Posts tagged with: Plug In

How do you solve number 2 on page 28 with the plugging in method?

Here’s the question (from the Plugging In chapter in the Math Guide):

The way to go here is to plug in values to evaluate choices, which makes this a bit of a hybrid between a plugging in question and a backsolving question. Use the answer choices to guide your plugging in.

For example, choice A tells us to try x < 0. That turns out to be a good idea. Say x = –2 and y equals, oh, I dunno, 3. That fits the conditions: xy and –3xy. Can you come up with any values of x greater than 0 where that would work? No? Well, maybe A is correct, then! But just for the moment, let’s try to eliminate the other choices to reassure ourselves.

What about choice B, though? Wouldn’t x = –2 still work if y = –1? Sure would, so we can eliminate that one.

Can we plug in values that would eliminate choice C? Sure can! In fact, x = –2 and y = 3, which we just used above when we were considering choice A, show that choice C doesn’t need to be true.

Does choice D have to be true? Well, it’s true when x = –2 and y = 3, but what if x = –2 and y = 5 instead? Then xy would be true, and –3xy would be true, but (–2)^2 is not greater than 5, so choice D doesn’t need to be true.

By trying a few quick numbers informed by the answer choices, we can eliminate B, C, and D, leaving only choice A standing. We’re done!

Let’s work backwards, and use hours instead of distance for ease (assume the plane travels at a constant speed for our purposes).

Let’s say when she woke up, she had 1 hour left in her flight. That’s half of the time she was sleeping so she must’ve slept for 2 hours.

She first fell asleep halfway through the flight, so she was awake for the first 3 hours, then slept for 2, then was awake for the last 1.

She was asleep for 2/6 = 1/3 of the trip.

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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.

Practice test 8 Calculator #13

First, you can plug in on this one, so if you feel rusty on your exponent rules at all, that’s a good move. Especially on the calculator section. Say, for example, that you plug in 4 for a. Just enter it all into your calculator (you may need to be careful with parentheses in the exponent depending on the kind of calculator you have):

    \begin{align*}4^{-\frac{1}{2}}&=x\\0.5&=x\end{align*}

Now that you know x, plug 0.5 into each answer choice to see which one gives you 4.

A) \sqrt{0.5}\approx 0.707

B) -\sqrt{0.5}\approx -0.707

C) \frac{1}{0.5^2}=4

D) -\frac{1}{0.5^2}=-4

Obviously, C must be the answer.

To solve this algebraically, first start by squaring both sides. Raising a power to a power is the same as multiplying the powers, so that’ll get rid of the 1/2 on the left:

    \begin{align*}\left(a^{-\frac{1}{2}}\right)^2&=x^2\\a^{-1}&=x^2\end{align*}

Now raise both sides to the –1 power to get a truly alone. Remember that a negative exponent is the same as 1 over the positive exponent, so you can transform the right hand side from x^{-2} to \frac{1}{x^2} to finish the problem.

    \begin{align*}\left(a^{-1}\right)^{-1}&=\left(x^2\right)^{-1}\\a&=x^{-2}\\a&=\frac{1}{x^2}\end{align*}

A triangle’s base was increased by 15%. If its area is increased by 38%, what percent was the height of the triangle increased by?

The easiest way to get this question is to plug in! Say the base and height of the original triangle are each 10. The formula for finding the area of a triangle is A=\frac{1}{2}bh, where b and h are the base and height, so the area of our original triangle is A=\frac{1}{2}(10)(10)=50.

Increasing the base by 15% brings it from 10 to 10\times 1.15=11.5. Increasing the area by 38% brings it from 50 to 50\times 1.38=69. Plug those back into the formula to solve for the new height:

    \begin{align*}69&=\frac{1}{2}11.5h\\138&=11.5h\\12&=h\end{align*}

If the original height was 10 and the new height is 12, then the height increased by 20%.

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:

image

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

image

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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1/2 x = a

x + y = 5a

In the system of equations above, a is a constant such that 0 <a <1/3. If (x,y) is a solution to the system of equations, what is one possible value of y?

(Answer: 0 < x < 1)

The easiest way to go here is to plug in! Let’s say a = 0.1, which is certainly between 0 and 1/3.

The first equation tells us:

    \begin{align*}\frac{1}{2}x&=0.1\\x&=0.2\end{align*}

From there, we can solve for y in the second equation:

    \begin{align*}0.2+y&=5(0.1)\\0.2+y&=0.5\\y&=0.3\end{align*}

So there you go—grid in 0.3 as a possible value of y and move along.

 

To see the whole range of possible answers, plug in the endpoints you’re given for a.

If a = 0, then the first equation tells us that x = 0, too.

If x and a both equal zero, then the second equation becomes 0 + y = 5(0), which of course means y = 0, too!

OK, now plug in 1/3 for a. The first equation tells us:

    \begin{align*}\frac{1}{2}x&=\frac{1}{3}\\x&=\frac{2}{3}\end{align*}

Easy enough to solve for y in the second equation:

    \begin{align*}\frac{2}{3}+y&=5\left(\frac{1}{3}\right)\\\frac{2}{3}+y&=\frac{5}{3}\\y&=\frac{3}{3}\\y&=1\end{align*}

So if a = 0, y = 0, and if a = 1/3, y = 1. That’s why the full range of possible answers is 0 < y < 1.

Hi. Would it be possible for you to explain #13 on SAT practice test 5 on calculator inactive (the tea bag problem)? I think the main reason I found it confusing was the wording, and the SAT explanation for the answer was also a little bit wordy.
Thanks

I’ll try to be as un-wordy as possible! 🙂

I think plugging in numbers helps here. Say the restaurant wants to make 20 cups of tea one night, so n=20. The equation t=n+2 tells us that they’d use t=22 tea bags to make 20 cups of tea.

Now, what if they wanted to make one more cup of tea (in our example, 21 cups instead of 20 cups)? Well, the equation tells us they’d use 23 tea bags to do that:

    \begin{align*}t&=n+2\\t&=21+2\\t&=23\end{align*}

So they needed 22 bags to make 20 cups, and 23 bags to make 21 cups. That’s one extra bag for one extra cup.

The other way to think about this is that they’re just giving you a linear equation and asking for the rate of change (i.e., the slope). Think of the equation t=n+2 as y=x+2. That’s a line with a slope of 1 (and a y-intercept of 2, but we don’t need that for this question). A slope of 1 means for every change in x, you get an equal change in y. In other words, for every change in cups of tea, you get an equal change in tea bags.

Do either of those explanations help?

If f and g are functions, where f(x) = x^3 -10x^2 +27x – 18 and g(x) = x^3 – x^2 – 6x, which of the following gives a relationship between f and g?

A) g(x) = 3 f(x)
B) g(x) = f(x) – 3
C) g(x) = f(x) + 3
D) g(x) = f(x – 3)
E) g(x) = f(x + 3)

Can you solve w/o graphing?

Yes!

Note that the first term in both functions is x^3, so the relationship isn’t multiplication. Eliminate choice A.

Note also that it’s obvious that g(x) isn’t just 3 bigger or 3 smaller than f(x), so you can also eliminate choices B and C without any real work.

From there, it gets a little tricky. Focus on the constant terms, though. Note that f(x) has a –18 in it that goes away in g(x). Note also that when we plug x-3 or x+3 in for x, we’ll have a bunch of binomials to expand, but the only way that –18 goes away is if those binomial expansions spit out a positive 18. Without doing all the math, you should be confident that choice E is the one that will do that. To test it quickly, though, set x=0 (since this does need to be true when x=0).

    \begin{align*}g(0)&=f(0+3)\\g(0)&=f(3)\\0^3 - 0^2 - 6(0)&=3^3 -10(3)^2 +27(3) - 18\\0&=27-90+81-18\\0&=0\end{align*}

Yep! That worked nicely.

Here are a couple questions from the old official SAT Subject Test Math I practice exam:

The function f is defined by f(x) is x^4 – 4x^2 + x + 1 for -5 ≤ x ≤ 5. In which of the following intervals does the minimum value of f occur?
A) -5 ≤ x ≤ -3
B) -3 ≤ x ≤ -1
C) -1 ≤ x ≤ 1
D) 1 ≤ x ≤ 3
E) 3 ≤ x ≤ 5

Can you solve w/o graphing?

Yes, but just before beginning I think it’s important to stress HOW USEFUL a graphing calculator is for the math Subject Tests. Anyone who’s prepping for those that doesn’t own one and can’t borrow one from school should find another way to get their hands on one they know how to use for test day.

To get this without graphing, you’re going to want to plug in values Just the integers should work. Note that x^4 will always be 0 or positive and -4x^2 will always be 0 or negative, but x will be both positive and negative, you should start by plugging in negative numbers.

(-5)^4-4(-5)^2-5+1=521

(-4)^4-4(-4)^2-4+1=189

(-3)^4-4(-3)^2-3+1=43

(-2)^4-4(-2)^2-2+1=-1

(-1)^4-4(-1)^2-1+1=-3

(0)^4-4(0)^2-0+1=-1

From there, you probably see that the minimum is around –1. So plug in a couple more values to see whether you want to choose B or C.

(-1.5)^4-4(-1.5)^2-1.5+1=-4.4375

(-0.5)^4-4(-0.5)^2-0.5+1=-0.4375

Yeah…gonna want to go with B.

Again, the graph is SO helpful here:

With that, you immediately see that the minimum is between –1 and –2.

Test 8 Section 4 #23

I think the easiest way to get this one is to plug in some numbers. We know that every 4 quarters is a year, so the correct answer will be the one that gives us the same result for 1 year as 4 quarters, 2 years as 8 quarters, 3 years as 12 quarters, etc.

At 1 year, the original equation works out thusly:

    \begin{align*}M&=\num{1800}(1.02)^t\\M&=\num{1800}(1.02)^1\end{align*}

Which answer choice(s) result in the same expression when q = 4? Only choice A does:

    \begin{align*}M&=1800(1.02)^\frac{q}{4}\\M&=1800(1.02)^\frac{4}{4}\\M&=1800(1.02)^1\end{align*}

If you’re concerned that another one might also work, just check with your calculator!

    \begin{align*}1800(1.02)^1&=1836\\1800(1.02)^\frac{4}{4}&=1836\\1800(1.02)^{4(4)}&\approx 2471\\1800(1.005)^{4(4)}&\approx 1950\\1800(1.082)^4&\approx 2467\end{align*}

Sure enough, only choice A gives you the same result after 4 quarters that the original equation gives you for 1 year.

Hi Mike
For question 13 test 1 no calculator, I used plugging in. I made x = 4 and solved to get 42/13. then I plugged 4 into my answer choices and B gave me 42/13.
I am curious as to why you did not use plug in for your answer and explanation.

(Link to the referenced solution.)

Good question. I definitely have this one categorized as plug-innable in my book, but because there’s some algebraic manipulation required whether you plug in or not, I guess I chose to take it as an opportunity to push that skill. Your solution (elaborated below) is also totally valid, and would definitely be my preferred method on a section where calculators were allowed.

    \begin{align*}\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}}\\\\=\dfrac{1}{\dfrac{1}{4+2}+\dfrac{1}{4+3}}\\\\=\dfrac{1}{\dfrac{1}{6}+\dfrac{1}{7}}\\\\=\dfrac{1}{\dfrac{7}{42}+\dfrac{6}{42}}\\\\=\dfrac{1}{\left(\dfrac{13}{42}\right)}\\\\=\dfrac{42}{13}\end{align*}

Now set x=4 in each answer choice to see which one simplifies to \dfrac{42}{13}:

A) \dfrac{2(4)+5}{4^2+5(4)+6}=\dfrac{13}{42}

B) \dfrac{4^2+5(4)+6}{2(4)+5}=\dfrac{42}{13}

C) 2(4)+5=13

D) 4^2+5(4)+6=42

It might be a stupid question, but on exercise 6 about plugging in, I plugged in 3 for x, and none of the answer choices were correct. Letter c was the closest, but still, 3 raised to 6 equals 729; minus 9 equals 720. okay, but the result for letter c, if y=9, is 72.
So, I wanted to ask you if, on this specific exercise, the only option to get the right aswer is plugging in 2 for x.
(sorry if I said any terms in the wrong way, engish is not my first language)

Here’s the question for everyone else playing along:

I think the problem is that you’re squaring x instead of cubing it. If you’re plugging in x = 3, then you should have x^3=3^3=27 for y and x^6-x^3 should be 3^6-3^3=729-27=702.

So you’re looking for an answer choice that equals 702 given y = 27. Choice C works: 27(27 - 1) = 702.

For the question 8 about plugging in
I plugged in 30 for the number of desks, and so e=25 and c=48. But, it didn’t match any of the answer choices. I don’t understand what went wrong. I checked it over and over again. Pleease help aaaaa

Thanks for the question, but I think you mean #10? This one?

Let’s go step by step, starting with your choice of 30 for the number of desks. Five desks are not occupied, so you’re correct that there are 25 employees. So far so good.

All but two employees have two chairs. So 23 employees have 2 chairs each—46 chairs. Here’s the tricky part: all other desks, whether occupied or not, have one chair. That means the 2 employees that don’t have two chairs plus the 5 empty desks each have a chair at them. So there are 46 + 2 + 5 = 53 chairs.

Which expression is equal to 53? Choice D works: 2e + 3 = 2(25) + 3 = 53.

Could you please explain Test 2 Calculator Active #25

Sure thing. The best way to see what’s going on with this one is to plug in. When they tell you that ab = 0 and a ≠ b, they’re telling you that a and b must be nonzero opposites (i.e., that a = –b and that a ≠ 0). So just pick any nonzero number and rock and roll!

I’ll pick 3 for a, which means I’m picking –3 for b. Therefore, the graph passes through (3, 0) and (0, –3). What do we know about its slope? (Don’t just imagine it in your head—draw it in your test book!)

That’s a positive slope! Conveniently, finding a positive slope for even one set of values of a and b is enough to eliminate choices B, C, and D: a positive slope is not negative, zero, or undefined!

If you’re nervous though, maybe see what happens if you set a = –3 and b = 3 instead. Then your line would go through (–3, 0) and (0, 3). That’s still a positive slope!