Word Problems on 2 × 2 Linear Systems

Calculators: Calculators

Prerequisites:
(1.) Numbers and Notations
(2.) Fractions, Decimals, and Percents
(3.) Expressions and Equations

Notable Notes About Word Problems
(1.) Word problems are written in standard British English.
(2.) Some word problems are very lengthy and some are unnecessary. Those ones are meant to discourage you from even trying.
(3.) Word problems in Mathematics demonstrate the real-world applications of mathematical concepts.
(4.) Embrace word problems. See it as writing from "English to Math".
Take time to:
(a.) Read to understand. Paraphrase and shorten long sentences as necessary.
(b.) Re-read and note/underline the vocabulary Math terms written in English.
(c.) Translate/Write important sentences one at a time.
(d.) Review what you wrote to ensure correctness.
(e.) Solve the math, and check your solution in the word problem.
Does that solution makes sense?
If it does, you may be correct.
If it does not, please re-do it. For example, if you were asked to calculate the length of something and you get a negative number, then you will need to re-do it.

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students: Unless specified otherwise:
Any question labeled WASSCE-FM is a question from the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Use appropriate variables.
Indicate the method(s) used.
Show all work.
Interpret your solutions.

(1.) NYSED An ice cream shop sells small and large sundaes.
One day, 30 small sundaes and 50 large sundaes were sold for $420.
Another day, 15 small sundaes and 35 large sundaes were sold for $270.
Sales tax is included in all prices.

If x is the cost of a small sundae and y is the cost of a large sundae, write a system of equations to represent this situation.

Peyton thinks that small sundaes cost $2.75 and large sundaes cost $6.75.
Is Peyton correct?

Using your equations, determine algebraically the cost of one small sundae and the cost of one large sundae.

$ (a.) \\[3ex] 30x + 50y = 420 ...eqn.(1) \\[3ex] 15x + 35y = 270 ...eqn.(2) \\[3ex] (b.) \\[3ex] \underline{Based\;\;on\;\;Peyton's\;\;Thoughts} \\[3ex] One\;\;day: \\[3ex] 30\;\;small\;\;sundaes\;\;@\;\;\$2.75\;\;each = 30(2.75) = 82.5 \\[3ex] 50\;\;large\;\;sundaes\;\;@\;\;\$6.75\;\;each = 50(6.75) = 337.5 \\[3ex] Total\;\;cost = 82.5 + 337.5 = \$420.00 \\[3ex] Peyton\;\;is\;\;correct\;\;for\;\;this\;\;day \\[3ex] Another\;\;day: \\[3ex] 15\;\;small\;\;sundaes\;\;@\;\;\$2.75\;\;each = 15(2.75) = 41.25 \\[3ex] 35\;\;large\;\;sundaes\;\;@\;\;\$6.75\;\;each = 35(6.75) = 236.25 \\[3ex] Total\;\;cost = 41.25 + 236.25 = \$277.50 \\[3ex] Peyton\;\;is\;\;not\;\;correct\;\;for\;\;this\;\;day \\[3ex] (c.) \\[3ex] \underline{Simplify\;\;before\;\;solving} \\[3ex] 30x + 50y = 420 ...eqn.(1) \\[3ex] 3x + 5y = 42 ...simplified\;\;eqn.(1) \\[3ex] 15x + 35y = 270 ...eqn.(2) \\[3ex] 3x + 7y = 54 ...simplified\;\;eqn.(2) \\[3ex] Our\;\;new\;\;equations\;\;are: \\[3ex] 3x + 5y = 42 ...eqn.(1) \\[3ex] 3x + 7y = 54 ...eqn.(2) \\[3ex] \underline{Elimination-by-Subtraction\;\;Method} \\[3ex] To\;\;find\;\;y;\;\;eliminate\;\;x \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] (3x + 7y) - (3x + 5y) = 54 - 42 \\[3ex] 3x + 7y - 3x - 5y = 12 \\[3ex] 2y = 12 \\[3ex] y = \dfrac{12}{2} \\[5ex] y = 6 \\[3ex] \underline{Elimination\;\;by\;\;Addition\;\;Method} \\[3ex] To\;\;find\;\;x;\;\;eliminate\;\;y \\[3ex] -7 * eqn.(1) + 5 * eqn.(2) \implies \\[3ex] -7(3x + 5y) + 5(3x + 7y) = -7(42) + 5(54) \\[3ex] -21x - 35y + 15x + 35y = -294 + 270 \\[3ex] -6x = -24 \\[3ex] x = \dfrac{-24}{-6} \\[5ex] x = 4 \\[3ex] $ The cost of one small sundae is $4.00 and the cost of one large sundae is $6.00

(2.) ACT Tanisha, a manager at a state park, counted the money in the cash register at the end of her shift, and then she deposited the money in the bank.
When she went back to her office, she accidently shredded the deposit slip.
She remembered that there were only $5 and $10 bills.
She also recalled that there were 27 bills totaling $205.
How many $5 bills were in Tanisha's cash register at the end of her shift?

$ F.\;\; 13 \\[3ex] G.\;\; 14 \\[3ex] H.\;\; 16 \\[3ex] J.\;\; 23 \\[3ex] K.\;\; 32 \\[3ex] $

Let:
the number of $5 bills = x
the number of $10 bills = y

She also recalled that there were 27 bills

$ x + y = 27 ...eqn.(1) \\[3ex] $ totaling $205

$ 5x + 10y = 207 ...eqn.(2) \\[3ex] \underline{Elimination\;\;Method} \\[3ex] 10 * eqn.(1) - eqn.(2) \implies \\[3ex] 10(x + y) - (5x + 10y) = 10(27) - 205 \\[3ex] 10x + 10y - 5x - 10y = 270 - 205 \\[3ex] 5x = 65 \\[3ex] x = \dfrac{65}{5} \\[5ex] x = 13 \\[3ex] $ There were 13 $5 bills in Tanisha's cash register at the end of her shift.

(4.) Ming and Carlos are selling cookie dough for a school fundraiser.
Customers can buy packages of chocolate chip cookie dough and packages of gingerbread cookie dough.
Ming sold 8 packages of chocolate chip cookie dough and 12 packages of gingerbread cookie dough for a total of $364.
Carlos sold 1 package of chocolate chip cookie dough and 4 packages of gingerbread cookie dough for a total of $93.
Find the cost each of one package of chocolate chip cookie dough and one package of gingerbread cookie dough.

Let:
the cost of one package of chocolate chip cookie dough = c
the cost of one package of gingerbread cookie dough = g

$ Ming:\;\; 8c + 12g = 364 ...eqn.(1) \\[3ex] Carlos:\;\; c + 4g = 93 ...eqn.(2) \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(2) \\[3ex] c = 93 - 4g ...eqn.(3) \\[3ex] Substitute\;\;(93 - 4g)\;\;for\;\;c\;\;in\;\;eqn.(1) \\[3ex] 8(93 - 4g) + 12g = 364 \\[3ex] 744 - 32g + 12g = 364 \\[3ex] 744 - 364 = 32g - 12g \\[3ex] 380 = 20g \\[3ex] 20g = 380 \\[3ex] g = \dfrac{380}{20} \\[5ex] g = 19 \\[3ex] Substitute\;\;19\;\;for\;\;g\;\;in\;\;eqn.(3) \\[3ex] c = 93 - 4(19) \\[3ex] c = 93 - 76 \\[3ex] c = 17 \\[3ex] $ The cost each of one package of chocolate chip cookie dough is $17.00
The cost of one package of gingerbread cookie dough is $19.00

(5.) The local amusement park is a popular field trip destination.
This year the senior class at High School A and the senior class at High School B both planned trips there.
The senior class at High School A rented and filled 16 vans and 8 buses with 752 students.
High School B rented and filled 5 vans and 5 buses with 380 students.
How many students can a van carry?
How many students can a bus carry?

Let:
the number of students that a van can carry = v
the number of students that a bus can carry = b

$ \underline{High\;\;School\;A} \\[3ex] 16v + 8b = 752 ...eqn.(1) \\[3ex] \underline{High\;\;School\;B} \\[3ex] 5v + 5b = 380 ...eqn.(2) \\[3ex] \underline{Elimination-by-Addition\;\;Method} \\[3ex] To\;\;find\;\;v,\;\;eliminate\;\;b \\[3ex] -5 * eqn.(1) + 8 * eqn.(2) \implies \\[3ex] -5(16v + 8b) + 8(5v + 5b) = -5(752) + 8(380) \\[3ex] -80v - 40b + 40v + 40b = -3760 + 3040 \\[3ex] -40v = -720 \\[3ex] v = \dfrac{-720}{-40} \\[5ex] v = 18 \\[3ex] To\;\;find\;\;b,\;\;eliminate\;\;v \\[3ex] -5 * eqn.(1) + 16 * eqn.(2) \implies \\[3ex] -5(16v + 8b) + 16(5v + 5b) = -5(752) + 16(380) \\[3ex] -80v - 40b + 80v + 80b = -3760 + 6080 \\[3ex] 40b = 2320 \\[3ex] b = \dfrac{2320}{40} \\[5ex] b = 58 \\[3ex] $ A van can carry 18 students.
A bus can carry 58 students.

(6.) ACT At Gussie's Pizzeria, Odetta spends $8.75 (before tax) on the purchase of 2 appetizers and 3 slices of pizza.
The price of each slice of pizza is p dollars.
The price of each appetizer is twice the price of a slice of pizza.
Which of the following systems of equations, when solved, gives the price, a dollars, of an appetizer and the price, p dollars, of a slice of pizza at Gussie's?

$ F.\;\; 2a + 3p = 8.75 \\[1ex] \hspace{4.5em}a = 2p \\[3ex] G.\;\; 2a + p = 8.75 \\[1ex] \hspace{4em}3a = 2p \\[3ex] H.\;\; 2p + a = 8.75 \\[1ex] \hspace{4em}3p = 2a \\[3ex] J.\;\; 3a + 2p = 8.75 \\[1ex] \hspace{4.5em}p = 2a \\[3ex] K.\;\; 3p + a = 8.75 \\[1ex] \hspace{4.5em}a = 2p \\[3ex] $

Odetta spends $8.75 (before tax) on the purchase of 2 appetizers and 3 slices of pizza.
2a + 3p = 8.75

The price of each appetizer is twice the price of a slice of pizza.
a = 2p

Correct option is F.

(8.) The school that Lisa goes to, is selling tickets to the annual talent show.
On the first day of ticket sales the school sold 4 senior citizen tickets and 5 student tickets for a total of $102.
The school took in $126 on the second day by selling 7 senior citizen tickets and 5 student tickets.
What is the price of one senior citizen ticket and one student ticket?

Let:
the price of one senior citizen ticket = c
the price of one student ticket = d

$ \underline{First\;\;Day} \\[3ex] 4c + 5d = 102...eqn.(1) \\[3ex] \underline{Second\;\;Day} \\[3ex] 7c + 5d = 126...eqn.(2) \\[3ex] \underline{Elimination-by-Subtraction\;\;Method} \\[3ex] To\;\;find\;\;d,\;\;eliminate\;\;c \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] (7c + 5d) - (4c + 5d) = 126 - 102 \\[3ex] 7c + 5d - 4c - 5d = 24 \\[3ex] 3c = 24 \\[3ex] c = \dfrac{24}{3} \\[5ex] c = 8 \\[3ex] To\;\;find\;\;c,\;\;eliminate\;\;d \\[3ex] 7 * eqn.(1) - 4 * eqn.(2) \implies \\[3ex] 7(4c + 5d) - 4(7c + 5d) = 7(102) - 4(126) \\[3ex] 28c + 35d - 28c - 20d = 714 - 504 \\[3ex] 15d = 210 \\[3ex] d = \dfrac{210}{15} \\[5ex] d = 14 \\[3ex] $ The price of one senior citizen ticket is $8.00
The price of one student ticket is $14.00

(11.) ACT Ms. Johnson purchased 1000 prizes for the school carnival for $90.
Each prize costs either $0.05 or $0.25.
How many of the less expensive prizes did she buy?

$ F.\:\: 200 \\[3ex] G.\:\: 360 \\[3ex] H.\:\: 500 \\[3ex] J.\:\: 640 \\[3ex] K.\:\: 800 \\[3ex] $

Let the:
Less expensive prize be x
More expensive prize be y
Ms. Johnson purchased 1,000 prizes

$x + y = 1000 ...eqn.(1)$

Less expensive prize goes with $\$0.05$ and the More expensive prize goes with $\$0.25$
This is because $0.25 \gt 0.05$
Ms. Johnson purchased these prizes for $\$90$

$0.05x + 0.25y = 90 ...eqn.(2)$

The two equations are:

$ x + y = 1000 ...eqn.(1) \\[3ex] 0.05x + 0.25y = 90 ...eqn.(2) \\[3ex] \underline{Substitution\:\:Method} \\[3ex] From\:\:eqn.(1) \\[3ex] y = 1000 - x...eqn.(3) \\[3ex] Substitute\:\:eqn.(3)\:\:into\:\:eqn.(2) \\[3ex] 0.05x + 0.25(1000 - x) = 90 \\[3ex] 0.05x + 250 - 0.25x = 90 \\[3ex] -0.2x = 90 - 250 \\[3ex] -0.2x = -160 \\[3ex] x = \dfrac{-160}{-0.2} \\[5ex] x = 800 \\[3ex] $ Ms. Johnson purchased $800$ prizes of the $\$0.05$ prize

(12.) GCSE A theme park sells entrance tickets.
2 adult tickets and 3 child tickets would cost a total of £72.
3 adult tickets and 1 child ticket would cost a total of £66.
A family ticket costs £45 and allows entry for 2 adults and 2 children.
How much cheaper is it to buy a family ticket than it would be to buy 2 adult and 2 child tickets?

$ Let\;\;the: \\[3ex] cost\;\;of\;\;an\;\;adult\;\;ticket = x \\[3ex] cost\;\;of\;\;a\;\;child\;\;ticket = y \\[3ex] \implies \\[3ex] 2x + 3y = 72 ...eqn.(1) \\[3ex] 3x + y = 66 ...eqn.(2) \\[3ex] \underline{Elimination-by-Subtraction\;\;Method} \\[3ex] Finding\;\;x \\[3ex] eqn.(1) - 3 * eqn.(2) \implies \\[3ex] (2x + 3y) - 3(3x + y) = 72 - 3(66) \\[3ex] 2x + 3y - 9x - 3y = 72 - 198 \\[3ex] -7x = -126 \\[3ex] x = \dfrac{-126}{-7} \\[5ex] x = 18 \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(2) \\[3ex] y = 66 - 3x ...eqn.(3) \\[3ex] Substitute\;\;18\;\;for\;\;y\;\;in\;\;eqn.(3) \\[3ex] y = 66 - 3(18) \\[3ex] y = 66 - 54 \\[3ex] y = 12 \\[3ex] Cost\;\;for\;\;2\;\;adult\;\;and\;\;2\;\;child\;\;tickets \\[3ex] = 2x + 2y \\[3ex] = 2(18) + 2(12) \\[3ex] = 36 + 24 \\[3ex] = £60 \\[3ex] Cost\;\;of;\;family\;\;ticket = £45 \\[3ex] Difference = 60 - 45 = 15 \\[3ex] $ It is £15 (15 pounds) cheaper to purchase a family ticket (2 adults and 2 children tickets) than to purchase the individual tickets for 2 adults and 2 children.

(14.) Do you drink? Please do not get drunk!
For those who care to know, I do not drink.

Timothy likes to drink. However, he does not like strong drinks.
During his birthday party, he received several alcoholic drinks as gifts.
One of the gifts was a vodka containing 40% alcohol by volume (40% ABV).
Another gift was a wine containing 10% ABV.
He wants a 5-liter solution containing 16% alcohol. So he mixes the wine with the vodka.
How many liters of each drink (wine and vodka) should he mix?

Here is my video on Mixture Problems
Two factors are important here: Volume and Concentration.
In this case, the two drinks (10% alcohol wine and 40% alcohol vodka) and the mixture are alcohol solutions.

Let the volume of the 10% alcohol wine be x
and the volume of the 40% alcohol vodka be y
This means that the volume of the mixture is x + y

Volume
$x + y = 5 ...eqn.(1)$

Concentration
10% of x + 40% of y = 16% of the mixture

$ 0.1 * x + 0.4 * y = 0.16 * 5 \\[3ex] 0.1x + 0.4y = 0.8 ...eqn.(2) \\[3ex] $ The two equations are:

$ x + y = 5 ...eqn.(1) \\[3ex] 0.1x + 0.4y = 0.8 ...eqn.(2) \\[3ex] x + 4y = 8 ...modified\;\;eqn.(2) \\[3ex] \underline{Elimination\;\;Method} \\[3ex] To\;\;find\;\;y,\;\;eliminate\;\;x \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] (x + 4y) - (x + y) = 8 - 5 \\[3ex] x + 4y - x - y = 3 \\[3ex] 3y = 3 \\[3ex] y = \dfrac{3}{3} \\[5ex] y = 1 \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(1) \\[3ex] x = 5 - y ...eqn.(3) \\[3ex] Substitute\;\;y = 1 \;\;into\;\;eqn.(3) \\[3ex] x = 5 - 1 \\[3ex] x = 4 \\[3ex] $ Timothy should mix 4 liters of the 10% alcohol wine with 1 liter of the 40% alcohol vodka in order to obtain 5 liters of the 16% alcohol solution.

(16.) Continuation from Question (14.)

Timothy wants to dilute the remaining liters of vodka.
Rather than mixing it with wine, he intends to mix it with pure water.
Please NOTE: that pure water is 0% alcohol. It has no alcohol.
He still wants a 5-liter solution containing 16% alcohol.
How many liters of water and vodka should he mix?

Here is my video on Mixture Problems
Two factors are important here: Volume and Concentration.
In this case, the two drinks are different.
But the second drink (40% alcohol vodka) and the mixture are alcohol solutions.
The first drink is pure water. It is 100% water. It is 0% alcohol.
The second drink is vodka, 40% alcohol.
The mixture is also an alcohol solution.
So, alcohol is the main factor.
In that sense, we shall use pure water as 0% alcohol.

Let the volume of the pure water be x
and the volume of the 40% alcohol vodka be y
This means that the volume of the mixture is x + y

Volume
$x + y = 5 ...eqn.(1)$

Concentration
0% of x + 40% of y = 16% of the mixture

$ 0 * x + 0.4 * y = 0.16 * 5 \\[3ex] 0.4y = 0.8 ...eqn.(2) \\[3ex] y = \dfrac{0.8}{0.4} \\[5ex] y = 2 \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(1) \\[3ex] x = 5 - y ...eqn.(3) \\[3ex] Substitute\;\;y = 2 \;\;into\;\;eqn.(3) \\[3ex] x = 5 - 2 \\[3ex] x = 3 \\[3ex] $ Timothy should mix 3 liters of the pure water with 2 liters of the 40% alcohol vodka in order to obtain 5 liters of the 16% alcohol solution.

(18.) Two planes travel toward each other from cities that are about 1725km apart.
They travel at rates of 360km/hr and 330km/hr.
They started at the same time.
In how many hours will they meet?

The two planes started at the same time.
Hence, their times are equal.
Let the two planes be Plane A and Plane B
Let the speed of Plane A = 360 km/hr
and the speed of Plane B = 330 km/hr
Let the time be $t$

$ Speed = \dfrac{distance}{time} \\[5ex] \therefore distance = speed * time \\[3ex] $ Distance traveled by Plane A = $360 * t = 360t$
Distance traveled by Plane B = $330 * t = 330t$
The sum of the distance should be the total distance

$ 360t + 330t = 1725 \\[3ex] 690t = 1725 \\[3ex] t = \dfrac{1725}{690} \\[5ex] t = \dfrac{5}{2}\: hours $

(19.) ACT Kojo has an internet site where his classmates can sell items in online auctions.
For each item, a student pays Kojo a listing fee, based on the item's starting price, and a selling fee calculated as a percent of the selling price, as shown in the tables below.

Starting price	Listing fee
$0.01 — $4.99 $5.00 — $19.99 $20.00 — $49.99 $50.00 and up	$0.25 $0.50 $1.00 $2.00

Selling price	Selling fee
$0.01 — $49.99 $50.00 and up	5% of selling price 3% of selling price

Erick sold 2 items on Kojo's site.
The sum of selling prices for the 2 items was $116.00.
The sum of the selling fees for the 2 items was 4.34.
The system of equations below can be used to obtain the selling price for each item.
What was the total listing fee for the 2 items given that the starting price was equal to the selling price for each of the 2 items? $$ x + y = 116.00 \\[3ex] 0.03x + 0.05y = 4.34 $$ $ A.\;\; \$2.00 \\[3ex] B.\;\; \$2.25 \\[3ex] C.\;\; \$2.50 \\[3ex] D.\;\; \$3.00 \\[3ex] E.\;\; \$4.00 \\[3ex] $

$ x + y = 116.00 ...eqn.(1) \\[3ex] 0.03x + 0.05y = 4.34 ...eqn.(2) \\[3ex] 3x + 5y = 434 ...modified\;\;eqn.(2) \\[3ex] \underline{Elimination-by-Addition\;\;Method} \\[3ex] Finding\;\;y \\[3ex] -3 * eqn.(1) + eqn.(2) \implies -3(x + y) + (3x + 5y) = -3(116) + 434 \\[3ex] -3x - 3y + 3x + 5y = -348 + 434 \\[3ex] 2y = 86 \\[3ex] y = \dfrac{86}{2} \\[5ex] y = 43 \\[3ex] \underline{Substitution\;\;Method} \\[3ex] Finding\;\;x \\[3ex] From\;\;eqn.(1) \\[3ex] x = 116 - y ...eqn.(3) \\[3ex] Substitute\;\;43\;\;for\;\;y\;\;in\;\;eqn.(3) \\[3ex] x = 116 - 43 \\[3ex] x = 73 \\[3ex] Selling\;\;price\;\;for\;\;item:\;x = \$73.00 \\[3ex] Selling\;\;price\;\;for\;\;item:\;y = \$43.00 \\[3ex] Because: \\[3ex] $ the starting price was equal to the selling price for each of the 2 items

$ Starting\;\;price\;\;for\;\;item:\;x = \$73.00 \\[3ex] Listing\;\;fee\;\;for\;\;item:\;x = \$2.00...on\;\;the\;\;table \\[3ex] Starting\;\;price\;\;for\;\;item:\;y = \$43.00 \\[3ex] Listing\;\;fee\;\;for\;\;item:\;y = \$1.00...on\;\;the\;\;table \\[3ex] \therefore\;the\;\;total\;\;listing\;fee = \$2.00 + \$1.000 = \$3.00 $

(20.) GCSE The diagram below shows triangle ABC

(a) Use the angles in the triangle to complete the following equation. $$ 3x + 11y = ..................... $$ (b) Use the angles on the straight line at C to obtain a further equation.

Use the two equations to show that ABC is an isosceles triangle.
You must use an algebraic method.

$ 3x + 11y = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] At\;\;Point\;C:\;\;6x + 7y = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \implies \\[3ex] 3x + 11y = 180 ...eqn.(1) \\[3ex] 6x + 7y = 180 ...eqn.(2) \\[3ex] \underline{Elimination-by-Addition\;\;Method} \\[3ex] Finding\;\;y \\[3ex] -2 * eqn.(1) + eqn.(2) \implies \\[3ex] -2(3x + 11y) + (6x + 7y) = -2(180) + 180 \\[3ex] -6x - 22y + 6x + 7y = -360 + 180 \\[3ex] -15y = -180 \\[3ex] y = \dfrac{-180}{-15} \\[5ex] y = 12 \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(2) \\[3ex] 6x = 180 - 7y \\[3ex] x = \dfrac{180 - 7y}{6} ...eqn.(3) \\[5ex] Substitute\;\;12\;\;for\;\;y\;\;in\;\;eqn.(3) \\[3ex] x = \dfrac{180 - 7(12)}{6} \\[5ex] x = \dfrac{180 - 84}{6} \\[5ex] x = \dfrac{96}{6} \\[5ex] x = 16 \\[3ex] \angle A = 4y^\circ = 4(12) = 48^\circ \\[3ex] \angle B = 3x^\circ = 3(16) = 48^\circ \\[3ex] \angle C = 7y^\circ = 7(12) = 84^\circ \\[3ex] $ Because ∠A = ∠B = 48°, ▵ABC is an isosceles triangle.

(22.) ACT Susan completed 20 courses in college and her grade point average was 3.15 (A = 4.0 and B = 3.0).
If all of her grades were A's or B's, and each course grade carried equal weight, what is the number of A's that Susan received?

$ F.\;\; 1 \\[3ex] G.\;\; 3 \\[3ex] H.\;\; 4 \\[3ex] J.\;\; 9 \\[3ex] K.\;\; 15 \\[3ex] $

This sentence should have been worded correctly as:
...and each course ~~grade~~ carried equal weight
So, we shall assume such for this question.
In other words, we shall assume that each course is a 1-credit unit course.
So we have 20 credit units for the 20 courses.

Let:
the number of A's courses = x
the number of B's courses = y

$ \underline{Number\;\;of\;\;courses} \\[3ex] x + y = 20 ...eqn.(1) \\[3ex] \underline{GPA} \\[3ex] \dfrac{4x + 3y}{20} = 3.15 \\[5ex] 4x + 3y = 20(3.15) \\[3ex] 4x + 3y = 63 ...eqn.(2) \\[3ex] We\;\;want\;\;to\;\;find\;\;x \\[3ex] \underline{Elimination\;\;Method} \\[3ex] eqn.(2) - 3 * eqn.(1) \implies \\[3ex] (4x + 3y) - 3(x + y) = 63 - 3(20) \\[3ex] 4x + 3y - 3x - 3y = 63 - 60 \\[3ex] x = 3 \\[3ex] $ Susan made A's in 3 courses.

(24.) A boat traveled 210 miles each way going downstream and back.
The trip downstream took 10 hours.
The trip back took 70 hours.
(a.) What is the speed of the boat in still water?
(b.) What is the speed of the current?

Let:
the speed of the boat in still water = x
the speed of the current = y

$ Speed = \dfrac{distance}{time} \\[5ex] \underline{Downstream} \\[3ex] speed = \dfrac{210}{10} = 21\;mph \\[5ex] Working\;\;with\;\;the\;\;current \\[3ex] \implies \\[3ex] x + y = 21 ...eqn.(1) \\[3ex] \underline{Trip\;\;back:\;\;Upstream} \\[3ex] speed = \dfrac{210}{70} = 3\;mph \\[5ex] Working\;\;against\;\;the\;\;current \\[3ex] \implies \\[3ex] x - y = 3 ...eqn.(2) \\[3ex] \underline{Elimination-by-Addition\;\;Method} \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] (x + y) + (x - y) = 21 + 3 \\[3ex] x + y + x - y = 24 \\[3ex] 2x = 24 \\[3ex] x = \dfrac{24}{2} \\[5ex] x = 12\;mph \\[3ex] -1 * eqn.(1) + eqn.(2) \implies \\[3ex] -1(x + y) + (x - y) = -1(21) + 3 \\[3ex] -x - y + x - y = -21 + 3 \\[3ex] -2y = -18 \\[3ex] y = \dfrac{-18}{-2} \\[5ex] y = 9\;mph \\[3ex] $ The speed of the boat in still water is 12 miles per hour.
The speed of the current is 9 miles per hour.

(25.) ACT A community theater group performed at 5 local schools.
For each school, the table below shows the total number of tickets sold and the total dollar amount collected from ticket sales.

School	Number of tickets sold	Ticket sales
A B C D E	200 250 300 150 275	$1,400 $1,650 $1,800 $1,350 $1,625

At School A, only 2 types of tickets were sold: premium tickets for $10 each and value tickets for $6 each.
How many value tickets were sold at School A?

$ F.\;\; 22 \\[3ex] G.\;\; 50 \\[3ex] H.\;\; 100 \\[3ex] J.\;\; 150 \\[3ex] K.\;\; 178 \\[3ex] $

Let:
the number of premium tickets sold at School A = x
the number of value tickets sold at School A = y

$ 200\;\;tickets\;\;sold\;\;at\;\;School\;A \\[3ex] x + y = 200 ...eqn.(1) \\[3ex] \$1400\;\;ticket\;\;sales\;\;at\;\;School\;A\;\;@\;\;\$10\;premium\;\;tickets\;\;and\;\;\$6\;\;value\;\;tickets \\[3ex] 10x + 6y = 1400 \\[3ex] 5x + 3y = 700 ...eqn.(2) \\[3ex] \underline{Elimination-by-Addition\;\;Method} \\[3ex] To\;\;find\;\;y;\;\;eliminate\;\;x \\[3ex] -5 * eqn.(1) + eqn.(2) \implies \\[3ex] -5(x + y) + (5x + 3y) = -5(200) + 700 \\[3ex] -5x - 5y + 5x + 3y = -1000 + 700 \\[3ex] -2y = -300 \\[3ex] y = \dfrac{-300}{-2} \\[5ex] y = 150 \\[3ex] $ 150 value tickets were sold at School A

(26.) The state fair is a popular field trip destination.
This year the senior class at High School A and the senior class at High School B both planned trips there.
The senior class at High School A rented and filled 8 vans and 8 buses with 240 students.
High School B rented and filled 4 vans and 1 bus with 54 students.
Find the number of students in each van and in each bus.

Let:
the number of students in each van = v
the number of students in each bus = b

$ \underline{High\;\;School\;A} \\[3ex] 8v + 8b = 240 ...eqn.(1) \\[3ex] \underline{High\;\;School\;B} \\[3ex] 4v + b = 54 ...eqn.(2) \\[3ex] \underline{Substitution\;\;Method} \\[3ex] From\;\;eqn.(2) \\[3ex] b = 54 - 4v...eqn.(3) \\[3ex] Substitute\;\;(54 - 4v)\;\;for\;\;b\;\;in\;\;eqn.(1) \\[3ex] 8v + 8(54 - 4v) = 240 \\[3ex] 8v + 432 - 32v = 240 \\[3ex] -24v = 240 - 432 \\[3ex] -24v = -192 \\[3ex] v = \dfrac{-192}{-24} \\[5ex] v = 8 \\[3ex] Substitute\;\;8\;\;for\;\;v\;\;in\;\;eqn.(3) \\[3ex] b = 54 - 4(8) \\[3ex] b = 54 - 32 \\[3ex] b = 22 \\[3ex] $ There were 8 students in each van.
There were 22 students in each bus.

(28.) A boat traveled 336 miles each way going downstream and back.
The trip downstream took 12 hours.
The trip back took 14 hours.
(a.) What is the speed of the boat in still water?
(b.) What is the speed of the current?

Let:
the speed of the boat in still water = x
the speed of the current = y

$ Speed = \dfrac{distance}{time} \\[5ex] \underline{Downstream} \\[3ex] speed = \dfrac{336}{12} = 28\;mph \\[5ex] Working\;\;with\;\;the\;\;current \\[3ex] \implies \\[3ex] x + y = 28 ...eqn.(1) \\[3ex] \underline{Trip\;\;back:\;\;Upstream} \\[3ex] speed = \dfrac{336}{14} = 24\;mph \\[5ex] Working\;\;against\;\;the\;\;current \\[3ex] \implies \\[3ex] x - y = 24 ...eqn.(2) \\[3ex] \underline{Elimination-by-Addition\;\;Method} \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] (x + y) + (x - y) = 28 + 24 \\[3ex] x + y + x - y = 52 \\[3ex] 2x = 52 \\[3ex] x = \dfrac{52}{2} \\[5ex] x = 26\;mph \\[3ex] -1 * eqn.(1) + eqn.(2) \implies \\[3ex] -1(x + y) + (x - y) = -1(28) + 24 \\[3ex] -x - y + x - y = -28 + 24 \\[3ex] -2y = -4 \\[3ex] y = \dfrac{-4}{-2} \\[5ex] y = 2\;mph \\[3ex] $ The speed of the boat in still water is 26 miles per hour.
The speed of the current is 2 miles per hour.