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DSC 10’s only prerequisite is high school algebra. In order to gauge your preparedness for the type of math you’ll see in this class, you should work through the following practice problems on your own. You will not submit it, and it will not be graded, but it will help you identify any gaps in your background knowledge and it will teach you some useful test-taking skills.


Problem 1

Express \frac{42}{9} \cdot \frac{36}{21} as an integer without using a calculator and without long division or multiplication.

Answer: 8

The idea is to look for common factors in the numerator and denominator and then cancel. We notice that 42 and 21 both share a factor of 7, and that 36 is a multiple of 9.

\begin{align*}\frac{42}{9} \cdot \frac{36}{21} &= \frac{6 \cdot 7}{9} \cdot \frac{4 \cdot 9}{3 \cdot 7} \\ &= \frac{6 \cdot \sout{7}}{\sout{9}} \cdot \frac{4 \cdot \sout{9}}{3 \cdot \sout{7}} \\ &= \frac{6}{1} \cdot \frac{4}{3} \\ &= \frac{2 \cdot \sout{3}}{1} \cdot \frac{4}{\sout{3}} \\ &= 2 \cdot 4 \\ &= 8 \end{align*}


Problem 2

Choose the answer below that is closest to 412 \cdot 289, without using a calculator and without performing long multiplication.

Answer: 120{,}000

We can estimate the answer by solving a similar problem, 400\cdot300, which has numbers close to those given, but are much easier to work with. Since 400\cdot300 = 120{,}000, we would estimate that 412\cdot289 is similar. If you’re curious, a calculator will tell you that 412\cdot289 = 122{,}776.


Problem 3

Express \frac{16}{40} as a percentage without using a calculator and without long division or multiplication.

Answer: 40%

There are many ways to do this. One way is to simplify the fraction. The numerator, 16, and the denominator, 40, are both divisible by 4, so the fraction is equivalent to \frac{4}{10}, which is same as \frac{40}{100}, or 40%.

Another way is to solve the equation \frac{16}{40} = \frac{x}{100}. We can solve this equation for x by multiplying both sides by 100, to get x = \frac{1600}{40}. Since both the numerator and denominator end in 0, we can divide both by 10, to get x = \frac{160}{4}, or x = 40.

We can also solve this by recognizing that 40 \cdot 2.5 = 100. Therefore if we also multiply the numerator of 16 by 2.5, we will convert \frac{16}{40} to its corresponding fraction out of 100, which is \frac{16 \cdot 2.5}{40 \cdot 2.5} = \frac{40}{100}, or 40%.


Problem 4

I just sent my friend $50 on Venmo and that used up 25% of my Venmo account balance. How much money is left in my Venmo account? Solve this problem without a calculator.

Answer: $150

If $50 was 25%, or one fourth, of my account balance, then my original account balance was \$50 \cdot 4 = \$200. After I pay the $50 to my friend, my account balance drops to \$200 - \$50 = \$150.


Problem 5

A bicycle shop is going out of business and has a 30% discount on all bikes. I also have a coupon for an additional 40% off any bike. As compared to the original price, what discount will I get if I use my coupon on top of the going-out-of-business discount? Solve this problem without a calculator.

Answer: 58%

The going-out-of-business discount is a 30% discount, which means I will pay 70% of the original cost of a bicycle. If I use my coupon to get 40% off, this means I will pay only 60% of that already-discounted amount. 60% of 70% is 42% of the original cost, which we obtain by multiplying 0.6 \cdot 0.7 = 0.42. This corresponds to a 58% discount off the original price. So I can get the mattress for less than half its original price when combining both discounts!

Notice that 40% off on top of 30% off does not equal 70% off. Percentages don’t add in this way, they multiply. An easy way to see this is if you took 50% off on top of 50% off. That would be half price of half price, which is a quarter of the original price. Your discount would be 75%, not 50% + 50% = 100%. That would be too good to be true!


Problem 6

You plan to drive to campus for your Monday, Wednesday, Friday classes and you are interested in knowing how many parking spaces are available in each of three parking structures (Gilman, Hopkins, and Pangea). We’ll assume for this problem that these are the only three parking options available. The table below shows how many unoccupied spaces there are in each parking structure at 10am on Monday, Wednesday, and Friday of Week 1.

Gilman Hopkins Pangea
Monday 180 840 190
Wednesday 150 850 200
Friday 165 835 220


Problem 6.1

What proportion of available spaces on Wednesday are in Gilman? Fully simplify your answer without using a calculator and without performing long division.

Answer: \frac{1}{8}

There are 150 + 850 + 200 = 1200 available parking spaces on Wednesday. 150 of these are in Gilman. So the proportion we’re looking for is \frac{150}{1200}. To simplify this, we look for common factors in the numerator and denominator. Both 150 and 1200 end in 0, which means they are both divisible by 10. If we cancel out the common factor of 10 from the numerator and denominator, we can simplify the proportion to \frac{15}{120}. This can be further simplified. Since 15 has factors of 3 and 5, let’s look for factors of those numbers in the denominator of 120. It turns out that both are factors, but suppose we first notice that 3 is a factor of 120. This simplifies the proportion to \frac{15}{120} = \frac{5}{40}. Then we could notice that 40 is a multiple of 5, and simplify the proportion to \frac{1}{8}.


Problem 6.2

On average, Gilman is 80% full at 10am on Monday, Wednesday, and Friday of Week 1. What is the capacity of Gilman? Solve this problem without a calculator.

Answer: 825

First, we use the Gilman column of the table to compute that on average, there are 165 available spaces. We can compute this average quickly without calculation by noticing that 150 and 180 are both equally far from 165 on a number line, so the average of 180, 150, and 165 must be 165. Since 80% of Gilman is full on average, this means 20% of Gilman is not full on average. We have figured out that 20% of capacity is 165 parking spaces, which we can write as 0.2x = 165, or equivalently, \frac{1}{5}x = 165, where x is the capacity of Gilman. Writing 0.2 as the fraction \frac{1}{5} makes it easier to solve for x by hand without a calculator. We can solve for x by multiplying both sides by 5, so the answer is 165 \cdot 5, which can do using long multiplication to get 825.

If we want to avoid using long multiplication, we can use a trick: to multiply by five, multiply by ten and divide by two, both of which are pretty easy to do. 165 \cdot 10 = 1650 and half of that is 825. So, there are 825 parking spaces in Gilman.


Problem 6.3

On Monday, the percentage of occupied spaces in Pangea is twice the percentage of occupied spaces in Hopkins. If Pangea has 950 parking spaces, how many parking spaces does Hopkins have? You may use a calculator for this part.

Answer: 1400

Let’s start with Pangea since we know more information about that facility. On Monday, 190 out of 950 parking spaces in Pangea are available, which means 950 - 190 = 760 parking spaces are occupied. The percentage of occupied spaces in Pangea is \frac{760}{950} \cdot 100\% = 80\%. We’re told that is twice the percentage of occupied spaces in Hopkins, so that means Hopkins must be 40% full. There are 840 available spaces in Hopkins on Monday. Since Hopkins is 40% full, these 840 unoccupied spaces correspond to the remaining 60% of spaces in Hopkins. If x is the capacity of Hopkins, we have 0.6x = 840. We can solve for x by dividing both sides by 0.6 to obtain x = 1400. So, there are 1400 parking spaces in Hopkins.



Problem 7

Every year, San Diego’s Regional Task Force on Homelessness conducts a “point-in-time” count of the number of homeless individuals in San Diego County. They send out a team of over a thousand volunteers to locate and count homeless individuals throughout the county’s shelters, streets, encampments, cars, etc. In 2023, newspapers reported that the count increased by 22% from the 2022 count. Part of that increase was explained by the fact that for the first time, volunteers were able to access and count people on Caltrans property. It was also reported that if we were to exclude the people who were counted on Caltrans property, the count would have still increased by 14%, meaning that the additional area covered could not explain all of the increase from one year’s count to the next. The 2023 point-in-time count was 10,264 individuals.


Problem 7.1

The point-in-time count is an underestimate of the true number of homeless individuals in San Diego County. Why is it always an underestimate?

Answer: Volunteers may miss some people who should have been counted, since it can be hard to locate every single homeless person.


Problem 7.2

How many individuals in 2023 were counted on Caltrans property? You may use a calculator.

Answer: 673

We are told that 10,264 homeless individuals were counted in 2023 and that this was a 22% increase from the 2022 count. We can use this information to find the 2022 count, which we will call x. Since 1.22x = 10{,}264, we can divide both sides by 1.22 to find that the 2022 count was 8413. The 2023 count of people excluding those counted on Caltrans property is 14% higher than this number, which is 1.14 \cdot 8413 = 9591. The difference between the total 2023 count and the count excluding those on Caltrans property is therefore 10{,}264 - 9591 = 673.

Here is another way to solve this problem without actually calculating the 2022 count. Again, let x represent the 2022 count. We want to find the difference between the full 2023 count, which is 1.22x, and the 2023 count without those counted on Caltrans property, which is 1.14x. Therefore, we need to find 1.22x - 1.14x = 0.08x. We know 1.22x = 10{,}264 as before. So, x = \frac{10{,}264}{1.22} and 0.08x = 0.08 \cdot \frac{10{,}264}{1.22} = 673.


Problem 7.3

The actual number of individuals counted on Caltrans property in 2023 was 661, which should be close to your answer to the previous question, but not exactly the same. Did the newspapers make a mistake or is there another explanation for the discrepancy?

Answer: The discrepancy is most likely due to rounding. Notice that when the newspaper reported the percent increases (22 and 14), it gave them as whole numbers even though the actual percent increases could not have been whole numbers.

If x represents the 2022 count, then based on the informtion given in the question, we have 1.22x = 10{,}264. The exact solution to this equation is actually x = 8413.1147541, but since x represents a count of people, it must be a whole number. This means the 1.22 number in this equation is an approximation, because it’s easier (both to report and to interpret) a percent increase that is a whole number and not a decimal.



Problem 8

At a furniture store, suppose every item has a “true value” that represents its worth. Every item in the store has a ticketed price, which is the amount that is printed on the item’s price tag that customers will pay, that is 25% higher its true value. The store wants its employees to be able to buy items at their true value. What percent discount does it need to give its employees off the ticketed price to accomplish this?

In general, if a store marks up prices by p%, what percent discount does it need to give to employees off the ticketed price so that they can buy things at the true value?

Solve this problem without a calculator.

Answer: 20%, or more generally, \frac{100p}{100+p}\%

Let v represent the true value of an item in the furniture store. The store sells the item for v + 25% of v, which can be expressed as 1.25v. We’ll call this the ticketed price, which is the amount that would be printed on the item’s price tag.

Now, just to demonstrate the kind of calculation involved, let’s say the store provides employees with a 25% discount. This would mean that employees pay 100\% - 25\% = 75\% of the ticketed price, which is 0.75 \cdot 1.25v. This simplifies to 0.9375v, which is less than the true value v. This means a 25% employee discount is too much of a discount, since it allows employees to buy items at less than the true value. Since the ticketed price is 25% more than the true value, 25% of the ticketed price is larger than 25% of the true value; since the employee discount is taken off of the ticketed price, it needs to be less than 25%.

The question, then, is what percentage discount should the store provide to employees so that the post-discount price is equal to v? Suppose they provide employees with a discount of d, where d is a decimal between 0 and 1 (in the previous example, for instance, it would have been 0.25, which corresponds to 25%). Then, employees pay (1 - d) \cdot 1.25v. Now, we need to pick the value of d that makes (1 - d) \cdot 1.25v equal to v. Let’s set up an equation:

(1 - d) \cdot 1.25v = v

We need to solve this for d, and fortunately, if we divide both sides by v, this becomes an equation with just one variable, which we can solve with algebra:

\begin{align*}(1 - d) \cdot 1.25 &= 1 \\ 1 - d &= \frac{1}{1.25} \\ d &= 1 - \frac{1}{1.25} \end{align*}

Since 1.25 = \frac{5}{4}, we have:

d = 1 - \frac{1}{1.25} = 1 - \frac{1}{\frac{5}{4}} = 1 - \frac{4}{5} = \frac{1}{5}

This means that the store needs to give employees a discount of d = \frac{1}{5} = 0.2, or 20%, in order for their post-discount price to be equal to the true value of the item they’re buying.

More generally, if a store sells an item for p% more than its true value, v, then the ticketed price is \left(1 + \frac{p}{100} \right)v, or \left(\frac{100 + p}{100}\right)v. For instance, if p = 25, then the ticketed price is \frac{100 + 25}{100}v = 1.25v, which we saw before. Then, the store would need to offer employees a discount of d (where d is a decimal between 0 and 1) such that (1 - d) \cdot \frac{100 + p}{100} = 1. Solving for d gives:

\begin{align*} (1 - d) \cdot \frac{100 + p}{100} &= 1 \\ 1 - d &= \frac{100}{100 + p} \\ d &= 1 - \frac{100}{100 + p} \\ d &= \frac{100 + p}{100 + p} - \frac{100}{100 + p} \\ d &= \frac{p}{100 + p} \end{align*}

\frac{p}{100 + p} is a decimal between 0 and 1, so to convert it to a percentage we can multiply it by 100\%. To conclude, if a store sells an item for p% more than its true value, v, then they must offer employees a discount of \frac{p}{100 + p} \cdot 100\% in order for their post-discount price to be equal to v.

This is a scary-looking formula, but let’s try it in action: if we let p be 25, as it was in the first part of this question, we then have a discount of \frac{25}{100 + 25} \cdot 100\%, or 0.2 \cdot 100\%, or 20\%.


Problem 9

One of your classes this quarter, PTS 1: Principles of Taylor Swift, has the following grading scheme:

There are 4 homework assignments, but your lowest homework score is dropped from your overall grade calculation. Each homework assignment is worth the same amount in your overall grade calculation, even though they have different numbers of available points. Your scores on all parts of the course, before the Final Exam, are as follows:

What is the minimum possible score you can earn on the Final Exam, as a percentage, to guarantee that you finish with at least a B grade (80% overall) in PTS 1? You may use a calculator (tip: you can type arithmetic expressions into Google and it will perform calculations for you!).

Answer: 87.5%

Let f be your Final Exam score in PTS 1, as a decimal between 0 and 1. (We’ll treat f as a decimal because all of your other scores are provided to us as decimals, or fractions, as well.) If we knew f, we could calculate your overall grade right now:

20\% \cdot \text{Midterm 1} + 20\% \cdot \text{Midterm 2} + 30\% \cdot \text{Homeworks} + 30\% \cdot f

We don’t know f, but we do know the other variables in the above expression. Your Midterm Exam scores are given to us directly, but your average score on homeworks, after dropping your lowest homework, requires a bit of calculation to find. First, we need to find which of the four homeworks you did the worst on, so that we can exclude it when calculating your average homework score. That happens to be Homework 2, which you scored a 72% on; we can find this without a calculator by noticing that 25 \cdot 4 = 100, so \frac{18}{25} = \frac{18 \cdot 4}{25 \cdot 4} = \frac{72}{100} = 72\%. This is lower than Homework 1, which you scored a \frac{3}{4} = 75\% on; Homework 3, which you scored a \frac{9}{10} = 90\% on, and Homework 4, which you scored a \frac{12}{13} = 92.3\% on (\frac{12}{13} is a tough fraction to simplify without a calculator, but you don’t need to, since has to be greater than \frac{9}{10} which you know is 90%, and thus it has to be greater than 72%).

This means that the three homework scores that we’re going to include in your average homework score calculation are \frac{3}{4}, \frac{9}{10}, and \frac{12}{13}. Note that we’re using the fractional forms rather than something like 92.3% because there’s no need to round at this stage. The average of these three fractions is \frac{1}{3} \cdot \left(\frac{3}{4} + \frac{9}{10} + \frac{12}{13}\right); the result is your average homework score. We’ll leave it like this for now so that you only need to put one (big) expression in the calculator.

Now that we have your average homework score, we know most parts of your overall grade:

20\% \cdot \frac{98}{100} + 20\% \cdot \frac{16}{38} + 30\% \cdot \frac{1}{3} \cdot \left(\frac{3}{4} + \frac{9}{10} + \frac{12}{13}\right) + 30\% \cdot f

Simplifying everything before 30\% \cdot f in a calculator gives us 53.75%.

It’s worth commenting on how to evaluate such an expression in a calculator. It’s easier to treat percentages, like 20%, as decimals, like 0.2, when using a calculator. Here you can see the exact expression we typed into a Google search (no need to use a physical calculator!); 0.5375… is the same as 53.75..%.

Now, we have that your overall grade is

53.75\% + 30\% \cdot f

We need to find the smallest value of f such that your overall grade is at least 80%, or in other words, the smallest value of f that satisfies the inequality

53.75\% + 30\% \cdot f \geq 80\%

Let’s solve:

\begin{align*} 53.75\% + 30\% \cdot f &\geq 80\% \\ 30\% \cdot f &\geq 26.25\% \\ f &\geq \frac{26.25}{30} \\ f &\geq 0.875 \end{align*}

So, to earn a B in PTS 1, you’ll need at least an 87.5% on the Final Exam. Good luck!


Problem 10

Your professor for PTS 1: Principles of Taylor Swift can grade 150 homeworks in 50 minutes. The TA can grade 150 homeworks in 75 minutes. If the professor and TA work together, how many minutes will it take them to grade 150 homeworks? Solve this problem without a calculator.

Answer: 30 minutes

To actually solve this problem requires a bit of math, but it turns out we can narrow it down to the correct answer using the process of elimination and some common sense. First of all, some of these answers are tempting because they come naturally from the numbers given in the problem. For example, 125 minutes is the sum of 50 minutes and 75 minutes. Likewise, 62.5 minutes is the average of 50 minutes and 75 minutes. But does it make sense to sum or average the times? No! Think of it this way: there is a fixed amount of work to be done (grade 150 homeworks). Working alone, the professor can knock out that task in 50 minutes. With help, the professor will finish the task even faster. So the answer must be less than 50 minutes.

That leaves two options: 25 minutes and 30 minutes. Notice that 25 minutes is half the time it would take for the professor to grade the homeworks alone. If the professor were able to clone themself, then the two professors could grade the homeworks in 25 minutes, working double the speed of one professor. But the TA is not quite as fast as a second professor. So it will take more than 25 minutes for one professor and one TA to complete the grading. This leaves only one answer that can make sense, so the answer must be 30 minutes. We can solve the problem without doing any math!

It’s always good to double-check that our answer is correct. Let’s actually solve this problem using math now. To start, we know that the professor grades at a rate of 150 homeworks per 50 minutes, which is 3 homeworks per minute because \frac{150}{50} = \frac{3}{1}. The TA grades 150 homeworks per 75 minutes, or 2 homeworks per minute. This means that in one minute, the professor can grade 3 homeworks and the TA can grade 2 homeworks, for a total of 5 homeworks.

When working together, the professor and TA grade at a rate of 5 homeworks per minute. So how much time do they need to grade 150 homeworks? 5 homeworks per minute is equivalent to 150 homeworks per 30 minutes, since \frac{5}{1} = \frac{150}{30}. Therefore, it takes them 30 minutes to complete their grading task, the same result we obtained through process of elimination.


Problem 11

You are working in a spreadsheet editor and you highlight several consecutive rows of your spreasheet. The first highlighted row is row 543 and the last highlighted row is row 897. How many rows are highlighted?

In general, if rows n through m are highlighted, with n<m, how many rows is that?

Solve this problem without a calculator.

Answer: 355, or more generally, m-n+1

It might help to visualize the problem, as in the image below.

One way to solve this problem is to count the rows and look for a pattern. Row 543 is row 1 of the highlighted section. Row 544 is row 2 of the highlighted section. Row 545 is row 3 of the highlighted section. The pattern seems to be that row x is row x - 542 of the highlighted section. Therefore, with x = 897, row 897 is row 355 of the highlighted section. Since this is the last row of the highlighted section, there are 355 rows highlighted.

Let’s also think about the problem a different way. Imagine we had highlighted all the rows up to and including row 897. That would mean we had highlighted 897 rows. But we actually highlighted 542 fewer rows than this (since the first highlighted row was row 543, the first 542 rows were not highlighted). So the total number of highlighted rows is 897 - 542 = 355.

In general, imagine highlighting all m rows up to an including row m. That would be m highlighted rows. But if we only start highlighting at row n, this includes n-1 rows that should not be counted. So there are m-(n-1) = m - n + 1 rows highlighted when we highlight rows n through m.


Problem 12

There are six different colors of M&M candies: yellow, blue, green, red, orange, and brown. You and your friend are playing a game. You will pick five M&Ms from a bag with your eyes closed. Your friend will give you a prize if none of the candies you picked are yellow, green, or blue. Unfortunately, when you play this game, you don’t win the prize. What does this say about the candies you chose? For each statement below, say whether the statement must be true, may be true, or cannot be true.

  1. All five M&M’s were the same color, either yellow, blue, or green.
  2. All five M&M’s were the same color, either red, orange, or brown.
  3. All five M&M’s were one of the following colors: yellow, blue, green.
  4. All five M&M’s were one of the following colors: red, orange, brown.
  5. One or more M&M’s was yellow, blue, or green.
  6. One or more M&M’s was red, orange, or brown.
  7. None of the M&M’s were yellow, blue, or green.
  8. None of the M&M’s were red, orange, or brown.

To win the prize, you must select five M&Ms and end up with no yellow, no green, and no blue M&Ms. Since you didn’t win the prize, among your five selected M&Ms, there must be at least one M&M that is yellow, blue, or green. That’s all we know for sure about the five M&Ms we selected.


1. All five M&M’s were the same color, either yellow, blue, or green. May be true.

You could have gotten either of the groups of candies shown below, because both of them would fail to win you the prize. It’s possible that all five candies were green (or yellow, or blue) but it’s not necessarily the case.


2. All five M&M’s were the same color, either red, orange, or brown. Cannot be true.

If this were true, the condition for winning the prize would be satisfied: none of the candies you picked would be yellow, green, or blue. So this cannot be true.


3. All five M&M’s were one of the following colors: yellow, blue, green. May be true.

You could have gotten either of the groups of candies shown below, because both of them would fail to win you the prize. All we know is that at least one of the candies has to be yellow, blue, or green. They might all be those colors, or they might not.


4. All five M&M’s were one of the following colors: red, orange, brown. Cannot be true.

If all five candies were red, orange, or brown, such as in the picture below, you would have actually won the prize. Since you didn’t win the prize, that tells us this can’t be true.


5. One or more M&M’s was yellow, blue, or green. Must be true.

This has to be true, because if it were not true, you would have won the prize. The opposite of “one or more” is “none” and the condition to win the prize was that none of your candies could be yellow, blue, or green.


6. One or more M&M’s was red, orange, or brown. May be true.

It could be the case that you had a red, orange, or brown candy among your five, as in the first picture below. But you may have also not had any red, orange, or browns, like in the second picture below. For both groups of candies shown below, you would have failed to win the prize.


7. None of the M&M’s were yellow, blue, or green. Cannot be true.

This is exactly the winning condition, but we know you did not win the prize, so this cannot be true.


8. None of the M&M’s were red, orange, or brown. May be true.

You failed to win the prize because you had at least one candy that was yellow, blue, or green. You may have had some red, orange, or brown candies included among your five, or you may not have. Either of the two groups of five below is possible, because you would not have won the prize either way.


Once you’ve worked through the above problems, watch the following video. It summarizes key concepts covered in the pretest and discusses important test-taking strategies you should keep in mind moving forward, both in DSC 10 and your other classes.


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