ISEE Upper Level QR Study Guide with Solutions

The ISEE is the Independent School Entrance Exam and is required by many middle and high schools for the admissions process.

This guide goes through problems from the Upper Level Quantitative Reasoning section of the publicly available What to Expect on the ISEE practice test (scroll to the bottom of the pdf file to see the practice test), written by the ERB and found at This is one of the best ways to study for the exam, because the problems are written by the same people who write the actual exam and will be as close to the actual test as possible.

I wanted to put down, in simple language, my thoughts on each problem after using this guide for a number of years while tutoring the ISEE. The guide is very straight forward. Use this as a tool to understand each problem and what the test makers are thinking when they are writing. If you can see that, the ISEE will be a piece of cake.

I have been tutoring the ISEE for nearly 10 years and have been on both sides of the aisle — both as an admissions officer and as a tutor. I know the ins and outs of the exam and what admissions officers are looking for in terms of scoring and essay.

For more information check out my website.

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Photo by Changbok Ko on Unsplash

Each number below corresponds to the number problem in the Quantitative Reasoning section of the ISEE Upper Level Sample Exam (exam at bottom of pdf file). Please attempt each problem first on your own and then consult the guide below.


Note: The ISEE will use this type of problem frequently. They define some ‘function’, in this case the function is represented by n*, that is not typical mathematical function that you may have seen before. So the way I would read this in words is “some number starred is four times that number plus three”. So now if you need to plug in a number, which they want you to use the number 8, you can just replace “some number” with 8.

(A) Correct. 4*8+3=32+3=35


Note: there are a couple ways of thinking about this problem. In my opinion, the simplest way is just to do the two algebraic manipulations necessary to get the answer, i.e.

x − y = 3
x = 3 + y
x − 3 = y

Done. Another way to do this is to think about which expression from the answers you could plug into y and get a true statement.

x − y = 3
x − (x − 3) = 3
x − x + 3 = 3
3 = 3

Done. Since we got 3 = 3, which is a true statement, x − 3 is correct.

(B) Correct.


Note: how much is left off by only going to 998? Just 1000 and 999.

(A) Correct. We lose 1000 + 999 = 1999 by removing the last two numbers from the sum.


Note: There are two ways of solving this problem (1) the math-y way and (2) the non-math-y way. First the math-y way. An increase of 10% in the base is equivalent to writing 1.1b. A 20% decrease in the height is equivalent to writing 0.8h. The area of a triangle is 12 bh, so the proportional decrease in the area is just 1.1*0.8, which we know 8*11 = 88 so 1.1*0.8 = 0.88 or a 0.12 decrease. Now the non math-y way. To get a feel for this problem let’s consider a square. Let’s say we increase the base of the square by 20%. Then the area increases by 20% (1.2b * h = 1.2 * Area). Separately, if we increase the length of the height by 10% the area increases by 10% (b * 1.1h = 1.1 * Area). But, if we increase the base by 20% and the height by 10%, we get a 32% increase in Area (1.2b ∗ 1.1h = 1.32 * Area).

(See Figure 1.)

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Figure 1

The same idea works for a triangle, as we have in the problem. Just because the shape is a triangle does not mean the proportional increase or decrease in area is somehow halved or different than the square. Because it is a proportion the decrease is the same for the square and triangle. When we have a simultaneous 10% increase in base and 20% decrease in height, the decrease in height removes a portion of the increase in base. The same strange thing occurs at that top right corner, but in this case we are losing a small portion rather than gaining it. If we only decreased the height by 20%, we would lose 20% of the area. If we only increased the base by 10%, we would gain 10% in area. But when we do both, we gain only 8% from the increase because 2% of the original area is lost from decreasing the height. (See figure 2.)

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Figure 2

If you can see this result, you are sitting pretty for this type of problem on exam day. The reason for this is because there’s only one answer choice that is greater than 10%, so we know it must be 12% (without even doing any of the math). If we only need to see that a simultaneous 10% increase in base and 20% decrease in height will result in some decrease in area greater than 10%, we are done. The ISEE test makers purposefully design the answers in this way so if you can recognize this premise you already have the answer without calculating any results.

(D) Correct.


Note: A quadratic function can be written in many ways. We have two of those ways here. Factored form on the left hand side (LHS) of the equation, and general form on the RHS. General form always looks like: y = ax² +bx+c. And if we say the factored form is (x+d)², then c = d² and b = 2d. This result comes directly from the distributive property. The last term in general form is always the product of the factors, and the middle term is the sum of the factors. But when we have both factors are the same, it’s just 7²=49 and 7+7=14.

(B) Correct.


Note: So our new sum is 370 + 85 = 455. Then we just divide that by 5.

(C) Correct. 455/5= 91


Note: The first thing I thought of with this problem is the longest perimeter possible. The longest perimeter comes from the easiest Area to calculate, l = 110, w = 1. So it makes sense that the shortest perimeter would be the opposite of this method; to spread out the perimeter evenly across the length and width. So we need to determine two numbers that multiply to 110, and are as close as possible in value. These numbers are 10 and 11.

(A) This must be the most common wrong answer, because once you determine 10 and 11 are the length and width, you add them and then you’re done, right? Wrong. You only have half of the perimeter. Try multiplying that number by 2 and you should have the correct answer.

(C) Correct. 10+10+11+11=42


Note: This question is verging on silly to me. But good news for us, it’s fairly easily solvable. The first part of the question says the potato is cool to start, so that rules out (B and C). A potato that is over 300 degrees F is not cool. So between (A and D), the potato can’t become hotter than the temperature in the oven, so (D) is not possible.

(A) Correct.


Aside: Similar triangles have the property that corresponding sides are proportional in their ratio.

Note: So this problem does us the favor of showing us the proportional increase in the sides. The increase from QR to TU is by a factor of 3/2. So the increase in size of QS to TV will also be by a factor of 3/2.

(A) This is not an increase in size at all, so this can be ruled out.

(B) Correct. One and a half times the size of 4cm is 6cm.


Note: I think the best way to go about this problem is to change everything to powers of 3. We know that 9 = 3², so why not just write everything in this manner? If we do that we see,

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(B) Correct.


Note: this is just a matter of noticing when Jane stops moving, then reading the graph. So part (1), Jane stops moving when here distance doesn’t change as time continues. This occurs at 0.50 hours. Part (2), read the graph for how far Jane is from home. She is 0.25 miles from home.

(A) Correct.


Aside: ‘mean’ is the same as average. To calculate the mean we add up all of the numbers and divide by the number of numbers. But, we can also think of mean like a balance point. If all of the numbers were weights on a ‘number line seesaw’, the point at which you would have to place the base of the seesaw for it to not tip left or right is the mean.

Note: My instincts for this problem is that 3 of answer choices will change and one will not. This is just because we are increasing all of the values by the same amount. So many metrics will also just increase proportionally, but I’d bet that one stays relative to the size of the data set.

(A) If every value is increasing by the same amount, we can see that this idea of the ‘balance point’ (i.e. the mean) will just increase by the same amount.

(B) If the median is the ‘middle number’ when the list is written from least to greatest, this number will certainly increase when all of the numbers increase.

(C) The mode is the most frequent number. If all of the values are increasing, then certainly this value will increase as well.

(D) Correct. The range is the greatest number minus the smallest number. If all of the values increase by the same amount, this value will stay the same. The range is just relative to the greatest and least values, so as long as they stay the same distance apart the range won’t change.

Note: We didn’t even look at the dataset to determine this answer choice. I always skip over the data first and get to what the question is asking before taking the time to look through each entry.


Aside: number cubes are dice.

(B) Correct. If Maud and Jim both only won when they received a sum of 6, then their probability of getting a point would be the same. But Maud has another way of winning, so her probability is greater.


If the data are symmetric about 7, then 8’s frequency will be the same as 6’s. And 9’s the same as 5’s. And so on.

(D) Correct. So 8’s frequency is 7, 9’s frequency is 4, 10’s is 2, and 11’s is 1. So 7+4+2+1=14.


Aside: If we have a function where the biggest exponent on any x-value is 2, then this is a quadratic function and looks like a parabola.

Note: This function is a parabola and is ‘stretched’ by a factor of 2 and moved up by 1 unit, but has no horizontal shifts. This means that the function is symmetric about y = 0 and opens upwards because the leading coefficient is positive (i.e. the number in front of the x² term is positive). So this means that the highest point achieved on this interval of −2 ≤ x ≤ 1 occurs the furthest away from the vertex (or y=0 in this case). So this is when x=−2.

(C) Correct. 2(−2)² +1=2*4+1=9.


Note: The intuition here is key. Squaring a decimal makes it smaller, and square rooting a decimal makes it bigger. Then, squaring a number greater than 1 makes it bigger, and square rooting a number greater than 1 makes it smaller.

(B) Correct. This one shows what was stated in the note. g(0.9) < f(0.9) because squaring a decimal makes it smaller. And f(1.1) < g(1.1) because square rooting a number greater than 1 makes it smaller.


(D) Correct. The difference between the two speeds is exactly what we need. If they are both running at constant speeds, the relative speed they are moving together is just the difference between the two of them. If John was running at 5m/s and Erin running at 8m/s, then the difference, 3m/s, is what would allow us to see how long it would take for Erin to catch up with John because she would be gaining on him at a rate of 3 meters every second.


Note: Pick a side a stick to it. I chose the upside down triangle on the reference cube. Now I’m looking for which net will give me an upside down triangle if I fold the cube so that it appears in front.

(A) Correct. This one shows the upside down triangle in front, and if I fold the entire net back on itself, I get a circle on top and a triangle to the right face.

End of word problems
Begin quantitative comparison problems

Aside: the problems will progressively get more difficult up until the quantitative comparison problems, then the difficulty level loops back where the problems become easy again and then become more difficult until the end of the section.

Note: in quantitative comparison problems the answer choices are always exactly the same. So you should have the answer choices memorized by the time of the exam.

(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.


Aside: Order of Operations: PEMDAS (Parentheses Exponents Multiplication Division Addition Subtraction)

(C) Correct. Parentheses: (4 + 3) = 7. Multiplication: 2 * 7 = 14. Addition: 5 + 14 = 19.


Aside: If the question states Note: Figures not drawn to scale, they aren’t.

Area of a rectangle is A=l*w and perimeter is the sum of the side length sp=l+l+w+w. So in this case, Area=2x*x=2x² and P =2y+2y+y+y=6y. If 2x² =18 then x=3. And if 6y=30 then y = 5.

(B) Correct. y > x.


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(C) Correct. They are the same.


Note: we only care if the total value of the quarters is more than $3. So to determine this we just try to make the quarters add up to as much as we possibly can. So start adding quarters until it doesn’t fit into $4.50 anymore (including the dimes). Let’s see, 4 quarters gives us a dollar and then we would need 8 dimes, that’s $1.80. 8 quarters and 16 dimes gives us $3.60. So we’re getting closer. We need 90 cents left. Luckily that could be 2 quarters and 4 dimes. So 10 quarters and 20 dimes gives us $4.50. So the max value of the quarters is $2.50 which is still less than $3.00.

(B) Correct.


Aside: Parallel lines have the same slope.
Aside: Slope intercept form: y = mx + b has slope of m.
Note: So given these two asides, the slopes of both lines are 3, and this is greater than -3.

(A) Correct.


Note: The greatest area we can make is one where the sides multiply to be the largest value. This occurs when the sides are closer in length to one another. Let’s see this. If I make the sides as far apart in magnitude as I can, I would have one side as 24 and the other side as 1. The area of that rectangle is 24 but the perimeter is still 50. Now let’s see what happened if I make the sides as close in length as I can. We’ll set one side as 12 and the other as 13, this gives a perimeter of 50, but has an area of A=l*w=12*13=156. This area is now greater than the 144 in column B. So we have one instance where the area is less than and another instance where the area is greater than, so the relationship cannot be determined.

(D) Correct.


Note: Whenever we must determine the area of an irregular figure, or some shaded portion of a larger figure. We can typically take the area of the larger figure, then the area of the smaller removed portion, and take the difference between those two areas. So in this case we have the area of the large square is x*x=x². And the area of the smaller removed portion is y*y=y². So the area of the shaded portion is x²− y².

(A) Correct. It doesn’t matter what the values are, if we are subtracting an extra xy in column B, it will be less than if we didn’t subtract that quantity.


Aside: Consecutive means, ‘in a row’.
Note: So we only care if the greatest of the 3 consecutive integers is greater than 10. So why note just make the three consecutive integers with a greatest value equal to 10? 8 *9 *10 = 8 * 90 = 720.
(B) Correct. The three numbers must be less than 8, 9, and 10, because 720 is larger than 210. So the greatest of them would be certainly less than 10.


Note: I would first rewrite these in a similar fashion (aka distribute the 25 in column B). So if I do that, Column A stays the same, and Column B becomes 25n − 25. Now I want to ask myself is there any situation where I can think of a number for n that when I subtract 25 from it I get a value greater than when I only subtract 1. The three ‘types’ of numbers I need to try out are, positive, zero, and negative. Positive numbers for n, certainly when I subtract 25 I get a smaller value than when I subtract 1. Zero, I still get smaller values for subtracting 25. And negative numbers, I still am getting more negative.

(A) Correct.


Note: We have no information about what x and y could be so there is really no way of solving this problem. We cannot determine the relationship.

(D) Correct.


Note: These are independent events. Flipping a coin and rolling a die are completely independent of each other. This means that flipping the coin has no bearing over what happens with rolling the die. The probability of flipping a coin and getting heads or tails is still 1/2 no matter what happens with the roll of the die.

(C) Correct.


Aside: an interesting note about finding the median on a histogram is that if you think of the histogram as an area of some irregular shape, then if you shade in half of the area starting from either the left side or right side, the point at which you stop shading is the median.

Note: We don’t really know what the exact range is. We can find bounds on the range though. The smallest a value could be in this dataset is 51 and the largest is 100, this gives a value of 49 for the range. But the range could also occur where the smallest value is a 60 and the largest value is 91, this would give a range of 31. Then the median is that middle number when the data are arranged least to greatest. Well that part is done for us, so we can count that there are 11 numbers in this dataset, so the median will occur on the 6th number. This value falls in the histogram bin between 71–80.

(A) Correct. Since even 71 is still greater than the greatest range value, we know the median must be greater than the range.


Note: This is almost exactly like problem 13. In this problem, If columns A and B both were simply the probability of reaching in and getting a green candy, then the probabilities would be the same. But in column A now it’s asking us to reach in again and grab another green candy. The chances of this happening are certainly less than just grabbing once.

(B) Correct.


Note: this problem is slightly similar to problem 4. A 10% increase followed by a 10% decrease is not the same value that you started with. If you increase some number by 10%, and then decrease that new value by 10%. You are taking 10% of a larger number the second time (when you decrease). So you will end up with a smaller number than you started with.

(B) Correct. 1.50 * 1.1 = 1.65 and 1.65 * 0.9 = 1.485. But don’t worry too much about the numbers, all that matters is that when you take 10% of the bigger number you are decreasing by more than you just increased by.

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