in reality, weighted averages are not any more complicated than the plain-ol’ average, and should only slow you down by 10-15 seconds as you rapidly go through a little more number crunching than usual.

we’ll review the concept of averages first, and then head on to tackle weighted averages. the average, also known as the mean, is the sum of a group of numbers divided by the amount of numbers added together.

definition of the average for act math

average = (sum of numbers) / (total amount of numbers)

for example, the average of 3, 3, 5, 6, 4, 2, and 5 is 28/4 or 7.

definition of weighted averages for act math

when some numbers in a group carry more ‘weight’ than other numbers, you need to take that weight into account before plugging the numbers into the formula. this is called a weighted average.

in order to convert the numbers into their weighted counterparts, you need to multiply each number by its weight before adding them together. also, instead of dividing by the total amount of numbers, you divide by the total weight of the numbers.

you’ll be given the weight of each number, so all you really need to do is figure out which numbers to plug in so that you can get to the answer.

let’s take a look at this example and break it down step-by-step.

in a class of 2 boys and 3 girls, the boys’ average test score was 75 and the girls’ average test score was 80. what was the average score for the entire class?

step 1: first we need to recognize that there are an unequal amount of boys and girls. the girls’ scores carry more weight than the boys’ scores, so we need to multiply their scores by their weight.

boys’ weighted score = 75*2 = 150
girls’ weighted score = 80*3 = 240
total weighted score = 150 + 240 = 390

step 2: we need to find out the total weight to divide by. in this case, there are 2 boys and 3 girls, so the total weight is 5.

step 3: now we simply divide the answers we got from number 1 and number 2 to get the solution. 390/5 = 78

now that you have an understanding of how to solve weighted averages problems, check out our posts on other act math concepts, such as how to solve probability problems and how to solve circle problems.

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new sat math: basic circle skills review

if you’re a little shaky or foggy on your knowledge of circle concepts, check this section out. it’s a great refresher on what you’ve probably already learned before during one of your math classes.

if you already have a strong grasp on this, go ahead and skip to the next section!

value of pi

the value of pi is about 3.14. you may already have a pi button on your calculator, but you should have this value memorized anyway.

circumference of a circle

the circumference of a circle is diameter*pi. the circumference tells you the length of the perimeter of a circle.

it takes about 3.14 diameters to equal the circumference of a circle.

degrees in a circle

a circle consists of 360 degrees. half of the circle is 180 degrees, and a quarter of a circle is 90 degrees.

radians

radians are an angle measure equivalent to an arc length of one radius of a circle. it is a relative measurement. 1 radian equals about 57.3 degrees.

remember to check your calculator settings carefully to make sure it is set properly for either degrees or radians. you can bet that test makers will throw in a trap answer here or there for unsuspecting students.
 

new sat math: arc length of a circle

now that you’ve thoroughly brushed up on the basics of circles, let’s get right into figuring out how to calculate the arc length of a circle. the new sat contains problems that really dig deep into seeing how well you understand basic concepts, so take extra care to know the fundamentals forwards and backwards.

circle arc length

the total circle arc length is the circumference, or the perimeter of a circle. when we want to find out the length of a small part of that circumference, we need to multiply it by a fraction between 0 and 1.

we can use either degrees or radians in order to represent this fraction. an entire circle equals 360 degrees or 2pi radians, so our formula looks like this:

circle arc length = circumference * (x degrees / 360)
or
circle arc length = circumference * (x radians / 2pi)
 
looking for more help with circles on the sat? check out our videos about unit circle basics and tangents to a circle.

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in the most basic terms possible, functions relate a set of inputs with a set of outputs. usually, you will be able to spot a function when you see something like f(x) or g(x) instead of the usual y variable on one side of the equation.

the key thing to remember about functions is that each input can only give you one output. for example, you can’t plug in 3 into a function and get two different answers. however, to different values of inputs can yield the same output.

on a graph, an easy way to test whether or not a curve is a function is to use the vertical line test. if you draw an imaginary vertical line anywhere on the graph and it crosses the curve twice, then it is not a function.

most linear equations will pass the vertical line test. the only situation that you have to look out for is a line that is perfectly vertical. in that case, it is not a function.

the standard notation for a linear function is f(x) = kx + f(0).

the f(x) is similar to what the y-variable represents in a linear equation. test makers can also call it the range or the output of the function.

the k represents the slope. take a moment to recall the slope-intercept form and note the similarities between the two.

the f(0) is the y-intercept, or where the line passes through the y-axis when x=0.

the x-variable can be called the domain or the input. the x-variable is what you can control, and the y is the value that you get based on what you put in.

don’t be overwhelmed by the verbiage. just remember that you’re dealing with basically the same thing as a linear equation. the key here is to get used to all the terms that the test makers can throw at you.
 

new sat math: linear function graphs

many problems you will encounter ask you to find a value of a certain coordinate point on a graph. in this case, f(x) corresponds to the y-axis and the x corresponds to the x-axis.

all you have to do is go to a given spot on the curve see what the other coordinate equals. for example, if you see an ‘f(5)’, that means that you need to find what the y-coordinate is on the curve labeled “f(x)” where x = 5.

new sat math: linear function tabular notation

sometimes, you will be given a chart of a function and be expected to interpret it. the chart will only give you a few values to work with.

you will easily be able to see whether or not there is a linear relationship between the input and outputs. by comparing a few values, there should be a linear trend going upward or downward.

the key here is to make sure that you keep your variables, inputs, and outputs straight. if you do that, you can handle any function table that comes your way.

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ratio basics

a ratio tells you the proportional quantity of one thing relative to another.

make sure not to get ratios confused with fractions. fractions tell you the proportional quantity of something relative to its whole. ratios expressed as fractions do not tell you the whole.

one instance where you need to use the concept of ratios involves baking. if you want to make double the amount of cookies that a recipe will yield, then you need to double the quantity of each ingredient.

act math: dealing with ratios

you might see act ratios written in fraction form, colon form, or in plain english. whatever the case may be, you can treat them all the same way. in the case of the fraction form, do not get it confused with a regular fraction! the denominator of a ratio is not necessarily equivalent to the denominator of a ratio.

for example, the ratio 12/8, 12:8, and 12 to 8 are all the same. like fractions, you should reduce ratios down to simplest terms – in this case, it is 3/2. keep your numbers manageable, especially when you need to look for the lowest common multiple later on in the multi-step ratio section.

on the test, ratios will be clearly spelled out for you. if you are looking at a ratio problem, you’ll know it because the test makers will make it obvious.

the important part lies in knowing how to manipulate ratios to get to your answer. the two main things you need to know are proportions and multi-step ratios.

proportions

you’ll find that these are very common on the act. thankfully, they are also easy to solve.

you will usually be given a ratio along with a hypothetical quantity of one of the things on the original ratio. the key is to set up two ratios and cross-multiply as you would two fractions to solve for the missing fourth quantity.

if you have a ratio of 3 cats to 2 dogs, how many cats do you have if you have 20 dogs? you could use mental math or set up two fractions to get 30 cats as your answer.

multi-step ratios

these are a little bit more involved, but shouldn’t pose much of a threat to your act math score one you learn about how to go about solving them.

here you’ll be given two ratios and three different types of quantities: a, b, and c. the two ratios given compare a to b and b to c. you’ll then be asked to figure out the ratio of a to c.

in order to solve, you need to figure out the least common multiple of the two b’s and multiply the respective a’s accordingly. now with your b’s equal to each other, simply take the values of your a and c and create a new ratio.

for example, if a:b is 2:3 and b:c is 6:9, then what is a:c? here we need to multiply our first ratio by 2 in order to get 4:6. since our b from the first and second ratio match, we can take our a and c and form a new ratio.

our answer is 4:9.

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here you will be tested on figuring out missing side lengths, and you’ll have to brush up on some triangle knowledge before we begin:

• the sum of interior angles in a triangle add up to 180 degrees.
• the pythagorean theorem states that the sum of the squares of the two shorter sides of a right triangle equals the square of the longest side (the hypotenuse).

new sat math: similar triangles

similar triangles have the same angle measure for all angles, but they don’t necessarily have equal side lengths. all triangles with the same angle measures are similar.

the side lengths of similar triangles are proportional to each other, so we can set up a ratio in order to figure out missing side lengths:

photo by jwilson

a/e = b/f = c/g

new sat math: congruent triangles

congruent triangles are the same size, so they have the same angle measurements and equal side lengths. in other words, they are basically the same triangle.

new sat math: sample triangle problem

let’s take a look at a sample problem:

triangle adg (above) has an area of 6 square units. if ad = af = 3, then what is the length of ef?

a) 12/5
b) 2
c) 8/3
d) 1/2

1. the first thing we want to do here is label our given measurements. whenever we have a problem that asks us to use or draw a reference picture, we always want to write and draw everything out. let’s mark down the length of sides ad and af.

2. since the area of the entire triangle is given, we can go ahead and solve for the missing side length dg using area = (½)base*height. in this case, we should get
6 = .5*3*height
4 = height

side length dg is equal to 4.

3. now we can either use the pythagorean theorem to solve for side length ag or see that we have a 3-4-5 right triangle. either way, we should get 5 for the length of side ag.

4. from here, recognize that since both triangle aef and adg share an angle (angle a) and are both right triangles, they must be similar. therefore, we can set up a ratio between the two triangles in order to solve for length ef.

we will use the two known hypotenuse lengths and the length of dg.

3/5 = x/4

x = 12/5

our answer here is a!

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for simple games, such as calling heads or tails on a coin flip, it’s easy to figure out what the chances are of winning. physics and other variables aside, you have a 50% of getting it right.

when it comes to calculating more involved games, such as the chance of calling two or three coin flips correctly, then things get a bit more complicated. sure, you could map out and think about every possibility, but that would cost you a lot of time.

what you need is a method for figuring out probability problems. however, it’s important to know that it’s not absolutely necessary to know a formal way to get the answer. the key here is to familiarize yourself with enough act-style questions so that you have a good idea of what to expect when you take the real test.
 

act math: what is probability?

in simple terms, probability is the likelihood of a particular event happening.

probability = (favorable outcomes) / (total outcomes)

therefore, a probability of 0 means that it will never happen. a probability of 1 means that it will always occur. your answer will usually be somewhere between 0 and 1.

to find the probability of something not happening, simply subtract the probability from 1.

for multiple probabilities

to find the probability that two events will occur, multiply the probabilities together.

to find the probability of either event occurring, add the probabilities together.
 

act math: solving probability problems

the act math section does not dive deep into probability concepts, so you’ll be fine knowing just the basics. take a look at the example problem below for some practice:

a bag contains 2 purple marbles, 5 black marbles, and 5 blue marbles. you choose two marbles out of the bag at random. after the first marble is chosen, it is not replaced in the bag. what is the probability of choosing two purple marbles?

a. 1/66
b. 1/6
c. 1/36
d. 1/16
e. 1/26

okay, there’s a lot of things going on here, so let’s break it down.

1. first of all, let’s take stock of how many purple marbles and total marbles we have here.

we have 2 purple marbles, and
2+5+5 = 12 marbles total.

the probability of choosing the first purple marble is 2/12.

2. now let’s figure out the value of our second probability.

since the first marble does not go back into the bag once it is chosen, we only have 11 marbles total for the second draw.

whether or not we draw a purple marble the second time around only matters if we chose a purple marble the first time. what this means is that in order to figure out our second probability, we should assume that we picked a purple marble on our first draw.

therefore, our second probability comes out to be 1/11.

3. since the problem is asking for the probability of drawing two purple marbles, we need to multiply the two values together.

(2/12) * (1/11) = 2/132

all we need to do now is simplify, and our answer is a.
 

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the basics of exponents

there are two parts to an exponent: the base and the value of the exponent itself. the base tells you what is being multiplied by itself, and the value of the exponent tells you how many times the base gets multiplied.

for example, 5⁴ is equivalent to 5*5*5*5. the 5 is the base, and the 4 is the exponent. laid out like this, it is easy to see why exponents can be simplified. since we are combining large amounts of the same numbers being multiplied together, we can easily group them.

most problems that have to do with exponents on the act test you on your ability to simplify correctly. therefore, it is vital that you know the rules on how to combine exponents in order to turn complicated expressions into something more manageable.

with careful practice and repetition, you won’t have to rely on memorizing rules. instead, it’ll become almost second nature, allowing you to concentrate on deconstructing more difficult problems and concepts.

how to tackle exponents on the act

dividing exponents with the same base

in order to divide two terms containing the same base, subtract the exponents and keep the base.

for example, 3⁵ – 3³ = 3².

multiplying exponents with the same base

in order to multiply two terms containing the same base, add the exponents and keep the base.

for example, 3⁵ + 3³ = 38^.

raising exponents to a power

in order to raise an exponent to a power, multiply the exponents and keep the base.

for example, (3⁵)⁵ = 3²⁵.

evaluating negative exponents

in order to solve for negative exponents, take the reciprocal of the base and exponent. after that, change the sign of the exponent from negative to positive.

for example, x^-5 = 1/x⁵.

evaluating fractional exponents

evaluate the numerator of the exponent like normal. evaluate the denominator of the exponent as you would a radical.

for example, x^(5/2) = the square root of x⁵.

squares and cubes

squaring a number means to multiply the value by itself.

cubing a number is taking it to the third power.
 

the post act math: how to solve problems with exponents appeared first on magoosh blog | high school.

photo by wecometolearn

quadratic equations giving you trouble on the sat? check out this post for all you need to know about solving quadratic equations on the new sat math.

solving quadratic equations on the new sat math: the basics

quadratic equations consist of a variable raised to the power of 2 as the highest-powered term. on a graph, they usually produce a “u” shape.

the standard form of a quadratic equation is ax² + bx + c = 0. we always want to set the equation equal to 0 so that we can find out what x is when y = 0.

because they contain multiple solutions, quadratic equations are a bit more involved than linear equations. quadratics have either 0, 1, or 2 real solutions – you will generally not need to find the imaginary solutions.

there are a few common ways to say “find the solution” to a quadratic problem:

• find the root.
• find the x-intercept.
• find the zeros.

solving quadratic equations on the new sat math: use algebra

multiplying binomials

remember foil? they stand for first, outer, inner, and last. it’s a handy acronym for remembering how to distribute everything correctly.

(a + b)(c + d) = ac + ad + bc + bd

from here, you’ll usually need to simplify a couple of terms in order to get it to look like one of the answer choices on the test.

factoring

this is basically the opposite of foil. you are trying to figure out what two binomials, when multiplied together, equal the expression. remember to set the entire equation given equal to 0 first before starting on factoring.

this method works best when you have a 1 for the coefficient of your quadratic variable (x²). we just need to find out what our b and d is to create (a + b)(c + d).

let’s take a look back at ax² + bx + c.

the first way to narrow down your options is to think about what two numbers multiply together to equal the value of the constant c. once you’ve written out the possible options, figure out which pair of numbers, when added together, equal the coefficient of the variable raised to the power of 1.

now you can solve for x! all you have to do is set x equal to whatever will make each binomial 0.

once you are good at this, you will be able to get to the answer more quickly and not have to think about every single possible choice when guessing numbers.

completing the square

this method is a bit more involved but is usually faster than using the quadratic equation. note that you can only use this option if the coefficient of the variable raised to the power of 1 is an even number.

let’s try solving x² – 2x – 1 = 0.

1. the first step is to move the constant over to the right side. this gives us

x² – 2x = 1

2. next, figure out what binomial, when multiplied by itself, matches the left side of the equation.

(x – 1)² = 1

3. since (x – 1)² gives us x² – 2x + 1 on the left side, we need to add that 1 to the right side as well.

(x – 1)² = 2

4. square root both sides and solve for x.

x = 2.41 and x = -0.41

solving quadratic equations on the new sat math: use a calculator

with your graphing calculator, you can just punch the numbers in and see where the graph intersects with the x-axis. make sure you enter in the numbers correctly so that you don’t get a weird answer.

the best way to save time while doing this is to plug everything into the calculator, and then move on to the next question while waiting for the calculator to load so that you don’t lose precious time.

solving quadratic equations on the new sat math: use the quadratic formula

if you’re really stuck and nothing’s working out for you, you can try using the quadratic formula. here it is in all its glory:

it’s not a super difficult process, but it can eat up a lot time to list out the equation and plug in the numbers. this route should only be considered as a last resort.

the post new sat math: solving quadratic equations appeared first on magoosh blog | high school.

the formulas themselves are not too difficult to remember. what might trip you up, however, are the vast amount of key terms you are expected to know.

lastly, since act test makers have a tendency to combine circles, triangles, and other possibe shapes together in one problem, you’ll need to utilize skills from all parts of your geometry toolkit to handle them.

act math: definition of a circle

a circle is a set of all the points that are equidistant from a point (the center of the circle). you probably already know what a circle looks like, but it’s important to know the formal definition so that you can confidently handle any conceptual problems that come up.

act math: glossary of circle terms

radius

a line drawn from the center of a circle out to the edge of the circle.

formula: r

diameter

a line drawn from one edge of a circle out to the other side of the circle. it goes through the center of the circle, cutting it in half.

formula: 2r

circumference

the length of the perimeter of the circle.

formula: 2*(pi)*r

area

the size of the enclosed region of a circle.

formula: (pi)r²

arc length

the length of a particular fraction of the circumference of a circle.

formula: (x/360)*circumference

x equals the number of degrees of the arc’s central angle.

minor arc

the shorter distance along a circle between two points on the edge of the circle.

major arc

the longer distance along a circle between two points on the edge of the circle.

sector

the area of a circle enclosed by an arc and two radii.

formula: (n/360)*(area)

n is the degree measure of the central angle of the sector.

concentric circles

circles that have the same center and are of a different size.

semicircle

half of a circle is called a semicircle.

chord

a line that goes from one edge of the circle to the other side. unlike the diameter, chords don’t necessarily have to go through the center.

a diameter is a chord, but a chord is not necessarily a diameter.

tangent

a line on the outside of the circle that touches the circle at only one point. the radius of the circle that touches that point forms a right angle with the tangent line.

central angle

the angle that is formed by two radii.

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solving linear equations with one variable

these are simple to solve once you get the hang of it.

1. for linear equations that consist of only one variable, the key is to isolate the variable on one side.

2. next, get all the constants to the other side of the equation.

3. lastly, simplify both sides of the equation until you get to the answer.

new sat math: linear equation example

let’s take a look at a common question type that the new sat test makers will throw at you.

if 6x = 42 and xk = 2, what is the value of k?

a. 2/7
b. 1/6
c. 7
d. 1/7

let’s take a look at both equations and think about it for a second. in the first equation, there is only one variable, and it is raised to the power of 1. since there is more than one variable in the second equation (x and k), we can’t know for sure what k is unless we solve for x first.

we should go ahead and solve the first equation for x before moving on to the next equation.

1. in order to isolate the variable on one side, we need to divide both sides by 6. that yields x = 7.

2. now we can plug in x = 7 into the second equation. this gives us 7k = 2.

3. in order to isolate the k on one side, divide both sides by 7. this leaves us with k = 2/7.

in this case, the answer is a.

new sat math: linear equation example #2

james is budgeting his time to think about the number of classes c he will take this year. for every class that he takes, he believes that he’ll spend 2 1/2 hours each week working on homework. he believes that he’ll spend an additional 6 1/2 hours each week completing the reading work for all of his classes together. if james has 19 hours free every week to finish homework and reading work for his classes, which equation best models this situation?

a. 2.5c – 6.5 = 19
b. 2.5c + 6.5 = 19
c. 6.5c – 2.5 = 19
d. 6.5c + 2.5 = 19

this type of question looks much more involved than the first one, but don’t worry! the key here is to circle keywords so that we can construct an equation quickly and move on to the next question without getting bogged down too much from all the reading we have to do.

1. circle all the keywords – the variables involved and the numbers that we have to work with. in this case, we have 2.5 hours per class per week, 6.5 hours for reading work per week, and 19 hours of free time.

2. look at what the question is asking for. here we want to figure out how many classes james can take in a week given his time constraint (19 hours per week).

3. put the numbers together and create the equation. we don’t know how many classes james is going to take each week, so we should put a variable after 2.5 to indicate the unknown number. in this case, c = the number of classes.

2.5c

since james believes that he will spend 6.5 hours total each week for reading work regardless of the number of classes taken, we should add that to 2.5x to represent the total number of hours spent on homework and reading work.

2.5c + 6.5

james has 19 hours total to complete all of his work, so let’s set 2.5x + 6.5 equal to 19 to figure out the maximum number of classes he can take.

2.5c + 6.5 = 19

choice b matches our answer! a job well done.
 

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