AIEEE Questions

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Let the length of the stationary train Y be LY
Given that length of train X, LX = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 ---> ( 1 )
(300 + LY) / V = 60 ---> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + LY)/12 = 60
=> 300 + LY = 60 (12) = 720
=> LY = 720 - 300 = 420 m
Length of the stationary train = 420 m

Let the principle be Rs. P
As the amount double itself the interest is Rs. P too
So P = P x r x 15/100
=> r = 100/15 = 20/3 % = 6.66 %.

(45 x 9)/40 = (M x 7)/56 => M = 81

=> 81 – 45 = 36

The smallest unit of data storage in a computer is called Bit (Binary Digit). Half a byte (four bits) is called a Nibble.

1 bit (binary digit) = the value of 0 or 1

8 bits = 1 byte

1024 bytes = 1 kilobyte

1024 kilobytes = 1 megabyte

1024 megabytes = 1 gigabyte

1024 gigabytes = 1 terabyte

1024 terabytes = 1 petabyte

Hence, Computer storage units smallest to largest i.e, in ascending order is

 1 bit  <  1 byte  <  1 KB  <  1 MB  <  1 GB  <  1TB  <  1 PB.

Let work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.

First send 17 and 7. After 7 minutes, 7 is at finish and 17 needs 10 more minutes to complete his travel. now send 13. After 10 minutes 17 reaches finish and 13 needs 3 more minutes to complete. now send 3. both 3 and 13 reach destination at the same time after 3 minutes. hence the time taken is 7+10+3=20 minutes.

 ZA = 88  ; ZB = 52

So formula of compound is AB52 

i.e., A2B5.

Let there are L men

job requires LK man hours.

job completed in first 4 hrs = L x 4 = 4L

job completed in next 4 hrs = 4 x L/2 = 2L

job completed in next 4 hrs = 4 x L/4 = L

job completed in last 4 hrs = 4 x L/8 = L/2

4L + 2L + L + L/2 = KL

K = 7+1/2 = 7.5 hours.