ACT Math: Five Solutions (3/13/15)

Hopefully you were able to find some time and a quiet place this weekend to try our Five Problems for Your Weekend (3/13/15). Below you’ll find the fully worked-out solutions to those problems.

11.  C
Begin by replacing the variables in the expression with the given values. Then simplify the resulting expression paying careful attention to order of operations.

2x² + 4y
2(3)² + 4(5)
2(9) + 4(5)
18 + 20
38

38.  H
You just need to carefully solve the inequality using opposite operations. In the final step there will be two important things that you need to be very careful with.

x + 8 > 14
x + 8 – 8 > 14 – 8
x > 6
(-2)(-½x) < -2(6)
x < – 12

Two important ideas from the last step:

  • Dividing by a fraction is the same as multiplying by its reciprocal. So dividing by -½ is the same as multiplying by -2.
  • When solving an inequality, if you multiply or divide by a negative you must switch the direction of the inequality symbol.

35.  D
When a line crosses the x-axis, the y-coordinate of that point (the x-intercept) is always zero. You can substitute 0 for y in the equation and then solve for x.

6x – 8y = 24
6x – 8(0) = 24
6x – 0 = 24
6x = 24
x = 4

The line crosses the x-axis at the point (4, 0), so the value of a is 4. Alternatively, you could have put the equation into slope-intercept form (y = 3/4 x – 3), graphed it on your calculator and then looked at where it crosses the x-axis.

48.  J
The determinant of the matrix is given by (x)(x) – (7)(x) and we are told that this is equal to -12. Now you can write and solve an equation to find the value(s) of x for which this is true.

(x)(x) – (7)(x) = -12
x² – 7x = -12
x² – 7x + 12 = 0
(x – 4)(x – 3) = 0
x = 4; x = 3

You also could have plugged each of the answer choices into the matrix and evaluated the determinant until you found one that gave you a determinant equal to -12. Be careful with answer choice F — even though 3 works, it might not be the only answer. You would need to try any other answer that also includes 3.

55.  E
In order to get this problem correct, you need to be very familiar with the log rules (summarized in the table below).

Name of Rule Rule
Product log A·B = log A + log B
Quotient log (A/B) = log A – log B
Power log (A^B) = B·log A

Start by simplifying log a²b³ using the Product and Power rules.

log a²b³
log a² + log b³
2log a + 3log b
2x + 3y

In the final step, recall that it was given that log a = x and log b = y.

If you have questions about these problems or anything else to do with the ACT, leave a comment below or send me an email at info@cardinalec.com.

One thought on “ACT Math: Five Solutions (3/13/15)

Leave a Reply

Your email address will not be published. Required fields are marked *