How to Analyze an SHSAT Math Problem:

Welcome to the SHSATAcademy math course. Let’s learn some basic techniques for solving math problems on standardized math exams like the SHSAT or even the SAT. Every student should know the following approaches and understand the weaknesses of each. Subscribers to the full course will gain access to some additional advanced strategies.

**Understanding Methods and Strategies To Solve Math Problems**

There are only two general approaches — and several sub-strategies — to solve a math problem on the SHSAT or SAT or PSAT. Successful test takers will want to learn to do both.

1) Apply your knowledge of relevant mathematics to solve the problem directly from the question prompt without reference to the answer choices.

2) Look at the answer choices and apply a strategy that leverages the answer choices to find the correct answer.

**Solve the problem before referring to the answer choices:**

Over 90% of the problems in the math section can be solved before referring to the answer options, and that requires knowing the correct theorems or concepts and how to execute them. In essence, you must know your subject material. There is no substitute for gaining comfort and familiarity with the math. Students who have stronger prior experience in topics like algebra or geometry, two prevalent topics on the SHSAT, tend to perform substantially better on the math exam. That is why we provide extensive key topic lessons and related exercises in later chapters to help students identify and fill their math content knowledge gaps. There are abundant concepts and theorems required in the core curriculum for the SHSAT math exam, and this course includes over 100 lessons with videos and related exercises organized into 10 lesson chapters. It is the most extensive SHSAT math content toolbox we know of — applicable to students who need to build their skills from the ground up to experienced students who only need to access select chapters to fill the gaps in their content knowledge. In either case, almost every student is sure to learn something new from our lessons.

Although most every problem can be solved formally without referring to the answer choices, there are often faster, more efficient ways to solve problems on standardized math exams. Remember, the SHSAT does not require you to show your work like a math teacher in school. For example, word problems almost always can be solved by translating the problem into equivalent algebra equations and solving over many steps, but often there are more effective approaches. The SHSAT rewards clever approaches to unlock the single correct multiple choice answer with less work, fewer calculations, and less risk for error. Every SHSAT math problem is designed to be solved in 30 seconds or less with the right approach, but the question becomes, “How do I find the most efficient approach to each problem?” There is a good reason why one step in the Math Path to the correct solution above includes a review of the answer choices. A lot of useful methods are available to unlock the correct answers more quickly, and they are only limited by the creativity of the student. For all the potential deficiencies of standardized exams, this is one reason the SHSAT and the SAT have and likely will stand the test of time. Both provide a great way to test math problem-solving creativity, not just application of formulas. The following provides insight into some of the ways to leverage the answer options on the standardized math exam.

**Look at the answer choices:**

**a. Backsolving** – This approach entails plugging the different answer options back into the formula or conditions of the question prompt to determine which answer satisfies the question conditions. *Example:*

Find the greatest common factor (GCF) between 150, 225, and 750.

A. 15

B. 25

C. 75

D. 125

One formal, robust method to solve this problem is to calculate the prime factorizations of all three numbers and take the product of only the common prime factors to each. On the other hand, some students will look at the answer options and, starting with the greatest, plug each back in to check for divisibility, which will determine if the answer is a factor. 125 does not divide into 150, so that is incorrect. Moving to the next largest answer option, 75 divides into 150, 225, and 750. It must be the correct answer. Time is saved, possible calculation errors are avoided, and the student can be confident they have the correct answer. This approach works well when the calculations required to test answer choices are simple and can be limited. An example where this approach will not work too well is the following.

The average of 15, 23, x and 8, 14, 22, and 2x are the same. What is the value of x?

A. 5

B. 8

C. 10

D. 12

In this example, the calculation required for each answer is more complicated. It is easier to set up the equation and solve for x.

**b. Trial and Error (Guess and Check)** – In many cases the answer options include variables. In these cases, guessing a reasonable value for the variable(s) and checking the results for each answer choice can be effective. *Example:*

If R + S is even and S – T is odd, then which of the following must be odd?

A. R

B. S

C. R + T

D. S – R

In lieu of general scenario analysis, students might guess numbers for each variable. Suppose R = 1. A guess of S = 3 would satisfy the first condition: 3 + 1 = 4 — an even result. Turning to the second condition, the student might guess T = 2 because S – T = 3 – 2 = 1 — an odd result. Once the conditions are satisfied, the student can check the answer choices. S – R = 3 – 1 = 2 — an even result. Option D is not always odd. On the other hand, R + T = 1 + 2 = 3 — an odd number. Option C works, and it is the correct answer.

The problem with substituting guessed values is the potential for false positives. For example, if the student checked option A or B, they would realize either of those is an odd result also based on the guessed numbers. R = 1 is odd. S = 3 is odd. Could those also be correct answers? In this case, option A and B are not because we could have just as easily chosen each variable to be even. In fact, the student must run a second scenario substituting an even value for R, like R = 4, to be completely confident option C is the correct answer.

**c. ****Estimates** – The ability to round numbers and make effective estimates can help students solve some questions rapidly and accurately. *Example:*

Which of the following is closer to 0.19?

A.

B.

C.

D.

In lieu of a proper long division calculations, students could estimate the equivalent fraction for 0.19; 0.19 ~ 0.20 = and compare that on an apples to apples basis with the answer options by estimating the answer choices in the format . A ~, B ~ , C ~ , D ~. All the answers are smaller — they have larger denominators. As a result, the correct answer is option A where the denominator is closest to 5. As a result, is closest to 0.19.

The shortcoming of this strategy is that easy estimates are not always available and often not accurate enough to differentiate some or all of the answer options, but the method can be useful where it applies.

**d. Reasonableness Test/Common Sense Checks** – In many problems it is useful to assess the scale or range of values to determine which answers can be quickly eliminated. *Example:*

Determine the value of ?

A. 11

B. 100

C. 101

D. 1010

The correct answer is approximately 7,000 ÷ 7 = 100. Students can quickly eliminate 11 and 1,010 because they are off by factors of 10. Note, both answers are traps designed to catch students who make typical long division mistakes. 11 results when students fail to insert a 0 for the ten’s place in the long division and 1,010 results when students divide by 7 rather than 70. The correct answer must be option B or option C, but the estimate of 100, in this case, must be low because the calculation above rounded 7,070 down to 7,000. Hence, the correct answer can only be option C, 101.

The method is not always effective at eliminating all wrong answers, which means other approaches and calculations must still be made in many cases.

**e. Patterns** – SHSAT and SAT math answer options frequently demonstrate patterns by design, and these patterns can often be used to help students more carefully or more quickly determine the correct answer. Typical patterns include a) multiples b) opposites c) imitations of the correct answer and more. Multiples, for example, often present answers that are doubled and halved from one another, which indicates the answer depends on doubling or halving a certain value and students must make sure they do that the correct number of times. Likewise, opposites present a positive and negative option for the same value(s), which should alert students to pay careful attention to signs. Also, many answers try to imitate the correct answer. For example, x^{2} may be an expected part of the result, so that term appears in several answer choices. Test takers must take care that the other parts of their calculation are correct. *Example:*

5×6×7×8 is divisible by which of the following?

A. 2^{4}

B. 2^{5}

C. 2^{6}

D. 2^{7}

Note the pattern among the answer choices. That can help students resolve the problem. Let’s look at a few possible approaches. The first approach is to complete the multiplication calculations and try to backsolve. The product in the question is 1,660. Divide that value by each of 16, 32, 64, and 256 to see which results in a whole number quotient. This solution, using backsolving, may be the longest and result in the most calculations and possibilities for errors. Another simpler approach is to take advantage of the fact that all the answers are in a particular pattern — they are all powers of 2. Some students might count the factors of 2 in the numbers included in the question prompt to see how many powers of 2 are in the product. Six has one factor of 2 — 2 × 3. 8 has three factors of 2 — 2 × 2 × 2. In all, that amounts to 4 factors of 2 in the product, so the correct answer must be option A, 2^{4}. No multiplication calculations were required. In this approach, the student is leveraging their knowledge of factors and the pattern of answer choices to find the correct answer—an abstraction. There is no limit to the potential abstractions students can make to find faster, easier solutions. The SHSAT and SAT math sections are designed to reward this informal creative thinking.

**Start to Train Your Abstract Brain**

The SHSAT, and the SAT for that matter, reward abstract solutions in math that prove to be faster and more effective. These solutions are abstract because they result from test takers observing some aspect of the problem that is tangentially related; the methodologies are not the direct solutions made by the conventional methods. For example, the idea that every squared real number must also be positive helped to solve a problem in the last chapter. Practice and familiarity will improve these insights. Let’s begin to train your abstract brain.

**If a + b = 9, then what is the value of 3a + 3b?**

Resist the temptation to solve every equation for a single variable like you have done hundreds of times over in math class. That won’t work in this case because there are two variables and only one equation. You cannot know the value of a and the value for b individually, but that is okay because the question does not ask for the value of a or the value of b. Trained test takers will make certain they are answering the question asked. In this case, the goal is to express the desired result, 3a + 3b, in terms of the known value, a + b. As you probably already surmised, the expressions look alike, so the solution might come easily as it often does on SHSAT math problems. The desired expression to answer the question is just three times the known expression, so it must also be three times the known value: 3a + 3b = 3(a + b) = 3(9) = 27. Let’s look at a slightly more complicated variation.

**If a + 2b = 9 and 2a + b =15, then what is the value of a + b?**

In this case, it is possible to find a value for a and b individually by solving the system of two equations. However, it is a lot easier if you recognize this is similar to the previous problem in reverse. Add the two equations to get 3a + 3b = 24. Extract the common factor 3(a + b) = 24 and solve: a + b = 8. There is no need to find a or b individually even if you can. The effort will only take more time and take more calculations, which translates into more opportunities to make a mistake. Let’s take things a step further.

**If 15x – 15 = 99, then what is 30x + 40?**

The goal is to manipulate one expression to look like another, not solve for the variable. You could solve for x, but x is not even an integer, so that solution gets messy. It is easier to recognize the similarities between the expressions and utilize that observation. The second expression looks similar to twice the first. 2(15x – 15) = 30x – 30. In fact, the second expression in the question prompt is +40 – (-30) = 70 more. 3x + 40 = (3x – 30) + 70. Add 70 to twice the known expression value. 3x + 40 = 2(99) + 70 = 198 + 70 = 268. Math becomes more fun when you can make it easier. Try another.

**If a + 2b + 3c = 25 and 3a + 2b + c = 35, what is the value of a + b + c?**

Many students begin with an explanation why this problem cannot be solved. Recall, the goal is not to solve for a single variable. Add the equations again. 4a + 4b + 4c = 60. 4(a + b + c) = 60. a + b + c = 15. The answer to the question is not only possible; it is quick. Let’s shift gears and look at a word problem, a large percentage of the question formats on the exam.

**If 3 hamburgers and 2 orders of fries cost $5.80, and 2 hamburgers and 3 french fries cost $4.30, then how much more do hamburgers cost?**

An insightful student might realize one more hamburger less one more french fries order is equivalent to the difference between $5.80 and $4.30. In other words, hamburgers cost $1.50 more, which answers the question. Set up the equivalent equations to see this example is much like the previous problems. 3H + 2F = 5.80 and 2H + 3F = 4.30 where H and F are the prices of one hamburger and one order of french fries respectively. Similar to before, subtract one equation from the other. 1H – 1F = 5.80 – 4.30 = 1.50. Another example follows.

**If three-quarters of a house gets c****ompleted in 60 days, then how many more days will it take to finish the house?**

If represents 60 days of effort, then the remaining must represent one-third of 60 or 20 days more. The more formal approach is to set up a corresponding proportion equation and solve, and that does not take too much longer in this case. Let’s try it again with another question.

**If five machines produce five widgets in five minutes, then how long will it take for 100 machines to make 100 widgets?**

The corresponding proportion equation for this work problem is a little trickier, so many students will fall back on their pattern recognition skills and assume 5-5-5 corresponds to 100-100-100. Unfortunately, 100 minutes is wrong, and that is the wrong easy approach – something to keep in mind when attempting abstract methods. The logic must be sound. In this case, it appears the per unit rate is 1 machine makes 1 widget every 5 minutes. That is the same as 5 machines make 5 widgets in 5 minutes. At that rate, any N machines will make N widgets in just 5 minutes. There are just more of them producing more overall in the same time frame. As a result, the correct answer is 5 minutes. 100 machines make 100 widgets in the same 5 minutes. Once you begin thinking more abstractly, it may be true you will find the abstract wording of problems easier to translate.

**If X widgets can be purchased for Y dollars, then what is the cost of Z widgets?**

Many students will be unsure where to begin in the absence of numbers. They feel lost in the woods without a compass. The approach and the logic, however, should not change. The per unit cost of a widget is the overall cost divided by the amount, . If instead, we buy Z objects then we multiply the per unit cost by the number of objects Z. The result is . Or perhaps…

**What is the greatest common factor between every two-digit number decreased by the sum of its digits?**

What does this question mean? Where do we begin? There are 90 two-digit numbers. How can we easily find the GCF? Think about what is going on. Try an example: the lowest two-digit number 10. The number minus the sum of digits is 10 – 1 – 0 = 9. Increase the value by one. 11 – 1 – 1 = 9. The result is still 9 because every time the number increases by one the units digit that is subtracted increases by one to offset the change. When the tens digit increases by one the number still goes up by one, but the subtracted digits decrease by 9 – 1 = 8. The result is one more multiple of 9: 20 – 2 – 0 = 18. Thinking in the abstract frame of mind as the above problems, the expression to calculate for 20 is two times the expression for 10. It should be no surprise the answer is two times the value as well. In all cases, the result will be divisible by 9, so that must be the greatest common factor. Try this question.

**The sum of an odd number of even numbers must be**

…a headache. Just kidding. A trained test taker will actually say, “An easy correct answer!” Try an example of an even number, 2. An odd number of those might be three 2s. The sum of three 2s is 2 + 2 + 2 = 6. Change the numbers, but the result will still be an even integer multiplied by an odd integer, which must be even. Abstract approaches are not limited to word problems or ratios. Let’s see an example in geometry.

**If the sides of a triangle with area A are doubled, then what is the new area?**

Most students want to apply a formula like . In the absence of values for base or height, they may be unsure how to proceed. Most believe the new area is double the old area without a justifiable reason. It may be easy to realize that doubling all sides doubles the base, but only some realize it also doubles the height.

The new triangle area is 4 times the old area, and there is no need to know the particular details. Trained test takers will know the area is always the square of lengths, so if the lengths are scaled by a factor, F, then the area will be scaled by F^{2}. It does not matter what the shape is. Likewise, double the sides for a three-dimensional shape, and the new volume will be F^{3} = 2^{3} = 8 times the original volume.

These are but a few examples of abstract approaches that are limited only by your creativity and experience. Hopefully, they serve as a warm-up for your brain as we progress forward to gain more experience. The coming chapters will provide short lessons covering all the material on the SHSAT math exam. Each lesson chapter will include related skill exercises and finally SHSAT level problems that incorporate the same concepts. Recall, there are many more lessons and video tutorials, over 100 in all, available online at testprepshsat.com.

**How to Analyze an SHSAT Math Problem – The 7 Step Math Path**

- Read the question carefully, and consider all the words in the question; this is the most important step.
- Assess what information is given including diagrams and charts.
- Look at the answer choices to identify any relationships or patterns that may be useful.
- Determine which math topic(s) will help solve the problem. In this effort, diagram the problem to help visualize the problem or guess a value that may be appropriate. Action, not inaction, helps more often than not at this stage.
- Can you find a solution that would take about 30 seconds? If you cannot find it in 30 seconds, then you may want to skip the problem and return in a later pass.
- Execute your solution.
- Before marking your answer, check that you are answering the exact question asked. Too often careless errors — answering the wrong questions or using intermediate solutions by mistake — are the downfall of test takers.