Question

1. the drama club raises money by selling tickets and snacks at a school play. tickets bring in $8 and snacks bring in $2.50 each. the club needs to raise more than $500 to cover expenses. write an inequality that represents the number of tickets t and snacks s sold.
2. the band is hosting a concert fundraiser. they must raise at least $2,000 to pay for the event.

• youth tickets cost $10 each.
• adult tickets cost $16 each.

part a: write an inequality that represents the number of youth tickets, y, and adult tickets, a, that must be sold to meet the fundraising goal.
part b: which of the following ticket sales are viable solutions to the inequality? select all that apply.
a. 200 youth tickets and 0 adult tickets
b. 100 youth tickets and 75 adult tickets
c. 75 youth tickets and 100 adult tickets
d. 50 youth tickets and 90 adult tickets
e. 25 youth tickets and 150 adult tickets
f. 0 youth tickets and 125 adult tickets

Explanation:

Part A

Step1: Determine revenue from each ticket

Youth tickets: $10 each, so revenue from \( y \) youth tickets is \( 10y \).
Adult tickets: $16 each, so revenue from \( a \) adult tickets is \( 16a \).

Step2: Set up the inequality

Total revenue must be at least $2000, so \( 10y + 16a \geq 2000 \).

For each option, substitute \( y \) and \( a \) into \( 10y + 16a \) and check if it's \( \geq 2000 \).

Step1: Option A

\( y = 200 \), \( a = 0 \): \( 10(200) + 16(0) = 2000 \). \( 2000 \geq 2000 \): valid.

Step2: Option B

\( y = 100 \), \( a = 75 \): \( 10(100) + 16(75) = 1000 + 1200 = 2200 \). \( 2200 \geq 2000 \): valid. Wait, no—wait, 1675=1200, 10100=1000, total 2200? Wait, no, the goal is 2000. Wait, no, 10100 +1675=1000+1200=2200 ≥2000: valid? Wait, but let's recheck. Wait, maybe I miscalculated. Wait, 1675: 1670=1120, 165=80, total 1200. 10100=1000. 1000+1200=2200 ≥2000: yes. But wait, let's check other options.

Wait, no, let's do all:

Step3: Option C

\( y = 75 \), \( a = 100 \): \( 10(75) + 16(100) = 750 + 1600 = 2350 \). \( 2350 \geq 2000 \): valid.

Step4: Option D

\( y = 50 \), \( a = 90 \): \( 10(50) + 16(90) = 500 + 1440 = 1940 \). \( 1940 < 2000 \): invalid.

Step5: Option E

\( y = 25 \), \( a = 150 \): \( 10(25) + 16(150) = 250 + 2400 = 2650 \). \( 2650 \geq 2000 \): valid.

Step6: Option F

\( y = 0 \), \( a = 125 \): \( 10(0) + 16(125) = 2000 \). \( 2000 \geq 2000 \): valid.

Wait, but wait—wait, Option B: 10100 +1675=1000+1200=2200 ≥2000: valid. Option A: 2000, valid. Option C:2350, valid. Option E:2650, valid. Option F:2000, valid. Wait, but let's recheck Option B: 100 youth (1000) +75 adult (1200) =2200, which is ≥2000. So valid. Wait, but maybe I made a mistake. Wait, the problem says "at least $2000". So:

A: 2000 ≥2000: yes.
B: 2200 ≥2000: yes.
C:2350 ≥2000: yes.
D:1940 <2000: no.
E:2650 ≥2000: yes.
F:2000 ≥2000: yes.

Wait, but let's check the options again. Wait, the original problem's Part B options:

A. 200 youth, 0 adult: 10*200=2000: yes.
B. 100 youth, 75 adult: 10100 +1675=1000+1200=2200: yes.
C.75 youth, 100 adult:1075 +16100=750+1600=2350: yes.
D.50 youth,90 adult:1050 +1690=500+1440=1940: no.
E.25 youth,150 adult:1025 +16150=250+2400=2650: yes.
F.0 youth,125 adult:16*125=2000: yes.

So valid options are A, B, C, E, F? Wait, but maybe I miscalculated B. Wait, 1675: 7510=750, 756=450, so 750+450=1200. 10100=1000. 1000+1200=2200. Correct. So B is valid.

Wait, but let's check the problem again. Maybe the options have a typo, but based on calculation:

A: valid.
B: valid.
C: valid.
E: valid.
F: valid.

Wait, but let's confirm:

A: 200*10=2000 ≥2000: yes.
B:10010 +7516=1000+1200=2200 ≥2000: yes.
C:7510 +10016=750+1600=2350 ≥2000: yes.
D:5010 +9016=500+1440=1940 <2000: no.
E:2510 +15016=250+2400=2650 ≥2000: yes.
F:010 +12516=2000 ≥2000: yes.

So the valid options are A, B, C, E, F.

Answer:

\( 10y + 16a \geq 2000 \)

Part B