(1) Consider the sequence 2, 6, 18, 54, 162, ... If the 48th term is [tex]A[/tex] and the 51st term is [tex]B[/tex], what is the value of [tex]\frac{B}{A}[/tex]?

(2) At Central High School, 50 girls play intramural basketball and 40 girls play intramural volleyball. If 10 girls play both sports, what is the ratio of the number who play only volleyball to those who play only basketball?

(3) Maria is 6 times as old as Tina. In 20 years, Maria will be only three times as old as Tina. How old is Maria now?

(4) If a secretary types 50 words per minute, how many minutes will he take to type 330 words?

(5) If Scott can mow [tex]\frac{2}{5}[/tex] of a lawn each hour, how many lawns can he mow in a given number of hours?

(6) If it is now June, what month will it be 100 months from now?

(7) In a group of 100 students, more students are on a team than are members of a club. If 70 are in clubs and 20 are neither on a team nor in a club, what is the minimum number of students who could be both in a club and on a team?

(1) The sequence given is 2, 6, 18, 54, 162, ... . This is a geometric sequence where each term is multiplied by 3 to get the next term. The general formula for the nth term of a geometric sequence is given by:

[tex]a_n = a_1 imes r^{(n-1)}[/tex]

where [tex]a_1[/tex] is the first term, [tex]r[/tex] is the common ratio, and [tex]n[/tex] is the term number.

For this sequence, [tex]a_1 = 2[/tex] and [tex]r = 3[/tex]. So,

[tex]a_n = 2 \times 3^{(n-1)}[/tex]

To find the 48th term [tex]A[/tex] and the 51st term [tex]B[/tex]:

[tex]A = 2 \times 3^{47}[/tex]
[tex]B = 2 \times 3^{50}[/tex]

The ratio [tex]\frac{B}{A}[/tex] is:

[tex]\frac{B}{A} = \frac{2 \times 3^{50}}{2 \times 3^{47}} = \frac{3^{50}}{3^{47}} = 3^{3} = 27[/tex]

Therefore, the value of [tex]\frac{B}{A}[/tex] is 27.

(2) At Central High School, 50 girls play basketball and 40 play volleyball. 10 play both sports. To find how many play only volleyball, subtract the 10 who play both:

[tex]40 - 10 = 30 \, \text{girls play only volleyball.}[/tex]

To find how many play only basketball, subtract the 10 who play both:

[tex]50 - 10 = 40 \, \text{girls play only basketball.}[/tex]

The ratio of girls who play only volleyball to those who play only basketball is:

[tex]\frac{30}{40} = \frac{3}{4}[/tex]

So the ratio is 3:4.

(3) Let Tina's age now be [tex]x[/tex]. Then Maria is [tex]6x[/tex]. In 20 years, Tina will be [tex]x + 20[/tex] and Maria will be [tex]6x + 20[/tex].

The equation relating their ages in 20 years is:

[tex]6x + 20 = 3(x + 20)[/tex]
[tex]6x + 20 = 3x + 60[/tex]
[tex]6x - 3x = 60 - 20[/tex]
[tex]3x = 40[/tex]
[tex]x = \frac{40}{3} = 13.33[/tex]

Since Maria is [tex]6x[/tex]:

[tex]6x = 6 \times 13.33 = 80[/tex]

Maria is 80 years old now.

(4) If the secretary types 50 words per minute, to type 330 words, calculate the time taken:

[tex]\text{Time} = \frac{330 \, \text{words}}{50 \, \text{words/minute}} = 6.6 \approx 7 \, \text{minutes}[/tex]

It will take approximately 7 minutes.

(5) If Scott can mow [tex]\frac{2}{5}[/tex] of a lawn each hour, then in [tex]h[/tex] hours, he can mow:

[tex]\text{Lawns in } h \, \text{hours} = \left(\frac{2}{5}\right)h = \frac{2h}{5}[/tex]

This shows that in [tex]h[/tex] hours, Scott can mow [tex]\frac{2h}{5}[/tex] lawns.

(6) To find the month 100 months from June, note that a year has 12 months. Divide 100 by 12 to find complete years:

[tex]\frac{100}{12} \approx 8 \, \text{remainder} \, 4[/tex]

This means 8 years pass, and 4 months beyond June is October.

The month 100 months from now is October.

(7) In a group of 100 students:

70 are in clubs,

20 are neither,

Hence, 80 are in clubs or teams, or both.

Find the minimum number in both by subtracting those in clubs and neither from total:

If more students are in teams, then the remaining after clubs is max number possible for team and club.

Assume all 80 in clubs or teams; subtract those in clubs only:

[tex]80 - 70 = 10[/tex]

So, the minimum number in both clubs and teams is 10.