rightarrow x\in \left[\mathrm{1,2}\right)\cup [-3,-2)\end{array}$

A fluctuating trend can be seen in the highest package from 2024 to 2024. The highest package recorded in the recent years at  Xavier Institute of Management and Entrepreneurship  is present in the table below:

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatVariance\left({\sigma }^2\right)=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2\\ =\frac{18000}{60}-{\left(\frac{960}{60}\right)}^2=300-256=44\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

The placement rate for Xavier Institute of Management and Entrepreneurship has not been disclosed yet. However, Xavier Institute of Management and Entrepreneurship provides good placement to its students. The data tabulated below provides XIME Chennai placement highlights for 2024:

Yes, in order to secure BCom admission at ICFAI University Mizoram, students are required to make payment of the fees. Usually, to confirm admission, students must pay a one-time fee of INR 10,000. Along with this, students may also be required to make payment of the first installment of tuition fees. As per the structure, the total tuition fee for BCom is INR 65,500.

Note: The above-mentioned fee is taken from various official sources. However, it is subject to change.

To crack XIME Chennai placement interview students can follow the steps given below:

·       Review your resume

·       Practice common interview questions

·       Showcase your strengths

·       Be confident and articulate

If you miss VSAT counselling date, it may result in you losing your chance at preferred courses. Hence students are requested to keep track of the official website as well shiksha.com to stay updated on the dates.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}First10positiveintegersare1,2,3,4,5,6,7,8,9,10\\ onmultiplyingeachnumberby-1,weget\\ -1,-2,-3,-4,-5,-6,-7,-8,-9,-10\\ onadding1toeachofthenumber,weget\\ 0,-1,-2,-3,-4,-5,-6,-7,-8,-9\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=0-1-2-3-4-5-6-7-8-9\\ =-45\\ and{\displaystyle \sum _{}^{}{x}_i^2}=0^2+{\left(-1\right)}^2+{\left(-2\right)}^2+{\left(-3\right)}^2+{\left(-4\right)}^2+\dots +{\left(-9\right)}^2\\ =\frac{9\times 10\times 19}{6}=285\left[?{\displaystyle \sum _{}^{}{n}^2=}\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\\ \therefore S.D=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2}=\sqrt[]{\frac{285}{10}-{\left(\frac{-45}{10}\right)}^2}\\ =\sqrt[]{\frac{285}{10}-\frac{2025}{100}}=\sqrt[]{\frac{2850-2025}{100}}=\sqrt[]{8.25}\\ \therefore Variance={\left(SD\right)}^2={\left(\sqrt[]{8.25}\right)}^2=8.25\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

The average package has been fluctuating over the past three years. Refer to the data tabulated below to know more:

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givennumbersare1,2,3,4,5,6,7,8,9,10\\ Numbersobtainedwhen1isaddedtotheabovenumbersis2,3,4,5,6,7,8,9,10and11.\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=2+3+4+5+6+7+8+9+10+11\\ =\frac{10}{2}\left[2\times 2+\left(10-1\right).1\right]=5\left[4+9\right]=5\times 13=65\\ Now{\displaystyle \sum _{}^{}{x}_i^2}=2^2+3^2+4^2+\dots +11^2\\ =\left(1^2+2^2+3^2+4^2+\dots +11^2\right)-{\left(1\right)}^2\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}-1=\frac{11\times 12\times 23}{6}-1\\ =22\times 23-1=506-1=505\\ \therefore Variance\left({\sigma }^2\right)=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\right)}^2=\frac{505}{10}-{\left(\frac{65}{10}\right)}^2\\ =50.5-{\left(6.5\right)}^2=50.5-42.25=8.25\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

Oxford University placement rate is 92%. As per Oxford University reviews, is has been rated 4.3 stars for the quality of its faculty. The students rated the Oxford University for placement is-

Various companies visit Xavier Institute of Management and Entrepreneurship for placements some of them are given below:

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatSDoffirstnnaturalnumbers\sqrt[]{\frac{n^2-1}{12}}\\ Heren=10\\ \therefore SD=\sqrt[]{\frac{{\left(10\right)}^2-1}{12}}=\sqrt[]{\frac{99}{12}}=\sqrt[]{8.25}=2.87\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

Xavier Institute of Management and Entrepreneurship average package stood at INR 8.60 LPA. To know more about the average package at Xavier Institute of Management and Entrepreneurship check the table:

ICFAI University Mizoram offers a comprehensive BCom course. This course runs for a duration of three years. Moreover, students enrolled in this course are offered various exit options. Check below to know the exit options:

• After first year/two semesters: Certificate in Commerce
• After second year/four semesters: Diploma in Commerce

${\sum }_{r=0}^{20}?{}^{50-r}{\mathrm{C}}_6={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+{}^{30}{\mathrm{C}}_6$

$\begin{array}{rr}& ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{30}{\mathrm{C}}_6+{}^{30}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{31}{\mathrm{C}}_6+{}^{31}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{50}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\\ & ={}^{51}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat{w}_i=x_i+k,\overline{x_i}=48,SD\left(x_i\right)=12,\\ w_i=55andSD\left(w_i\right)=15\\ then\overline{w_i}=\overline{x_i}+k\left(\overline{w_i}=meanof{w}_i\text{'}sand\overline{x_i}isthemeanof{x}_i\text{'}s\right)\\ \Rightarrow 55=48+k\dots \left(i\right)\\ SDof{w}_i=SDof{x}_i\\ 15=l\times 12\\ \Rightarrow l=\frac{15}{12}=1.25\dots \left(ii\right)\\ fromeqn.\left(i\right)and\left(ii\right)wehave\\ k=\overline{w_i}-\overline{x_i}=55-1.25\times 48=55-60=-5\\ Hence,thecorrectoptionis\left(a\right).\end{array}$