Word Problems Involving Algebraic Models – 50 SAT Practice Questions (No Calculator)
Word problems are a crucial part of the SAT Math section, testing not only your ability to manipulate algebraic expressions but also your skill at translating real-world scenarios into mathematical models. In the no-calculator portion of the SAT, it’s essential to be able to handle numbers, variables, and algebraic manipulations efficiently, all while paying careful attention to the context of the question.
Below are 50 carefully designed word problems that require creating and solving algebraic models. The topics cover a wide range of classic scenarios, including distance-rate-time problems, mixture and solution calculations, consecutive integer puzzles, age relationships, pricing and budget constraints, system of linear equations in word contexts, proportional relationships, and more. Each question is followed by a detailed solution that explains step-by-step how to set up the equations, how to solve them, and how to interpret the results.
By working through these examples, you will strengthen your problem-solving techniques and develop strategies to tackle any scenario on test day. You will learn to identify relevant variables, translate statements into equations, carefully handle units, and keep an eye out for extraneous or nonsensical results. This practice will help you become more comfortable under the SAT’s time constraints and sharpen your mental math skills for the no-calculator segment.
Remember to practice systematically: read each problem carefully, identify what is being asked, denote variables clearly, form equations, and solve carefully without a calculator. Check that your answers make sense in the given context. Practice truly is the key to success. Good luck, and let’s begin with the questions!
Question 1
A bakery sells cookies and brownies. Each cookie costs \$2, while each brownie costs \$3. If Michael buys a total of 12 items and spends \$29 altogether, how many cookies and how many brownies does he purchase?
Solution
Let c = number of cookies, and b = number of brownies. We have two conditions:
- He buys 12 items in total: c + b = 12.
- The total cost is \$29: 2c + 3b = 29.
From the first equation, c = 12 - b. Substitute into the cost equation:
2(12 - b) + 3b = 29
24 - 2b + 3b = 29
24 + b = 29
b = 5
Since b = 5, then c = 12 - 5 = 7. Therefore, Michael buys 7 cookies and 5 brownies.
Question 2
A cyclist travels to a park at a constant speed of 12 miles per hour, then returns along the same route at a speed of 8 miles per hour. The total trip takes 5 hours. How far is the park from the cyclist’s starting point?
Solution
Let d = distance to the park (in miles). Time = distance / speed. The time to go to the park is d/12, and the time to return is d/8. We know the total time is 5 hours, so:
(d / 12) + (d / 8) = 5
Find a common denominator for 12 and 8, which is 24:
(2d / 24) + (3d / 24) = 5
5d / 24 = 5
Multiply both sides by 24:
5d = 5 × 24 = 120
d = 120 / 5 = 24
Thus, the park is 24 miles away from the cyclist’s starting point.
Question 3
A rectangular garden has a perimeter of 54 meters. Its length is 3 meters longer than twice its width. What are the dimensions of the garden?
Solution
Let w = width (meters), and let l = length (meters). We know:
- The perimeter, 2l + 2w = 54.
- The length is 3 meters more than twice the width, l = 2w + 3.
Substitute l = 2w + 3 into the perimeter equation:
2(2w + 3) + 2w = 54
4w + 6 + 2w = 54
6w + 6 = 54
6w = 48
w = 8
Then l = 2(8) + 3 = 16 + 3 = 19. Thus, the width is 8 meters and the length is 19 meters.
Question 4
Noah is 7 years older than his sister Emma. Together, their ages sum to 21. How old is each sibling?
Solution
Let N = Noah’s age and E = Emma’s age. We have:
- N = E + 7 (Noah is 7 years older than Emma).
- N + E = 21 (the sum of their ages is 21).
Substitute N = E + 7 into N + E = 21:
(E + 7) + E = 21
2E + 7 = 21
2E = 14
E = 7
Then N = 7 + 7 = 14. So, Emma is 7 years old and Noah is 14 years old.
Question 5
A train traveling at 60 miles per hour leaves a station. Two hours later, a second train leaves the same station on the same track at 80 miles per hour. How long after the second train departs will it catch up to the first train?
Solution
Let t = time (in hours) that the second train travels until it catches up. In that time, the second train covers 80t miles. The first train has a 2-hour head start. By the time the second train leaves, the first train has traveled 60 × 2 = 120 miles. After the second train travels for t hours, the first train has traveled an additional 60t miles. For them to be at the same place:
distance of first train = distance of second train
120 + 60t = 80t
Subtract 60t from both sides:
120 = 20t
t = 120 / 20 = 6
Therefore, the second train will catch up 6 hours after it departs.
Question 6
The sum of three consecutive integers is 72. What are the integers?
Solution
Let the smallest integer be n, the next consecutive integer is n + 1, and the next is n + 2. Their sum is 72:
n + (n + 1) + (n + 2) = 72
3n + 3 = 72
3n = 69
n = 23
So, the integers are 23, 24, and 25. Therefore, the three consecutive integers are 23, 24, and 25.
Question 7
Tim invests \$5000 in a savings account. The principal grows by 4% per year, and no additional deposits or withdrawals are made. Let t = number of years since the initial deposit. Write an expression for the account balance after t years and then find the balance after 2 years.
Solution
The balance grows by 4% each year. The growth model for an initial principal P with an annual interest rate r (in decimal form) is:
Balance = P(1 + r)^t
Here, P = 5000 and r = 0.04. The expression for the balance after t years is:
B(t) = 5000(1.04)^t
After 2 years, t = 2:
B(2) = 5000(1.04)^2 = 5000 × (1.04 × 1.04)
1.04 × 1.04 = 1.0816 (though we typically might leave it in exact form if no calculator is used).
B(2) = 5000 × 1.0816 = 5408 (approx.)
So, the balance after 2 years is approximately \$5408.
Question 8
A rectangular rug has an area of 88 square feet. Its length is 5 feet more than its width. Find the width and length of the rug.
Solution
Let w = width, and let l = length. The area is given by w × l = 88, and the length is 5 feet more than the width: l = w + 5. Substitute into the area equation:
w(w + 5) = 88
w^2 + 5w - 88 = 0
Solve this quadratic by factoring or by inspection. Factors of -88 that add up to 5 are 11 and -8:
(w + 11)(w - 8) = 0
So w = -11 or w = 8. A negative width is not physically meaningful, so w = 8. Then l = 8 + 5 = 13.
The rug’s dimensions are 8 feet (width) by 13 feet (length).
Question 9
Abby and Bobby are solving a jigsaw puzzle. Working alone, Abby can finish the puzzle in 3 hours, while Bobby can finish the puzzle in 6 hours. If they work together at their respective rates, how long will it take for them to complete the puzzle?
Solution
Abby’s rate = 1 puzzle / 3 hours = 1/3 puzzle per hour.
Bobby’s rate = 1 puzzle / 6 hours = 1/6 puzzle per hour.
Together, their combined rate = (1/3 + 1/6) = 1/2 puzzle per hour. If they complete 1 puzzle at a rate of 1/2 puzzle/hour, the time needed is:
time = 1 ÷ (1/2) = 2 hours
Therefore, working together, Abby and Bobby will finish the puzzle in 2 hours.
Question 10
Lisa and Maria decide to share the cost of renting a boat for a day. Lisa pays \$14 more than Maria. If the total cost is \$70, how much does each pay?
Solution
Let M = amount Maria pays. Then Lisa pays M + 14. The total is \$70, so:
M + (M + 14) = 70
2M + 14 = 70
2M = 56
M = 28
Hence, Maria pays \$28, and Lisa pays 28 + 14 = \$42.
Answer: Maria: \$28, Lisa: \$42.
Question 11
A rectangular swimming pool is 4 times as long as it is wide. If its perimeter is 150 meters, find its dimensions.
Solution
Let w = width. Then length l = 4w. The perimeter is given by 2l + 2w = 150:
2(4w) + 2w = 150
8w + 2w = 150
10w = 150
w = 15
Then l = 4(15) = 60. Therefore, the dimensions are 15 meters wide and 60 meters long.
Question 12
The difference between two positive integers is 12. Their sum is 80. Find these two integers.
Solution
Let the integers be x and y, with x > y. We have:
- x - y = 12
- x + y = 80
Add the equations:
(x - y) + (x + y) = 12 + 80
2x = 92
x = 46
Then using x + y = 80, we get 46 + y = 80, so y = 34.
Thus, the numbers are 46 and 34.
Question 13
A boat goes 36 miles downstream in the same time it takes to go 24 miles upstream. If the speed of the current is 2 miles per hour, find the speed of the boat in still water.
Solution
Let v = speed of the boat in still water. The downstream speed is (v + 2) mph, and the upstream speed is (v - 2) mph (because the current assists downstream and opposes upstream). The times for each journey are equal. Time = distance / speed, so:
(36 / (v + 2)) = (24 / (v - 2))
Cross-multiply:
36(v - 2) = 24(v + 2)
36v - 72 = 24v + 48
36v - 24v = 48 + 72
12v = 120
v = 10
Hence, the speed of the boat in still water is 10 mph.
Question 14
A clothing store marks up every item by 25% of the store’s cost. If a pair of jeans has a cost to the store of \$48, what is the retail price of the jeans?
Solution
The store’s cost is \$48. Markup at 25% means we add 25% of \$48 to the cost:
25% of \$48 = 0.25 × 48 = \$12
Therefore, the retail price = \$48 + \$12 = \$60.
The pair of jeans will retail for \$60.
Question 15
Jenny has \$100. She spends \$x on a birthday present for her friend. Then she spends twice as much on a new pair of shoes. Finally, she has \$24 left. Write an equation to model this situation and solve for x.
Solution
She starts with \$100. She spends \$x on a birthday present and \$2x on shoes. The total spent is x + 2x = 3x. After these purchases, she has \$24 left, so:
100 - 3x = 24
Subtract 100 from both sides:
-3x = -76
x = (-76)/(-3) = 76/3
x = 25⅓ if we convert to a mixed number, or 25.333...
Therefore, Jenny spends \$25⅓ (approximately \$25.33) on the birthday present.
Question 16
The sum of four consecutive even integers is 100. Find the integers.
Solution
Let the smallest of the four consecutive even integers be n. Then the next three are n+2, n+4, n+6. Their sum is 100:
n + (n + 2) + (n + 4) + (n + 6) = 100
4n + 12 = 100
4n = 88
n = 22
The four integers are 22, 24, 26, and 28. Therefore, the consecutive even integers are 22, 24, 26, 28.
Question 17
A chemist needs 25 liters of a 40% acid solution. She has a 60% acid solution and a 30% acid solution on hand. How many liters of each should she mix to get 25 liters of 40% acid solution?
Solution
Let x = liters of 60% solution. Then (25 - x) = liters of 30% solution, because we want a total of 25 liters. The acid from the 60% solution is 0.60x liters, and from the 30% solution is 0.30(25 - x). We want this total acid to be 40% of 25 liters, i.e. 0.40 × 25 = 10 liters of acid.
So:
0.60x + 0.30(25 - x) = 10
Distribute:
0.60x + 7.5 - 0.30x = 10
0.30x + 7.5 = 10
0.30x = 2.5
x = 2.5 / 0.30 = 25/3 = 8.333...
So x ≈ 8⅓ liters (60% solution). Then 25 - x ≈ 16⅔ liters (30% solution).
The chemist should mix about 8.33 liters of 60% solution and 16.67 liters of 30% solution.
Question 18
A car rental company charges a fixed daily fee of \$20 plus \$0.15 for each mile driven. If someone rents a car and pays \$47 in total, how many miles did the person drive?
Solution
Let m = number of miles driven. The total cost is given by:
20 + 0.15m = 47
Subtract 20 from both sides:
0.15m = 27
m = 27 / 0.15 = 27 ÷ 0.15
0.15 = 15 cents, which is 3/20. So dividing 27 by 3/20 is like 27 × 20/3 = 27 × (20/3) = 180
So, m = 180. The person drove 180 miles.
Question 19
Margaret has 30 coins in her purse, consisting of only dimes and quarters. If she has a total of \$5.10, how many of each coin does she have?
Solution
Let d = number of dimes, and q = number of quarters. Then we know:
- d + q = 30 (she has 30 coins total).
- The total value is 10d + 25q = 510 cents (because \$5.10 = 510 cents).
From d + q = 30, we get d = 30 - q. Substitute into the value equation:
10(30 - q) + 25q = 510
300 - 10q + 25q = 510
300 + 15q = 510
15q = 210
q = 14
Then d = 30 - 14 = 16. Margaret has 16 dimes and 14 quarters.
Question 20
Kendra has \$650 in a checking account. She writes a check for \$c, then later deposits \$150. Afterwards, her balance is \$780. Write and solve an equation to find \$c, the amount of the check.
Solution
Kendra starts with \$650, then subtracts the check amount \$c, and then adds \$150. The final result is \$780:
650 - c + 150 = 780
Combine like terms:
800 - c = 780
-c = -20
c = 20
The check amount is \$20.
Question 21
A group of 12 people shares the cost of renting a large vacation home equally. If there were 2 fewer people, each person would have to pay \$25 more. What is the total cost of renting the home?
Solution
Let T = total cost of renting the home, in dollars. If 12 people share the cost equally, each pays T/12. If there were only 10 people, each pays T/10. We are told that T/10 is \$25 more than T/12:
(T / 10) - (T / 12) = 25
Find a common denominator, 60:
(6T / 60) - (5T / 60) = 25
(6T - 5T) / 60 = 25
T / 60 = 25
T = 25 × 60 = 1500
Therefore, the total cost is \$1500.
Question 22
Max has a part-time job walking dogs. He charges \$12 for the first dog and \$3 for each additional dog in a day. If he walked 10 dogs total and earned \$39, how many “first dog” charges did he have? In other words, how many separate jobs with different owners did he have?
Solution
Let n = number of separate jobs. For each job, Max charges \$12 for the first dog. If for the nth job, he walks a certain number of additional dogs, each additional dog is \$3. However, the total dogs for all jobs is 10, and the total money is \$39.
Another way: Each job includes at least 1 dog at \$12. Past that, all other dogs are \$3. Let’s suppose he had n first-dogs at \$12 each, so total from these is 12n. Then the total additional dogs across all jobs is (10 - n), each at \$3, so 3(10 - n). The total is \$39:
12n + 3(10 - n) = 39
12n + 30 - 3n = 39
9n + 30 = 39
9n = 9
n = 1
This implies Max only had 1 separate job overall. But let’s check if that’s feasible with the total number of dogs and cost.
If n=1, Max has 1 first dog at \$12, and 9 additional dogs at \$3 each for a total of 12 + 27 = 39. Indeed, that fits.
So, he had 1 separate job (or one client) with a total of 10 dogs, if we interpret the problem strictly this way.
Question 23
In one day, a factory makes three times as many small widgets as large widgets. If the factory produces 380 widgets in total (small plus large) in a day, how many of each size are made?
Solution
Let L = number of large widgets, and S = number of small widgets. From the problem:
- S = 3L (three times as many small as large).
- S + L = 380.
Substitute S = 3L into S + L = 380:
3L + L = 380
4L = 380
L = 95
Then S = 3 × 95 = 285. The factory makes 95 large widgets and 285 small widgets.
Question 24
Julian is reading a book with 240 pages. He reads 20 pages per day for the first few days, then he reads 30 pages per day for the remaining days. If he finishes the book in 10 days total, for how many days did he read 20 pages per day?
Solution
Let x = number of days Julian reads 20 pages/day. Then (10 - x) = number of days reading 30 pages/day. The total pages read is 240:
20x + 30(10 - x) = 240
Distribute:
20x + 300 - 30x = 240
-10x + 300 = 240
-10x = -60
x = 6
Julian read 20 pages/day for 6 days and 30 pages/day for the remaining 4 days.
Question 25
The length of a rectangle is 2 inches more than its width. The area of the rectangle is 24 square inches. Find the length and width.
Solution
Let w = width. Then length l = w + 2. The area is w × (w + 2) = 24:
w^2 + 2w - 24 = 0
Factor if possible: We look for factors of -24 that add up to 2. This is 6 and -4:
(w + 6)(w - 4) = 0
So w = -6 or w = 4. A negative width is not valid, so w = 4. Then l = 4 + 2 = 6.
The rectangle’s dimensions are 4 inches by 6 inches.
Question 26
A punch mixture is 30% fruit juice. You have 15 liters of a 50% fruit juice concentrate. How much of the 30% punch must you add to the concentrate to end up with a 40% fruit juice mixture?
Solution
Let x = liters of the 30% punch to be added. We have 15 liters of 50% concentrate, so that contains 0.50 × 15 = 7.5 liters of juice. The 30% punch has 0.30x liters of juice. The total volume after mixing is (15 + x) liters, and we want a 40% mixture of fruit juice, meaning 0.40(15 + x) liters of juice in total.
So, set up the equation:
(Juice from concentrate) + (Juice from punch) = (Total juice in final mix)
7.5 + 0.30x = 0.40(15 + x)
Expand the right side:
7.5 + 0.30x = 6 + 0.40x
Subtract 0.30x and also subtract 6 from both sides:
7.5 - 6 = 0.40x - 0.30x
1.5 = 0.10x
x = 1.5 / 0.10 = 15
Therefore, you must add 15 liters of the 30% punch.
Question 27
Three consecutive odd integers have a sum of 111. What are they?
Solution
Let the smallest odd integer be n. Then the next two consecutive odd integers are n+2 and n+4. Their sum is 111:
n + (n + 2) + (n + 4) = 111
3n + 6 = 111
3n = 105
n = 35
The three consecutive odd integers are 35, 37, and 39.
Question 28
A teacher has \$120 to spend on notebooks. Each notebook costs \$2.50. If the teacher also needs \$20 left over for shipping, how many notebooks can the teacher buy?
Solution
Let n = number of notebooks. Each costs \$2.50, so total cost for n notebooks is 2.50n. The teacher has \$120 in total, but must save \$20 for shipping. Hence, the money available for notebooks is (120 - 20) = \$100:
2.50n = 100
n = 100 / 2.50 = 40
Therefore, the teacher can buy 40 notebooks.
Question 29
Trevor drives from his home to a distant city, averaging 45 miles per hour. He returns along the same route at 55 miles per hour. If the total driving time is 10 hours, how far apart are the two cities?
Solution
Let d = distance between Trevor’s home and the city (one-way). Time to go = d / 45 hours, and time to return = d / 55 hours. The total driving time is 10 hours:
d/45 + d/55 = 10
Find a common denominator: 45 and 55 have a least common multiple of 495. Multiply everything by 495 to clear denominators:
495(d/45) + 495(d/55) = 495(10)
(495 / 45)d + (495 / 55)d = 4950
11d + 9d = 4950
20d = 4950
d = 4950 / 20 = 247.5
The distance between the two cities is 247.5 miles.
Question 30
Anna can mow the lawn in 4 hours. Her younger sister Daisy can mow the same lawn in 6 hours. If they work together at their constant individual rates, how long will it take them to finish the lawn?
Solution
Anna’s rate = 1/4 lawn per hour. Daisy’s rate = 1/6 lawn per hour. Together, their combined rate is 1/4 + 1/6 = 3/12 + 2/12 = 5/12 lawn per hour. The time needed to complete 1 lawn is:
time = 1 ÷ (5/12) = 12/5 = 2.4 hours
Hence, working together, they will finish the lawn in 2.4 hours (2 hours and 24 minutes).
Question 31
The sum of two numbers is 30, and their difference is 6. Find the two numbers.
Solution
Let the two numbers be x and y with x ≥ y. Then we have:
- x + y = 30
- x - y = 6
Add these two equations:
(x + y) + (x - y) = 30 + 6
2x = 36
x = 18
Then, from x + y = 30, we get 18 + y = 30, so y = 12.
The two numbers are 18 and 12.
Question 32
A certain airplane can travel 840 miles with a tailwind in the same amount of time it takes to travel 720 miles against the same wind. If the speed of the wind is 30 mph, find the speed of the airplane in still air.
Solution
Let v = speed of the plane in still air. With a 30 mph tailwind, the effective speed is (v + 30). Against the wind, it is (v - 30). The times are equal:
840 / (v + 30) = 720 / (v - 30)
Cross-multiply:
840(v - 30) = 720(v + 30)
840v - 25200 = 720v + 21600
840v - 720v = 21600 + 25200
120v = 46800
v = 46800 / 120 = 390
Thus, the plane’s speed in still air is 390 mph.
Question 33
A rectangular piece of paper is 4 inches longer than it is wide. Its perimeter is 28 inches. Find the width and length of the paper.
Solution
Let w = width, and l = length = w + 4. The perimeter is 2w + 2l = 28:
2w + 2(w + 4) = 28
2w + 2w + 8 = 28
4w = 20
w = 5
Then l = 5 + 4 = 9.
The paper’s dimensions are 5 inches wide and 9 inches long.
Question 34
A painter charges a flat rate of \$50 for a consultation plus \$16 per hour for labor. If a customer is billed \$130, how many hours did the painter work?
Solution
Let h = number of hours the painter worked. The total cost is:
50 + 16h = 130
Subtract 50:
16h = 80
h = 80/16 = 5
The painter worked 5 hours.
Question 35
Together, Jessica and Karen earn \$56 per hour tutoring. Jessica earns \$6 more per hour than Karen. If their total is \$56 per hour, how much does each earn?
Solution
Let K = Karen’s hourly rate, and J = Jessica’s hourly rate. We have:
- J + K = 56
- J = K + 6
Substitute J = K + 6 into J + K = 56:
(K + 6) + K = 56
2K + 6 = 56
2K = 50
K = 25
Then J = 25 + 6 = 31.
Karen earns \$25 per hour and Jessica earns \$31 per hour.
Question 36
The sum of two positive integers is 100. One number is 4 times the other. Find the numbers.
Solution
Let x and y be the two positive integers. Suppose x = 4y. We also know x + y = 100:
4y + y = 100
5y = 100
y = 20
x = 4(20) = 80
Therefore, the numbers are 80 and 20.
Question 37
A landscaper can plant 24 bushes in 6 hours. Assuming he works at the same rate, how many hours does it take him to plant 40 bushes?
Solution
Rate = (24 bushes) / (6 hours) = 4 bushes/hour. We want to plant 40 bushes. At 4 bushes/hour:
time = 40 / 4 = 10 hours
It takes 10 hours to plant 40 bushes.
Question 38
A total of \$1200 is split between two investments, one paying 4% annual interest and the other paying 7% annual interest. The annual interest from the two investments is \$66. How much is invested at each rate?
Solution
Let x = amount invested at 4%, and (1200 - x) = amount at 7%. The total interest is \$66 for the year. The interest from the 4% portion is 0.04x, and from the 7% portion is 0.07(1200 - x).
0.04x + 0.07(1200 - x) = 66
Distribute:
0.04x + 84 - 0.07x = 66
-0.03x + 84 = 66
-0.03x = -18
x = (-18)/(-0.03) = 600
Thus, \$600 is invested at 4%, and 1200 - 600 = \$600 at 7%. Both amounts are \$600 each.
Question 39
Mary has a coupon for \$5 off any purchase plus a 20% discount on the remaining amount. She buys an item for \$50. How much does she pay after these two deductions?
Solution
First, subtract the \$5 coupon from the \$50 purchase:
50 - 5 = \$45
Then apply the 20% discount on the remaining \$45. A 20% discount means we pay 80% of that amount:
0.80 × 45 = \$36
So Mary pays \$36 in total.
Question 40
A chef has a 50% vinegar solution and an 80% vinegar solution. She needs 10 liters of a 65% solution for a recipe. How many liters of each solution should she mix?
Solution
Let x = liters of the 50% solution, and (10 - x) = liters of the 80% solution. The total vinegar in the final 65% mixture (which is 10 liters) is 0.65 × 10 = 6.5 liters. The amount of vinegar from the 50% solution is 0.50x, and from the 80% is 0.80(10 - x):
0.50x + 0.80(10 - x) = 6.5
Distribute:
0.50x + 8 - 0.80x = 6.5
-0.30x + 8 = 6.5
-0.30x = -1.5
x = (-1.5)/(-0.30) = 5
So x = 5 liters of the 50% solution, and 10 - 5 = 5 liters of the 80% solution.
The chef should mix 5 liters of 50% solution and 5 liters of 80% solution.
Question 41
Nate and Natalie are working on assembling gift baskets. Nate can finish assembling 100 baskets in 10 hours, while Natalie can finish the same 100 baskets in 8 hours. If they work together, how many hours does it take them to assemble 100 baskets?
Solution
Nate’s rate = 100 baskets / 10 hours = 10 baskets/hour.
Natalie’s rate = 100 baskets / 8 hours = 12.5 baskets/hour.
Together, combined rate = 10 + 12.5 = 22.5 baskets/hour.
Time to complete 100 baskets at this rate = 100 / 22.5 ≈ 4.44 hours (about 4 hours and 26 minutes).
Therefore, working together, they finish in 4.44 hours (approximately).
Question 42
Henry needs to fence a rectangular field adjacent to a long wall. He only needs to fence three sides (two widths and one length) because the wall covers the other length side. If the area of the field should be 500 square meters, and the length is along the wall, find the dimensions that minimize the amount of fencing used. (Hint: express the fencing in terms of width and length, use area constraint.)
Solution
Let w = width (meters), and l = length along the wall (meters). The area constraint: w × l = 500. The fencing needed is: F = 2w + l (since only three sides need fencing).
From w × l = 500, we have l = 500 / w. Substitute into F:
F = 2w + (500 / w)
Minimizing F typically is a calculus approach, but we can also reason about symmetrical shapes. For a rectangle of fixed area, the perimeter is smallest when it is as close to a square as possible. But here, we only need 2 widths plus 1 length.
A shortcut approach: If we do use optimization (though the SAT might not require advanced derivatives), set derivative of F w.r.t. w = 0:
dF/dw = 2 - 500 / w^2 = 0 => 2 = 500 / w^2 => w^2 = 250 => w = √250 = 5√10
Then l = 500 / (5√10) = 100 / √10 = 100√10 / 10 = 10√10.
So the dimensions that minimize fencing:
w = 5√10 meters and l = 10√10 meters.
(Approximately w ≈ 15.81 m, l ≈ 31.62 m.)
Question 43
Maria swims 50 laps in the pool in the same time it takes Robin to swim 30 laps. If Maria swims 2 laps per minute faster than Robin, find the lap rate of each swimmer in laps per minute.
Solution
Let r = Robin’s rate in laps per minute. Then Maria’s rate is (r + 2) laps per minute. We are told the time for Maria to swim 50 laps is the same as the time for Robin to swim 30 laps.
Time = distance / rate. So:
50 / (r + 2) = 30 / r
Cross-multiply:
50r = 30(r + 2)
50r = 30r + 60
20r = 60
r = 3
So, Robin swims at 3 laps/minute, and Maria swims at 3 + 2 = 5 laps/minute.
Question 44
A hardware store is having a “Buy One, Get One at 50% Off” sale on light bulbs. If the regular price of each light bulb is \$2.40, and Andrew buys 2 bulbs under this offer, how much does he pay in total?
Solution
The first bulb is \$2.40 at regular price. The second bulb is 50% off \$2.40, which is \$1.20.
So total = 2.40 + 1.20 = \$3.60
Andrew pays \$3.60 in total for 2 bulbs.
Question 45
An airplane traveling with a tailwind covers 1500 miles in 2.5 hours. The return trip (the same distance) against the wind takes 3 hours. Find the speed of the airplane in still air and the speed of the wind.
Solution
Let p = speed of the plane in still air, and w = wind speed. With tailwind, speed = p + w; with headwind, speed = p - w.
For the tailwind trip:
distance = 1500, time = 2.5 hours, so speed = 1500 / 2.5 = 600 mph. Hence, p + w = 600.
For the headwind trip:
distance = 1500, time = 3 hours, so speed = 1500 / 3 = 500 mph. Hence, p - w = 500.
Add the two equations:
(p + w) + (p - w) = 600 + 500
2p = 1100
p = 550
Then p + w = 600 => 550 + w = 600 => w = 50.
The plane’s speed in still air is 550 mph, and the wind speed is 50 mph.
Question 46
A sum of \$6000 is divided into two bank accounts, one paying 3% annual interest and the other paying 5%. If the total annual interest is \$240, how much is invested at 3% and how much is invested at 5%?
Solution
Let x = amount invested at 3%. Then (6000 - x) = amount at 5%. The total interest from the two accounts after one year is \$240. The interest from the 3% portion is 0.03x, and from the 5% portion is 0.05(6000 - x).
0.03x + 0.05(6000 - x) = 240
Distribute:
0.03x + 300 - 0.05x = 240
-0.02x + 300 = 240
-0.02x = -60
x = 3000
So, \$3000 is invested at 3%, and 6000 - 3000 = \$3000 at 5%.
Question 47
The ratio of the measures of two supplementary angles is 3:7. Find the measure of each angle.
Solution
Supplementary angles add up to 180 degrees. Let the angles be 3x and 7x (since their ratio is 3:7). Then:
3x + 7x = 180
10x = 180
x = 18
The angles are 3(18) = 54 degrees and 7(18) = 126 degrees.
Question 48
A mother is three times as old as her daughter. In 12 years, she will be twice as old as her daughter at that time. How old is each now?
Solution
Let M = mother’s current age, and D = daughter’s current age. We have:
- M = 3D (mother is three times the daughter’s age).
- In 12 years, mother’s age = M + 12, daughter’s age = D + 12. At that time, M + 12 = 2(D + 12).
Substitute M = 3D into the second equation:
(3D + 12) = 2(D + 12)
3D + 12 = 2D + 24
3D - 2D = 24 - 12
D = 12
Then M = 3(12) = 36.
So currently, the mother is 36 years old and the daughter is 12 years old.
Question 49
A driving school charges a fixed registration fee plus \$30 per lesson. If 12 lessons cost \$540 total, while 8 lessons cost \$390 total, find the registration fee and the cost per lesson.
Solution
Let R = registration fee, and C = cost per lesson. We are told it is \$30 per lesson, but let's confirm the problem statement. Actually, the problem states “\$30 per lesson,” yet the second part of the data indicates something else might be going on. Let’s do it systematically:
For 12 lessons, total = R + 30 × 12 = 540.
For 8 lessons, total = R + 30 × 8 = 390.
Let’s verify each scenario:
1) R + (12 × 30) = R + 360 = 540 => R = 180
2) R + (8 × 30) = R + 240 = 390 => R = 150
We have a discrepancy. Possibly, the problem means the driving school charges some unknown amount per lesson C, not necessarily \$30. Let’s rename it:
For 12 lessons: R + 12C = 540
For 8 lessons: R + 8C = 390
Subtract the second equation from the first:
(R + 12C) - (R + 8C) = 540 - 390
4C = 150
C = 37.5
Then from R + 8C = 390, substitute C = 37.5:
R + 8(37.5) = 390
R + 300 = 390
R = 90
The registration fee is \$90, and each lesson costs \$37.50.
(Note: The original mention of \$30 per lesson might be a confusion in the wording. The algebraic approach clarifies the actual cost.)
Question 50
The altitude of a triangle is 4 cm shorter than its base. If its area is 30 cm2, find the base and altitude of the triangle.
Solution
Let b = base (cm). Then the altitude is (b - 4). The area of a triangle is (1/2)(base)(altitude) = 30:
(1/2)(b)(b - 4) = 30
b(b - 4) = 60
b^2 - 4b - 60 = 0
Solve this quadratic. Look for factors of -60 that add to -4. We might try -10 and 6, which gives -4, but that yields -10 × 6 = -60. So b^2 - 10b + 6b - 60 = 0 => b^2 - 4b - 60 = 0. Let’s do the quadratic formula:
b = [4 ± √(16 + 240)] / 2 = [4 ± √256] / 2 = [4 ± 16] / 2
So b = (4 + 16) / 2 = 20 / 2 = 10, or b = (4 - 16) / 2 = -12 / 2 = -6 (not physically meaningful).
Hence, b = 10 cm. The altitude is 10 - 4 = 6 cm.
The triangle’s base is 10 cm and its altitude is 6 cm.
Conclusion and Final Tips
Well done on making your way through these 50 SAT-level algebraic word problems designed for the no-calculator portion of the test. These types of problems require a careful blend of reading comprehension, variable assignment, equation formation, and arithmetic skill. Here are some final insights to solidify your preparation:
- Define Variables Clearly: Often the biggest challenge in word problems is translating the real-world scenario into correct algebraic expressions. Setting up variables in a concise, step-by-step way can prevent confusion later.
- Check the Context: After solving, make sure your answer makes sense. For instance, a negative width, or a negative time, usually indicates a need to discard that solution or re-check your equations.
- Use Systematic Methods: For mixture problems, carefully track the concentration or percentage. For distance-rate-time, remember that time = distance ÷ speed. For money problems, convert to cents if that makes arithmetic easier. For consecutive integers, define them systematically as n, n+1, etc.
- Practice Mental Arithmetic: Since this is the no-calculator section, time spent doing simple arithmetic steps can add up. Learning to do fraction manipulation and quick multiplication or division is crucial for efficiency.
- Stay Organized: Label all steps clearly, including equations and final answers, to keep track of your logic under test conditions.
By consistently practicing these techniques, you will gain speed and accuracy in solving word problems without a calculator. Remember to double-check your work for potential errors, and go into test day feeling confident in your algebraic modeling skills!