#1
July 1st, 2014, 09:01 AM
 
 
Eligible to apply for Infosys, IBM, HCL after 87.6%, 77.6% and 69.3% in 10th, 12th and BE (IT) upto 3rd year
Am I Eligible for Infosys, IBM, HCL after 87.6%, 77.6% and 69.3% in 10th, 12th and BE (IT) upto 3rd year r??? Yes, you are Eligible for Infosys, 87.6%, 77.6% and 69.3% in 10th, 12th and BE (IT) upto 3rd year. the basic requirements for these companies are as following: Eligibility: The applicant should have 60% marks in 10th, 12th and B.Tech till 6th semester and 65% marks. 60% in xth 60% in xiith and 60% in graduation Clear all backlogs before interview. not more than 1 year gap Selection process: Written test Interview (technical question) Group discussion placement paper Question 1 In a bus stand, there are two services namely A and B. Every 10 minutes buses will leave from A and this service works from 6.10 am to 2 pm. The service at B starts at 2.20 pm and for every 20 minutes buses will leave from the bus stand. Find the probability of getting bus from service B between 2.20 pm to 2.50 pm, if service A is late by 1 hour. a) 1/2 b) 1/3 c) 1/4 d) 1/5 Answer : b) 1/3 Solution : Usually service A starts at 6.10 am. Since the service is late by 1 hour, the first and last bus will leave by 7.10 am and 3 pm respectively. Note that, there is a bus for every 10 minutes. Number of buses leaving between 2.20 pm to 2.50 pm is 4 and the timings are 2.20 pm, 2.30pm, 2.40pm and 2.50pm. There is a bus for every 20 minutes from service B. Number of buses leaving between 2.20 pm to 2.50 pm is 2 and they start at 2.20 pm and 2.40 pm. Therefore, 4 buses from A and 2 buses from B are available. The probability of getting bus from B = buses from B / total number of buses from A and B. = 2/6 = 1/3. Question 2 From a railway station, trains leave for every 15 minutes and 25 minutes to city A and city B respectively. First train to city A and city B start at 9 am and 10.15 am respectively. If a man arrives to the station in between 11.25 am and 12.25 pm then the probability of getting train for city A is: a) 1/4 b) 4/7 c) 3/5 d) 2/5 Answer : b) 4/7. Solution : The man wants to go to city A and he arrives station in between 11.25 am and 12.25 pm. First train to city A is at 9 am and there is a train for every 15 minutes. Trains for city A will leave at the following times : 9 am, 9.15 am, 9.30 am,…,11.30 am, 11.45 am, 12 pm, 12.15pm, and so on. Number of trains for city A between 11.25 am and 12.25 pm is 4. First train to city B is at 10.15 am and there is a train for every 25 minutes. Trains for city B will leave at the following times: 10.15 am, 10.40 am, 11.05 am, 11.30 am, 11.55 am, 12.20 pm, and so on. Number of trains for city B between 11.25 am and 12.25 pm is 3. Probability of getting train for city A between 11.25 am and 12.25 pm = Number trains for city A from 11.25 am to 12.25 pm / Total number of trains for city A and B from 11.25 am to 12.25 pm = 4/7. Question 3 There are two bus stands, namely X and Y. Buses leave from X for every 30 minutes and its first bus starts at 8.05 am. Every hour number of buses leaving from Y increases by 1 and its first bus starts at 7 am. From Y there is only 1 bus for the 1st hour. Any bus from either of the bus stations takes 15 minutes to reach a nearby bus stop. Suppose a person reaches the stop in between 12.15 pm and 1.15 pm. The probability that the person will get a bus from Y is: a) 3/4 b) 1/3 c) 1 d) 1/4 Answer : a) 3/4 Solution : From bus stand X : The first bus will leave by 8.05 am and reach the bus stop in 15 minutes, i.e. at 8.20 am Second bus will leave after 30 minutes i.e. at 8.35 am and will reach the stop at 8.50 am Therefore, buses will reach the stop at the following times: 8.20am, 8.50am, 9.20 am,…,12.20 pm, 12.50 pm, 1.20 pm and so on. Between 12.15 pm and 1.15 pm, two buses will reach the stop at 12.20 pm and 1.20 pm. Therefore, the person will get 2 buses from X. From bus stand Y : The first bus will leave by 7 am and reach the bus stop in 15 minutes, i.e. at 7.15 am. There is only one bus for first 1 hour. i.e., the second bus will leave after 8 am. Note that, the number of buses leaving from Y is increased by 1 per hour. From 8 am to 9 am, two buses will leave from Y and reach the stop between 8.15 am to 9.15 am. And from 9 am to 10 am, 3 buses will leave from Y and reach the stop between 9.15 am to 10.15 am. Proceeding like this, we have, From 12 pm to 1 pm, 6 buses will leave from Y and reach the stop between 12.15 pm to 1.15 pm. Therefore, the person will get 6 buses from Y between 12.15 pm to 1.15 pm. Probability of getting bus from Y between 12.15 pm to 1.15 pm = Number buses from Y in between 12.15 pm to 1.15 pm / Total number of buses from X and Y in between 12.15 pm to 1.15pm = 6/(2+6) = 6/8 = 3/4. Last edited by Neelurk; April 23rd, 2020 at 04:08 PM. 
