**SAT Level 5 Passport to Advanced Math Problem**

with Solution

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*)=*x*^{2} + 4*x* *–* 1

*h*(*x*)=2*x*^{3} + 3*x*^{2} + *x*

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

A) *k*(*x*) = *g*(*x*) *–* *h*(*x*)

B) *k*(*x*) = 12*g*(*x*) *–* *h*(*x*)

C) *k*(*x*) = *g*(*x*) *–* 10*h*(*x*)

D) *k*(*x*) = 12*g*(*x*) *–* 10*h*(*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 2*x* *– *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*) = 12*g*(*x*) *– *10*h*(*x*), then *k*(1/2) = 0, and it follows that 2*x* *– *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 2*x* *– *1 as 2(*x* *– *1/2). We see that 2*x* *– *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.

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**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*)=*x*^{2} + 4*x* *–* 1

*h*(*x*)=2*x*^{3} + 3*x*^{2} + *x*

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

A) *k*(*x*) = *g*(*x*) *–* *h*(*x*)

B) *k*(*x*) = 12*g*(*x*) *–* *h*(*x*)

C) *k*(*x*) = *g*(*x*) *–* 10*h*(*x*)

D) *k*(*x*) = 12*g*(*x*) *–* 10*h*(*x*)

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**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

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**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

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**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, *m*∠*DFE *= 60°, *EF *= 3, and *CF *= 4. What is the area of square *ABCD* ?

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

We can now use the Pythagorean Theorem to get

*DC*^{2} = *DF*^{2 }*–* *CF*^{2 }= 6^{2 }*–* 4^{2} = 36 *–* 16 = **20**.

**Notes: **(1) The area of a square is *A* = *s*^{2}, 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 *DC*^{2}, 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.

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**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, *m*∠*DFE *= 60°, *EF *= 3, and *CF *= 4. What is the area of square *ABCD* ?

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**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* = 2*x* + 7 and *k* = 2*x* *–* 7, and write *km* = *cx*^{2} + *d*, where *c* and *d* are constants. What is the value of *c – **d* ?

*** Solution using the difference of two squares: **

*km *= (2*x* + 7)(2*x* *–* 7) = 4*x*^{2} *– *49.

So, we have 4*x*^{2} *– *49 = *cx*^{2} + *d*.

Thus, *c* = 4 and *d* = *–*49.

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

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**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* = 2*x* + 7 and *k* = 2*x* *–* 7, and write *km* = *cx*^{2} + *d*, where *c* and *d* are constants. What is the value of *c – **d* ?

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**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 = 0.75*T*, 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 + 18*x*) ≥ 441

480 + 18*x* ≥ 441/0.75

480 + 18*x* ≥ 588

18*x* ≥ 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 18*x* dollars.

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

(4) We are given that 75% of the total earned must be at least 441. So, 0.75*T* = 441, where *T* = 12 ⋅ 40 + 18*x* (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 + 18*x –* 720 ≥ 441/0.75

18*x –* 240 ≥ 588

18*x *≥ 828

*x *≥ 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 *x* = **40**, the total earnings is 12 ⋅ 40 + 18 ⋅ **0**

If *x* = **41**, 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.

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