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SAT Math Prep Course 22 Grid Ins Questions | SAT Online Classes AMBiPi

Welcome to AMBiPi (Amans Maths Blogs). SAT (Scholastic Assessment Test) is a standard test, used for taking admission to undergraduate programs of universities or colleges in the United States. In this article, you will get SAT 2022 Math Prep Course 22 Grid Ins Questions with Answer Keys | SAT Online Classes AMBiPi.

SAT 2022 Math Prep Course 22 Grid Ins Questions with Answer Keys

SAT Math Practice Online Test Question No 1:

The table above shows information about the tickets sold for a recent performance by a theater troupe. The total revenue in ticket sales for this performance was $15,000.

Before the tickets for this performance went on sale, a consultant for the theater had predicted that n, the number of tickets sold per section, would vary with p, the price in dollars for a ticket in that section, according to the formula n = 2,800/p. By how many tickets did this model underestimates the actual total number of tickets sold?

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Correct Answer: 25

Problem Solving (extended thinking) HARD

The mathematical model n = 2,800/p has embedded in it the predicted revenue per section: np = revenue per section = $2,800. Notice that this prediction is $200 less than the actual average revenue per section of $3,000, so clearly, the model underestimated the number of tickets sold per section.

If we want to analyze this situation in detail, we can compare the predicted tickets sold to the actual tickets sold by adding a new column to the table entitled “predicted sold,” which we can fill in using the calculations from our model. Also, it might be helpful to add columns for “total revenue” for each situation.Images

You might notice that the predicted number of tickets sold in the Front Orchestra and the Third Mezzanine are fractions, which seems strange. (Of course, we can’t sell a fraction of a ticket!) But even if we round these predictions to the nearest whole numbers, 47 and 93, the total number of tickets is the same: 346, which underestimates the number of tickets sold by 25.

SAT Math Practice Online Test Question No 2:

The table above shows a set of ordered pairs that correspond to the function h(x) = x2/2 + k. What is the value of k?

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Correct Answer: 3/2 or 1.5

Advanced Mathematics (quadratics) MEDIUM

Original function: h(x) = x2/2 + k

Substitute h(3) = 6 (from table): 6 = 32/2 + k

Simplify: 6 = 4.5 + k

Subtract 4.5: 1.5 = k

To check your answer, you can plug in the second row of the table to verify that 52/2 + 1.5 = 14

SAT Math Practice Online Test Question No 3:

If 0 < x < 2π and cosx = √5, what is the value of (sinx/3)2?

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Correct Answer: 4/45 or. 088 or. 089 

Special Topics (trigonometry) MEDIUM-HARD

Given equation: 5 cosx = √5

Divide by 5: cosx = √5/5

This gives us the value of cos x, but we are asked to evaluate (sinx/3), which of course is in terms of sin x. For all real numbers x, sin2 x + cos2 x = 1.

Pythagorean Identity: sin2 x + cos2 x = 1

Subtract cos2 x: sin2 x = 1 – cos2 

Expression to be evaluated: (sinx/3)2

Simplify: sin2x/9

Substitute sin2 x = 1 – cos2 x: 1 – cos2 x/9

Substitute cosx = √5/5 : (1 – [√5/5]2)/9

Simplify: (1 – 5/25)/9

Simplify: (4/5)/9

Simplify by multiplying 5/5: 4/45

SAT Math Practice Online Test Question No 4:

 An Internet service provider offers three different plans for residential users. Plan A charges users $500 for the first year of service and $80 per month thereafter. Plan B charges users $68 per month. Plan C is a “high speed” plan that offers 200% higher speeds for $92 per month.

Isabelle has been using Plan A for over a year. She recently reviewed her plan and realized that if she had been using Plan B for the same amount of time, she would have saved $104 for Internet service over the entire period. At the time of her review, how many months had Isabelle been on Plan A?

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Correct Answer: 47

Problem Solving (extended thinking) HARD

Let n equal the number of months that Isabelle has been on Plan A. If she has been on Plan A for over a year, then n > 12. This means that she has been on Plan A for n – 12 months beyond the first year. Since Plan A costs $500 for the first year and $80 per month thereafter, the total cost for her n months of service is $500 + $80(n – 12). If she had been on Plan B, the cost would have been $68 per month or a total of $68n. If Plan B would have saved her $104 over this period, 500 + 80(n – 12) – 104 = 68n

Distribute and simplify: 396 + 80n – 960 = 68n

Simplify: 80n – 564 = 68n

Add 564: 80n = 68n + 564

Subtract 68n: 12n = 564

Divide by 12: n = 47

SAT Math Practice Online Test Question No 5:

A state environmental study determines that the coastal regions of the state lose 24.5 acres of wetlands per month. At this rate, how many months will it take these coastal regions to lose a total of 343 acres?

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Correct Answer: 14

Problem Solving and Data Analysis (proportions) EASY

If x is the number of months it takes until the regions lose 343 acres, then 24.5 = 343/x

Cross-multiply: 24.5x = 343

Divide by 24.5: x = 14

SAT Math Practice Online Test Question No 6:

V(t) = 1000(1 + k)m

An analyst wants to use the formula above to estimate the value, in dollars, of a $1,000 initial investment in a mutual fund after m quarters have passed. If a $1,000 initial investment in this fund is worth $1,102.50 after 2 quarters, what number should the analyst choose for k?

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Correct Answer: .05 or 1/20 

Problem Solving (analyzing formulas) HARD

If the investment is worth $1,102.50 after 2 quarters, 1,000(1 + k)2 = 1,102.5

Divide by 1000: (1 + k)2 = 1.1025

Take the square root: 1 + k = 1.05

Subtract 1: k = .05

SAT Math Practice Online Test Question No 7:

If I = FV/(1 + r)n

The formula above indicates the initial investment, I, that must be made in an account with an annual interest rate of r to ensure a future value of FV after n years.

What value of r, to the nearest thousandth, would ensure that the value of an investment would increase by 69% in 2 years?

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Correct Answer: 0.3

Problem Solving and Data Analysis (formula analysis) HARD

If the value of the investment increases by 69% in 2 years, this means that FV = 1.691 when n = 2.

Substituting into the formula gives I = 1.69I/(1 + r)2.

Divide by I: 1 = 1.69/(1 + r)2.

Cross-multiply: (1 + r)2 = 1.69 

Take the square root: 1 + r = 1.3

Subtract 1: r = 1.3 – 1 = 0.3

SAT Math Practice Online Test Question No 8:

A new color copier purchased for $8,500 is expected to depreciate (lose value) according to the equation y = –1,250x + 8,500, where y is the value of the copier x years after it was purchased. The company that bought the copier plans to sell it when the value is $1,000 and upgrade to a new one. How many years after the copieris purchased will the company sell it?

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Correct Answer: 6 

Difficulty: Medium

Category: Heart of Algebra / Linear Equations

Strategic Advice: In a word problem, being able to match each variable to the quantity that it represents is often the key to getting started.

Getting to the Answer: The copier will be worth $1,000 when its value (y) is equal to 1,000. Plug 1,000 in for y and solve for x using inverse operations: y = -1,250x + 8,500

1,000 = -1,250x + 8,500

-7,500 = -1,250x

6 = x

The company will sell the copier 6 years after its purchase.

SAT Math Practice Online Test Question No 9:

If (2/3)j – (1/4)k = 5/2, what is the value of 8j – 3k?

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Correct Answer: 30

Difficulty: Medium

Category: Heart of Algebra / Linear Equations

Strategic Advice: Don’t panic when you see two variables and only one equation. Instead, look for a way to manipulate the given equation, or at least one side of it, so that it looks like the desired expression.

Getting to the Answer: The desired expression (8j – 3k) doesn’t have fractions, so multiply both sides of the given equation by the least common denominator (12) to eliminate the fractions: (2/3)j – (1/4)k = 5/2

12 ⦁ [(2/3)j – (1/4)k] = 12 ⦁ (5/2)

8j – 3k = 30

Notice that the coefficients of j and k are 8 and -3, respectively, which is exactly what you’re looking for. The number on the other side of the equation, 30, is the correct answer.

SAT Math Practice Online Test Question No 10:

A street vendor sells two types of newspapers, one for $0.25 and the other for $0.40.If she sold 100 newspapers for $28.00, how many newspapers did she sell at $0.25?

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Correct Answer: 80

Difficulty: Medium

Category: Heart of Algebra / Systems of Linear Equations

Strategic Advice: Use the Kaplan Strategy for Translating English into Math to build a system of equations. Be sure to use consistent units throughout. Use cents (rather than dollars) to avoid having to work with decimals.

Getting to the Answer: Call the number of 25-cent newspapers x and the number of 40-cent newspapers y. Because the vendor sold 100 newspapers in total, x + y = 100. Also, the total cost of these newspapers was $28.00, or 2,800 cents. So 25x + 40y = 2,800. The question asks for x, the number of 25-cent newspapers sold. From the first equation, y = 100 – x. Substitute this value for y into the second equation, and solve for x.

25x + 40y = 2,800

25 + 40(100 – x) = 2,800

25x + 4,000 – 40x = 2,800

-15x + 4,000 = 2,800

-15x = -1,200

x = -1200 ÷ (-15)

x = 80

The question asks for the number of 25-cent newspapers sold (which is x), so you’re all done. Grid in 80 and move on to the next question.

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