Combine like terms: 56 t = 40
↓
Multiply both sides by^65 to solve for t: t = 40 × (^65 ) = 48- B To answer this question, you need to determine the net increase in the
population during the year. If 500 organisms were born each day and 300
organisms died each day, then the population increased at a constant rate of
200 organisms/day. For the entire year, the population increase will be 200
organisms/day × 365 days = 73,000. The approximate total population will
thus be 50,000 + 73,000 = 123,000. - 101 Use the r × t = d table. Let j = Jack’s rate and j + 2 = Bob’s rate:
rate (mi/hr) × time (hr) = distance (miles)
↓ ↓ ↓
Jack: j × t = jt
Bob: (j + 2) × t = (j + 2)(t)
If they end up^15 of a mile apart, then Bob’s distance must be^15 mile more than
Jack’s. Expressed algebraically, you arrive at (j + 2)(t) =^15 + jt. Solve for t:
(j + 2)(t) =^15 + jt
jt + 2t =^15 + jt
2 t =^15
t = 101 - 12 Use the r × t = w formula, where x is the rate of each scientist:
rate (beakers/min) × time (min) = work (beakers)
↓ ↓ ↓
2 x × 10 = 8
Thus
(2x)10 = 8
2 x = 108
x = 208
x =^25
Now that you know the rate of each scientist, you need to determine how
many scientists are needed to fill 24 beakers in 5 minutes. Let y = the number
of scientists. Substitute the given information into the r × t = w table:
rate (beakers/min) × time (min) = work (beakers)
↓ ↓ ↓
(^25 )y × 5 = 24
CHAPTER 12 ■ FROM WORDS TO ALGEBRA 35304-GRE-Test-2018_313-462.indd 353 12/05/17 12:04 pm