the name of our book for catalyst prep
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| The digit all the way to the right in a whole number. In the number 345, five is the units digit. | ||
| The digit immediately to the left of the units digit. In the number 345, four is the tens digit. | ||
| The digit immediately to the left of the tens digit. In the number 345, three is the hundreds digit. (This applies all the way up, to thousands digit, ten-thousands digit, hundred-thousands digit, etc.) | ||
| The digit immediately to the right of the decimal point. In the number 0.345, three is the tenths digit. (Again, this applies all the way down, to hundredths digit, thousandths digit, etc.) | ||
| All whole numbers and the number zero. | ||
| Whole numbers greater than zero. | ||
| Whole numbers less than zero. | ||
| All positive numbers AND the number zero. | ||
| Whole numbers divisible by 2 (including zero!) | ||
| Whole numbers NOT divisible by 2 (NOT including zero!) | ||
| Whole numbers that differ by one and can be lined up in a row, for instance: 3, 4, 5, 6, 6, 8. Consecutive integers can be negative numbers too. They can be listed in ascending or descending order. | ||
| Whole numbers that differ by two. (so either 2, 4, 6, 8...or 1, 3, 5, 7). Consecutive odd/even integers can be negative numbers too. | ||
| Unlike integers, number do NOT have to be whole. The fraction 1/8 is a number, as is 0.325. Think of it this way, only ONE integer exists between 1 and 3 (and that would be 2), but an INFINITE amount of numbers exist between 1 and 3 (1.003, 5/2, 2.999, etc.) | ||
| A fraction with a denominator of 100. To convert any decimal into a percentage, simply shift the decimal point two places to the right. For example, the decimal .05 is 5%. The number 1.1 is 110%. | ||
| Any of the numbers (generally, integers) that can be multiplied to form a product. 1, 2, 3, 4, 6, and 12 are all factors of the number 12. | ||
| All numbers that are products of a number. 24, 36, 48, and 144 are all multiples of the number 12. | ||
| Any number greater than one, whose only factors are one and itself. | ||
| A group of elements or members that share a common characteristic. For example, {6,9,18,54} is a set whose elements are all multiples of 3. | ||
| The two end points in a range are included. If a problem wants to know how many integers are between 10 and 20 inclusive, it's always the difference of the two end points PLUS ONE. There are 11 integers between 10 and 20 inclusive, and 101 integer between 100 and 200 inclusive. | ||
| Identified by variables in both question and answer. Plug-in a value for "x" and solve accordingly. (Try not to choose 1 or 2, since they might have unique properties...go for a nice number, like 10!) | ||
| A. 280k | ||
| A problem asking you to determine how many different ways a number of things could be chosen or combined. A combination is when the options do NOT cancel each other out. Let's suppose we're wondering how many cup/plate/fork combinations we can get with 3 different types of cups, 5 different plates, and 6 different forks. First, list out each option. (A cup, a plate, a fork.) Second, list the amount or number of each option above it. (3/cups, 5/plates, 6/forks). Third, multiply the NUMBER of options together. (3*5*6) The answer, which would be the number of combinations, is the product of that multiplication. (90 options) | ||
| A problem asking you to determine how many different ways a number of things could be chosen or combined. A permutation is when the options do cancel each other out. Let's suppose that we're wondering how many different varieties of eye shadow color, glitter color, and nail color you could get for Homecoming, where every color would be different! And you have 4 color options for each: Green, Blue, Red, Yellow. First, list out each option without the quantity. (Eye Shadow, Glitter, Nail) Second, for the FIRST option, list out the quantity of options above the option. (4/Eyeshadow) Third, for the SECOND option, list out the quantity of options above the option, IF the first option had taken one of the color options. (4/Eyeshadow, 3/Glitter...because if eyeshadow had been green, then there would only be 3 other options [red, blue, yellow] to choose from.) Fourth, repeat this step for the THIRD option, listing out the quantity of the options above the option, if the SECOND option had taken one of the color options. (4/Eyeshadow, 3/Glitter, 2/Nail....because if eyeshadow had been green, and glitter had been blue, that would only leave TWO options [red, yellow] for nail color.) Fifth, multiply the quantity of options. (4*3*2) Sixth, this product is the number of combinations possible based on options that cancel each other out. (24). | ||
| Typically represented by a fraction. The best way to solve any probability problems on the PSAT is to use the formula: (WHAT YOU WANT) / (THE TOTAL POSSIBLE NUMBER OF DIFFERENT OUTCOMES) | ||
| Underline the key word "each." This means that BOTH Karen and Tina will get a corner office. First, solve if Karen gets the office. Karen has a 2/4 opportunity of getting a corner office, which simplifies to 1/2 chance. Second, solve for Tina as if Karen got a corner office. This means there are 3 rooms left, and only one of them is a corner office. So Tina has a 1/3 chance. Third, since we're solving for BOTH Tina and Karen, we'll multiply their probabilities together (1/2 * 1/3) in order to the probability of each of them getting a corner office, which is c. 1/6 | ||
| Every PSAT features at least one function problem that uses strange symbols like Δ or ☺. They use these strange symbols to confuse you, but they are NO different than any other functions. Like all function problems on the PSAT, they involve one basic strategy: substitution. Simply if x ☺ y = 3x +y , and 1 ☺ 2, then you would replace 1 for x, and 2 for y, which would giving you 3(1) + (2), and the answer 5. | ||
| Replace x with 3, and y with 1. (3)^2 + (3)(1) + (1)^2 = 9 + 3 + 1= 13 Now, since this is double-symboled, it reads (13)♦1. Solve again, with 13 as x, and 1 as y. (13)^2 + (13)(1) + (1)^2 = 169 + 13 + 1 = 183 So e. 183 | ||
| The typical formulate for solving problems involving percent increase or decrease is (THE DIFFERENCE) / (THE ORIGINAL). | ||
| This problem will best be solved using plug&chug, and what we know about percent increase/decrease. We know it is a rectangle, so it is okay for both the length and width to be the same measurement. Since this is a percent problem, let's use 100 for each length of the side. To find the area, you'd times 100 by 100, which would give you 10,000. Now let's reduce each side. Since the length is INCREASED by 30%, let's add 30 to 100 (since 30 is 30% of 100), and we'll get 130. Since the width is DECREASED by 30%, let's SUBTRACT 30 from 100 (since 30 is 30% of 100), and we'll get 70. Now we have a rectangle with a length of 130, and a width of 70. In order to find the area, we multiply the side, which gives us 9,100. Now in the common percent increase/decrease problem, we put the DIFFERENCE over the original. So in order to find the difference, we must subtract 9,100 from the original, which would give us 900. Now put the difference of 900 over the original of 10,000, and you'll get 900/10,000 which is equivalent to 9/100, which is e. 9% | ||
| In order to determine the area of perimeter of a tough-to-measure shape, just divide & conquer the shape into smaller, easier-to-measure pieces, then add up their components. (Unfortunately, since Quizlet is just text, I cannot put up a shape and explain this process.) | ||
| Pretty straightforward. Just watch out for boobytraps from the PSAT test writers. For instance, it may give you Tom's budget of $240 for a weekend vacation. And say he spent 30% on his hotel costs, splitting it with 3 other buddies. Then it may ask you how much the total hotel bill was. Do not just calculate 30% of his $240 budget ($72), but also remember that he split the cost with 3 other buddies, so the hotel room cost is 4 times $72 (3 buddies + Tom), so $288. | ||
| A remainder is the whole number left over after you divide a number by a smaller number. For instance, 3 is the remainder when 18 is divided by 5. Let's use this problem: When the positive integer m is divided by 5, the remainder is 3. What is the remainder when 20m is divided by 25? First, pick a number for m, based on the conditions. An easy number to pick is to take the remainder and add it to the divisor (5+3), so we'll use 8. Now, for the new set of conditions (when 20m is divided by 25) substitute your value for m. So 20(8) = 160, which when divided by 25, has 6 equal parts, with a remainder of 10. So the answer is 10. | ||
| A sequence is a list of number that follow a specific pattern. To guide us through explaining this math problem, let's use this problem. In the decimal 0.12678, where the digits 12678 repeat, which digit is 6,003 places to the right of the decimal? It'd be brain numbing to write this out to the 6,003rd digit, so we'll simplify it. Count the number of digits, in this case we have 5 digits (1,2,6,7,8). Take 6003 and DIVIDE it by 5. 5 goes into 6003, 1200 times. However, there is a remainder of 3. It is this remainder that is important, NOT how many times 5 goes into 6003. Take this remainder of 3, and count 3 spots into the sequence, (1, 2, 6). It ends on 6, which indicates that the 6003rd digit of that sequence will be 6. So the answer is 6. |
