The Hardest Official SAT Math Questions of the Day by Student Response Rate and by Difficulty Level

The dates referred to in this post will be different if you have a newer version of the app. However, the app just cycles through the same questions, so everything in this post will still be relevant to you. You just won't be able to track down questions according to their date.

I've been taking a closer look at the math questions of the day available on The College Board SAT app recently. It's quite helpful and serves as great practice. I highly recommend it.

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As you can see, the app tells you the percentage of responses that were correct for each question. It also tells you the official difficulty level.

On the whole, the correct response rate was about 62%, which means on average, each question was answered correctly 62% of the time.

After going through all the math questions in the archive, I noticed something peculiar. Some of the questions that were designated as hard (difficulty level 3) had correct response rates that were actually quite high.

For example, the November 14th 2015 question (supposedly difficult) had a correct response rate of 63%, which is higher than the average for all questions. Similarly, the November 24th 2015 and January 15th 2016 questions, both labeled difficult by The College Board, had correct response rates of 59%.

On the other hand, quite a few medium questions had correct response rates as low as 32%!!

This discrepancy could be due to several factors:

The top students only bother with the most difficult questions.
What The College Board considers difficult is different from what most students would consider difficult.
Because the exam was new when The College Board wrote some of these questions, they didn't have an accurate idea of their actual difficulty.

All factors are probably at play.

In any case, it's worthwhile to look at all the tough questions, whether it's by student response rate or by the official difficulty level, so I'm presenting them all in this post. Enjoy!

Top 5 Toughest Questions according to Correct Student Response Rates

1. September 27th, 2015 (Only 36% were Correct)

Tonya hits a golf ball from an initial height of 10 feet. The height of the golf ball, $h$, in feet above sea level, $t$ seconds after the ball was hit, can be modeled by a quadratic equation. If the golf ball reaches its maximum height of 74 feet exactly 2 seconds after it has been hit, which of the following equations represents this relationship?

A) $h = 10(t - 2)^2 + 74$
B) $h = -10(t + 2)^2 + 74$
C) $h = -16(t - 2)^2 + 74$
D) $h = 16(t + 2)^2 + 74$

2. November 2nd, 2015 (Only 35% were Correct)

A potato chip company produces the same number of snack-sized bags of potato chips every month. A manager at the company randomly selects a sample of snack-sized bags of potato chips to be weighed in a routine quality-assurance check each month. The January sample had a mean weight of 0.95 ounces and a margin of error of 0.06 ounces. The February sample had a mean weight of 0.98 ounces and a margin of error of 0.04 ounces. Based on these findings, which of the following is an appropriate conclusion?

A) Most of the snack-sized bags of potato chips produced by the company in January each had a weight of at least 0.95 ounces, while most of the snack-sized bags of potato chips produced by the company in February each had a weight of at least 0.98 ounces.

B) The mean weight of all the snack-sized bags of potato chips produced by the company in January must have been 0.03 ounces more than the mean weight of all the snack-sized bags of potato chips produced by the company in February.

C) The mean weight of all the snack-sized bags of potato chips produced by the company in January was 0.03 ounces less than the mean weight of all the snack-sized bags of potato chips produced by the company in February.

D) The number of snack-sized bags of potato chips in the January sample was less than the number of snack-sized bags of potato chips in the February sample.

3. November 16th, 2015 (Only 32% were Correct)

The temperature $T$ in degrees Celsius of a chilled drink after the drink has been sitting on a table for $m$ minutes is represented by the function below.

\[T(m) = 32 - 28\cdot 36^{-0.05m}\]

What is the best interpretation of the number 32 in this context?

A) The drink is originally 32 degrees Celsius.
B) Every 32 minutes, the temperature warms by 3 degrees Celsius.
C) After 32 minutes, the drink will fully warm to the ambient temperature.
D) After the drink has been sitting for a very long time, the temperature of the drink will approach 32 degrees Celsius.

4. February 8th, 2016 (Only 37% were Correct)

\[6(1 - a) = 3(a - b) + 1\] \[4(b - 2) = 3a\]

If $(a, b)$ is the solution to the system above, then what is the value of $a\cdot b$?

A) $-\dfrac{1,276}{243}$

B) $\dfrac{29}{9}$

C) $\dfrac{44}{27}$

D) $\dfrac{1,276}{243}$

5. January 1st, 2016 (Only 37% were Correct)

\[Q(x) = \dfrac{P(2x)}{2}\]

In the equation shown above, $Q$ and $P$ are functions. If $(x_0, y_0)$ is a point on the graph of $y = Q(x)$, which of the following is a point on the graph of $y = P(x)$?

A) $(x_0, y_0)$

B) $\left(2x_0, \dfrac{y_0}{2}\right)$

C) $\left(\dfrac{x_0}{2}, 2y_0\right)$

D) $(2x_0, 2y_0)$

For more challenging questions, I highly recommend my SAT Math book. Check it out.

Answers and Explanations

Top 3 Toughest Questions according to Official Difficulty Levels

D — The standard deviation is a measure of how spread out the data is. It says nothing about how the mean or the median compares with that of another data set. Because the standard deviation is lower for Juan's math class, the heights of the students in that class are less variable (less spread out).

B — Let $t$ be the time it takes for Lisa to catch up to Sam. In that time, Lisa covers a distance of $8t$. Sam covers a distance of $5t$, but his head start gave him $15\times 5 = 75$ meters on Lisa. So, Lisa has to cover the head start distance as well as the distance Sam runs up until the point she catches up to him: \[8t = 5t + 75\] \[3t = 75\] \[t = 25\]

D — From the formula, \[\text{mass} = \text{density} \times \text{volume}\] \[\text{mass} = 1.5 \times 10.5 = 15.75\] Now because the density is less than or equal to 1.5 grams per milliliter, the mass we just calculated is actually the maximum. \[\text{mass} \leq 1.5 \times 10.5 = 15.75\]

Top 5 Toughest Questions according to Correct Student Response Rates

C — When $t = 2$, $h = 74$. Because of this condition, it should be obvious that the answer is between (A) and (C). When $t = 0$, $h = 10$. The answer choice that also meets this condition is (C).

D — Answer (A) is wrong because the margin of error doesn't tell us anything about where most of the data are. We don't know whether most of the bags are above the mean or below the mean. Answer (B) is wrong because we can't infer from the samples that the mean for all the January bags was 0.03 ounces more than the mean for all the February bags. In fact, the sample means suggest the opposite. In any case, what we got from the samples were estimated means. You can't know the exact mean for all bags unless you weigh all of them. Answer (C) is wrong for this same reason. Answer (D) is correct because the higher the sample size, the lower the margin of error.

D — As $m$ gets to be a very very large number, $36^{-0.05m}$ gets very very small. It approaches zero (feel free to try some values on your calculator to check this). As a result, $28\cdot 36^{-0.05m}$ also approaches 0, and the expression \[32 - 28\cdot 36^{-0.05m}\] approaches 32. What does this mean? The temperature of the drink approaches 32 degrees Celsius after the drink has been sitting for a very long time.

D — The first equation gives \[6 - 6a = 3a - 3b + 1\] \[5 = 9a - 3b\] The second equation gives \[4b - 8 = 3a\] \[-8 = 3a - 4b\] So now we have \[ \begin{cases} 9a - 3b &= 5\\ 3a - 4b &= -8 \end{cases} \] Multiply the second equation by $-3$ and the resulting equation to the first to get, \[9b = 29\] \[b = \dfrac{29}{9}\] To solve for $a$, plug this result into the first equation: \[9a - 3\left(\dfrac{29}{9}\right) = 5\] \[9a = \dfrac{44}{3}\] \[a = \dfrac{44}{27}\] Finally, \[a\cdot b = \dfrac{44}{27}\cdot \dfrac{29}{9} = \dfrac{1,276}{243}\]

D — Don't be confused by the notation. Plug in $x_0$ as if it were a number: \[Q(x_0) = \dfrac{P(2x_0)}{2}\] Since $Q(x_0) = y_0$, \[y_0 = \dfrac{P(2x_0)}{2}\] \[2y_0 = P(2x_0)\]
Stop and take a look at what we have. What does this mean? When we plug in $2x_0$ into $P(x)$, we get $2y_0$. That means $2x_0, 2y_0$ is a point on $y = P(x)$.
By the way, don't get confused by the notation $y = P(x)$. It just refers to the function being a graph in the $xy$-plane, much like $y = x^2 + 1$ is a function that can be graphed in the $xy$-plane, with $x$ representing the input and $y$ representing the output. Here, $x_0$ and $y_0$ are the actual constants you actually want to be working with. It would have been pointless to do something like \[Q(x) = \dfrac{P(2x)}{2}\] \[y = \dfrac{P(2x)}{2}\] where you're substituting $Q(x)$ with $y$. While it's true that $y$ in this case would represent the output of $Q(x)$, it doesn't really lead anywhere. Think of $y = P(x)$ and $y = Q(x)$ as more of notational thing to designate that they are functions that have graphs in the $xy$-plane.