Posts tagged with: heart of algebra

a = 4800 – 6t
b = 5400 – 8t

In the system of equations above, a and b represent the distance, in meters, two marathon runners are from the finish line after running for four hours and t seconds. How far will runner a be from the finish line when runner b passes her?

A. 300 meters
B. 500 meters
C. 100 meters
D. 3000 meters

 

Runner b will pass runner a when they are the same distance from the finish line, so set those distance expressions equal.

4800-6t=5400-8t
2t=600
t=300

But be careful, t is not a measure of distance, it’s time! The runners will meet at 300 seconds from the 4 hour mark. To find the distance they’ll be from the finish line at that time, plug 300 back into one of the original equations.

a=4800-6(300)
a=3000

So the answer is D.

Hi. Please, can you explain test 6,section 3,number 14. I get confused by “at least twice as many containers of detergents as many softeners.” I was thinking it was 2d< or = s. Please explain why it is not so. Thanks

“At least” is the same as “greater than or equal to.” Looks like you had that part. The rest of the syntax, though, might have been the issue. When one says “at least twice as many boys as girls,” that means the first thing listed, the number of boys, is the bigger number.

To figure out where to put the 2 that comes from the “twice,” plug in: If there are 3 girls, and at least twice as many boys, then there are at least 6 boys. Therefore, put the 2 on the girls: b\geq 2g.

To return to the question you’re asking, “at least twice as many containers of detergent as containers of fabric softener” means you write d\geq 2s.

That right there tells you the answer must be A or C. From there, you just need to identify that choice A is the one that gets the weights of the containers right.

Hi! Can you please show me how problem 8 is solved in test 2 section 3?

Sure! First, don’t get bogged down in the wall of text. You should read it, of course, but you should read it while remembering that the formula you’re going to need has already been given to you at the top of the question. A simplified version of the question is this:

nA = 360. If A > 50, what’s the greatest integer value of n?

I’ll suggest two ways to solve. First, you can backsolve. The question asks for the greatest number of sides, so start with the largest answer choice, 8, and see if you get a value of A that’s greater than 50 when you set n = 8.

8A = 360
A = 45

Nope, not quite. So try n = 7.

7A = 360
A = 51.4…

That works! So choice C is correct.

The mathier way is to use an inequality. First, get A by itself in the given equation:

A=\dfrac{360}{n}

Now, remembering that we know A must be greater than 50 and therefore \dfrac{360}{n} must be greater than 50, make an inequality.

50<\dfrac{360}{n}

50n<360

n<7.2

Since the question asked for the largest possible value of n, which can’t be a decimal because n represents a number of sides and a polygon can’t have a decimal number of sides, we go with 7.

Test 1 Section 4 #34

Lots of people miss this one because they don’t read carefully enough. The question asks for the total number of slots the station can sell on Tuesday AND Wednesday—two days! So we need to figure out how many 30-minute slots there are in a day, and then multiply by two.

There are 2 slots in an hour, and 24 hours in a day, so there are 48 slots in a day. Multiply that by 2 to cover Tuesday and Wednesday and you get the answer: 96. Below is how you’d set it up if you were being very careful with your units, which, incidentally, I strongly recommend.

\dfrac{2\text{ half-hour slots}}{1\text{ hour}}\times\dfrac{24\text{ hours}}{1\text{ day}}\times 2\text{ days}=96\text{ half-hour slots}

Test 3 Section 3 #19

Let’s say h is the number of calories in each hamburger, and f is the number of calories in each order of fries. The question gives us enough information to write two equations:

hf + 50 (each hamburger has 50 more calories than each order of fries)
2h + 3f = 1700 (2 hamburgers and 3 orders of fries have a total of 1700 calories)

Let’s solve that system by substitution, since we already have h in terms of f.

2(f + 50) + 3f = 1700
2f + 100 + 3f = 1700
5f + 100 = 1700
5f = 1600
f = 320

The question asks us for h, so now we solve for h:

hf + 50
h = 320 + 50
h = 370

T2 Sec3 #16

To solve this question, you basically need to make $3000 out of $250 and $750 increments. The thing to note right away is that $3000 and $750 are both multiples of $250, so as long as you choose a number of $750 bonuses that add up to less than $3000, you will be able to land on a whole number of $250 bonuses.

For example, let’s say only one $750 bonus is given out. Then there must be $2250 in $250 bonuses, which means there were \dfrac{2250}{250}=9 $250 bonuses.

Other acceptable answers:

  • Two $750 bonuses –> \dfrac{1500}{250}=6 $250 bonuses.
  • Three $750 bonuses –> \dfrac{750}{250}=3 $250 bonuses.

t2 s4 #35 please!

a = 18t + 15

The key to correctly interpreting questions like this one is paying attention to where the variable (t, in this case) is. The variable will be multiplied by the number that’s added every time. The variable will not be attached to the starting point. You want Jane’s initial deposit here—the starting point. That’s going to be 15. She started by depositing $15, and then every week after that she deposits $18.

I think it’s useful to note that this is really just the slope-intercept form of a line, usually shown as ymxb. In that equation, the variable is attached to m, which you know is the slope, or the rate of change. Another way of thinking about this is that every time x increases by 1, the y value increases by m. The starting point, b, is the value of y when x = 0 (also known as the y-intercept).

Test 2 Section 4 #34 Thanks!!

Let’s say that d = the number of hours Doug spent in the lab, and l = the number of hours that Laura spent in the lab. Here are the equations we can write from the question:

ld = 250
dl + 40

This question is begging for a little substitution. Let’s take the expression for d we have in the second equation and plug it into the first.

l + (l + 40) = 250
2l +40 = 250
2l = 210
l = 105

Q#22, S#4, SAT Test#3

This one is tricky! Here’s why: they talk about three numbers, but they really only give you enough information to write two equations, and they only ask for the value of one of the three numbers. So it looks way more complicated than it really is.

Let’s come up with some equations. First, the sum of three numbers is 855. Easy enough:

x+y+z=855

And then the second thing they tell you: x is 50% more than the sum of the other two numbers.

(y+z)\times\dfrac{50}{100}+(y+z)=x

That simplifies to (or maybe you prefer shorthand and just started by writing this) the following:

1.5(y+z)=x

Now we can solve by substituting or eliminating. I think substitution is a little easier here since we already have an x = equation:

1.5(y+z)+y+z=855

1.5(y+z)+1(y+z)=855

2.5(y+z) = 855

y+z=\dfrac{855}{2.5}

y+z=342

Now that you’ve got y + z, all you need to do is solve for x.

x+342=855

x=513

(Note that, while I didn’t do this in my explanation because I think it might have been confusing, you can also just use a single letter (like a) instead of yz to represent the sum of the other two numbers, since we never need to split them apart.)

Please explain question #8, section #4, SAT test #3! Thanks Mike!

The equation you’re given is y = 0.56x + 27.2, where y is the average number of students per classroom and x is the number of years since 2000. You’re asked what the number 0.56 represents in the equation.

Before you get too lost in the details of the particular model, recognize that y = 0.56x + 27.2 is really just a line in ymxb form! So without even thinking about average students per classroom or years, we know that 0.56 represents the line’s slope, or rate of change. Which answer choices refer to rates of change?

Really, only choice C does. A talks about a total number, B talks about an average number, and D talks about a difference (which is just one number minus another). Only C talks about an INCREASE, which is the kind of word we’re looking for to describe slope. So the answer must be C.

Test 1 Section 3 number 18. Thanks

I love this question—it’s a great argument for making the elimination method of solving linear systems a major part of your SAT approach. The question asks you for x, but the fastest way to get x is to solve for y first.

x+y=-9\\x+2y=-25

See how there’s an x in each equation? That means that you can eliminate x by subtracting the equations from each other.

(x+2y)-(x+y)=(-25)-(-9)\\x+2y-x-y=-25+9\\y=-16

Of course, once you know that y = –16, it’s easy to plug back into the first equation to solve for x.

x+(-16)=-9\\x=7

If n/5 + y/10 =1 what is the value of 2n+y?

Multiply everything by 10!

10\left(\dfrac{n}{5} + \dfrac{y}{10}\right)=10(1)

2n+y=10

A tank initially contains 7 gallons of water. A faucet is opened and water begins pouring into the tank at a rate of 1.5 gallons per minute until the tank is full. Which of the following represents the volume V of water in gallons in the tank as a function of time t in minutes that has elapsed since the faucet is opened?

A) v(t)=1.5+t
B)v(t)=8.5+t
C)v(t)=1.5t
D)v(t)=1.5t-7
E)v(t)=1.5t+7
Thanks

The correct answer needs to have 1.5t, because 1.5 gallons go into the tank every minute, and there are t minutes that the faucet is running. It also needs to have a +7, withoutt, because that original 7 gallons in the tank is the starting point—it doesn’t change with the amount of time the faucet has been running.

A good mental roadmap for a question like this is to start by noting how much was in the tank before the faucet was turned on, which is time 0. The question tells us that at time 0, there were 7 gallons in the tank, so we should be able to plug 0 in for t in the right answer and get 7. Only choice E does that, so we’re done. If that didn’t eliminate every choice, though, then the next step would be to figure out that after 1 minute of the faucet running, there should be 8.5 gallons in the tank. Of the choices not yet eliminated, which gives you 8.5 when t = 1? Etc.

If x-y+z=10 and -x+y=3 what is the value of z?

A) 13
B) 7
c) 10/3
D) -7
E) -13

xy = 3 can be rewritten: –(xy) = 3. That means that x – y = –3.

So xy + z = 10 can be rewritten: –3 + z = 10. That means z = 13.

The new SAT places a heavy emphasis on the “Heart of Algebra,” which is a bizarre and tortured euphemism for, mostly, working with linear equations. One of the kinds of questions you know you’re going to see, probably more than once, on your SAT is solving systems of linear equations. For example:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

There are a bunch of good ways to solve such a problem, and I think you should know all of them. So, I’m going to talk about all of them. That’s what you come here for, after all.

Substitution

To solve by substitution, first you’ll want to get one of the variables in one of the equations alone. It doesn’t matter which variable, and it doesn’t matter which equation, so pick the one that looks easiest to isolate. I’m gonna get the x from the first equation by itself. First, subtract the y-term from each side…

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{3}x=12-\dfrac{1}{6}y

Then multiply by 3 to get x alone…

3\left(\dfrac{1}{3}x\right)=3\left(12-\dfrac{1}{6}y\right)

x=36-\dfrac{3}{6}y

x=36-\dfrac{1}{2}y

Once I’ve got x alone, I’m going to substitute what I now know x equals into the other equation. This is important! You won’t get anywhere if you substitute back into the equation you just manipulated—you’ll eventually just end up at 0 = 0. Right, so:

x=36-\dfrac{1}{2}y

2x+5y=21

2\left(36-\dfrac{1}{2}y\right)+5y=21

Then I’m going to solve that equation for y:

72-y+5y=21

72+4y=21

4y=-51

y=-\dfrac{51}{4}

Once I’ve got my y I can actually look at the answer choices and see that I’m able to eliminate every choice but A. If I were in a hurry, I’d bubble that and move on. Since we’re practicing, though, I’m going to finish this process by putting the y I just found into one of the previous equations to get x. I like to use the equation I already solved for x to save me a step or two:

x=36-\dfrac{1}{2}y

x=36-\left(\dfrac{1}{2}\right)\left(-\dfrac{51}{4}\right)

x=36+\dfrac{51}{8}

x=\dfrac{288}{8}+\dfrac{51}{8}

x=\dfrac{339}{8}

So, there you have it. Choice A is for sure the answer: \left(\dfrac{339}{8},\dfrac{-51}{4}\right) is a solution for that system of equations.

Elimination

Substitution is the classic, the mainstay. Most people know how to do it and are comfortable with it. Elimination, to my constant amazement, seems less universally known. I say to my amazement because elimination is awesome.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

To solve the same problem with elimination (I’ve pasted it in above so you don’t have to scroll to see it), what you’ll want to do is instead of trying to isolate any variables, find a way to multiply one of the equations by something so that either the x-coefficients in each equation are equal, or the y-coefficients are equal. I see an easy way to do that:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

6\left(\dfrac{1}{3}x+\dfrac{1}{6}y\right)=6(12)

\dfrac{6}{3}x+\dfrac{6}{6}y=72

2x+y=72

That’s what I’m talking about! I’ve got a 2x there, and a 2x in the other equation: 2x+5y=21. Now all I need to do is subtract one equation from the other, eliminating the 2x making it very easy to solve for y:

     \begin{align*} 2x+y &=72\\ -(2x+5y&=21)\\ -4y&=51\\ y&=-\dfrac{51}{4} \end{align*}

Easy, right? Now to find x, we just substitute -\dfrac{51}{4} in for y in either equation:

2x+5y=21

2x+5\left(-\dfrac{51}{4}\right)=21

2x-\dfrac{255}{4}=21

2x-\dfrac{255}{4}=\dfrac{84}{4}

2x=\dfrac{339}{4}

x=\dfrac{339}{8}

Cool, right? Unsurprisingly, because math works, we landed on the same answer with both methods. But we’re not done yet, folks. There are still more ways to solve a system of linear equations!

Graphing

Always remember, as you’re taking the SAT, that when two graphs intersect, they do so at a point (x,y) that is a solution set for the two equations that make those graphs. If you’re in the section where calculators are allowed (and with the numbers in this question we’re playing with, you would be), then you have still more weapons for solving systems of equations at your disposal.  Let’s look at that question again.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

To solve a system of equations by graphing, first get each equation into y= form:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{6}y=12-\dfrac{1}{3}x

y=72-2x

2x+5y=21

5y=21-2x

y=\dfrac{21}{5}-\dfrac{2}{5}x

Pop those into your calculator and graph! If your window is currently set to standard zoom, you’ll probably only see one line. You should expect this to happen from time to time—the writers of the SAT would love to see you doing the algebra, not taking graphing shortcuts. Try zooming out a bunch by setting your window to go from –100 to 100 in both axes, instead of the standard –10 to 10.

Now just use your calculator’s intersect function! On a TI-83 or TI-84, you’re going to hit [2nd][TRACE] to open up the CALC menu, and then select “intersect.” Now just hit ENTER on the first line (your calculator asks for the “first curve,” but that’s only because this same function will also find intersections of non-linear functions), then hit ENTER on the second line. The calculator asks for a guess next—you don’t need to do anything here but hit ENTER. You should see something like this:

Of course, those are the decimal values of the answers we already know are correct:

\left(\dfrac{339}{8},-\dfrac{51}{4}\right)=(42.375,-12.75)

Backsolving

Finally, a multiple choice question like this can be solved by backsolving. All you have to do is try answer choices by making sure they work in both equations. Only the correct answer choice will result in true outcomes when substituted into the given equations.

\dfrac{1}{3}x+\dfrac{1}{6}y=12

2x+5y=21

Which ordered pair (x,y) satisfies the system of equations above?

A) \left(\dfrac{339}{8},-\dfrac{51}{4}\right)

B) \left(18, 36\right)

C) \left(\dfrac{1}{2},4\right)

D) \left(-3,\dfrac{29}{5}\right)

Here’s what happens when you backsolve with a wrong answer choice (I’ll use choice B):

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\dfrac{1}{3}(18)+\dfrac{1}{6}(36)=12

6 + 6=12

12=12

That worked for the first equation. But let’s see what happens in the second equation:

2x+5y=21

2(18)+5(36)=21

36 + 180 =21

216=21

Nope—that’s not true at all!

The right answer, though, will work beautifully:

\dfrac{1}{3}x+\dfrac{1}{6}y=12

\left(\dfrac{1}{3}\right)\left(\dfrac{339}{8}\right)+\left(\dfrac{1}{6}\right)\left(-\dfrac{51}{4}\right)=12

\dfrac{113}{8}-\dfrac{17}{8}=12

\dfrac{96}{8}=12

12=12

Yep!

2x+5y=21

2\left(\dfrac{339}{8}\right)+5\left(-\dfrac{51}{4}\right)=21

\dfrac{339}{4} - \dfrac{255}{4} =21

\dfrac{84}{4}=21

21=21

Yep again! That’s a good answer.

Of course, I’m showing a lot of intermediate steps you won’t need to do yourself. All you’ll want to do is type the substituted left-side of the equation into your calculator and see what you get. For example, you might enter

2(339/8)+5(-51/4)

When you hit ENTER and get 21, which is what the question said you should get, then you know that ordered pair works for that equation.

Now you try

Think you’ve got all this? I bet you do. But just so that we can both sleep a little better tonight, why don’t you try a few more questions. Because this is practice and all, you might consider trying to solve all of these questions all 4 ways.

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