I'm downloading a big image. The size of the image is 25 Mb. The site says me that I've completed 2 5 % of the download and it will finish in 5 minutes.
Knowing that 1 Mb is 1024 Kb, what is the speed of my connection in Kb/s?
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My mystake was the remain of the downloading
I made a calculation mistake.
vry goooooooooood
Total size of image= 25 Mb = ( 2 5 × 1 0 2 4 ) Kb. Now, let us take the downloading speed in Kb/s be x and time remaining= 5 min = ( 5 × 6 0 ) sec.
Remaining size to be downloaded in Kb = ( x × ( 5 × 6 0 ) )
⟹ ( 2 5 × 1 0 2 4 ) − 1 0 0 2 5 × ( 2 5 × 1 0 2 4 ) = 3 0 0 x
⟹ 1 0 2 4 ( 2 5 − 4 2 5 ) = 3 0 0 x
⟹ 1 0 2 4 × 4 7 5 = 3 0 0 x
⟹ x = 1 0 2 4 × 4 7 5 × 3 0 0 1 ⟹ x = 6 4
25/100 ??? from where??
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Since it is said that 25% of the download is complete, so I subtracted 25% of the download to get the remaining data to be downloaded i.e., 1 0 0 2 5 part of the original download
1024*25=25600 5 minutes = 300 seconds so whole 25mb takes 400 sec here we divide 25600 to 400sec = 64
Firstly we convert Mb to Kb, so 25MB is 25600kb. According to problem statement already completed 25% that means 6400Kb download completed out of 25600Kb,so remaining data size is 19200Kb .According to problem statement download completed after 5 minutes and problem requirement is what is the speed in Kb/s?. So we should convert minutes to second so we have 5*60=300second. So download speed is 19200/300=64Kb/s
yes
25Mb= 25 1024 Kb= 25600Kb 25% of 25mb= 25600 25/100Kb=19200Kb but, it will finish in 5min= 5*60s=300s so in 300s 19200Kb will be downloaded so in 1s the speed of his connection is 19200/300Kbps = 64Kbps
size = 25 mb 25mb x 75% divided by 100 = 18.75mb left 18.75mb x 1024kb = 19200 kbs 19200kb divided by(in 5 min) 300 seconds = 64 kb/s
remaining data = 75 % of 25 = 18.75 Mb = 19200 Kb speed = 19200 / (5*60) = 64 Kb/s
Hello....
25% of 25 Mb=25/4 Mb speed(kbps)=25(1024)-(25/4 x 1024) / 60(5)= 19 200/300=64 kbps....thanks
Size of image=25mb (first convert it into kb) =25 1024 = 25600kb completed download =25% Remaining download=total download-complete download =25600-25% =19200 time remaining = 5min (As we need speed in seconds so we convert it into seconds) =5 60 =300 NOW speed in kb/second= speed in kb divided by time in seconds =19200/300 =64kb/sec
25 * 1024*(3/4)=
19200Kb/(5*60s)=
64Kb/s
total picture size=25,so when download give signal that 25% complete and it take further 5 minute so it mean that out of 100% picture is download and 75% remaining and it vl taking 5 minute..... so according to statement in one minute down load cmplt=75/5=15% it mean that the remaining download vl complet in 15% in on minute 25% size hv MB=25 .25=6.25MB 15% size hv MB=25 .15=3.75MB it mean that in per minute 3.75 MB are downloading in remaining minute speed=kb/s=3.75*1024kb/60sec=64
Image Size Is 25 Mb =25 Times 1024= 25600 Kb........25% Downloading Is Done Which Implies 3/4 Is Left=19200 Kb Which Is Estimated To be Done In 5 Minutes = 300 Seconds..........Speed = 19200/300=64 Kb/s
25Mbx1024 =25600 bits
one fourth of 25600 = 25600/4 = 6400 bits are over
remaining 25600 - 6400 = 19200 bits
time remaining = 5 minutes = 300 seconds
speed in KB/s = 19200/300 = 64
He have completed 25% of the download. So he have 75% remain of the download. It equal 18,75 Mb = 19.200 Kb. And it will finish in 5 min = 300 s. So the speed connection = 19.200 / 300 = 64 Kb/s
25% means 6.25 MB have been completed . so (25-6.25) = 18.75 have remain . Then 18.75MB=19200KB and 5min=300sec . so, In 1 sec (19200/300) or 64 KB is downloading ........
25 Mb * 1024 = 25600 kb and 25 % of 25600 is 6400 kb so 25600-6400= 19200 kb is remain to download and 5*60= 300 second time is required. Therefore, 19200/300= 64 kb/s. is speed of downloading.
given the image size to be 25mb(25600kb), given 25%already complete so.....25600-(25%of25600) which comes to 19200 which is to completed in 5 min so now for speed it is going to be 19200kb/5*60sec coming to 64
25600 kb = 100 %
19200 kb = 75 %
6400 kb = 25 %
64 kb = 1 %
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First, convert 25 MB into KB. 1 2 5 M B ⋅ 1 M B 1 0 2 4 K B = 2 5 6 0 0 K B Next, multiply by 75% to find out the remaining number of KB to be downloaded. 2 5 6 0 0 K B ⋅ 4 3 = 1 9 2 0 0 K B Now, divide by 5 minutes to get 5 m i n 1 9 2 0 0 K B = 1 m i n 3 8 4 0 K B ⋅ 6 0 s 1 m i n = 6 4 s K B