• Proofreaders Wanted for New SAT Math Book
December 11, 2017
• Proofreaders Wanted for New SAT Math Book
• SAT Level 5 PAM Problem with Solution
December 8, 2017

### SAT Level 5 Passport to Advanced Math Problem with Solution

Today I would like to provide a solution for the difficult Passport to Advanced Math SAT problem I recently posted. Here is the original post: SAT Level 5 PAM Problem

### Level 5 Passport to Advanced Math

g(x)=x2 + 4x 1
h(x)=2x3 + 3x2 + x

The polynomials g and h are defined above. Which of the following polynomials is divisible by 2x 1 ?

A) k(x) = g(x) h(x)
B) k(x) = 12g(x) h(x)
C) k(x) = g(x) 10h(x)
D) k(x) = 12g(x) 10h(x)

First recall that the factor theorem says that p(r) = 0  if and only if x – r is a factor of the polynomial p.

* Solution using the factor theorem: We note that 2x – 1 = 2(x – 1/2), and use r = 1/2. We have

g(1/2) = (1/2)2 + 4(1/2) – 1 = 1/4 + 2 – 1 = 1/4 + 1 = 5/4

h(1/2) = 2(1/2)3 + 3(1/2)2 + 1/2 = 2(1/8) + 3(1/4) + 1/2 = 1/4 + 3/4 + 1/2 = 3/2

Since 12(5/4) – 10(3/2) = 15 – 15 = 0, we see that if k(x) = 12g(x– 10h(x), then k(1/2) = 0, and it follows that 2x – 1 is a factor of k. So, the answer is choice D.

Note: To use the factor theorem, we need to divide by a linear polynomial of the form x – r. So, we rewrite 2x – 1 as 2(x – 1/2). We see that 2x – 1 is a factor of the polynomial if and only if x – 1/2  is a factor of the polynomial. So, we use r = 1/2.

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• SAT Level 5 PAM Problem
December 6, 2017

### SAT Level 5 Passport to Advanced Math Problem

Here is a difficult Passport to Advanced Math problem for the SAT. I will provide a solution soon.

### Level 5 Passport to Advanced Math

g(x)=x2 + 4x 1
h(x)=2x3 + 3x2 + x

The polynomials g and h are defined above. Which of the following polynomials is divisible by 2x 1 ?

A) k(x) = g(x) h(x)
B) k(x) = 12g(x) h(x)
C) k(x) = g(x) 10h(x)
D) k(x) = 12g(x) 10h(x)

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• SAT Level 5 Heart of Algebra Problem with Solution
December 1, 2017

### SAT Level 5 Heart of Algebra Problem

Here is a hard Heart of Algebra problem for the SAT. I will provide several different solutions to this problem over the next few days.

### Level 5 Heart of Algebra

A worker earns \$12 per hour for the first 40 hours he works in any given week, and \$18 per hour for each hour above 40 that he works each week. If the worker saves 75% of his earnings each week, what is the least number of hours he must work in a week to save at least \$441 for the week?

A) 6
B) 8
C) 46
D) 47

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• SAT Level 5 Heart of Algebra Problem
November 30, 2017

### SAT Level 5 Heart of Algebra Problem

Here is a hard Heart of Algebra problem for the SAT. I will provide several different solutions to this problem over the next few days.

### Level 5 Heart of Algebra

A worker earns \$12 per hour for the first 40 hours he works in any given week, and \$18 per hour for each hour above 40 that he works each week. If the worker saves 75% of his earnings each week, what is the least number of hours he must work in a week to save at least \$441 for the week?

A) 6
B) 8
C) 46
D) 47

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• Hard SAT Geometry Problem with Solution
November 27, 2017

### Hard SAT Geometry Problem with Solution

Today I would like to solve the SAT geometry problem I posted yesterday. You can see the original post here: Hard SAT Geometry Problem

### Level 5 Geometry

ABCD shown above is a square, mDFE = 60°, EF = 3, and CF = 4. What is the area of square ABCD ?

Solution: DF is the hypotenuse of a 30, 60, 90 triangle (ΔDEF). Since EF = 3, we have DF = 6.

We can now use the Pythagorean Theorem to get

DC2 = DF CF= 6 42 = 36 16 = 20.

Notes: (1) The area of a square is As2, where s is the length of a side of the square.

(2) There is no need to find DC here because the area of the square is equal to DC2, and that’s what we found.

(3) The 30, 60, 90 triangle and the Pythagorean Theorem are both given to you at the beginning of each SAT math section.

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• Hard SAT Geometry Problem
November 26, 2017

### Hard SAT Geometry Problem

Here is a hard SAT Geometry problem. Come back tomorrow for a solution.

### Level 5 Geometry

ABCD shown above is a square, mDFE = 60°, EF = 3, and CF = 4. What is the area of square ABCD ?

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• SAT Level 4 Passport to Advanced Math Problem with Solution
November 21, 2017

### SAT Level 4 Passport to Advanced Math Problem with Solution

Today I would like to solve the last SAT math problem I posted. You can see the original post here: SAT Level 4 Passport to Advanced Math Problem

Level 4 Passport to Advanced Math

Let m = 2x + 7 and k = 2x 7, and write km = cx2 + d, where c and d are constants. What is the value of c – d ?

* Solution using the difference of two squares:

km = (2x + 7)(2x 7) = 4x2 – 49.

So, we have 4x2 – 49 = cx2 + d.

Thus, c = 4 and d49.

It follows that c – d = 4 – (49) = 4 + 49 = 53.

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• SAT Level 4 Passport to Advanced Math Problem
November 19, 2017

### SAT Level 4 Passport to Advanced Math Problem

Here is a hard Passport to Advanced Math SAT problem. Come back tomorrow for a solution.

### Level 4 Passport to Advanced Math

Let m = 2x + 7 and k = 2x 7, and write km = cx2 + d, where c and d are constants. What is the value of c – d ?

If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books.

• Inequality Word Problem for the SAT with Solutions
November 17, 2017

### Inequality Word Problem for the SAT with Solutions

Today I would like to give a solution to the SAT problem from last Wednesday on Inequalities. You can see the original post here: Inequality Word Problem for the SAT

### Level 5 Heart of Algebra

A worker earns \$12 per hour for the first 40 hours he works in any given week, \$18 per hour for each hour above 40 that he works each week. If the worker saves 75% of his earnings each week, what is the least number of hours he must work in a week to save at least \$441 for the week?

A) 6
B) 8
C) 46
D) 47

* Informal solution: If \$441 represents 75% of the worker’s earnings, then the worker’s total earnings is  441/0.75 = \$588.

For the first 40 hours, the worker earns 12 ⋅ 40 = 480 dollars. So, the remaining amount that the worker needs to earn is 588 – 480 = 108 dollars. So, the number of additional hours above 40 that the worker will work is  108/18 = 6.

The total number of hours that the worker must work is therefore 40 + 6 = 46, choice C.

Notes: (1) We change a percent to a decimal by moving the decimal point to the left 2 places. The number 75 has a “hidden” decimal point at the end of the number (75 = 75. or 75.0). When we move this decimal point to the left two places we get .75 or 0.75.

(2) We can find the worker’s total earnings formally as follows:

We are given that 441 is 75% of the worker’s total earnings. So, we have 441 = 0.75T, where T is the worker’s total earnings. We divide each side of this equation by 0.75 to get T = 441/0.75 = 588.

(3) Be careful that you do not accidentally choose 6 as the answer. 6 is the number of hours above 40 that the worker must work. The question is asking for the total number of hours, which is 40+6.

Algebraic solution 1: Let x be the number of hours that the worker works above 40 hours. We need to solve the following inequality for x:

0.75(12⋅40 + 18x) ≥ 441
480 + 18x ≥ 441/0.75
480 + 18x ≥ 588
18x ≥ 108
x ≥ 6

So, the least number of hours the worker needs to work is 40 + 6 = 46, choice C.

Notes: (1) For the first 40 hours, the worker earns 12 ⋅ 40 = 480 dollars.

(2) For x hours above 40, the worker earns 18x dollars.

(3) using notes (1) and (2), we see that he worker earns a total of 12 ⋅ 40 + 18x = 480 + 18x dollars.

(4) We are given that 75% of the total earned must be at least 441. So, 0.75T = 441, where T = 12 ⋅ 40 + 18x (see note (2) in the previous solution).

(5) Remember that we let x represent the number of hours the worker works above 40. So, at the end, we need to add 40 to the result.

Algebraic solution 2: This time we let x be the total number of hours that the worker works, where x must be at least 40. We need to solve the following inequality for x:

0.75(12⋅40 + 18(x – 40)) ≥ 441
480 + 18x – 720 ≥ 441/0.75
18x – 240 ≥ 588
18≥ 828
≥ 46

So, the least number of hours the worker needs to work is 46, choice C.

Notes: (1) This time we are letting x represent a number greater than or equal to 40.

If x40, the total earnings is 12 ⋅ 40 + 18 ⋅ 0

If x41, the total earnings is 12 ⋅ 40 + 18 ⋅ 1

If x = 42, the total earnings is 12 ⋅ 40 + 18 ⋅ 2

Observe the relationship between x and the last number in the expression (both in bold). We subtract 40 from the x-value to get that number.

This shows that the total earnings is 12 ⋅ 40 + 18(x – 40).

(2) This time, the value that we get for x is the answer to the question.