Greetings!

So, CA is the toughest course in India. After your PU you can take addmission in B.Com, BBM or BBA. For the colleges you can take admission in government colleges or m.sc College of Bangalore or government first grade college of davangere.

For CA there will be three levels

1. Foundation level
2. Intermitted level
3. Final

In CA the subjects will be typically include accounting, taxation, auditing, business laws, etc

In foundation level there will be four subject.

In intermediate there will be two groups each group consists of four papers.

In final level there will be two groups each level consists of four papers.

You have to take 40% in each paper. And your percentage should be 50 or else you will be fail and you have to give all the papers again.

Hello Waziha

You can do D.M.L.T Course after completing PUC .

DMLT is Diploma in laboratory technology . It is two years Course including internship .

Eligibility for the Course is to have passed class xii or equivalent  course with  minimum 50% marks from any recognized university.

Course fees ranges from 50,000 to 1,50,000 depending upon jnstitute .

Top colleges offering DMLT Course

KP paramedical Institute

Bangalore medical and research indtitute

DPMI , Delhi paramedical and management Institute

Job opportunities

You can work as lab technician in multispeciality hospitals.

You can open your own laboratory .

Work as lecturer .

https://www.careers360.com/courses/diploma-in-medical-laboratory-technology

Correct Answer: 236 : 325 : 263

Solution : Let's check each option –
First option: 236: 325 : 263; In the second number digits are different from the the first and the second number.
Second option: 189: 198 : 819; In all three numbers same digits are repeated.
Third option: 761: 617 : 176; In all three numbers same digits are repeated.
Fourth option: 416 : 641 : 461; In all three numbers same digits are repeated.

So, the first option is different from the others. Hence, the first option is correct

Hello,

The 13 digit 'Student Number' issued from PUE Department has to be entered by the candidates who appeared for the 12th PUC exam.

To find your 12th/PUC student/Reg number for the KCET application form in 2024, you'll need to check your 12th/PUC registration documents or reach out to your school or college. They should be able to provide you with the specific number you need to enter in the KCET application form.

Hope this helps,

Thank you

Correct Answer: $\frac{4}{5}$

Solution : Given: $a^2+b^2+c^2=16$----(equation 1)
$x^2+y^2+z^2=25$-----(equation 2)
$ax+by+cz=20$-----(equation 3)
Substitute the values of $b=0$ and $c=0$ in the given equation 1, we get,
$a^2+0^2+0^2=16$
$⇒a^2=16$
$\therefore a=4$
Substitute the values of $y=0$ and $z=0$ in the given equation 2, we get,
$x^2+0^2+0^2=25$
$⇒x^2=25$
$\therefore x=5$
Substitute the values in the given equation 3, we get,
$4\times5+0\times0+0\times0=20$
$⇒20=20$
So, the above equation is satisfied.
Substitute the values in the given below equation, we get,
$\therefore\frac{a+b+c}{x+y+z}=\frac{4+0+0}{5+0+0}=\frac{4}{5}$
Hence, the correct answer is $\frac{4}{5}$.

Correct Answer: National Capital Territory of Delhi

Solution : The correct option is the National Capital Territory of Delhi.

As of the 2011 Census of India, the National Capital Territory of Delhi had the highest population density among all states and union territories in India. Delhi is a densely populated urban area, and its population density is significantly higher compared to other regions in the country.