Arithmetic - PSAT Math
Card 1 of 4032
A TV show lasts 30 minutes, what fraction of the show is left after 12 minutes have passed?
A TV show lasts 30 minutes, what fraction of the show is left after 12 minutes have passed?
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After watching 12 minutes of the show 18 remain. 18 is 60% of the total 30 minutes. As a fraction it can be expressed as 3/5.
After watching 12 minutes of the show 18 remain. 18 is 60% of the total 30 minutes. As a fraction it can be expressed as 3/5.
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A bag contains red, orange, and yellow marbles only. The marbles occur in a ratio of 5 red marbles: 4 orange marbles: 1 yellow marble. If one-third of the red marbles, one-half of the orange marbles, and one-fourth of the yellow marbles are removed, then what fraction of the remaining marbles in the bag is red?
A bag contains red, orange, and yellow marbles only. The marbles occur in a ratio of 5 red marbles: 4 orange marbles: 1 yellow marble. If one-third of the red marbles, one-half of the orange marbles, and one-fourth of the yellow marbles are removed, then what fraction of the remaining marbles in the bag is red?
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Marty drove 40 mi/hr for 3 hours, then 60 mi/hr for 1 hour, and finally 70 mi/hr for the last 2 hours. What was Marty's average speed?
Marty drove 40 mi/hr for 3 hours, then 60 mi/hr for 1 hour, and finally 70 mi/hr for the last 2 hours. What was Marty's average speed?
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Marty's total driving time was 3 + 1 + 2 = 6 hours. He drove 40 mi/hr for 3 hours, or 3/6 = 1/2 of the time. He drove 60 mi/hr for 1 hour, or 1/6 of the drive. Lastly, he drove 70 mi/hr for 2 hours, or 2/6 = 1/3 of the drive.
To find the average speed, we need to multiply the speeds with their corresponding weights and add them up.
Average = 1/2 * 40 + 1/6 * 60 + 1/3 * 70 = 53.33... ≈ 53 mi/hr
Marty's total driving time was 3 + 1 + 2 = 6 hours. He drove 40 mi/hr for 3 hours, or 3/6 = 1/2 of the time. He drove 60 mi/hr for 1 hour, or 1/6 of the drive. Lastly, he drove 70 mi/hr for 2 hours, or 2/6 = 1/3 of the drive.
To find the average speed, we need to multiply the speeds with their corresponding weights and add them up.
Average = 1/2 * 40 + 1/6 * 60 + 1/3 * 70 = 53.33... ≈ 53 mi/hr
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If the length of a rectangle is increased by thirty percent, which of the following most closely approximates the percent by which the rectangle's width must decrease, so that the area of the rectangle remains unchanged?
If the length of a rectangle is increased by thirty percent, which of the following most closely approximates the percent by which the rectangle's width must decrease, so that the area of the rectangle remains unchanged?
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If the price of a TV was decreased from $3,000 to $1,800, by what percent was the price decreased?
If the price of a TV was decreased from $3,000 to $1,800, by what percent was the price decreased?
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The price was lowered by $1,200 which is 40% of $3,000.
The price was lowered by $1,200 which is 40% of $3,000.
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Simplify:
Simplify:
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With this problem the first thing to do is cancel out variables. The x2 can all be divided by each other because they are present in each system. The equation will now look like this:
Now we can see that the equation can all be divided by y, leaving the answer to be:
With this problem the first thing to do is cancel out variables. The x2 can all be divided by each other because they are present in each system. The equation will now look like this:
Now we can see that the equation can all be divided by y, leaving the answer to be:
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There is a 5 to 3 ratio of red marbles to green marbles in a bucket. How many red marbles can be in a bucket of 40 marbles?
There is a 5 to 3 ratio of red marbles to green marbles in a bucket. How many red marbles can be in a bucket of 40 marbles?
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In this problem, we use a proportion to solve for the number of red marbles in the bucket. Since the ratio is 5 to 3 we can set up a proportion of the red over total where the total is (red marbles + green marbles). We then set this proportion equal to 
over our new total, which is 40. Then we cross multiply and divide to get the number of red marbles in a bucket of 40 total marbles.
In this problem, we use a proportion to solve for the number of red marbles in the bucket. Since the ratio is 5 to 3 we can set up a proportion of the red over total where the total is (red marbles + green marbles). We then set this proportion equal to
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There is a 2 to 3 ratio of dogs to cats in a park. How many cats are there if there are 12 dogs?
There is a 2 to 3 ratio of dogs to cats in a park. How many cats are there if there are 12 dogs?
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We set up the following ratio of dogs to cats and set it equal to the new ratio we are interested in.
Then we cross multiply and divide to solve for 
.
We set up the following ratio of dogs to cats and set it equal to the new ratio we are interested in.
Then we cross multiply and divide to solve for
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You have a bag filled with red and green apples. If there are 9 red apples and 12 green apples, what is the probability of choosing a red apple out of the bag?
You have a bag filled with red and green apples. If there are 9 red apples and 12 green apples, what is the probability of choosing a red apple out of the bag?
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To find the probability we need to set up a ratio. The ratio will be how many red apples are in the bag over the total number of apples (both red and green) in the bag.
We can now reduce this fraction:
To find the probability we need to set up a ratio. The ratio will be how many red apples are in the bag over the total number of apples (both red and green) in the bag.
We can now reduce this fraction:
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A deck of cards has 52 cards. What is the probability of dealing a card with a face value of 4?
A deck of cards has 52 cards. What is the probability of dealing a card with a face value of 4?
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For this problem we need to recall that that there are 4 different suits in a deck of cards. Each suit has numbers 1 through 10, jack, queen, king, and ace. Therefore, we can write a ratio of number of 4's over total cards in the deck.
From here we can reduce the fraction and we get
For this problem we need to recall that that there are 4 different suits in a deck of cards. Each suit has numbers 1 through 10, jack, queen, king, and ace. Therefore, we can write a ratio of number of 4's over total cards in the deck.
From here we can reduce the fraction and we get
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In a group of 20 children, 25% are girls. How many boys are there?
In a group of 20 children, 25% are girls. How many boys are there?
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Since 
of the children are girls, this totals to 
girls in the group.
boys.
Since
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If a rectangle's length decreases by fifteen percent, and its width decreases by twenty percent, then by what percent does the rectangle's area decrease?
If a rectangle's length decreases by fifteen percent, and its width decreases by twenty percent, then by what percent does the rectangle's area decrease?
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Let's call the original length and width of the rectangle 
and 
, respectively.
The initial area, 
, of the rectangle is equal to the product of the length and the width. We can represent this with the following equation:
Next, let 
and 
represent the length and width, respectively, after they have been decreased. The final area will be equal to 
, which will be equal to the product of the final length and width.
We are asked to find the change in the area, which essentially means we want to compare 
and 
. In order to do this, we will need to find an expression for 
in terms of 
and 
. We can rewrite 
and 
in terms of 
and 
.
First, we are told that the length is decreased by fifteen percent. We can think of the full length as 100% of the length. If we take away fifteen percent, we are left with 100 – 15, or 85% of the length. In other words, the final length is 85% of the original length. We can represent 85% as a decimal by moving the decimal two places to the left.
= 85% of 
= 
Similarly, if we decrease the width by 20%, we are only left with 80% of the width.
= 80% of 
= 
We can now express the final area in terms of 
and 
by substituting the expressions we just found for the final length and width.
= (
)(
) = 
Lastly, let's apply the formula for percent of change, which will equal the change in the area divided by the original area. The change in the area is equal to the final area minus the original area.
percent change = 
(100%)
=
(100%)
=
(100%) = –0.32(100%) = –32%
The negative sign indicates that the rectangle's area decreased. The change in the area was a decrease of 32%.
The answer is 32.
Let's call the original length and width of the rectangle
The initial area,
Next, let
We are asked to find the change in the area, which essentially means we want to compare
First, we are told that the length is decreased by fifteen percent. We can think of the full length as 100% of the length. If we take away fifteen percent, we are left with 100 – 15, or 85% of the length. In other words, the final length is 85% of the original length. We can represent 85% as a decimal by moving the decimal two places to the left.
Similarly, if we decrease the width by 20%, we are only left with 80% of the width.
We can now express the final area in terms of
Lastly, let's apply the formula for percent of change, which will equal the change in the area divided by the original area. The change in the area is equal to the final area minus the original area.
percent change =
=
=
The negative sign indicates that the rectangle's area decreased. The change in the area was a decrease of 32%.
The answer is 32.
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Two two-digit numbers, 
and 
, sum to produce a three-digit number in which the second digit is equal to 
. The addition is represented below. (Note that the variables are used to represent individual digits; no multiplication is taking place).
What is 
?
Two two-digit numbers,
What is
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Another way to represent this question is:
In the one's column, 
and 
add to produce a number with a two in the one's place. In the ten's column, we can see that a one must carry in order to get a digit in the hundred's place. Together, we can combine these deductions to see that the sum of 
and 
must be twelve (a one in the ten's place and a two in the one's place).
In the one's column: 
The one carries to the ten's column.
In the ten's column: 
The three goes into the answer and the one carries to the hundred's place. The final answer is 132. From this, we can see that 
because 
.
Using this information, we can solve for 
.
You can check your answer by returning to the original addition and plugging in the values of 
and 
.
Another way to represent this question is:
In the one's column,
In the one's column:
The one carries to the ten's column.
In the ten's column:
The three goes into the answer and the one carries to the hundred's place. The final answer is 132. From this, we can see that
Using this information, we can solve for
You can check your answer by returning to the original addition and plugging in the values of
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Simplify:
Simplify:
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_x_2 – _y_2 can be also expressed as (x + y)(x – y).
Therefore, the fraction now can be re-written as (x + y)(x – y)/(x + y).
This simplifies to (x – y).
_x_2 – _y_2 can be also expressed as (x + y)(x – y).
Therefore, the fraction now can be re-written as (x + y)(x – y)/(x + y).
This simplifies to (x – y).
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Simplify the following expression:
Simplify the following expression:
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Following this equation, you divide 4 by 8 to get 1/2. When a variable is raised to an exponent, and you are dividing, you subtract the exponents, so 6 – 3 = 3.
Following this equation, you divide 4 by 8 to get 1/2. When a variable is raised to an exponent, and you are dividing, you subtract the exponents, so 6 – 3 = 3.
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Simplify the given fraction:
Simplify the given fraction:
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125 goes into 2000 evenly 16 times. 1/16 is the fraction in its simplest form.
125 goes into 2000 evenly 16 times. 1/16 is the fraction in its simplest form.
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Simplify the given fraction:
Simplify the given fraction:
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120 goes into 6000 evenly 50 times, so we get 1/50 as our simplified fraction.
120 goes into 6000 evenly 50 times, so we get 1/50 as our simplified fraction.
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Let 
= 
, where 
and 
are both positive integers whose greatest common factor is one. What is the value of 
?
Let
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First we want to simplify the expression: 
.
One way to simplify this complex fraction is to find the least common multiple of all the denominators, i.e. the least common denominator (LCD). If we find this, then we can multiply every fraction by the LCD and thereby be left with only whole numbers. This will make more sense in a little bit.
The denominators we are dealing with are 2, 3, 4, 5, and 6. We want to find the smallest multiple that these numbers have in common. First, it will help us to notice that 6 is a multiple of both 2 and 3. Thus, if we find the least common multiple of 4, 5, and 6, it will automatically be a multiple of both 2 and 3. Let's list out the first several multiples of 4, 5, and 6.
4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60
5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60
6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
The smallest multiple that 4, 5, and 6 have in common is 60. Thus, the least common multiple of 4, 5, and 6 is 60. This also means that the least common multiple of 2, 3, 4, 5, and 6 is 60. Therefore, the LCD of all the fractions is 60.
Let's think of the expression we want to simplify as one big fraction. The numerator contains the fractions 1/4, 1/3, and –1/5. The denominator of the fraction is 1/2, –1/6 and 1. Remember that if we have a fraction, we can multiply the numerator and denominator by the same number without changing the value of the fraction. In other words, x/y = (xz)/(yz). This will help us because we can multiply the numerator (which consists of 1/4, 1/3, and –1/5) by 60, and then mutiply the denominator (which consists of 1/2, –1/6, and 1) by 60, thereby ridding us of fractions in the numerator and denominator. This process is shown below:
= 
= 
= 
This means that a/b = 23/80. We are told that a and b are both positive and that their greatest common factor is 1. In other words, a/b must be the simplified form of 23/80. When a fraction is in simplest form, the greatest common factor of the numerator and denominator equals one. Since 23/80 is simplified, a = 23, and b = 80. The sum of a and b is thus 23 + 80 = 103.
The answer is 103.
First we want to simplify the expression:
One way to simplify this complex fraction is to find the least common multiple of all the denominators, i.e. the least common denominator (LCD). If we find this, then we can multiply every fraction by the LCD and thereby be left with only whole numbers. This will make more sense in a little bit.
The denominators we are dealing with are 2, 3, 4, 5, and 6. We want to find the smallest multiple that these numbers have in common. First, it will help us to notice that 6 is a multiple of both 2 and 3. Thus, if we find the least common multiple of 4, 5, and 6, it will automatically be a multiple of both 2 and 3. Let's list out the first several multiples of 4, 5, and 6.
4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60
5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60
6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
The smallest multiple that 4, 5, and 6 have in common is 60. Thus, the least common multiple of 4, 5, and 6 is 60. This also means that the least common multiple of 2, 3, 4, 5, and 6 is 60. Therefore, the LCD of all the fractions is 60.
Let's think of the expression we want to simplify as one big fraction. The numerator contains the fractions 1/4, 1/3, and –1/5. The denominator of the fraction is 1/2, –1/6 and 1. Remember that if we have a fraction, we can multiply the numerator and denominator by the same number without changing the value of the fraction. In other words, x/y = (xz)/(yz). This will help us because we can multiply the numerator (which consists of 1/4, 1/3, and –1/5) by 60, and then mutiply the denominator (which consists of 1/2, –1/6, and 1) by 60, thereby ridding us of fractions in the numerator and denominator. This process is shown below:
=
This means that a/b = 23/80. We are told that a and b are both positive and that their greatest common factor is 1. In other words, a/b must be the simplified form of 23/80. When a fraction is in simplest form, the greatest common factor of the numerator and denominator equals one. Since 23/80 is simplified, a = 23, and b = 80. The sum of a and b is thus 23 + 80 = 103.
The answer is 103.
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Simplify:
(2_x_ + 4)/(x + 2)
Simplify:
(2_x_ + 4)/(x + 2)
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(2_x_ + 4)/(x + 2)
To simplify you must first factor the top polynomial to 2(x + 2). You may then eliminate the identical (x + 2) from the top and bottom leaving 2.
(2_x_ + 4)/(x + 2)
To simplify you must first factor the top polynomial to 2(x + 2). You may then eliminate the identical (x + 2) from the top and bottom leaving 2.
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The expression 
The expression
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A negative exponent in the numerator of a fraction can be rewritten with a positive exponent in the denominator. The same is true for a negative exponent in the denominator. Thus, 
.
When 
is multiplied by 
, the numerators and denominators cancel out, and you are left with 1.
A negative exponent in the numerator of a fraction can be rewritten with a positive exponent in the denominator. The same is true for a negative exponent in the denominator. Thus,
When
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