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top 10 engineering college Tamilnadu

Tanya Gupta 5th Jan, 2024

Hello,

Not so sure about the top 10 but these are the top 4 colleges in tamil nadu for engineering in my opinion according to their rating, fees and packages.

Indian Institute of Technology, Madras

Amrita School of Engineering, Coimbatore    .

Anna University, Chennai

National Institute of Technology, Tiruchirappalli

Hope

40 Views

Annual fees for mechanical engineering.

Tanya Gupta 5th Jan, 2024

Hello,

One of the most diverse and versatile engineering fields, mechanical engineering is the study of objects and systems in motion.

The fees for the mechanical engineering course vary from institutions to institutions. The average fees of colleges are in the range of  Rupees 50,000 to Rupees 3,00,000 per annum.

292 Views

easiest engineering entrance exam

Sajal Trivedi 3rd Jan, 2024

Hello aspirant,

There is no such thing as an easy engineering entrance exam because the level of difficulty varies depending on how well you prepare. Therefore, an individual who has prepared well may find the same exam to be easier for them, while someone who has not may find it

12 Views

Question : Directions: Which answer figure will complete the pattern in the question figure?

Option 1:

Option 2:

Option 3:

Option 4:

Team Careers360 21st Jan, 2024

Correct Answer:


Solution : On comparing the question figure and all the answer figures, the following figure will combine and make the complete pattern –

Hence, the first option is correct.

351 Views

Question : Directions: CHMZ is related to WRMZ in a certain way based on the English alphabetical order. In the same way, NPWY is related to LJCA. Which of the given options is DGNO related to following the same logic?

Option 1: ONGD

Option 2: VSLK

Option 3: KECF

Option 4: CFMN

Team Careers360 11th Jan, 2024

Correct Answer: VSLK


Solution : Given:
CHMZ is related to WRMZ and NPWY is related to LJCA –

Find the opposite of each letter of CHMZ and then subtract 1 from the positional value of each letter to obtain the required code –
C→X; H→S; M→N; Z→A
Now, subtract 1

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