charge for the first minute is $0.75. The charge for all subsequent minutes will
be cost/minute × number of minutes. You are told that cost/minute is $0.25.
If l is the length of the entire phone call, then the number of minutes charged
at $0.25 will be l – 1. Thus you can construct the following equation:
0.75 + 0.25(l – 1) = 5.75. Solve for x:
0.75 + 0.25(l – 1) = 5.75
0.75 + 0.25l – 0.25 = 5.75
0.5 + 0.25l = 5.75
0.25l = 5.25
l = 21- B Assign variables: Let a = the price of stock A, and let b = the price of stock
B. The information in the problem provides two relationships:
“The price of stock A is $5 greater than the price of stock B”: expressed
algebraically: a = b + 5
“If the price of stock A increases by $10, the new price will be double the price
of stock B.”: a + 10 = 2b
Now use substitution to solve for b:
a = b + 5
a + 10 = 2b
Substitute b + 5 for a in the second equation: (b + 5) + 10 = 2b. Solve for b:
(b + 5) + 10 = 2b
b + 15 = 2b
15 = b - D Let h = the number of hamburgers Bob purchases, and let s = the number
of shakes Bob purchases. The amount Bob spends on hamburgers is 5h, and
the amount he spends on shakes is 4s. Since the maximum that he can spend is
$56, the sum of these two quantities must be less than or equal to 56: 5h + 4s ≤
56. Since the question asks for the maximum number of hamburgers that Bob
can purchase, you should minimize the number of shakes that he purchases by
letting s = 1:
5 h + 4(1) ≤ 56
5 h ≤ 52
h ≤^525
Since h must be an integer, the maximum value for h will be the greatest
integer less than^525.^525 = 10.4. Thus the greatest value for h will be 10. - A The final height of the tree will be the sum of the original height and the
increase after 8 years. If the height of the tree increases by x feet each year,
then the tree will have increased by 8x feet after 8 years. Thus you can create
the following equation: 10 + 8x = 60. Solve for x:
324 PART 4 ■ MATH REVIEW04-GRE-Test-2018_313-462.indd 324 12/05/17 12:03 pm