Data Analysis - PSAT Math
Card 1 of 1757
A number between 1 and 15 is selected at random. What are the odds the number selected is a multiple of 6?
A number between 1 and 15 is selected at random. What are the odds the number selected is a multiple of 6?
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In the set of 1 to 15, two numbers, 6 and 12, are multiples of 6. That means there are two chances out of 15 to select a multiple of 6.
2/15
In the set of 1 to 15, two numbers, 6 and 12, are multiples of 6. That means there are two chances out of 15 to select a multiple of 6.
2/15
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A big box of crayons contains a total of 120 crayons.
The box is composed of 3 colors; red, blue, and orange. 30 of the crayons are red, 40 of the crayons are blue and the rest are orange. If one picks a crayon randomly from the box, what is the probability that it will be orange?
A big box of crayons contains a total of 120 crayons.
The box is composed of 3 colors; red, blue, and orange. 30 of the crayons are red, 40 of the crayons are blue and the rest are orange. If one picks a crayon randomly from the box, what is the probability that it will be orange?
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To solve the problem one must calculate that there are 50 orange crayons in the box. So 50/120 are orange. If we simplify that fraction by 10 we get 5/12.
To solve the problem one must calculate that there are 50 orange crayons in the box. So 50/120 are orange. If we simplify that fraction by 10 we get 5/12.
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The average for 24 students on a test is 81%. Two more students take the test, averaging 74% between the two of them. What is the total class average (to the closest hundreth) if these two students are added to the 24?
The average for 24 students on a test is 81%. Two more students take the test, averaging 74% between the two of them. What is the total class average (to the closest hundreth) if these two students are added to the 24?
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The easiest way to solve this is to consider the total scores as follows:
Group 1: 81 * 24 = 1944
Group 2: 74 * 2 = 148
Therefore, the total percentage points earned for the class is 148 + 1944 = 2092. The new class average will be 2092/26 or 80.46. (For our purposes, this is 80.46%.)
The easiest way to solve this is to consider the total scores as follows:
Group 1: 81 * 24 = 1944
Group 2: 74 * 2 = 148
Therefore, the total percentage points earned for the class is 148 + 1944 = 2092. The new class average will be 2092/26 or 80.46. (For our purposes, this is 80.46%.)
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The average of 5, 10, 12, 15, and x is 11. What is the median?
The average of 5, 10, 12, 15, and x is 11. What is the median?
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You must first find what x is in order to find the median.
To find x, set up the following equation:
To solve the equation, first multiply both sides by 5:
5 + 10 + 12 + 15 + x = 55
Then, add up 5 + 10 + 12 + 15 to get 42:
42 + x = 55
x = 13
Now that you know what x is, you are ready to find the median. To find the median, order the numbers from lowest to highest. The median is the number in the middle.
5, 10, 12, 13, 15
You must first find what x is in order to find the median.
To find x, set up the following equation:
To solve the equation, first multiply both sides by 5:
5 + 10 + 12 + 15 + x = 55
Then, add up 5 + 10 + 12 + 15 to get 42:
42 + x = 55
x = 13
Now that you know what x is, you are ready to find the median. To find the median, order the numbers from lowest to highest. The median is the number in the middle.
5, 10, 12, 13, 15
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Looking at the table, what is the difference of the mean minus the mode.
Looking at the table, what is the difference of the mean minus the mode.
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The mode is 3 and subtracting 3 from 2.5 gives -0.5.
The mode is 3 and subtracting 3 from 2.5 gives -0.5.
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Find the mode of the following set of numbers:
1, 6, 3, 6, 5, 9
Find the mode of the following set of numbers:
1, 6, 3, 6, 5, 9
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Mode is the most frequently occuring number in a set. 6 appears most frequently since there are 2 sixes but the rest of the numbers only appear once. The mode is 6
Mode is the most frequently occuring number in a set. 6 appears most frequently since there are 2 sixes but the rest of the numbers only appear once. The mode is 6
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Find the mode in this set of numbers:
2, 100, 52, 97, 1, 7, 22, 19, 100
Find the mode in this set of numbers:
2, 100, 52, 97, 1, 7, 22, 19, 100
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The mode is the number that appears the most
The mode is the number that appears the most
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Find the mode from the following set of numbers
1, 4, 8, 17, 8, 8, 15, 21, 32, 17
Find the mode from the following set of numbers
1, 4, 8, 17, 8, 8, 15, 21, 32, 17
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The mode is the number that appears the most in a given set of numbers
The mode is the number that appears the most in a given set of numbers
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What is the mode of the following set of numbers?
24, 38, 26, 27, 7, 27, 34, 12, 13
What is the mode of the following set of numbers?
24, 38, 26, 27, 7, 27, 34, 12, 13
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The mode is the number that appears the most in a given set of numbers.
The mode is the number that appears the most in a given set of numbers.
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Find the range of the data set:
25, 83 , 51, 13, 37, 21, 52, 58
Find the range of the data set:
25, 83 , 51, 13, 37, 21, 52, 58
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44 is the median of the data. 42.5 is the arithmetic mean. 13 is the minimum value. 83 is the maximum value.
To find the range, subtract the minimum value from the maximum value:
83 - 13 = 70
44 is the median of the data. 42.5 is the arithmetic mean. 13 is the minimum value. 83 is the maximum value.
To find the range, subtract the minimum value from the maximum value:
83 - 13 = 70
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F, G, H are the only three numbers in a sequence in which each number is twice the number before it. The three number have an arithmetic mean of 21. What is the value of H?
F, G, H are the only three numbers in a sequence in which each number is twice the number before it. The three number have an arithmetic mean of 21. What is the value of H?
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We know that H = 2G = 4F, and G = 2F and that (F + G + H)/3=21. If we substitute each term with its F equivalent we get:
(F + 2F + 4F)/3 = 21
7F/3 = 21
7F = 63
F = 9
If we substitute F = 9 into H = 4F, we get H = 36.
The correct answer is 36.
We know that H = 2G = 4F, and G = 2F and that (F + G + H)/3=21. If we substitute each term with its F equivalent we get:
(F + 2F + 4F)/3 = 21
7F/3 = 21
7F = 63
F = 9
If we substitute F = 9 into H = 4F, we get H = 36.
The correct answer is 36.
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There is a special contest held at a high school where the winner will receive a prize of $100. 300 seniors, 200 juniors, 200 sophomores, and 100 freshmen enter the contest. Each senior places their name in the hat 5 times, juniors 3 times, and sophmores and freshmen each only once. What is the probability that a junior's name will be chosen?
There is a special contest held at a high school where the winner will receive a prize of $100. 300 seniors, 200 juniors, 200 sophomores, and 100 freshmen enter the contest. Each senior places their name in the hat 5 times, juniors 3 times, and sophmores and freshmen each only once. What is the probability that a junior's name will be chosen?
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The first thing to do here is find the total number of students in the contest. Seniors = 300 * 5 = 1500, Juniors = 200 * 3 = 600, Sophomores = 200, and Freshmen = 100. So adding all these up you get a total of 2400 names in the hat. Out of these 2400 names, 600 of them are Juniors. So the probability of choosing a Junior's name is 600/2400 = 1/4.
The first thing to do here is find the total number of students in the contest. Seniors = 300 * 5 = 1500, Juniors = 200 * 3 = 600, Sophomores = 200, and Freshmen = 100. So adding all these up you get a total of 2400 names in the hat. Out of these 2400 names, 600 of them are Juniors. So the probability of choosing a Junior's name is 600/2400 = 1/4.
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Michelle is randomly drawing cards from a deck of of 52 cards. What is the chance she will draw a black queen followed by a 5 of any color, without replacing the cards?
Michelle is randomly drawing cards from a deck of of 52 cards. What is the chance she will draw a black queen followed by a 5 of any color, without replacing the cards?
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There are 2 black queens in the deck, one of spades and one of clubs, so there is a 2/52 chance a black Queen will be drawn and 4/51 chance of drawing a 5 of any color, since the queen has already been removed from the deck. Thus: 2/52 * 4/51 = 8/2652 → 2/663.
1
There are 2 black queens in the deck, one of spades and one of clubs, so there is a 2/52 chance a black Queen will be drawn and 4/51 chance of drawing a 5 of any color, since the queen has already been removed from the deck. Thus: 2/52 * 4/51 = 8/2652 → 2/663.
1
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Zack has 10 green, 14 red, 2 blue, and 6 black marbles in a bag. What is the probability that Zack will not randomly pick a red or blue marble from the bag?
Zack has 10 green, 14 red, 2 blue, and 6 black marbles in a bag. What is the probability that Zack will not randomly pick a red or blue marble from the bag?
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To NOT choose a red or blue, leaves 6 black and 10 green to choose from. That leaves 16 marbles out of a total of 32 marbles, or a 1/2 chance.
To NOT choose a red or blue, leaves 6 black and 10 green to choose from. That leaves 16 marbles out of a total of 32 marbles, or a 1/2 chance.
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What is the probability of choosing three hearts in three draws from a standard deck of playing cards, if replacement of cards is not allowed?
What is the probability of choosing three hearts in three draws from a standard deck of playing cards, if replacement of cards is not allowed?
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The standard deck of cards has 52 cards: 13 cards in 4 suits.
Ways to choose three hearts: 13 * 12 * 11 = 1716
Ways to choose three cards: 52 * 51 * 50 = 132600
Probability is a number between 0 and 1 that is defines as the total ways of what you want ÷ by the total ways
The resulting simplified fraction is 11/850
The standard deck of cards has 52 cards: 13 cards in 4 suits.
Ways to choose three hearts: 13 * 12 * 11 = 1716
Ways to choose three cards: 52 * 51 * 50 = 132600
Probability is a number between 0 and 1 that is defines as the total ways of what you want ÷ by the total ways
The resulting simplified fraction is 11/850
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What is the arithmetic mean of all of the odd numbers between 7 and 21, inclusive?
What is the arithmetic mean of all of the odd numbers between 7 and 21, inclusive?
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One can simply add all the odd numbers from 7 to 21 and divide by the number of odd numbers there are. Or, moreover, one can see that 14 is halfway between 7 and 21.
One can simply add all the odd numbers from 7 to 21 and divide by the number of odd numbers there are. Or, moreover, one can see that 14 is halfway between 7 and 21.
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It takes Johnny 25 minutes to run a loop around the track. He runs a second loop and it takes him 30 minutes. If the track is 5.5 miles long, what is his average speed in miles per hour?
It takes Johnny 25 minutes to run a loop around the track. He runs a second loop and it takes him 30 minutes. If the track is 5.5 miles long, what is his average speed in miles per hour?
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The minutes must be converted to hours which gives 11/12 hours. The total distance he runs is 11 miles. 11/(11/12) = 12.
The minutes must be converted to hours which gives 11/12 hours. The total distance he runs is 11 miles. 11/(11/12) = 12.
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I recently joined a bowling team. Each night we play three games. During my first two games I scored a 112 and 134, what must I score on my next game to ensure my average for that night will be a 132?
I recently joined a bowling team. Each night we play three games. During my first two games I scored a 112 and 134, what must I score on my next game to ensure my average for that night will be a 132?
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To find the average you add all the games and divide by the number of games. In this case we have 112 + 134 + x = 246 + x. If we divide by 3 and set our answer to 132, we can solve for x by cross multiplying and solving algebraically. We can also solve this problem using substitution.
To find the average you add all the games and divide by the number of games. In this case we have 112 + 134 + x = 246 + x. If we divide by 3 and set our answer to 132, we can solve for x by cross multiplying and solving algebraically. We can also solve this problem using substitution.
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For the fall semester, three quizzes were given, a mid-term exam, and a final exam. To determine a final grade, the mid-term was worth three times as much as a quiz and the final was worth five times as much as a quiz. If Jonuse scored 85, 72 and 81 on the quizzes, 79 on the mid-term and 92 on the final exam, what was his average for the course?
For the fall semester, three quizzes were given, a mid-term exam, and a final exam. To determine a final grade, the mid-term was worth three times as much as a quiz and the final was worth five times as much as a quiz. If Jonuse scored 85, 72 and 81 on the quizzes, 79 on the mid-term and 92 on the final exam, what was his average for the course?
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The formula for a weighted average is the sum of the weight x values divided by the sum of the weights. Thus, for the above situation:
Average = (1 x 85 + 1 x 72 + 1 x 81 + 3 x 79 + 5 x 92) / ( 1 + 1 + 1 + 3 + 5)
= 935 / 11 = 85.
The formula for a weighted average is the sum of the weight x values divided by the sum of the weights. Thus, for the above situation:
Average = (1 x 85 + 1 x 72 + 1 x 81 + 3 x 79 + 5 x 92) / ( 1 + 1 + 1 + 3 + 5)
= 935 / 11 = 85.
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The average (arithmetic mean) of m, n and p is 8. If m + n = 15 then p equals:
The average (arithmetic mean) of m, n and p is 8. If m + n = 15 then p equals:
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If the arithmetic mean of the three numbers is 8, then the three numbers total 24. We are given m + n, leaving p to equal 24 – 15 = 9.
If the arithmetic mean of the three numbers is 8, then the three numbers total 24. We are given m + n, leaving p to equal 24 – 15 = 9.
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