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Question Explanations For

Intermediate Algebra (Intermediate Algebra Practice Questions)

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1.

The Correct Answer is (D) — To solve this question, isolate for x:

$\small \begin{align*} 3x - 4 &= -\frac{1}{2x} \\ 3x +\frac{1}{2x} &= 4 \\ \frac{7}{2}x &= 4 \\ x &= \frac{8}{7} \end{align*}$

2.

The Correct Answer is (F) —

To factor a quadratic where the second-order term has a coefficient of 1, find two numbers that multiply up to the constant, –7, and add up to the first-order term’s coefficient, 6. These two numbers are 7 and –1, so you can rewrite the quadratic equation as follows:

x² + 6x – 7 = (x + 7)(x – 1)

To find the solutions—or roots of the resulting quadratic function—set the equation equal to zero. You can see that this happens when x = –7 or x = 1. (F) is the only answer choice that lists one of these options.

If you chose (H) or (J), you probably forgot that a root is the negative of one of those two numbers you found, because it’s a value of x that makes one of the bracketed expressions equal to zero.

3.

The Correct Answer is (B) — You want to avoid writing out the entire sequence up to the 21^st term, thankfully, you can use the formula for an arithmetic sequence, a_n = a₁ + d(n – 1) , to find the nth term algebraically. Simply set a₁ = 11, d = –3, and n = 21 for the 21^st term in a sequence where the first term is 11 and it decreases by 3 each term, and solve:

$\small \begin{align*} a_{(21)} &= 11 + (-3)(21 - 1)\\ a_{21} &= 11 - 3(20)\\ a_{21} &= -49 \end{align*}$

If you got any of the other answer choices, you found a term from the right sequence, but not the 21^st term.

4.

The Correct Answer is (G) — You know r and q in terms of x, so you can write out the value of y also in terms of x:

$\small y = rq = x^2 \times 12x = 12x^3$

Now you can sub in $\small \frac{1}{8}$ x to see how the value of y changes with respect to x:

$\small \begin{align*} y &= 12\left ( \frac{1}{2}x \right )^3 \\ y &= \frac{1}{8} \times 12x^3 \end{align*}$

Therefore, when x is halved,y decreases by a factor of $\small \frac{1}{8}$ .

5.

The Correct Answer is (E) — Rewrite the first inequality so that it is in terms of x:

$\small \begin{align*} 2x &\leq 8 \\ x &\leq 4 \end{align*}$

Taking the second inequality, x > 1, into account, you know that the number line needs to show values between 1 and 4, including 4 and excluding 1. The only answer choice representing this is (E).

6.

The Correct Answer is (H) — To factor a quadratic equation in the form ax² + bx + c, you need to find two values that add to b and multiply to ac. In this case, two numbers that multiply to 2 × (–1) = –2 and add to –1 are –2 and 1. Split the middle term so that these numbers are the resulting coefficients, then factor as usual:

$\small \begin{align*} 2x^2 - x - 1 &= 2x^2 + (-2x + 1x) - 1\\ &= 2x^2 - 2x + x - 1\\ &= 2x(x -1 ) + 1(x -1 )\\ &= (2x + 1 )(x -1)\\ \end{align*}$

You can always check that your result is correct by expanding the brackets or by eliminating the other answer choices:

You can eliminate (G) because 2x × 2x does not equal 2x², a.

You can eliminate (J) and (K) because neither (–1) × (–1) nor 1 × 1 equals –1, c.

You can eliminate (F) because (1 × (–1)) + (1 × 2) = –1 + 2 = 1, and b should equal –1.

7.

The Correct Answer is (A) — Equate the expression to 0, collect like terms, and solve for the possible values of x:

$\small \begin{align*} 0&=x^2+2x-7x+1-7\\ 0&=x^2-5x-6\\ 0&= x^2-6x+x-6\\ 0&=x(x-6)+1(x-6)\\ 0&=(x-6)(x+1)\\ \end{align*}$

You have determined that the two possible values of x are –1 and 6. (A) is the only choice that lists one of these solutions.

8.

The Correct Answer is (J) — First, use y = x + 1 to plug in the value for y into the second equation:

$\small x+1=|x-7|$

Next solve for the two possible values of x:

$\small \begin{align*} x+1&=x-7\\ 0&=-8 ? \end{align*}$

This one doesn’t make any sense, so you can eliminate this result and try the second solution:

$\small \begin{align*} x+1&=-(x-7)\\ x+1&=-x+7\\ 2x&=6\\ x&=3 \end{align*}$

You can substitute this back into the equations to confirm your answer.

9.

The Correct Answer is (C) — At time T, the tank is being filled at a rate slower than it was previously being filled at. Hence, the only possible explanations are I and III, since II would increase the rate at which the tank is being filled.

Note that this is a bit tricky, because you have to see that I is possible even while the volume in the tank is increasing. The volume is not increasing as much as before, so I may be the result of this.

10.

The Correct Answer is (J) — The sum of 14 and 17 is 31, which is exactly half the number of students in the class. Doubling the whole ratio of boys to girls would therefore give the number of students in the class. The number of boys in the class is therefore the left side of the ratio multiplied by two: 14 × 2 = 28.

11.

The Correct Answer is (D) — All you need to do is to set the equation equal to 10 and solve, since b(w) is the number of blocks of pressed boards that the manager wishes to produce:

$\small \begin{align*} 10&=\sqrt{20+2w}\\ 100&=20+2w\\ 50&=10+w\\ w&=40 \end{align*}$

12.

The Correct Answer is (J) — First, find the number of boards that are produced with the new method:

$b(8)=8\text{ boards per hour}$

This means that after 5 hours, (8 boards per hour)(5 hours) = 40 boards.

Next, find the number of boards that would have been produced with the old method:

$\small \begin{align*} b\;(40)&=\sqrt{20+2(40)}\\&=\sqrt{20+80}\\&=\sqrt{100}\\&=10\;\text{boards} \end{align*}$

The difference between the two methods is 40 – 10 = 30 boards.

13.

The Correct Answer is (B) — The simplest way to solve the question is to rewrite the logarithm as an exponent:

2^x = 128

Solving for x, you will find that x = 7. You can confirm this by quickly plugging 2⁷ into your calculator.

14.

The Correct Answer is (J) — Substitute the given functions into the composite function and simplify:

$\small \begin{align*} \frac{f(x)}{g(x)}&=\frac{x^2-4x-4}{x^2+5x-14} \\ &=\frac{(x-2)(x-2)}{(x+7)(x-2)} \\&=\frac{x-2}{x+7} \end{align*}$

If you chose (K) you may have taken the reciprocal of the expression.

If you got any other result, check your algebra as it is most likely due to incorrect simplifying and elimination.

15.

The Correct Answer is (D) — Recall that the value of i raised to a power repeats itself in cycles of 4. i¹ = i, i² = –1, i³ = –1, i⁴ = 1… To find the value of i³⁵⁶, divide the exponent by 4: $\small \frac{356}{4}=89$ . Since there is no remainder, i³⁵⁶ is equivalent to i⁴ = 1.

16.

The Correct Answer is (J) — On the first day, Lucy has collected 2 cans. Every subsequent day, the number of cans collected doubles from the previous day.

On the second day, Lucy collects 4 cans; on the third, 8; on the fourth, 16; on the fifth, 32. Adding all the cans together gives 2 + 4 + 8 + 16 + 32 = 62 cans in total.

If you chose (H), you probably picked the number of cans collected on the fifth day alone, rather than the cumulative number of cans collected.

17.

The Correct Answer is (B) — To multiply matrices, multiply together the matching numbers from the first matrix’s row and the second matrix’s column, and add the products together. For these two matrices, the product is (7 × 2) + (0 × 2) + (3 × 7) = 14 + 0 + 21 = 35. Since there are no more rows or columns, the final product is [35].

18.

The Correct Answer is (G) — Substitute the given values into Euler’s equation and solve. You are given that V = F, and E = 6:

$\small \begin{align*} F-6+F=2\\ 2F=8\\ F=4 \\ \end{align*}$

Substituting V into the first equation:

$\begin{align*}4-2+F&=2\\F&=4 \end{align*}$

19.

The Correct Answer is (A) — If n > m, then $\small \frac{n}{m}$ must be greater than 1. Use the rule that any logarithm in the form $\small \text{log}_b\;b = 1$ . However, $\small 6^\frac{n}{m}$ must be greater than 6 since $\small \frac{n}{m}$ > 1. This means that $\small \text{log}_6\;6^{\left ( \frac{n}{m} \right )} > 1$ . You know then that p > 1.

If you chose another answer, make sure to review your log rules.

20.

The Correct Answer is (F) — First, expand and simplify the expression:

$\small \begin{align*} \frac{(i+a)(i-a)}{i^3-ia^2} &=\frac{i^2-a^2}{i(i^2-a^2 )} \\ &= \frac{1}{i} \end{align*}$

Next, you can see that none of the answer choices are fractions. What you need to do is to rationalize the expression so that i is in the numerator and not the denominator. You do this by multiplying both the numerator and denominator by the value i. This doesn’t change the value of the expression, but will shift the i to the numerator:

$\small \begin{align*} \frac{1}{i}\times \frac{i}{i} &= \frac{i}{-1}\\ &=-i \end{align*}$