Answer :
(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.
