Downloading

First, convert 25 MB into KB. $\frac{25 MB}{1}\cdot\frac{1024 KB}{1 MB}=25600 KB$ Next, multiply by 75% to find out the remaining number of KB to be downloaded. $25600 KB\cdot\frac{3}{4}=19200 KB$ Now, divide by 5 minutes to get $\frac{19200 KB}{5 min}=\frac{3840 KB}{1 min}\cdot\frac{1 min}{60 s}=\boxed{64\frac{KB}{s}}$

Total size of image= 25 Mb = ( $25 \times 1024)$ Kb. Now, let us take the downloading speed in Kb/s be x and time remaining= 5 min = $(5 \times 60)$ sec.

Remaining size to be downloaded in Kb = $(x \times (5 \times 60))$

$\implies (25\times1024) - \frac{25}{100} \times (25 \times 1024) = 300x$

$\implies 1024(25 - \frac{25}{4}) = 300x$

$\implies 1024 \times \frac{75}{4} = 300x$

$\implies x=1024 \times \frac{75}{4} \times \frac{1}{300} \implies x=\boxed{64}$

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

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 %