Correct Answer: 57,200 per quarter

Solution : Answer = 57,200 per quarter

Interest on drawings=Drawings$\times \frac{Rate}{100} \times \frac{Average time}{12}$.
8580=Drawings$\times \frac{10}{100} \times \frac{4.5}{12}$.
Drawings=2,28,800.
Hence, the correct option is 2.

Correct Answer: Harsh's loan account debited Rs 6000 , credited bank account Rs 5500 and credited realization account Rs 500

Solution : Answer = Harsh's loan account debited Rs 6000, credited bank account Rs 5500 and credited realization account Rs 500

Harsh's loan A/C Dr 6000
To Bank AIC 5500

Correct Answer: Only conclusions I and II follow

Solution : The possible Venn diagram according to the given statements is as follows –

Let's analyse the Conclusions –
Conclusion (I):  Some C is not X – As shown in the diagram, the line having a cross between the circle of

Correct Answer:

Solution : Both lions and foxes are carnivorous. Therefore, both the circles will be inside the circle of carnivorous. But there is no direct relationship between lion and fox. So, both the circles will not overlap with each other.

Based on the above conclusions, the following Venn diagram

Dear student !

Hope you are preparing well for the exam ! In the mht cet exam there are physics , chemistry, mathematics and biology . Class 12th and class 11th topics will be asked in the examination and 20% weightage is given to the Class 11th syllabus and 80%

With a MHT CET score of 150, you may have a chance to get into various colleges in Maharashtra, especially government colleges and some private institutions. However, top-tier colleges like COEP Pune, VJTI Mumbai, and MIT Pune might be challenging unless you belong to a reserved category or meet additional

Correct Answer: –2

Solution : $\frac{3\left(\operatorname{cosec}^2 26^{\circ}-\tan ^2 64^{\circ}\right)+\left(\cot ^2 42^{\circ}-\sec ^2 48^{\circ}\right)}{\cot \left(22^{\circ}-\theta\right)-\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\tan \left(\theta+68^{\circ}\right)+\tan ^2\left(28^{\circ}-\theta\right)}$
$= \frac{3\left(\operatorname{sec}^2 64^{\circ}-\tan ^2 64^{\circ}\right)+\left(\cot ^2 42^{\circ}-\operatorname{cosec} ^2 42^{\circ}\right)}{\tan \left(68^{\circ}+\theta\right)-\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\tan \left(\theta+68^{\circ}\right)+\cot ^2\left(62^{\circ}+\theta\right)}$
$= \frac{3\left(1\right)+\left(\cot ^2 42^{\circ}-1-\cot ^2 42^{\circ}\right)}{-(\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\cot ^2\left(62^{\circ}+\theta\right))}$
$= \frac{3-1}{-1}$
$= -2$
Hence, the correct answer is –2.