Chhattisgarh Pre Engineering Test
With a CRL rank of 92,216 and a General category rank of 9,418 in CAT, getting into a top National Institute of Technology (NIT) or Indian Institute of Information Technology (IIIT) for Computer Science Engineering (CSE) is highly unlikely, as these institutes typically have very competitive cutoffs for CSE, especially for the general category. You may still have a chance at other branches or colleges with slightly lower cutoffs, but it's important to explore other options or consider additional rounds of counseling and eligibility for specific state-level institutions.
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Question : O is the centre of the circle and $\angle AOB = 150°$, and the shaded portion is $x$ part of the circular region, then what will be the value of $x$?
Option 1: $\frac{1}{12}$
Option 2: $\frac{1}{9}$
Option 3: $\frac{1}{6}$
Option 4: $\frac{1}{4}$
Correct Answer: $\frac{1}{6}$
Solution :
Given: O is the centre of the circle and $\angle AOB = 150°$. We know that the vertically opposite angles are identical. ⇒ $\angle COD=150°$ Now, The angles between the shaded region and the two angles $\angle AOB$ and $\angle COD$ are making a complete circle. Let each shaded portion make an angle of $y°$ at the centre. ⇒ $150°+y°+150°+ y°=360°$ ⇒ $2y+300°=360°$ ⇒ $2y=60°$ ⇒ The shaded region covers 60° of the complete circle i.e., $x=\frac{60°}{360°}=\frac{1}{6}$ Hence, the correct answer is $\frac{1}{6}$.
Hi, with this rank it is not possible for your son to get any government engineer colleges like nit,IIIT,etc. With 47 percentile your son might get a private college but a 3rd or 4th tier college at most .
I hope this helps you.
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