Posts tagged with: translating words into math

Can you help me with Question #29 in Practice Test 3, Section 4? I found the answer by subtracting multiples of 9 from 122 to find one that was divisible by 5. Is there a better way?

 
The way I like to go here is to set up equations. If we say left-handed females are x and left-handed males are y, then we can fill in the table like this:

Now we have a system we can solve!

x+y=18
5x+9y=122

We eventually want a number of right-handed females, so let’s solve for x

5x+9(18-x)=122
5x+162-9x=122
40=4x
10=x

Remembering that x is the number of left-handed females, now we just need to multiply by 5 to get the right-handed females: 50.

The last trick here is that the question asks for a probability that a right-handed student being picked at random is female. There are 50 female right-handed students and 122 right-handed students overall, so the probability is \dfrac{50}{122}\approx 0.410.

d = 30 + 2(40 – s)

The machine begins the day with $30 inside, so that’s the “30 +” part. Easy enough.

The variable s is defined as how many sodas the machine has in it, but what we really care about is how many sodas are sold. We know the machine begins the day with 40, so 40 – s should give us the number of sodas sold. (When s = 40, no sodas have been sold; when s = 35, 5 sodas have been sold…)

For each soda that’s sold, the machine should have $2 more, so that’s why “2(40 – s)” is in there.

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You can make two equations here. First, you know the total number of marbles is 103, so:

The second equation is more complicated, so let’s do it in parts. First, he gives away 15 red marbles, so he should have r – 15 left. He gives away 2/5 of his blue marbles, so he should have b – 2/5b = 3/5b left.

So the ratio of red marbles he has left to blue marbles he has left (which the question tells us is 3/7) should be:

The question asks how many blue marbles he had originally, so let’s substitute and solve for b. First get r by itself in the first equation:

Now substitute that into the second equation and solve:

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This question comes from my own book, so my tips on how to deal with these can be found in the same chapter. The main key to getting it right is making sure you translate the words into math correctly.

Note that although the question tells you that Tariq makes brownies and Penelope makes cookies, in the end it only asks about “treats,” so we can lump cookies and brownies together.

Tariq makes 30 treats per hour and Penelope makes 48 treats per hour. Together, then, they make 78 treats per hour. We know they both worked for the same amount of hours.

The other key to getting this right is keeping track of the units of the numbers you know. In this case, we have treats and hours for units. We know the number of total treats, and we know the rate of treats per hour. We want the number of hours. How do we set up the equation we need to solve? We need to divide the total number of treats, 312, by the number of treats they made per hour, 78.

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Pilar is a salesperson at car company. Each car costs at least $15,000. For each car she sells, she gets 6% commission of the amount by which selling price exceeds $10,000. If Pilar sells a car at d dollars, which function gives her the commission in dollars on sale?
A) C(d)=0.06(d-10000)
B) C(d)=0.06(d-15000)
C) C(d)=0.06(10000-d)

Plugging in might help you think about this in a more concrete way. From what the question says, if Pilar sells a car for $18,000, for example, then we’d expect her to earn commission on $8,000—the amount of the car’s price above $10,000. A 6% commission on $8,000 is 0.06\times $8,000=$480. Which of the answer choices, when you plug in $18,000 for d, gives you $480?

Choice A is the only one that works.

The other way to think through this is to notice that all the choices have the same 0.06 in the beginning, so the 6% part of the problem is taken care of. Our job is to figure out which of the choices has the right thing in the parentheses. Which of those things will provide the amount that d, the selling price, exceeds $10,000? Well, translating the words into math, we’d have to say that “the amount d exceeds $10,000″ can be written as: d – 10,000.

One way to make sure you get questions like these right is to plug in some values to see which equation makes sense. For example, you might choose to plug in 0 for h here because you know that at zero feet above sea level the boiling point should be 212° F.

Choices C and D don’t give you 212 when h = 0, so they’re definitely wrong!

Now plug in 1000 for h. We should expect the right equation to do what the question says—the boiling point should be (212 – 1.84)° F = 210.16° F. Which remaining choice, A or B, does that when you plug in 1000 for h?

Choice A gives you a crazy low number: 212 – 1.84(1000) = –1628.

Choice B does exactly what you want: 212 – (0.00184)(1000) = 210.16

So the answer is B.

To get this without plugging in, you should think about the elements of the language you’re translating into math. You want to start at 212, and subtract 1.84 degrees for every thousand feet (h/1000), so you might write this to start:

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From there, a little manipulation lands you on the right answer choice:

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My recommendation, though: plug in. With a little practice you’ll get very fast at it, and then questions like this go from head scratchers to gimmies.

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Hello! Could you please explain how to answer Question #16 from Section #3 from official practice SAT test #5? Thank you in advance!

She only has to take the water safety course ($10) once, so you can really think about her budget as the $280 the question provides minus the cost of the safety course: $280 – 10 = $270.

The question is asking how many whole hours Maria can rent the boat, at $60 per hour, and stay within her budget. In other words, how many times does 60 go into 270?

60\times 4=240 <– Works.

60\times 5=300 <– Too big!

Therefore, Maria can afford to rent the boat for 4 hours—the answer is 4.

This no-calculator prompt is from the Daily Practice app:

Lisa gives her little brother Sam a 15-second (sec) head start in a 300-meter (m) race. During the race, Sam runs at an average speed of 5 m/sec and Lisa runs at an average speed of 8 m/sec, not including the head start.

Which of the following best approximates the number of seconds that had passed when Lisa caught up to Sam?

A. 5
B. 25
C. 40
D. 55

In 15 seconds at 5 m/sec, Sam will be 75 m ahead when Lisa starts running. Once she starts running, Lisa is going to gain 3 m/sec on Sam. How many seconds will it take to cover that 75 m gap at 3 m/sec?

    \begin{align*}\frac{75\text{ m}}{3\text{ m/sec}}=25\text{ sec}\end{align*}

The trick to this question is recognizing that you’re being asked (at least I think you’re being asked—the wording is not very precise) how many seconds have passed when Lisa catches up to Sam since the race started. You have to include Sam’s 15 second head start. Sam ran for 15 seconds, then both Lisa and Sam ran for 25 seconds, then Lisa caught up to Sam. Total elapsed time: 40 seconds.

Test 8 Section 3 #13

You’re translating from words into math here. Key words to note: “at a constant rate” means you’re dealing with something linear, that has a slope. And “decreased at a constant rate” means you’re looking for a negative slope.

Because the function is a function of t, years are your x-variable and and millions of barrels is your y-variable. Calculate the slope from the numbers the question provides: 4 million barrels in 2000 and 1.9 million barrels in 2013.

    \begin{align*}\text{Slope}&=\dfrac{4-1.9}{2000-2013}\\&=\dfrac{2.1}{-13}\end{align*}

Note that while no answer choice has that slope in that exact form, choice C is equivalent. \dfrac{2.1}{-13}=-\dfrac{21}{130}.

Because all the answer choices have the same y-intercept of 4, you need not think much about it, but it does make sense given the question. There were 4 million barrels in 2000 and t represents years after 2000, so t = 0 when the y-variable is 4.

Test 4 Section 3 #20 please. Also Mike, I still didn’t know how to answer the question even after I went and reviewed it from the official test breakdown section of your book.

There are a lot of words here, but when you strip that all away, this is basically a question about slope. You know you have a change in distance from the earth’s surface of 50 km to 80 km, and a change in temperature from –5° C to –80° C.

The big trick is that we’re told we need to find the degrees by which the temperature changes every 10 km, so we should think about our distance values in tens. That is, instead of 50 km and 80 km, we should use 5 and 8 and our units will be tens of km.

So let’s calculate the change in temperature in degrees Celsius per change in 10 km distance.

    \begin{align*}&\dfrac{\text{Change in Temperature in degrees Celsius}}{\text{Change in Distance in 10 kilometer units}}\\\\=&\dfrac{-5-(-80)}{5-8}\\\\=&\dfrac{75}{-3}\\\\=&-25\end{align*}

So there you go. Every 10 km further you go from the Earth’s surface, the temperature drops by 25° C.

Test 7 Section 3 #14

OK, so in this question we’re told that this company restricts package sizes such that the height of the box plus the perimeter of the base of the box cannot exceed 130 inches. The words “cannot exceed” can be translated to “is less than or equal to.” A peek at the answer choices shows that each one has a “0 <” at the beginning (because boxes can’t have negative dimensions), but to keep our work simple for now we’ll just incorporate that at the end.

    \begin{align*}(\text{Height})+(\text{Perimeter of Base})\le 130\end{align*}

Then we’re told that we have a box with a height of 60 inches and an unknown base perimeter. So the first step here is to recognize that we’ve got 70 inches to play with for the base perimeter.

    \begin{align*}60+(\text{Perimeter of Base})\le 130\\(\text{Perimeter of Base})\le 70\end{align*}

Now let’s deal with the perimeter. The perimeter of a shape is the sum of the lengths of its sides. For a rectangular box with width w and length l, the perimeter will be 2w + 2l. In this particular case, we’re told that the width is and the length is 2.5x, so the perimeter should be 2x + 2(2.5)x, which can be simplified.

    \begin{align*}(\text{Perimeter of Base})\le 70\\2x+2(2.5)x\le 70\\2x+5x\le 70\\7x\le 70\\x\le 10\end{align*}

Now include the “0 <” we discussed up top and you’re at choice A.

Test 3 Section 4 #30

This one is yucky. Let’s attack it by substituting. First, we’re told that b=c-\frac{1}{2}, so let’s reflect that (I’m going to use the decimal version of \frac{1}{2} for simplicity):

    \begin{align*}3x+(c-0.5)&=5x-7\\3y+c&=5y-7\end{align*}

Now, let’s solve both equations for c:

    \begin{align*}3x+(c-0.5)&=5x-7\\c-0.5&=2x-7\\c&=2x-6.5\\\\3y+c&=5y-7\\c&=2y-7\end{align*}

Because we now have two different equations with c isolated, we can substitute again here to get an equation that only contains xs and ys.

    \begin{align*}2x-6.5&=2y-7\\2x&=2y-0.5\\x&=y-0.25\end{align*}

That translates to “x is y minus \frac{1}{4},” so the answer is A.

 

Test 3 Section 4 #24

Of course, we could use algebra to solve this (and I will down below), but because the answer choices are numbers that can easily be dropped back into the word problem, my recommended strategy is backsolving.

Start by trying C; assume 23 students in the class. If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. 3(23)+5=74. For Mr. Kohl to give each student 4 ml, he’ll need an additional 21 ml. 4(23)=92. Is that 21 higher than 74? It is not—it’s only 18 more.

That suggests we need even more kids in the class, so the answer is almost certainly D, which says there are 26 students. If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. 3(26)+5=83. For Mr. Kohl to give each student 4 ml, he’ll need an additional 21 ml. 4(26)=104. Is that 21 higher than 83? Yes! D is the answer.

Now, the algebra. What we want to do here is come up with two expressions that are both equal to n, the amount of solution Mr. Kohl has. I’ll use x for the number of students he has.

If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. n=3x+5

In order to give each student 4 ml, he will need an additional 21 ml. n=4x-21

Now that we have those equations, we can solve by substitution.

    \begin{align*}3x+5&=4x-21\\5&=x-21\\26&=x\end{align*}

Can you help with 33 in test 7 section 4?

Yep! This one is all about setting up the equations—translating words into math. The question tells us that the score in the game is calculated by subtracting the number of incorrect answers from twice the number of correct answers. Let’s say x is the number of correct answers and y is the number of incorrect answers. “Subtracting the number of incorrect answers” is easy enough: you’re going to have a ” – y” in your score equation. “Twice the number of correct answers” is pretty straightforward, too: if the number of correct answers is x, then twice that is 2x. So we can say that:

Score = 2xy

We’re not done though. We need to write another equation before we can solve. The question tells us the total number of questions answered; we need to use that information. The total number of questions answered must be the number of questions answered correctly plus the number answered incorrectly, right?

Questions = xy

Now we can plug in he values we know and solve the system of equations.

50 = 2xy
40 = xy

The question asks for the number the player answered correctly, so we need to solve for x. Conveniently, we can solve for x by eliminating y—all we need to do is add the equations together!

    50 = 2x – y
+ (40 = xy)
    90 = 3x

Of course, if 90 = 3x, then 30 = x. 30 is the answer.

Can you do test 4 section 4 number 35 please?

This is a very SAT-ish question: the math is pretty trivial, but conceptually the question is still a bit tricky.

There are a couple things you need to have nailed down to get this one right.

First, you need to recognize that they’re asking you for a ratio of two dynamic pressures, so they’re asking you for a ratio of two q‘s. Let’s call them q_1 and q_2. We’ll say q_1 is the one that corresponds to a velocity of v and q_2 is the one that corresponds to a velocity of 1.5v.

Second, you need to make sure you’re providing the ratio that’s asked for: the q of the faster fluid to the q of the slower fluid. Which fluid is faster—the one with velocity v or the one with velocity 1.5v? 1.5v is always going to be a larger number than v, so that’s the faster fluid. Therefore, we need to calculate the ratio of q_2 to q_1.

    \begin{align*}\dfrac{q_2}{q_1}&=\dfrac{\frac{1}{2}n\left(1.5v\right)^2}{\frac{1}{2}nv^2}\\\\&=\dfrac{(1.5v)^2}{v^2}\\\\&=\dfrac{2.25v^2}{v^2}\\\\&=2.25\end{align*}