Solving Word Problems - SAT Math

If the bakery bakes 210 loaves of bread and sells 240 loaves of bread, then that means that, total, it loses 30 loaves of bread per day. Since you know that it starts with 450 loaves of bread, you can use this information to write a linear equation relating the number of loaves of bread left with how many days it has been since the bakery has closed for a baking day. It should look like:

B = 450 - 30d

Where B represents the number of loaves left and d represents the number of days since the bakery’s last baking day.

In order for the bakery to need to close for a baking day, the number of loaves left must equal 0. If you substitute in B = 0, you get: 0 = 450−30d.

Now you can solve: add 30d to both sides to get:

30d = 450

And then divide both sides by 30 to get d = 15

To turn this problem into an equation, you can start by putting Katharine's goal of $50,000 on one side of the equation, and then arranging all of the inputs to that goal (ways she earns money) and impediments (ways she loses money) on the other. If you use m to represent the number of months, an initial equation would look like:

$50,000 = $10,000 + $6,500m - $4,000m

Which, qualitatively, is that her goal of $50,000 is equal to the $10,000 she has plus the $6,500 she earns each month, minus the $4,000 she loses each month.

Now you can combine like terms to simplify the equation:

$40,000 = $2,500m

And then divide 40000 by 2500 (which simplifies to 400 divided by 25) to solve:

m = 16 months

To translate this word problem into equation form, note that your unknown is the capacity of the pool. You know that the pool is currently half full and that with the proposed addition it would be 2/3 full, but you need to assign those fractions to an actual value. So your variable, , will represent the total capacity of the pool. Then you can set up your problem as:

From here it's an algebra problem. To get your terms together, subtract from both sides:

Then you'll need to find a common denominator of 6, and rewrite what you have to reflect that common denominator:

You can then combine like terms on the right-hand side of the equation:

And then multiply both sides by 6 to finish the problem:

To translate this word problem into equations, first assign variables to the new beams. Since you know there will be a longer and a shorter beam you might use for longer and for shorter. You would then have:

(the total length is 48)

(the longer is 20 more than the shorter)

You can then substitute the second equation into the first, replacing with . This then gives you:

So and

But remember - the question asked you for the LONGER beam, so you'll need to plug that back in to one of the equations. Since you can say that , so 34 is the correct answer.

While it is tempting to set up and solve an equation for this problem, the algebra for this problem may be difficult to translate quickly and accurately. Because you are dealing with relatively simple numbers that add up to a total, you should see that this should be an easy problem to backsolve.

Start with a total of , $80. If Shay paid $7 less than half of the total, that means she paid , or .

And if Ashley paid $8 more than of the bill, that means that she paid , or $28. That means that Jenna must have paid the remainder, . However, since you know that Jenna paid $20, you should recognize that the total for answer choice $80 was too small and that the actual bill must have been slightly more.

You can go through the same process with answer choice $84. If Shay paid $7 less than half of the total, she paid .

And if Ashley paid $8 more than of the bill, that means that she paid .

Since you know that Jenna paid $20, you should recognize that . Since the original total and the final sum match up, you can conclude that $84 is your answer, pick it, and move on.

Alternatively, you could set up the algebra:

While the problem setup here may seem a bit involved, if you are quick to combine like terms you should see that the algebra is very manageable here. You're given two fractions of the bill ( and of the total) and three actual values (for Shay subtract 7, for Ashley add 8, and for Jenna add 20). So what you really have when you combine all of your like terms is:

(where represents the total bill). Since $21 represents the missing of the bill, the bill is 4($21) = $84.

This problem can be solved algebraically but is likely solved more quickly by simply back solving. Your algebra would set up as:

(eight times one-third of a number) (is one greater than) (five times one half of that number).

Those who are algebraically inclined could of course find common denominator, combine the terms, and solve (more on that below). But others might be quicker looking at the answer choices. Since one of the first steps the problem tells you is that the number has to be divided by three ("eight times one-third of a number"), you'll want to plug in a number divisible by three for an easy start. And since you'll also have to divide that number by 2 ("five times one half that number"), you'll want an even one.

So start with 6. Eight times one-third of 6 is 8(2) = 16. Is that one greater than five times one half of 6? It is. Five times one half of 6 is 5(3) = 15. The relationship works, so 6 is correct.

To begin on this problem, first take of 480 to determine that there are currently 120 hybrid cars. So if half were to be sold, then that would leave 60 hybrids.

Importantly, note that if 60 hybrids are sold, that means that 60 of the total of 480 cars are sold. That affects both the number of hybrids AND the number of total cars, which is now reduced to 420. So your new proportion of hybrids is the 60 hybrids out of the 420 total cars:

Age problems derive much of their difficulty from the fact that people are often careless in accounting for time shifts. The current age equation here should be easy to set up: . But the "5 years ago" equation takes a bit of thought: 5 years ago, they were BOTH five years younger, so you need to account for that on both sides of the equation. Juliet, five years ago, was three times as old as Will, five years ago:

If you then substitute in for , you have:

Adding 15 to both sides and subtracting from both sides leaves:

And since the problem asked for Juliet's age, not Will's, you'll plug in to to get .

Mathematically you can express the value of Devin’s investment using the equation , where represents the number of years. You can do the same for Anita’s using . And then to put them together, recognize that you’re looking for when Anita’s real estate is worth as much as Devin’s car, so you’ll want to use the equation

Then combine like terms by adding and subtracting on each side:

The math here reduces quickly, as you can first divide both sides by :

And you know that , so the answer then must be .

This problem provides you with a classic SAT Systems of Equations setup. You're given the sum of two entities - here that's Blin's Total + Alex's Total = $240,000 - and a relationship between those two entities. Here that's Blin raised $60,000 more than Alex did.

The sum equation is generally straightforward; you can express that as B + A = 240,000. For the relationship equation, you can say that Blin's total is $60,000 more than Alex's, which translates to B = 60,000 + A.

Now you have two equations:

B + A = 240,000

B = 60,000 + A

This structure sets up well for Substitution; you already have B in terms of A, so you can plug in the second equation for the first:

(60,000 + A) + A = 240,000

60,000 + 2A = 240,000

2A = 180,000

A = 90,000

Of course, at this point you need to check whether you solved for the proper variable. The question asks you about Blin's total, not Alex's, so you need to add $60,000 to get to Blin's total of $150,000. Then check your work: does Blin's total of $150,000 plus Alex's total of $90,000 arrive at the sum we know to be $240,000? It does, so you can confidently choose $150,000.

This problem provides you with a classic SAT Systems of Equations setup: you get a sum equation about a combined total of two entities, and then an equation that gives a relationship between the two. Here you know that Judy and Nancy combined for 42 miles, which you can turn into an equation as:

J + N = 42

Then you're told that Judy walked twice as many miles as Nancy did. That equation is J = 2N (be careful about putting the two in the wrong spot - it's best to think about "what do I have to do to Nancy's total to make it equal Judy's?? You'd have to double it to catch up, so twice Nancy's total is what it takes to equal Judy's --> J = 2N).

Now you can substitute the second equation for the first. If you plug in 2N where the J appears in the first, you have:

2N + N = 42

3N = 42

N = 14

And since this problem asks you for Nancy's total, you have it: Nancy walked 14 miles.

This problem provides you with two equations, setting up nicely for a Systems of Equations approach. The first equation is about their combined total, and you can set it up as:

T + D = 932

Then the second equation is:

T = D + 28. (you can think of this as "to equal Tamika's total, David would have had to add 28 hours")

One helpful tip on this problem is that while it sets up nicely for a Substitution Method approach (substitute D + 28 for T in the first equation), you want to solve for T, not for D. So you can with one quick algebra step set it up to use the Elimination Method. Just subtract D from both sides in the second equation, and it becomes T - D = 28

Now you can stack your equations and sum them:

T + D = 932

T - D = 28

These sum to:

2T = 960

Divide both sides by 2 and you get T = 480.

This problem provides you with two equations, allowing you to use Systems of Equations logic to solve once you've constructed the equations. Often on the SAT, these Systems problems will start by giving you a combined total as this one does. If the two dogs weigh a total of 87 pounds, you can express that as:

R + S = 87 (in which R = Rover's weight and S = Spot's weight)

Then if Rover weighs 9 pounds more than Spot, you have a few choices for how to express that. One is to say that "Rover's weight is 9 pounds more than Spot's weight" and then use the fact that "is" means "equals" so that the words translate directly to the equation: R = 9 + S. Another way is to see that Rover weighs more, and the difference is 9, meaning that R - S = 9. The equation with subtraction allows you to go straight into the Elimination Method as you have a + S in the first equation and a -S in the second, so that may be your preferred choice. If you structure it that way, you can stack the equations and sum them:

R + S = 87

R - 9 = 9

Which sums to:

2R = 96

That means that R = 48, but remember in Systems of Equations problems to always check that your algebra has answered the specific question, as a problem with two variables allows you to mistakenly solve for the wrong one! The question asks you about Spot's weight, so you can subtract 9 from 48 to see that Spot weighs 39 pounds.

This problem provides you with the opportunity to set up two equations: the total number of slices of pizza, and the total amount of money. If you use V to represent the number of vegetarian slices and P to represent the number of pepperoni slices, you can express the total number of slices as:

V + P = 20

Then for the money, the total revenue will be a combination of how much was made from selling vegetarian slices ($1.25 times the number of vegetarian) and how much was made from selling pepperoni ($1.50 times the number of pepperoni). This equation is then:

1.25V + 1.5P = 28.25

This problem sets up nicely for using the Substitution Method for systems of equations. Since V is likely easier to work with in the second equation (the substituted term is likely easier to multiply by 1.5 than by 1.25), you can leverage the first equation to get P in terms of V:

P = 20 - V

When you plug this in to the second equation, you get:

1.25V + 1.5(20 - V) = 28.25

This expands to:

1.25V + 30 - 1.5V = 28.25

Combine like terms to get:

-0.25V = -1.75

So V = 7, meaning that P = 13.

Now as with any systems of equations problem, it's imperative that you go back to double check what the question wants. It's asking for the difference between the two types of pizza (how many more pepperoni slices were sold than vegetarian), so your last step is to subtract 13 - 7 to get the correct answer, 6.

This problem provides you with the opportunity to set up two equations: one, the number of total tickets and two, the revenue from selling all of those tickets. If you use B to represent the number of balcony tickets and F to represent the number of floor tickets, your equations can look like:

Number of tickets:

B + F = 500

Revenue (the number of each type of ticket multiplied by the price of that type of ticket):

30B + 55F = 20000

If you multiply the entire first equation by -30, it sets you up nicely to be able to use the Elimination Method. That gives you:

-30B - 30F = -15000

30B + 55F = 20000

Which sums to:

25F = 5000

You can then solve for F, which is 200. Of course, on any systems of equations problem you should always double check the question being asked, since with two variables there are multiple different questions they could ask you. Here they ask about the number of main floor tickets so you can safely answer 200.

In setting up a word problem like this, be sure to use meaningful variables. Here your variables are lattes (which you can call ) and coffees (which you can call ).

You can set up an equation to account for the $60 total: $6 times the number of lattes plus $3 times the number of coffees will account for the $60, so . You can then simplify by dividing everything by : .

You can also set up an equation based on the fact that there were 2 more coffees than lattes. The number of coffees, , is equal to 2 plus the number of lattes, , so:

With two equations and two variables, you can now solve. A quick way to do that is to subtract from both sides of the second equation:

If you then multiply both sides by 2, you have:

Which sets up for the elimination method when you stack and add the two equations:

This yields , so . When you plug that back in to , you see that , meaning that the total number of drinks is .

Your job on this word translations problem is to assign variables ( for cupcakes and for brownies) and then set up equations. You know that one equation is , since the number of cupcakes plus the number of brownies must equal 8. The other equation deals with the price; Jeremy spends a total of $34, and that cost is made up of the number of cupcakes he buys times $5 for each, and the number of brownies he buys times $4 each brownie. So that equation is .

With those two equations, you should see that they stack up nicely so that you can use the Elimination Method to solve. You have:

And then if you multiply the entire bottom equation by -4, you'll have:

If you then sum the equations, you're left with . But remember: the question asked you about brownies, not cupcakes, so be careful with the temptation to select 2 as your answer! If and , that means that .

In this word problem, it's first helpful to assign meaningful variables. Since the matinee equation is the "cleaner" of the two, you might start with:

Meaning that the number of adult tickets to the matinee and the number of children's tickets to the matinee.

Then look at the second equation. Since is the same for each equation you're done there, but then unpack the relationship between matinee children's tickets and evening children's tickets. If, for the evening show, the number was "20 less than double," that means that you'd double the earlier number of children's tickets and subtract 20 to get . That means that your second equation is:

So you have two equations now:

So use the Elimination Method and subtract the simpler, matinee equation from the more-complex, evening equation and you'll be left with:

So

But remember: C is the number of children's matinee tickets, so you'll need to convert to the number of children's evening tickets. That number is , which equals 80.

This word problem starts you off with two equations. If you assign variables as = bats and = balls, you have:

, which reduces to

and

You can then use the Elimination Method to eliminate the terms and solve for :

, so . Plug that back in, and . So now you have the prices.

At this point, you know that you need the same number of and , and that the total can only be as high as $240. If you call that same number , you have:

. So you'll be able to buy 4 bats and 4 balls.