A,C, and D

To solve this question, plug in the answers.Choice (A) works: If the veggie burger costs $2.25 and the shake costs $1.50, then 2($2.25) + 2($1.00) = $6.50.Choice (B) does not work, since 2 veggie burgers and 2 beef burgers cost $7.00, but 2($2.25) + 2($1.75) = $8.00.Eliminate choice (B).Choice (C) works, since 2 veggie burgers and 2 shakes cost $6.50, and 2($2.00) + 2($1.25) = $6.50.Choice (D) works: If 2 veggie burgers and 2 beef burgers cost $7.00, and beef burgers cost $1.75, then 2(veggie) + 2($1.75) = $7.00, and veggie burgers cost $2.00.This works with the other equation, since 2 veggie burgers and 2 shakes cost $6.50, and 2($2.00) + 2($1.25) = $6.50. Finally, choice (E) does not work, since 2 veggie burgers and 2 shakes cost $6.50, but 2($2.25) + 2($1.25) = $7.00.Eliminate choice (E), and you’re left with choices (A), (C), and (D).

3

To solve this question, try Plugging In. Since the question is asking about a percentage, it is a Hidden Plug In, and—unless there’s a compelling reason not to—you should try 100. If there are 100 vehicles, then there are 75 buses and 25 train cars. Since 20% of 75 = \(\frac{20}{100}\times 75=15\) , you know 15 buses use alternative fuels; of those 15, 40% use biofuels, so \(\frac{40}{100}\times 15=6\) use biofuels, and 20% use hydrogen, so \(\frac{20}{100}\times 15=3\) use hydrogen.That leaves 6 that use natural gas.The number of train cars younger than 5 years old is 25%; that rate is half the rate for natural gas buses, so 50% of natural gas buses are less than 5 years old. 50% of 6 is 3, so 3 natural gas buses are less than 5 years old, and 3 are at least 5 years old. Finally, the question asks for what percentage of vehicles are natural gas buses at least 5 years old, which is 3 out of 100, or 3 percent.

D An algebraic equation and numbers for answer choices mean Plugging In the Answers, so start by listing out your answer choices.Try numbers near the middle first: If x = 3, then \(\frac{3^{2}}{4}+\frac{3}{2}-4=\frac{9}{4}+\frac{6}{4}-\frac{16}{4}=-\frac{1}{4}\) ; eliminate choice (C). Since choice (C) was too small, so are answer choices (A) and (B)—although it’ll only take a moment to verify if you’re not sure.Try choice (D) next: If x = 4, then \(\frac{4^{2}}{4}+\frac{4}{2}-4=4+2-4=2\) , so choice (D) is one of the correct answer choices.Keep going: If x = 5, then \(\frac{5^{2}}{4}+\frac{5}{2}-4=\frac{25}{4}+\frac{10}{4}-\frac{16}{4}=\frac{19}{4}=4\frac{3}{4}\) That’s too big, so eliminate choice (E)—and with it choice (F) as well. Only choice (D) works.

This problem is another hidden Plug In. There are no numbers provided and the question asks for the ratio of the perimeters, so plug in your own numbers according to the ratio you’ve been given. Make the length of the side of the square 8. Assume the sides of the small triangles in the corners are 2. That leaves 6 for the length of the rest of the side. The sides of the rectangle are the hypotenuses of the right triangles.The smaller right triangle is 2 by 2, so the hypotenuse is 2 .The larger triangles are 6 by 6, so the hypotenuses are 6√2 . Now, you just need to add up the sides to get the perimeters. The square is simply 8 + 8 + 8 + 8, which equals 32. The rectangle is 2√2 + 2√2 + 6√2 + 6√2 .This simplifies to 16√2.32 is times greater than 16√2 , making choice (A) correct.

Plug in the values in the answer choices and see which one works. Start with choice (C), which is 1.That makes the top of the fraction (3)(−4) and the bottom (−2)(5).This doesn’t equal 1.Try choice (D).The top of the fraction becomes (2.5)(−4.5) and the bottom becomes (−2.5)(4.5).This does equal 1, so choice (D) is correct.

Don’t attempt this problem without Plugging In some numbers. The problem wants to know “how many times” greater is the area of the circle than the area of the triangle. This is same as asking what the ratio of the two areas is. A ratio is a fraction, so plug in numbers and work from there. Make each side of the equilateral triangle 6. To find the area of the triangle, draw in the height to create a 30-60-90 triangle with base 6 and height √3. These values yield an area of 9√2 for the triangle. To find the radius, draw lines from the center of the circle to each vertex of the triangle. Thus, you have created two smaller 30-60-90 triangles. The side opposite the 60 degree angle is equal to 3, so the hypotenuse of the smaller triangle is twice the length of the side opposite the 30 degree angle.This shorter side is equal to \(\frac{3}{\sqrt{3}}\) (to move from the medium side of a 30-60-90 triangle to the shortest side, divide by ). The hypotenuse of the triangle is twice this value, or \(\frac{3}{\sqrt{3}}\). This is equal to the radius of the circle, so the area of the circle is π\(\frac{3}{\sqrt{3}}\) squared or 12π. Finally, the ratio of the areas is \(\frac{12\pi }{9\sqrt{3}}\) .This reduces down to choice (C)

Remember, a ratio is just another type of fraction. No numbers are provided, so plug in anything you want and see what happens. Make the side of each smaller square 4. To find the area of the square in the middle, you need the length of one of its sides. Get this by using the Pythagorean theorem. You have a right triangle formed by the length of one smaller square plus half of the length of the adjacent square (because points R, S, T, and U are all midpoints according to the problem). So if the length of a side of a smaller square is 4, you have a triangle with legs 6 and 2.By the Pythagorean theorem, the hypotenuse is \(\sqrt{40}\).This is the length of one of the sides of the square in the middle. The area of this inner square is therefore 40, which is \(\sqrt{40}\) squared. The area of the big square is going to be 64, or 8 squared.The ratio of areas is \(\frac{40}{64}\), which is equal to ⁵⁄₈, which is choice (E).

This problem is a great opportunity to plug in the answers: For any given side, simply find b (the area, or the square of the side) and n (the perimeter, or 4 times the length of the side), and then determine whether they meet the given condition of n = ^b⁄₃. Only choice (A) will give the desired results. If the side of the square is 12, then b = 144, n = 48, and 48 = \(\frac{144}{3}\)

Don’t be intimidated by the lack of numbers in the problem—that just means you can Plug In. The question asks for a fraction, so any numbers that you plug in will work, provided they make sense in the problem. The triangle is a right triangle, so use some familiar numbers for the legs: 6, 8, and 10. If l is 6, then the radius of the semicircle is 3 and the area is 4.5π (remember, it’s half of a circle, so you need to take half of the area). Similarly, m is 8, so the radius of the circle is 4 and the area of the semicircle is 8π. Lastly, n is 10, which means the radius is 5 and the area is 12.5π. If you add L + M, you get 12.5π. If you put this over N, you get a fraction equal to one, which is choice (C).

This problem is a bit tricky because at first it looks like a hidden Plug In.The question does not tell you the number of animals in the zoo and gives you a bunch of fractions.But, in fact, this is a Plugging In the Answers problem, because the question asks for the total number of non-mammals in the zoo and the answer choices are real numbers, not fractions or percents. Start by Plugging In choice (C), 54, for the number of non-mammals. Now, use the information in the problem to find the number of mammals. According to the problem, 24 mammals are allowed to interact with the public, and this is ²⁄₃ of all the mammals.Thus, there must be 36 total mammals in the zoo (because 24 is ²⁄₃ of 36). If there are 36 mammals and 54 non-mammals, then there are 90 animals in the zoo. Now, check this number against the information in the problem.The problem says that ²⁄₅ of all the animals are mammals and 36 is ²⁄₅ of 90.Choice (C) is the correct answer.