Hard ACT Math Problems

If youre wondering why hard ACT Math problems are so difficult, know that its not because they test crazy advanced topics like multivariable calculus or anything like that. Instead, its because they take some of the foundations of math topics youve already studied in school, like pre-algebra, algebra, and geometry, and turn them into multi-step processes that may combine concepts from different areas. Does this make them challenging? Yep. Does it mean you cant solve them? Absolutely not! To master these problems, youll need to refresh your understanding of the concepts, then put them into practice. Think youre ready to put your skills to the test? Check out the twelve ACT math challenge problems weve put together for you below. Answers and explanations follow! Want to make sure youve mastered the trickiest concepts in all areas? For ACT challenge problems from other sections, check out: Hard ACT English Problems Hard ACT Reading Problems Hard ACT Science Problems 1. Let x and y be nonzero real numbers such that 2(y+1)=2x. Which of the following expresses 2(y+2)in terms of x? A. 1/(2x3) B. 1/(2x+3) C. E. 15 9. If sin = , which of the following could be true about ? A. 0 /3 B. /. The Average Speed for the whole trip was 1 mph, which is different from simply taking the mean of the two speeds. If we had just averaged the two speeds (10mph and 20mph) we would have gotten 15mph. Average Speed is a weighted average. Since Suzanne spent more time in the problem going 10mph than 20mph, it makes sense that the Average Speed would be closer to 10mph. 8. ANSWER: D To find the average speed of the bus, we know we will need to find the Total Distance and the Total Time, so we can start by using another formula (Distance = Rate x Time) to help us find the pieces we’re missing for each part of the trip. For the first part of the field trip: 30 miles = 15mph x T, so we know that T = 2 hours. For the middle part of the trip, we know that D = 10mph x 3 hours, so we know that D = 30 miles. For the last part of the trip, we know that .28mph. Since the question used the word â€Å"approximately,† we can round to the nearest integer: 14. The correct answer is (D). 9. ANSWER: E This is a hard problem. First of all, notice that the angle must be in QIII or QIV, since the sine is negative. It has to be a value on the line = – . The first three choices are all in QI and QII, so the angle cant possibly be in any of those. We can easily eliminate (A) (B) (C). Lets think about the last two. First, option (D). This option is in QII and QIII, so the part in QIII might go down low enough to contain the angle. The lowest ray in that region is at . Technically, this ray isnt even included, because its the endpoint of an inequality, but we will just use this value. Sine is positive in QII, zero at the negative x-axis, and starts to get more and more negative as we go around through QIII. How low does this go by this last value? = = This does not go down as far as Heres a diagram with region (D) and the line = – . So, we can eliminate (D). This leaves (E) by the process of elimination, but lets verify that this works. Region (E) is in QIV. The sine is negative one at the bottom, where the unit circle intersects the negative y-axis. As we rise from this bottom, we get negative values of smaller and smaller absolute value, until it equals zero at the positive x-axis. We have to know how low the lowest values in this region are. Well, the limiting value is . We know that = Now, if you happen to have the decimal approximations memorized, you will see that: = If you dont have this decimal equivalent memorized, you could think about the triangles in QIV: Even if we cant directly compare the sizes of the two vertical legs, we definitely can compare the two horizontal legs, and 0.8 is definitely bigger than 0.5. That means the purple {0.6, .8, 1} triangle is further around in the counterclockwise direction from the green 30-60-90 triangle, and the point with a sine of – 0.6 would clearly be included in the arc from the 30-60-90 triangle up to the positive x-axis. Answer = (E) 10. ANSWER: D The graph clearly has a hole, a point discontinuity, at x = 3. This means that the function cannot be defined at x = 3. It must have (x – 3) in the denominator. On this basis, we can eliminate (A), (C), and (E). Notice that the graph is identical to the line y = 2x – 3 at every point except for the point discontinuity at x = 3. Also, to get a point discontinuity instead of an asymptote, we need to have a factor of (x – 3) in both the numerator and the denominator. Choice (B) at x = 3 has a zero denominator and a non-zero numerator, which is the condition for a vertical asymptote. We have eliminated the other four answers, so it would seem that (D) is the answer, but lets verify this. In order to get a graph that is identical to y = 2x – 3 at every point except for x = 3 and has a point discontinuity at x = 3, we would have to multiply (2x – 3) by (x – 3) over itself. So everything checks out and our answer is (D)! 11. ANSWER: E When we have functions inside functions, we call these â€Å"nested† functions – it’s important to always start from the innermost parentheses and work your way out. The expression m(m(n)) starts with an inntermost â€Å"m(n)† that we know equals -n3, so let’s start by substituting that in. Now the question is asking, what is the value of m(-n3). Just like we normally do for any f(x) function, we will plug whatever in inside the parentheses in for the variable in the function. Here, the testmaker is trying to confuse you by using the same variable â€Å"n,† but we’re too smart for the ACT’s tricks! 🙂 m(-n3) = -(-n3)3 According to the order of operations, or PEMDAS, we’d start with what is inside the parentheses. Since we don’t know the value of â€Å"n† we cannot simplify -n3 any further. Next, we simplify the exponents. Here’s where knowing your exponent rules really come in handy! When two exponents are separated by a parenthesis, we must always multiply them. m(-n3) = -(-n9) Remember that in number properties, any negative number raised to an odd exponent will stay negative, since it takes a pair of negatives to cancel each other out and become positive. -n9 will stay negative until it is multiplied by the -1 in front of the parentheses. Finally, we arrive at our answer: m(-n3) = n9 The correct answer is (E). 12. ANSWER: F The function values for p(x) vary directly as x for all real numbers. The graph of y=p(x) in a standard (x,y) coordinate plane is a line with a y-intercept at 0. Since y = p(x) doesnt have any coefficients, and p(x) varies directly with x (like p(x) = x, as opposed to something like p(x) = 2x), this line will cross the origin (0,0). The correct answer is (F).