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PSAT Review #1 *Questions adapted from the released Fall 2017 College Board PSAT/NMSQT.

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Presentation on theme: "PSAT Review #1 *Questions adapted from the released Fall 2017 College Board PSAT/NMSQT."— Presentation transcript:

1 PSAT Review #1 *Questions adapted from the released Fall 2017 College Board PSAT/NMSQT

2 #1 Lines that are perpendicular have opposite reciprocal slopes. To find the slope of the given line count the rise and the run. You get 3 2 , so the slope of a line perpendicular to this line would have a slope of − B is the only choice that has a slope of − 2 3 , so B must be the answer. ***Also, remember that parallel lines have the same slope because they rise, or fall, at the same rate.

3 #2 Since x and y are dollar amounts, we can eliminate answer choices B and C because 6 and 9 represent the x and y intercepts (which are dollar amounts not numbers of items purchased). Also, since x is a dollar amount we know the 3 in the equation stands for the number of bags of chips, not the cost of a bag, so we can eliminate answer choice A. The 3 and the 2 must represent the number of bags of chips purchased and the number of jars of salsa purchased, respectively. Since 3 divided by 2 is 1.5, D is the answer.

4 #3 Radicals can be rewritten as rational (fractional) exponents. The index on the radical is the denominator of the fraction and the power of the radicand is the numerator of the fraction.

5 #4 Use the information given to write two coordinate pairs: (2, 60) & (7, 75). Then find the equation of the line passing through these two points by finding the slope and the y-intercept. 𝑚= 75−60 7−2 = 15 5 =3 Plug either one of the given points and the calculated slope into the slope-intercept form of a linear equation, y = mx + b (or you can use point-slope form), to solve for b (the y-intercept). 60=3 2 +𝑏 Plug 3 in for slope and 54 in for the y-intercept 60=6+𝑏 to get 𝑇 𝐻 =3𝐻+54. 𝑏=54

6 #5 The easiest way to do this problem is to
graph the function on a graphing calculator and look at its graph. Domain includes the x-values. (left to right) Range includes the y-values. (bottom to top) Looking at the graph we see the x values start at 2 and go to the right.

7 #6 *This problem was on the non-calculator part of the PSAT, so you won’t be able to use your graphing calculator to help you. The graph is not linear, so we can eliminate answer choice C. The graph is not a parabola, so we can eliminate the quadratic function, choice B. To determine if A or D is the correct answer, look at the sign of the leading coefficient. The leading coefficient of answer choice D is -1. The graph shown rises to the right, so it must have a positive leading coefficient. The answer cannot be D, and therefore, must be A.

8 #7 Let length = 17 – 2x Let width = 11 – 2x Let height = x
V = lwh = (17 - 2x)(11 - 2x)(x) The area of the base would be length times width, so only multiply these two. V = ( x +4 𝑥 2 )x This matches answer choice C.

9 #8 Cross multiply to get: (x-20)(x-4) = (3x)(x-8)
𝑥 2 −24𝑥+80=3 𝑥 2 −24𝑥 − 2𝑥 2 +80=0 −2𝑥 2 =−80 𝑥 2 =40 You can stop here. The problem does not ask for x. It asks for 2𝑥 2 . Since 𝑥 2 =40, 2𝑥 2 =80.

10 #9 Plug the points (3, 17) and (10, 62.50) in for (n,C) to create a linear system which can be solved using elimination: 17 = (3) 2 a + b 62.50 = (10) 2 +b 17 = 9a b 62.50 = 100a + b Multiply the top equation by -1 and add the two equations to get: 45.50 = 91a Divide both sides by 91 to get a = Be careful here! The problem asks for b, not a. Plug 0.50 into either equation and solve for b: b =

11 #10 Write a linear inequality and solve for y:
Let x = number of hours worked during the week. Let y = number of hours worked on the weekend. So…. 10x + 15y ≥ 500 Plug 30 in for x (30) + 15y ≥ 500 and solve for y: y ≥ 500 15y ≥200 y ≥ Round up to the nearest whole number, 14.


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