Ask & Answer: India's Largest Education Community

There is no specific eligibility criteria for the scribe to be used in CAT exam. However, they need to be less qualified than the candidate in terms of academic degree. 

The number of students placed over the past three years during PSIT College of Higher Education placements is presented below:

No, you are not eligible to apply for CAT exam until you complete your CA course and clear ICAI CA exams. In case of absence of graduation degree, candidates are required to complete the professional degree before applying. The minimum score for applying for CAT on the basis of professional degree course in 50 per cent ( 45 per cent for SC/ST/PwD candidates).

Given, $2\left[\begin{array}{l}13\\ 0x\end{array}\right]+$ $\left[\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$ $=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\hfill 2\times 1\hfill & \hfill 2\times 3\hfill \\ \hfill 2\times 0\hfill & \hfill 2\times x\hfill \end{array}\right]$ $+\left[\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\hfill 2+y\hfill & \hfill 6+0\hfill \\ \hfill \hfill & \hfill 2x+2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

Equating the corresponding elements of the matrices we get,

2 + y = 5

y = 5 - 2 = 3

And 2x + 2 = 8

2x = 8 - 2

$\Rightarrow x=\frac{6}{2}$ $$

x = 3

∅x = 3, y - 3.

The list of leading recruiters that visited the campus during PSIT College of Higher Education placements 2025 is presented below:

Both TOEFL and IELTS are English language tests used for studying or working in English-speaking countries. Neither test is always harder than the other. It depends on your skills and what you are comfortable with. Some find TOEFL harder, if they are not used to computer tests. Others finds IELTS challenging, due to the face-to-face with an examiner. 

Yes, you are eligible to apply for CAT provided you have 50 per cent in first and second year of bachelor's degree programme. The aggregate marks are to be filled in the CAT 2025 application form and only those forms will be accepted that meet the minimum academic requirement fixed by the examination authorities.

Give, 2x + 4 = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]$

$\Rightarrow 2x=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]-$ y $=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]-\left[\begin{array}{ll}3\hfill & 2\hfill \\ 1\hfill & 4\hfill \end{array}\right]$ $=\left[\begin{array}{l}1-30-2\\ -3-12-4\end{array}\right]$

$\Rightarrow x=\frac{1}{2}\left[\begin{array}{ll}-2\hfill & -2\hfill \\ -4\hfill & -2\hfill \end{array}\right]=\left[\begin{array}{ll}-2\times \frac{1}{2}\hfill & -2\times \frac{1}{2}\hfill \\ -4\times \frac{1}{2}\hfill & -2\times \frac{1}{2}\hfill \end{array}\right]=$ $\left[\begin{array}{l}-1-1\\ -2-1\end{array}\right]$

There is no upper age limit to appear in CAT exam. Any individual can apply for CAT at any phase of their career. Candidates need to have only bachelor's degree with minimum 50 per cent aggregate (45 per cent for SC/ST/PwD) candidates. It means the CAT aspirant must be of at least 20 years of age at the time of appearing for the exam.

Candidates have to pay the registration fee amount in order to submit their applications on time. Motherhood University BA registration fee is INR 2,000. This fee amount is as per the official website. It is however, still subject to changes and hence, is indicative. 

PSIT College of Higher Education the highest package data is not released. However, as per the NIRF report 2025, the median package received during the past three placement drives is presented below:

(i) x + y = $\left[\begin{array}{ll}7\hfill & 0\hfill \\ 2\hfill & 5\hfill \end{array}\right]$

x - y = $\left[\begin{array}{l}30\\ 03\end{array}\right]$

$\Rightarrow Y=\frac{1}{2}\left[\begin{array}{ll}4\hfill & 0\hfill \\ 2\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{ll}\frac{1}{2}\times 4\hfill & \frac{1}{2}\times 0\hfill \\ \frac{1}{2}\times 2\hfill & \frac{1}{2}\times 2\hfill \end{array}\right]$ $=\left[\begin{array}{l}20\\ 11\end{array}\right]$

(ii) 2x + 3y = $\left[\begin{array}{ll}2\hfill & 3\hfill \\ 4\hfill & 0\hfill \end{array}\right]$ ___ (1)

2x + 2y = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill -1\hfill & \hfill 5\hfill \end{array}\right]$ _____ (2)

Multiplying  eqⁿ by 2 and $q^n(u)$

2 × (2x+ 3y) = 2 × $\left[\begin{array}{l}23\\ 40\end{array}\right]$

$\Rightarrow$ 4x + 6y = $\left[\begin{array}{ll}2\times 2\hfill & 2\times 3\hfill \\ 2\times 4\hfill & 2\times 0\hfill \end{array}\right]$

$\Rightarrow$ 4x + 6y = $\left[\begin{array}{ll}4\hfill & 6\hfill \\ 8\hfill & 0\hfill \end{array}\right]$ $\to (iii).$

3 × (3x+ 2y) = 3 × $\left[\begin{array}{l}2-2\\ -15\end{array}\right]$

$\Rightarrow$ 9x + 6y = $\left[\begin{array}{cc}\hfill 3\times (2)\hfill & \hfill 3\times (-2)\hfill \\ \hfill 3\times (-1)\hfill & \hfill 3\times 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 6\hfill & \hfill -6\hfill \\ \hfill -3\hfill & \hfill 15\hfill \end{array}\right]$ $\to (iv)$

Subtracting  eqⁿ  (iii) from (iv) we get,

4x+ 6y - (4x+ 6y) = $\left[\begin{array}{l}6-6\\ -315\end{array}\right]-$ $\left[\begin{array}{l}46\\ 80\end{array}\right]$

$\Rightarrow$ 5x = $\left[\begin{array}{l}6-4-6-6\\ -3-815-0\end{array}\right]=$ $\left[\begin{array}{l}2-12\\ -1115\end{array}\right]$

$\Rightarrow x=\frac{1}{5}\left[\begin{array}{rr}\hfill 2& \hfill -12\\ \hfill -11& \hfill 15\end{array}\right]=\left[\begin{array}{rr}\hfill 2\times \frac{1}{5}& \hfill -12\times \frac{1}{5}\\ \hfill -11\times \frac{1}{5}& \hfill 15\times \frac{1}{5}\end{array}\right]=$ $\left[\begin{array}{l}\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\raisebox{1ex}{$-12$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ \raisebox{1ex}{$-11$}\!\left/ \!\raisebox{-1ex}{$53$}\right.\end{array}\right]$

From eqⁿ  (u);

$3y=\left[\begin{array}{ll}2\hfill & 3\hfill \\ 4\hfill & 0\hfill \end{array}\right]-2x=\left[\begin{array}{ll}2\hfill & 3\hfill \\ 4\hfill & 0\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 2\times \frac{2}{5}\hfill & \hfill 2\times \left(-\frac{12}{5}\right)\hfill \\ \hfill 2\times (-11/5)\hfill & \hfill 2\times 3\hfill \end{array}\right]$

$=\left[\begin{array}{l}23\\ 40\end{array}\right]-$ $\left[\begin{array}{l}\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\raisebox{1ex}{$-24$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ \raisebox{1ex}{$-22$}\!\left/ \!\raisebox{-1ex}{$56$}\right.\end{array}\right]$

$=\left[\begin{array}{l}2\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.3-\left(\raisebox{1ex}{$-24$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)\\ 4-\left(\raisebox{1ex}{$-22$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)0-6\end{array}\right]$

$=\left[\begin{array}{cc}\hfill \frac{5\times 2-4}{5}\hfill & \hfill \frac{3\times 5+24}{5}\hfill \\ \hfill \frac{4\times 5+22}{5}\hfill & \hfill -6\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill \frac{10-4}{5}\hfill & \hfill \frac{15+24}{5}\hfill \\ \hfill \frac{20+22}{5}\hfill & \hfill -6\hfill \end{array}\right]$

y $=\frac{1}{3}\left[\begin{array}{cc}\hfill \raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$3q$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$42$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill -6\hfill \end{array}\right]=$ $\left[\begin{array}{l}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \frac{6}{5}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \frac{39}{5}\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \frac{42}{5}\frac{1}{3}\times \left(-6\right)\end{array}\right]$

$=\left[\begin{array}{l}\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ \raisebox{1ex}{$14$}\!\left/ \!\raisebox{-1ex}{$5-2$}\right.\end{array}\right]$

This is a Multiple Choice Questions as classified in NCERT Exemplar

54. Options (i) and (iii) are correct since nucleophiles are negatively charged or electron rich (as contains lone pair of electrons) species. 

The placement trend witnessed during PSIT College of Higher Education placements over the past three years is presented below:

$cos\theta \left[\begin{array}{cc}\hfill cos\theta \hfill & \hfill sin\theta \hfill \\ \hfill -sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right]+sin\theta \left[\begin{array}{cc}\hfill sin\theta \hfill & \hfill -cos\theta \hfill \\ \hfill cos\theta \hfill & \hfill sin\theta \hfill \end{array}\right]$

(E) $=\left[\begin{array}{l}co{r}^2\theta cot\theta sin\theta \\ -cos\theta sin\theta co{s}^2\theta \end{array}\right]$ $+\left[\begin{array}{l}sin2\theta -sin\theta cos\theta .\\ sin\theta cos\theta si{n}^2\theta \end{array}\right]$ $$

$=\left[\begin{array}{l}co{s}^2\theta +si{n}^2\theta cos\theta sin\theta -sin\theta cos\theta \\ -cos\theta sin\theta +sin\theta cos\theta co{s}^2\theta +si{n}^2\theta \end{array}\right]$

$=\left[\begin{array}{l}10\\ 01\end{array}\right]$

The key highlights of PSIT College of Higher Education placements 2024 are presented below:

3a - 5b = 3 × $\left[\begin{array}{ccc}\hfill \frac{2}{3}\hfill & \hfill 1\hfill & \hfill \frac{5}{3}\hfill \\ \hfill \frac{1}{3}\hfill & \hfill \frac{2}{3}\hfill & \hfill \frac{4}{3}\hfill \\ \hfill \frac{7}{3}\hfill & \hfill 2\hfill & \hfill \frac{2}{3}\hfill \end{array}\right]$ - $5\times \left[\begin{array}{ccc}\hfill \frac{2}{5}\hfill & \hfill \frac{3}{5}\hfill & \hfill 1\hfill \\ \hfill \frac{1}{5}\hfill & \hfill \frac{2}{5}\hfill & \hfill \frac{4}{5}\hfill \\ \hfill \frac{7}{5}\hfill & \hfill \frac{6}{5}\hfill & \hfill \frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 3\times \frac{2}{3}\hfill & \hfill 3\times 1\hfill & \hfill 3\times \frac{5}{3}\hfill \\ \hfill 3\times \frac{1}{3}\hfill & \hfill 3\times \frac{2}{3}\hfill & \hfill 3\times \frac{4}{3}\hfill \\ \hfill 3\times \frac{7}{3}\hfill & \hfill 3\times 2\hfill & \hfill 3\times \frac{2}{3}\hfill \end{array}\right]-$ $\left[\begin{array}{ccc}\hfill 5\times \frac{2}{5}\hfill & \hfill 5\times \frac{3}{5}\hfill & \hfill 5\times 1\hfill \\ \hfill 5\times \frac{1}{5}\hfill & \hfill 5\times \frac{2}{5}\hfill & \hfill 5\times \frac{4}{5}\hfill \\ \hfill 5\times \frac{7}{5}\hfill & \hfill 5\times \frac{6}{5}\hfill & \hfill 5\times \frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2-2\hfill & \hfill 3-3\hfill & \hfill 5-5\hfill \\ \hfill 1-1\hfill & \hfill 2-2\hfill & \hfill 4-4\hfill \\ \hfill 7-7\hfill & \hfill 6-6\hfill & \hfill 2-2\hfill \end{array}\right]$ = $\left[\begin{array}{l}000\\ 600\\ 000\end{array}\right]$

This is a long answer type question as classified in NCERT Exemplar

Alkali metals are ionic in nature due to their larger size. They have +1 oxidation states. Alkali metals forms three types of oxides such as peroxides, superoxides and normal oxides. The basic character of normal oxides increases from lithium oxide to caesium oxide.

The halides of alkali metals are also ionic except lithium halide. Lithium halide is covalent in nature because of small size and high polarizing power.

Oxosalts of alkali metal are solid water-soluble ionic compounds. Oxosalts of lithium show different properties due to small size of lithium.

A + B = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1+3\hfill & \hfill 2-1\hfill & \hfill -3+2\hfill \\ \hfill 5+4\hfill & \hfill 0+2\hfill & \hfill 2+5\hfill \\ \hfill 1+2\hfill & \hfill -1+0\hfill & \hfill 1+3\hfill \end{array}\right]$

$A+B=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 9\hfill & \hfill 2\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \end{array}\right]$