Difference between revisions of "1988 AHSME Problems/Problem 7"
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\textbf{(E)}\ 4\text{ hours}</math> | \textbf{(E)}\ 4\text{ hours}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | + | We want to figure out the number of chunks in <math>60</math> blocks, so we have <math>60\cdot 512 \approx 30000</math>. We divide this by <math>120</math> to determine the number of seconds necessary to transmit. <math>30000/120 \approx 250</math>, which means that it takes approximately <math>4</math> minutes to transmit. Thus, the answer is <math>\boxed{\text{D}}</math>. | |
| − | + | ==Solution 2== | |
| + | This solution is if you are running out of time and just want to write down an answer. So, this is quite unreliable. You can logic it out. It doesn't make sense for the first three options to be the answer since that is way too quick. The last option is way too long. That just leaves <math>\boxed{\text{D}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 20:15, 16 September 2021
Contents
Problem
Estimate the time it takes to send 
blocks of data over a communications channel if each block consists of 
"chunks" and the channel can transmit 
chunks per second.
Solution 1
We want to figure out the number of chunks in blocks, so we have 
. We divide this by to determine the number of seconds necessary to transmit. 
, which means that it takes approximately 
minutes to transmit. Thus, the answer is 
.
Solution 2
This solution is if you are running out of time and just want to write down an answer. So, this is quite unreliable. You can logic it out. It doesn't make sense for the first three options to be the answer since that is way too quick. The last option is way too long. That just leaves .
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
