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

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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]]>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

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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]]>Here is a hard SAT problem involving inequalities. A solution will be posted soon.

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

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]]>Today I would like to give a solution to the algebra SAT problem from last Wednesday. The original post can be found here: Algebraic Word Problem for the SAT

An ornithologist is in charge of a 400-acre bird sanctuary with only two types of birds: egrets and flamingos. There are currently 150 egrets and 200 flamingos living within the sanctuary. If 75 more egrets are introduced into the sanctuary, how many more flamingos must be introduced so that 5/6 of the total number of birds in the sanctuary are flamingos?

**Algebraic solution: **Let *x* be the number of flamingos that must be introduced. The number of flamingos will then be 200 + *x* and the total number of birds will be 150 + 75 + 200 + *x* = 425 + *x*. So, we must have (200 + *x*) = 5/6 (425 + *x*).

We multiply by 6 to get 6(200 + *x*) = 5(425 + *x*). Distributing on each side gives us 1200 + 6*x* = 2125 + 5*x*. Finally, we subtract 5*x* and subtract 1200 from each side of this last equation to get *x* = **925**.

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]]>Here is a hard SAT Heart of Algebra problem for you to try. A solution will be posted soon.

An ornithologist is in charge of a 400-acre bird sanctuary with only two types of birds: egrets and flamingos. There are currently 150 egrets and 200 flamingos living within the sanctuary. If 75 more egrets are introduced into the sanctuary, how many more flamingos must be introduced so that 5/6 of the total number of birds in the sanctuary are flamingos?

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]]>Today I would like to give a solution to the statistics problem I presented last Wednesday. The original post can be found here: Hard Statistics Question for the SAT

In Dr. Steve’s AP Calculus BC class, students are given a grade between 0 and 100, inclusive on each exam. Jason’s average (arithmetic mean) for the first 3 exams was 90. What is the lowest grade Jason can receive on his 4th exam and still be able to have an average of 90 for all 7 exams that will be given?

**Solution by changing averages to sums: **The sum of the first 3 exam grades was 90 ⋅ 3 = 270.

We want the sum of Jason’s 7 exam grades to be 90 ⋅ 7 = 630.

So, we need the sum of the grades on the last three exams to be

630 – 270 = 360.

The maximum grade is 100, and so the most Jason can score on the last 3 exams is 100 ⋅ 3 = 300.

Therefore, on the 4th exam he must score at least 360 – 300 = **60**.

**Note: **To see more details on the strategy used here, go to this page: Changing Averages to Sums

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]]>Here is a hard SAT Statistics problem for you to try. A solution will be posted soon.

In Dr. Steve’s AP Calculus BC class, students are given a grade between 0 and 100, inclusive on each exam. Jason’s average (arithmetic mean) for the first 3 exams was 90. What is the lowest grade Jason can receive on his 4th exam and still be able to have an average of 90 for all 7 exams that will be given?

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]]>The post Difficult Parabola Question for the SAT with Solution appeared first on GET 800.

]]>Let’s solve the parabola problem I posted on Wednesday. I’ll give the question one more time, followed by a solution.

The vertex of the parabola in the *xy*-plane above is (*h*, *k*). Which of the following is true about the parabola with the equation *y* = –*ax*^{2} + *k* ?

A) The parabola opens upward and the vertex is (0, *k*).

B) The parabola opens downward and the vertex is (0, *k*).

C) The parabola opens upward and the vertex is (0, –*k*).

D) The parabola opens downward and the vertex is (0, –*k*).

**Solution: **From the given picture, we see that the graph of the given equation is an upward facing parabola. It follows that *a* > 0. Therefore, –*a* < 0, and the graph of *y *= –*ax*^{2} + *k* is a downward facing parabola. So, we can eliminate choices A and C.

Now, the equation *y *= –*ax*^{2} + *k* can be written as *y *= –*a*(*x* – 0)^{2} + *k*. This is in standard form, and the graph of this equation is a parabola with vertex (0, *k*). So, the answer is choice **B**.

**Note: **Here is a graph of both parabolas drawn on the same set of axes.

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