7) Apti on Time speed distance

 https://www.geeksforgeeks.org/speed-time-distance-formula-and-aptitude-questions/

 Two stations B and M are 465 km distant. A train starts from B towards M at 10 AM with a speed of 65 km/hr. Another train leaves from M towards B at 11 AM at a speed of 35 km/hr. Find the time when both trains meet. 



The train leaving from B leaves an hour early than the train that leaves from M. 
=> Distance covered by train leaving from B = 65 km / hr x 1 hr = 65 km 
Distance left = 465 - 65 = 400 km 
Now, the train from M also gets moving and both are moving towards each other. 
Applying the formula for relative speed, 
Relative speed = 65 + 35 = 100 km / hr 
=> Time required by the trains to meet = 400 km / 100 km / hr = 4 hours 
Thus, the trains meet at 4 hours after 11 AM, i.e., 3 PM. 

Comments

Post a Comment

Popular posts from this blog

5) Apti on mathematical ability ( Time & Work )

 1) A+B+C can do a work in 12 days. B+ C can do work in 18 days. A+ B can do work un 15 days. If b do the work alone, then how many days needed We are given: A + B + C can do the work in 12 days B + C can do the work in 18 days A + B can do the work in 15 days We are to find how many days B alone will take to complete the work. We are given: A + B + C can do the work in 12 days B + C can do the work in 18 days A + B can do the work in 15 days We are to find how many days B alone will take to complete the work. Step 1: Convert to Work Rates Let the total work be LCM of 12, 18, 15 = 180 units (assume total work is 180 units for easier calculation). Now compute each group’s work per day : (A + B + C)’s rate = 180 / 12 = 15 units/day (B + C)’s rate = 180 / 18 = 10 units/day (A + B)’s rate = 180 / 15 = 12 units/day Step 2: Subtract to find individual rates From above: A + B + C = 15 B + C = 10 So, A = 15 − 10 = 5 units/day Also...
Read more

6. Apti on Mathematical ability ( LCM , HCF )

 1.  If the HCF of two numbers is 12 and their LCM is 360, find the numbers. We know that the product of the HCF and LCM of two numbers is equal to the product of the two numbers. So, for two numbers a and b with HCF = 12 and LCM = 360: HCF × LCM = a × b 12 × 360 = a × b 4320 = a × b Now, we need to find two numbers whose product is 4320 and HCF is 12. There can be multiple pairs of numbers that satisfy this condition, and one such pair is: a = 120 and b = 36 Because 120 × 36 = 4320 and the HCF of 120 and 36 is 12. If you want valid pairs, here are a few: 12 and 360 60 and 72 36 and 120 2.  Find the HCF of 36, 48, and 72. Prime factorization of 36: 2 2 × 3 2 Prime factorization of 48: 2 4 × 3 Prime factorization of 72: 2 3 × 3 2 Common factors: 2 2 and 3 (take the minimum power) So, the HCF of 36, 48, and 72 is 2 2 × 3 = 12. 3. What is the largest three-digit number that is exactly divisible by the HCF of 24 and 36? he HCF of 24 and 36 is 12. To find the ...
Read more

9) Total Collect

1) Inspection and walkthrough is done by ? walkthrough is done by Author/Developer Inspection is done by:  Formal Review Team including: Moderator (leads the session) Author (person who wrote the document/code) Reviewers (peers, QA, devs, domain experts) Scribe (documents defects) 2) if you found a a high priority, high seberity defect in production, what you will say, its defect or bug or failure ? answer : failure
Read more