2007-2008 Grade 8 Mathematics Item and Scoring Sampler
2007-2008 Grade 8 Mathematics Item and Scoring Sampler
2007-2008 Grade 8 Mathematics Item and Scoring Sampler
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MATHEMATICS<br />
<strong>Item</strong> #27<br />
Top <strong>Scoring</strong> Response:<br />
Part A Answer<br />
Train L : 3 (hours)<br />
Train M: 3.75 (hours) or<br />
3 3 } (hours) or<br />
4<br />
3 hours <strong>and</strong> 45 minutes<br />
Train L :<br />
Support<br />
d = r • t<br />
225 = 75 • t<br />
225<br />
}<br />
75 = 75t }<br />
75<br />
3 = t<br />
Train M :<br />
225 = 60 • t<br />
225<br />
}<br />
60 = 60t }<br />
60<br />
3.75 = tmi<br />
AND<br />
(2 score points)<br />
0.5 point for 2 correct times<br />
1.5 points for complete support<br />
Sample Explanation:<br />
I used the distance formula (distance equals rate times time) for each<br />
train. I put in 225 miles for the distance <strong>and</strong> then put in each train’s<br />
speed. I then solved each equation for t to determine each train’s time<br />
to travel 225 miles.<br />
OR equivalent<br />
Part B Answer<br />
810 (miles)<br />
Train L :<br />
Train M :<br />
Support<br />
d = r • t<br />
d = 75 • 6<br />
d = 450<br />
d = 60 • 6<br />
d = 360<br />
450 + 360 = 810<br />
AND<br />
(2 score points)<br />
0.5 point for correct answer<br />
1.5 points for complete support<br />
Sample Explanation:<br />
I put the trains’ time (6) <strong>and</strong> each train’s rate into the distance formula. By<br />
multiplying rate times time, I found the distance each train traveled in 6 hours.<br />
Then, since the trains are going in opposite directions, I added the distances together<br />
to determine the total distance between them.<br />
OR equivalent<br />
<strong>Grade</strong> 8 <strong>Mathematics</strong> <strong>Item</strong> <strong>Sampler</strong> <strong>2007</strong>–<strong>2008</strong> 39