The second period College Algebra class of 33 students could do n problems in x
hours. If 3 students were absent, how long would it take to do 100 problems???
33 students can do n problems in x hours.
Therefore, 1 student would take 33 times as long, So
1 student can do n problems in 33x hours.
Therefore, 1 student can do 1 problem in that time divided by n. So:
1 student can do 1 problem in hours.
Therefore 1 student can do 100 problems in 100 times as long. So:
1 students can do 100 problems in hours. So with 3
students absent:
33-3=30 students could do them in that time divided by 30. So
Solution:
30 students can do 100 problems in hours.
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Another way is to use the worker-time-job formula, which is:
where
W1 = the number of workers in the first situation.
T1 = the number of time units (hours in this case) in the first situation.
J1 = the number of jobs in the first situation.
W2 = the number of workers in the second situation.
T2 = the number of time units (hours in this case) in the second situation.
J2 = the number of jobs in the second situation.
W1 = 33 W2 = 30
T1 = x T2 = the unknown quantity
J1 = n J2 = 100
Cross-multiply
30nT2 = 3300x
Divide both sides by 30n
T2 = =
Answer: 30 students can do 100 problems in hours.
Edwin