jamesk486 wrote:In a sample of college students, 40% are third year students and 70% are not second year students. What fraction of those students who are not third-year students are second-year students?
A. 3/4
B. 2/3
C. 4/7
D. 1/2
E. 3/7

If we assume that students study only for 3 years, then the answer
should be D

3rd year: 40%
2nd year: x
1st year: y

y + 40 = 70 implies y = 30
If y = 30, x = 30 as 40 + x + y = 100

So, y / (x+y) = 30 / 60 = 1/2

jamesk486 wrote: Ms. Jiminize plans an automobile trip of 7000 to 9000 miles. The cost will be 87 to 95 cents per gallon and her automobile will average 20 to 30 miles per gallon. What is the maximum possible cost of the gasoline for the trip?
A. 485.00
B. 427.50
C. 382.50
D. 297.50
E. 256.00

Max cost is when Jiminize travels 9000 miles at max cents/gallon and
at min miles/gallon

20 miles -> .95
So, 9000 miles -> .95/20 * 9000 = 427.5$

In a sample of college students, 40% are third year students and 70% are not second year students. What fraction of those students who are not third-year students are second-year students?
A. 3/4
B. 2/3
C. 4/7
D. 1/2
E. 3/7

Let total students be x
4x/10 are III yr students
Thus 6x/10 are not III yr students
7x/10 are not II yr students
and 3x/10 are II yr students (x - 7x/10)
Other students (besides II and III yr students) = x - (4x/10 + 3x/10) = 3x/10

Ratio = (3x/10) / (6x/10)
= 1/2
Hence D

Ms. Jiminize plans an automobile trip of 7000 to 9000 miles. The cost will be 87 to 95 cents per gallon and her automobile will average 20 to 30 miles per gallon. What is the maximum possible cost of the gasoline for the trip?
A. 485.00
B. 427.50
C. 382.50
D. 297.50
E. 256.00

Maximum possible cost of gasoline is when Ms. Jiminize travels the maximum i.e. 9000 miles, pays the maximum price per gallon i.e. 95 cents and gets the least mileage i.e. 20 miles/gallon

9000 miles @ 20 miles/gallon = 9000/20 = 450 gallons consumption
Now 450 gallons @ 95 cents / gallon = 450*(95/100) = $427.5

Hence B[/img]