Amrita Entrance Examination - Engineering
Question : Divide 50 into two parts so that the sum of their reciprocal is $\frac{1}{12}$.
Option 1: 35 and 15
Option 2: 20 and 30
Option 3: 24 and 36
Option 4: 28 and 22
Correct Answer: 20 and 30
Solution : Given: The number is 50. Let the first part of 50 be $\text{x}$. The second part will be $50 - \text{x}$. The sum of their reciprocal is $\frac{1}{12}$ $\Rightarrow \frac 1 {\text {x} }+ \frac 1 {\text {50-x} }=\frac 1 {\text {12} }$ $\Rightarrow \frac {50+\text {x}-\text {x}} {\text {50x} -{\text{x}^{2}} }=\frac {1}{12}$ $\Rightarrow \text {50x} -{\text{x}^{2}} = 600$ $\Rightarrow \text {50x} -{\text{x}^{2}} - 600=0$ $\Rightarrow {\text{x}^{2}}-\text {50x} + 600=0$ After factorising, we get ⇒ (x-30)(x-20) = 0 ⇒ x = 20 and 30 Hence, the two parts are 20 and 30.
Question : Directions: Five professors A, B, C, D, and E are sitting in a row facing the North (not necessarily in the same order). A is second to the left of D. Only two professors are sitting between C and E. No professor is sitting to the right of E. How many professors are sitting to the left of D?
Option 1: 3
Option 2: 0
Option 3: 2
Option 4: 1
Correct Answer: 2
Solution : 1. No professor is sitting to the right of E. Only two professors are sitting between C and E.
2. A is second to the left of D.
So, the final arrangement will be as follows –
Thus, two persons are left of D. Hence, the third option is correct.
Question : Find the length of a tangent drawn to a circle with a radius 8 cm from a point 17 cm from the centre of the circle.
Option 1: 14 cm
Option 2: 12 cm
Option 3: 15 cm
Option 4: 10 cm
Correct Answer: 15 cm
Solution : Radius of circle, $PB$ = 8 cm Distance of a point from the centre from which the tangent is drawn to the circle, $PA$ = 17 cm Since tangent is perpendicular to the radius, $\angle PBA=90^{\circ}$ By Pythagoras in $\triangle PBA$, Length of the tangent, $AB$ = $\sqrt{17^2-8^2}$ = 15 cm Hence, the correct answer is 15 cm.
Hello,
Candidates who have appeared in Phase I of the AEEE 2024 examination can also appear for Phase II examination by paying an additional fee of INR 600. This is an opportunity given to the candidates to improve their scores and also it is not compulsory to apply again as it is just for improvement of score.
Hope this helps you,
thank you
Yes, you can definitely apply for the AEEE (Amrita Engineering Entrance Exam) in 2026 after passing out of 12th in 2024. The AEEE is conducted for admission to the engineering programs offered by Amrita Vishwa Vidyapeetham. Typically, the eligibility criteria include passing the 12th standard or its equivalent with the required subjects and minimum marks.
Thank you
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As far as I know, there is no specific limit on the number of attempts for the AEEE (Amrita Engineering Entrance Exam). You can appear for the exam as many times as you want to improve your score and secure admission to Amrita Vishwa Vidyapeetham.
https://engineering.careers360.com/exams/aeee/amp
Absolutely, you can definitely attend AEEE while preparing for JEE Main. The syllabus for both exams has some overlapping topics, so preparing for one can help you with the other. It is a great way to maximize your chances of getting into a good engineering college. Just make sure to manage your time effectively and create a study plan that allows you to cover all the necessary topics for both exams.
https://www.google.com/amp/s/engineering.careers360.com/exams/jee-main/amp
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