All SAT Math Resources
Example Questions
Example Question #131 : Statistics
Adam is comparing the prices of granola at different grocery stores in town. So far, these are the prices he's found:
If the average of these prices is then what is the range of all the prices?
First, we want to solve for x. Since we know the average of all the prices is $1.90, then we can say
Now, we can find the range of the prices. The range is represented by the highest value minus the lowest value in any set. With that in mind, our range is given by:
Example Question #133 : Statistics
A word game comprises 100 square tiles, each of which has a letter and a numeric value. Sixty of the tiles have value 1 each; twenty tiles have value 2 each; twelve of the tiles have value 4 each; four tiles have value 8 each; and two tiles have value 10 each. There are also two blank tiles with zero value.
What is the mean of the values of the tiles?
None of the other responses is correct.
The total value of the tiles is
There are 100 tiles, so the mean value is
Example Question #12 : Range
Find the range of the following set of numbers:
1,5,14,17,22,23,23
To find range of any set of data, simply subtract the smallest number from the largest number. It is easiest if you order the numbers first. Thus,
Example Question #1 : How To Find The Probability Of An Outcome
A bag of jellybeans has 20 watermelon jellybeans, 45 sour apple jellybeans, 30 orange jellybeans and 5 cotton candy jellybeans. If you reach in and grab one jelly bean, what is the probability that it will be watermelon flavored?
1/20
4/19
1/3
1/4
1/5
1/5
Add up the total number of jellybeans, 20 + 45 + 30 + 5 = 100.
Divide the number of watermelon jellybeans by the total: 20/100 and reduce the fraction to 1/5.
Example Question #1 : Probability
A circle is inscribed inside a square. If a point inside the square is selected at random, what is the probability that the point will also be inside the circle?
π/4
3/4
5/6
π/6
π/4
The probability of the point being inside the circle is the ratio of the area of the circle to the area of the square. If we suppose that the circle has radius r, then the square must have side 2r. The area of the circle is πr2 and the area of the square is 〖(2r)〗2=〖4r〗2, so the proportion of the areas is (πr2)/〖4r〗2 =π/4.
Example Question #1 : How To Find The Probability Of An Outcome
John has a bowl with 54 marbles in it. Half of the marbles are green and half of the marbles are blue. John takes 3 green marbles and 6 blue marbles from the jar. John then takes 13 additional marbles from the remaining marbles in the jar. What is the minimum number of these 13 marbles that must be green in order for there to be more blue marbles than green marbles left?
5
9
8
10
4
9
The bowl has 54 marbles, half green and half blue. This gives us 27 green and 27 blue marbles:
27 G / 27 B
John then takes 3 green and 6 blue from the bowl. This leaves the bowl with:
24 G / 21 B
If there are going to be more blue than green marbles after John's 13 marbles, he has to take at least 4 more green marbles than blue marbles, because right now there are 3 less blue marbles. Therefore, we need to take at least 9 green marbles, which would mean 4 or less of the marbles would be blue (8 green and 5 blue would leave us with equal green and equal blue marbles, so it would have to be more than 8 green marbles, which gives us 9 green marbles).
We can also solve this as an inequality. You take the difference in marbles, which is 3, which means you need the difference in green and blue marbles to be greater than 3, or at least 4. You have b + g = 13 and g - b > 3, where b and g are positive integers.
b + g = 13 (Subtract g on both sides of the equation)
b = 13 - g
g - b > 3 (Substitute above equation)
g - (13 - g) > 3 (Distribute negative sign in parentheses)
g - 13 + g > 3 (Add both g variables)
2g - 13 > 3 (Add 13 to both sides of the inequality)
2g > 16 (Divide both sides of the inequality by 2)
g > 8 so g has to be 9 or greater.
Example Question #1 : How To Find The Probability Of An Outcome
If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?
3/10
9/20
6/10
11/20
9/10
9/20
If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?
Here we have 5 possible choices for x and 4 possible choices for y, giving us 5 * 4 = 20 possible outcomes.
We know that odd times odd = odd; even times even = even; and even times odd = even. Thus we need all of the outcomes where x and y are odd. We have 3 possibilities of odd numbers for x, and 3 possibilities of odd numbers for y, so we will have 9 outcomes of our total 20 outcomes where xy is odd, giving us a probability of 9/20.
Example Question #2 : Probability
Mike has a bag of marbles, 4 white, 8 blue, and 6 red. He pulls out one marble from the bag and it is red. What is the probability that the second marble he pulls out of the bag is white?
1/6
4/17
3/18
4/18
4/17
There are 18 marbles in total. One of them is removed so now there are 17 marbles. This is our denominator. All of the original white marbles are still in the bag so there is a 4 out of 17 or 4/17 chance that the next marble taken out of the bag will be white.
Example Question #1 : Probability
Michael tosses three fair coins. What is the probability that at least one of these coins will land on heads?
3/4
1/2
1/4
1/8
7/8
7/8
Michael can toss either one head, two heads, or three heads.
If Michael tosses one head, then it could be on either the first, second, or third toss. We could model it like this, where H represents heads and T represents tails.
HTT, THT, or TTH
If Michael tosses two heads, then there are three possible combinations:
HHT, HTH, or THH
If Michael tosses three heads, then there is only one possible combination:
HHH
Thus, there are seven ways that Michael can toss at least one head. We must find the probability of each of these ways and then add them together.
The probability of rolling a head is ½ and the probability of rolling a tail is ½. Because each coin toss is independent, we can multiply the probabilities together.
For example, the probability of the combination HTT is (1/2)(1/2)(1/2) = 1/8
Probability of HTT = 1/8
Probability of THT = (1/2)(1/2)(1/2) = 1/8
Probability of TTH = (1/2)(1/2)(1/2) = 1/8
Probability of HHT = 1/8
Probability of HTH = 1/8
Probability of THH = 1/8
Probability of HHH = 1/8
So, there are seven possible ways that Michael can toss at least one head. The probability of each of these seven ways is equal to 1/8. Thus, the total probability of all seven events is 7/8.
ALTERNATE SOLUTION:
Michael can toss at least one head, or he can toss zero heads. The sum of these two probabilities must equal one, because they represent all of the ways that Michael could toss the coins. He could either toss at least on head, or he could toss no heads at all.
Probability of tossing at least one head + probability of tossing no heads = 1
The probability of tossing no heads is only possible with the combination TTT. The probability of tossing three tails is equal to (1/2)(1/2)(1/2) = 1/8
Probability of tossing at least one head + 1/8 = 1
Probability of tossing at least one head = 1 – 1/8 = 7/8 .
Example Question #1 : How To Find The Probability Of An Outcome
A jar of marbles has 125 marbles in it. 25 are blue marbles, 65 are red marble, 15 are green marbles and 20 are yellow marbles. What is the probability that the first three marbles picked are green or blue?
0.025
0.031
0.015
0.020
0.043
0.031
Probability of each event = (# green marbles + # blue marbles)/ Total # of Marbles
P1 = (15 + 25) / 125 = 40 / 125
Second event assumes a blue or green was chosen for first event so there is one fewer marble on top and also one fewer marble in the total number of marbles.
P2 = (14 + 25) / 124 = 39 / 124
Third event assumes a blue or green was chosen for first and second events so there are two fewer marbles on top and also two fewer marbles in the total number of marbles.
P3 = (13 + 25) / 124 = 38 / 123
Probability for multiple events = P1 x P2 x P3
(40 / 125) * (39 / 124) * (38 / 123)
( 40 * 39 * 38) / (125 * 124 * 123 ) = 59280 / 1906500 = 0.031