Upper Level: Quantitative Reasoning Study Guide for the ISEE

Geometry

Two and Three-Dimensional Figures

Two-dimensional objects are planes and surfaces of various shapes, including triangles, conic sections and polygons. You may be asked to solve for the area, circumference or perimeter, radius, interior and exterior angles, sum of angles, and other dimension or measure of these planes.

Cubes, prisms, pyramids, spheres, cones and cylinders are examples of three-dimensional figures or solids. A question may ask for their volume or capacity. They may be given in a quantitative comparison question, or in a word problem, such as this example:

A rectangular prism has square surfaces on both ends. The sides of the prism’s square surfaces are twice the sides of a smaller cube which has a volume of \(a^3\). If the prism has a length twice the length of its square sides, what is its volume?

The formula for the volume of a cube is \(V = s^{3}\), where \(s\) is the side length and \(V\) is the volume. Since the small cube has a volume of \(a^{3}\), let’s find the length of its sides:

\[V = s^{3}\] \[a^{3} = s^{3}\] \[s = a\]

So, the side length of the smaller cube is \(a\).

The sides of the square surface of the rectangular prism is twice the side length of the small cube. So, it is \(2a\).

Also, the length of the rectangular prism is twice the length of its square sides. So, the length is \(2 \times 2a = 4a\). Now, we can draw the rectangular prism:

To find the volume of a rectangular prism, \(V\), we take the product of its length, width, and height:

\[V = 4a \times 2a \times 2a\] \[V = 16a^{3}\]

Congruence and Similarity

Plane or solid figures are similar if they have the same shape, and their measurements or dimensions are in proportion. They are congruent if these measurements and dimensions are equal. The example given above can instead be stated this way:

A rectangular prism has square surfaces on both ends. The prism’s square surface is four times the area of a similar square with an area of \(a^2\). If the prism has a length twice the length of its square sides, what is its volume?

Given that the prism’s square surface is four times the area of a similar square with area \(a^{2}\), we can say that the area of the square surface of the prism is \(4 \times a^{2} = 4a^{2}\). The side length, \(x\), of the square surface is:

\[x^{2} = 4a^{2}\] \[x = \sqrt{4a^{2}}\] \[x = 2a\]

The length of the prism is twice the side length of the square side, so the length is \(2 \times 2a = 4a\).

This is the same rectangular prism from above, with length \(4a\), width \(2a\), and height \(2a\). So, its volume is:

\[V = 4a \times 2a \times 2a\] \[V = 16a^{3}\]

Trigonometric Relationships

Master the Pythagorean Theorem (shown below) because it’s the basic tool for solving countless geometry and trigonometry questions.

\[a^2 + b^2 = c^2\]

You must be able to recall the trigonometric functions automatically. They’re written in an easy-to-recall form below, but you must practice with them to really understand the concepts deeply:

Sine function: S = O/H

Cosine function: C = A/H

Tangent function: T = O/A

Where A is the side adjacent to the angle, O is the side opposite the angle, and H is the hypotenuse.

Supplement your study and practice with the trigonometric identities. Take note that the given relationships, so far, are applicable only to right-angled triangles. For triangles that are not right-angled, use the triangle identities. Find additional materials on this subject.

Cartesian Coordinates

The location of points is determined using their \(x\) and \(y\) positions in the Cartesian plane, and are normally written as \((x, y)\), also called the Cartesian coordinates. A point can be anywhere in the four quadrants of the \(xy\)-plane. A good background in locating points is essential when solving for lengths of line segments, distances, and slopes. It is also important when transforming functions or translating geometric objects.

Translations, Reflections, Rotations, and Dilations

Planes and solids can undergo various transformations, such as translation, reflection, rotation, and dilation (or resizing).

These transformations may change the position or orientation of an object, but its size, shape, angles, area, and lengths of sides will not change.

Translation—sliding an object from its original position to another location

Reflection—flipping an object so that the reflected object is the “mirror-image” of the original

Rotation—turning an object around a center or point

The above-mentioned transformations will result in an object that is congruent to the original object. Dilation or resizing will result in an object that is similar to the original but will either be increased or decreased in size.

Read further on the transformation of objects in the Cartesian plane and lines of symmetry of plane shapes.

Vertex Edge Graphs

Vertex-edge graphs use dots or circles (vertices) and lines (edges) to represent various situations and to help visualize situations and problems. The vertex is a point where lines meet. A degree-\(1\) vertex means that the vertex has one line connected to it. A vertex can have \(0\) to \(n\) degrees. An edge is a line that connects one vertex to another.

In the test, a vertex-edge graph may be shown, and you will be asked a question that measures how well you understand the representation.

Integrating Other Math Concepts

Your understanding of the previous concepts on equations, planes, solids, functions, and inequality must ultimately be integrated with geometry and the Cartesian coordinate system.

You must be able to visualize how a linear equation looks on the \(xy\)-axes, whether it slopes up or down, or where it crosses the axes.

What solid is formed by a second-degree function in two variables? How will these solids look on the \(xyz\)-axes? At this level, you are expected to be able to exercise spatial reasoning.

Find additional practice exercises. The only way to learn and internalize these concepts is to actually solve equations and plot the graphs yourself.

Measurement

In most questions in the Quantitative Reasoning section, the formulas will be given with emphasis on analysis and reasoning.

In a quantitative comparison question, for instance, the formula for the volume of a cylinder may be given. The student will then be asked to compare the volume of Cylinder \(1\) and Cylinder \(2\) if Cylinder \(1\) has twice the radius but half the height of Cylinder \(2\).

Answer:

Let the radius and height of Cylinder \(2\) be \(r\) and \(h\), respectively.

Using the formula for the volume of a cylinder, \(V = \pi r^2 h\), we can find its volume.

Volume of Cylinder \(2 =\pi r^2 h\)

Now, it is given that Cylinder \(1\) has twice the radius and half the height of Cylinder \(2\). Thus, the radius would be \(2r\) and height \(\frac{1}{2}h\). We can now come up with an expression for the volume of Cylinder \(1\):

\[V = \pi r^2 h\] \[V = \pi (2r)^2 \left(\frac{1}{2}h\right)\] \[V = \pi (4r^2) \left(\frac{1}{2}h\right)\] \[V = \pi (2r^{2} h)\] \[V = 2\pi r^{2} h\]

Thus, we can see that the volume of Cylinder \(1\) is twice the volume of Cylinder \(2\).

It is important to familiarize yourself with the various formulas for measuring lengths, areas, and volumes, but “thinking or reasoning through” a formula or concept is given more emphasis in these types of questions.

Formulas for Area, Surface Area, and Volume

Here are some of the formulas seen in ISEE tests, but this shouldn’t limit you from studying and problem-solving with other formulas:

Area:

Triangle: \(A = \frac{1}{2} bh\) where \(b\) = base, \(h\) = height

Square: \(A = s^2\) where \(s\) = length of side

Circle: \(A = \pi r^2\)

Rectangle: \(A = lw\) where \(l\) = length, \(w\) = width

Parallelogram: \(A = bh\) where \(b\) = base, \(h\) = vertical height

Also practice solving for the areas of polygons when their vertices are given as coordinate points, such as this question:

What is the area of a triangle with vertices at \((1, 6), (1, -2)\), and \((7, 3)\)?

First, we need to graph the triangle on the coordinate plane:

Now, let’s label the “base” and the “height” of the triangle:

From the graph, you can see that the base is \(8\) units, and the height is \(6\) units. The area of a triangle is given by the formula \(A = \frac{1}{2}bh\), where \(b\) is the base length and \(h\) is the height of the triangle. Using this formula, we can easily find the area, \(A\), of the triangle:

\[A = \frac{1}{2}bh\] \[A = \frac{1}{2}(8)(6)\] \[A = 24\]

The surface area of a solid is the sum of all the flat surfaces of the object. A cube, for instance, will have six square surfaces, such that its surface area is:

\[A = 6(s^2)\]

Volume

Sphere: \(V = \frac43 \pi r^3\)

Cube: \(V = s^3\)

Rectangular Prism: \(V = lwh\)

Cylinder: \(V = \pi r^2h\)

Cone: \(V = \frac{\pi r^2h}{3}\)

Pyramid: \(V = \frac{bh}{3}\)

Unit Analysis

Unit analysis, also referred to as dimensional analysis, is very useful when checking the correctness of answers. If a particular item asks for the area and you get an answer in feet, then you know that you need to check your solution for inaccuracies. (The answer should be in square feet.)

We know the distance equation, \(s = vt\), where \(s\) is the distance measured in meters, \(v\) is the speed measured in meters per second, and \(t\) is the time measured in seconds.

What if you didn’t know the distance equation properly and used the equation \(s = v \times t^2\) instead? You could use dimension analysis to check if this is correct.

Using the standard measurement units [distance is measured in meters (\(m\)), speed is measured in meters per second \(\left(\frac{m}{s}\right)\), and time is measured in seconds (\(s\))] of the quantities described, let’s do unit analysis:

\[s = v \times t^2\] \[m \stackrel{?}{=} \frac{m}{s} \times s^2\] \[m \stackrel{?}{=} m \times s\] \[m \neq m \cdot s\]

By doing unit analysis, we see that meters is not equal to meters-second. Now, we will perform the same analysis with the real distance equation to show that it’s correct:

\[s = vt\] \[m \stackrel{?}{=} \frac{m}{s} \times s\] \[m = m\]

So, meters are equal to meters. We get the same unit on both sides. This equation is true.

When answering math questions, it’s quite common for students to compute solely with the numbers, completely disregarding the units. A good way to check your answer is to perform unit analysis afterward. When units undergo conversion in the problem, this step becomes doubly important.

Accuracy and Precision in Measurement

Any deviation of a measured value from the actual value introduces inaccuracies to computations using this measured value. For example, the inaccurately measured width and length of a surface will affect the area determined from these measurements.

Your knowledge of such concepts as degree of accuracy, margin of error, acceptable error, upper and lower bounds, limits, and successive approximation should come in handy for increasing accuracy, not only for this test but also in actual situations involving measurement.

Analyze the statement below:

Several crates of fruit are being loaded for delivery. The weight of each crate of fruit should be \(40\) kg rounded to the nearest \(10\) kg, or it will not be loaded. What is the heaviest weight (upper bound) that a crate of fruit should weigh to be included in the loading?

Answer:

To round a weight to the nearest \(10\) kg, we either round up to the nearest multiple of ten or round down to the nearest multiple of ten.

A weight of \(38\) kg will be rounded up to \(40\) kg. A weight of \(42\) kg will be rounded down to \(40\) kg.

Similarly, a weight of \(44\) kg will be rounded down to \(40\) kg. A weight of \(45\) kg will be rounded up to \(50\) kg. So, the upper bound for this problem is a crate weighing in at \(44\) kg.

Anything over that will be excluded from the loading.

What is the least weight (lower bound)?

Answer:

Again, to round a weight to the nearest \(10\) kg, we either round up to the nearest multiple of ten or round down to the nearest multiple of ten.

A weight of \(37\) kg will be rounded up to \(40\) kg. A weight of \(35\) kg will be rounded up to \(40\) kg. A weight of \(34\) kg will be rounded down to \(30\) kg. Thus, the lower bound for this problem is a crate weighing in at \(35\) kg.

Anything below it will be excluded from the loading.

A word problem may state that a measuring instrument is “accurate to \(10\) centimeters” and ask you to determine the possible range for a measurement of \(110\) meters using this instrument. You would determine that it could be any length from \(109.95\) m to \(110.05\) m. By recognizing these inaccuracies, you are able to arrive at more accurate results.

Errors in Measurement

As mentioned above, errors do occur in measurement. Several factors contribute to this, including built-in imperfections and biases in measurement instruments, expansion due to temperature variations, and the like.

Math problems often use a plus/minus notation to represent errors. In the measuring tool given as an example above, the error was \(\pm 5 cm\). Take note that the same instrument was described in the question as “accurate to \(10\) centimeters”. Questions may also be phrased as “accurate to the nearest \(100\) grams”, “accurate to the nearest \(5\) degrees Centigrade”, “measures in whole numbers”, “measures to the nearest \(1\) inch”, so study more in this area to fully grasp the concept.

You may also encounter such terms as absolute error, relative error, or percentage error. In the example of the measuring tool, the absolute error was \(5\) cm (take note that we removed the plus/minus sign).

Absolute error = the difference between the maximum possible value and the measured value

In the example:

Absolute error = \(110.05 – 110 = 0.05 m = 5 cm\)

Relative error = Absolute error ÷ Measured value

Example: Relative error = \(0.05m ÷ 110m = 0.00045\)

Percentage error = Relative error \(\cdot 100\%\)

Example:

\[0.00045 \cdot 100\% = 0.045\%\]