Algebraic Fractions - PSAT Math
Card 1 of 644
Simplify: 
Simplify:
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Recall that dividing is equivalent multiplying by the reciprocal. Therefore, ((x - 4) / (1 / 2)) / (1 / (x + 4)) = ((x - 4) * 2) * (x + 4) / 1.
Let's simplify this further:
(2x – 8) * (x + 4) = 2x2 – 8x + 8x – 32 = 2x2 – 32
Recall that dividing is equivalent multiplying by the reciprocal. Therefore, ((x - 4) / (1 / 2)) / (1 / (x + 4)) = ((x - 4) * 2) * (x + 4) / 1.
Let's simplify this further:
(2x – 8) * (x + 4) = 2x2 – 8x + 8x – 32 = 2x2 – 32
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If 3x = 12, y/4 = 10, and 4z = 9, what is the value of (10xyz)/xy?
If 3x = 12, y/4 = 10, and 4z = 9, what is the value of (10xyz)/xy?
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Solve for the variables, the plug into formula.
x = 12/3 = 4
y = 10 * 4 = 40
z= 9/4 = 2 1/4
10xyz = 3600
Xy = 160
3600/160 = 22 1/2
Solve for the variables, the plug into formula.
x = 12/3 = 4
y = 10 * 4 = 40
z= 9/4 = 2 1/4
10xyz = 3600
Xy = 160
3600/160 = 22 1/2
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Find the inverse equation of:

Find the inverse equation of:
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To solve for an inverse, we switch x and y and solve for y. Doing so yields:

To solve for an inverse, we switch x and y and solve for y. Doing so yields:
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Evaluate
when x=11. Round to the nearest tenth.
Evaluate
when x=11. Round to the nearest tenth.
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Wherever there is an x, plug in 11 and perform the given operations. The numerator will be equal to 83 and the denominator will be equal to 46. 83 divided by 46 is equal to 1.804… and since the second decimal place is not greater than or equal to 5, the first decimal place stays the same when rounding so the final answer is 1.8.
Wherever there is an x, plug in 11 and perform the given operations. The numerator will be equal to 83 and the denominator will be equal to 46. 83 divided by 46 is equal to 1.804… and since the second decimal place is not greater than or equal to 5, the first decimal place stays the same when rounding so the final answer is 1.8.
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Solve 
Solve 
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The common denominator of the left side is x(x–1). Multiplying the top and bottom of 1/x by (x–1) yields




Since this statement is true, there are infinitely many solutions.
The common denominator of the left side is x(x–1). Multiplying the top and bottom of 1/x by (x–1) yields




Since this statement is true, there are infinitely many solutions.
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Mary walked to school at an average speed of 2 miles per hour and jogged back along the same route at an average speed of 6 miles per hour. If her total traveling time was 1 hour, what was the total number of miles in the round trip?
Mary walked to school at an average speed of 2 miles per hour and jogged back along the same route at an average speed of 6 miles per hour. If her total traveling time was 1 hour, what was the total number of miles in the round trip?
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Since Mary traveled 3 times as quickly coming from school as she did going to school (6 miles per hour compared to 2 miles per hour), we know that Mary spent only a third of the time coming from school as she did going. If x represents the number of hours it took to get to school, then x/3 represents the number of hours it took her to return.
Knowing that the total trip took 1 hour, we have:
x + x/3 = 1
3x/3 + 1x/3 = 1
4_x_/3 = 1
x = 3/4
So we know it took Mary 3/4 of an hour to travel to school (and the remaining 1/4 of an hour to get back).
Remembering that distance = rate * time, the distance Mary traveled on her way to school was (2 miles per hour) * (3/4 of an hour) = 3/2 miles. Furthermore, since she took the same route coming back, she must have traveled 3/2 of a mile to return as well.
Therefore, the the total number of miles in Mary's round trip is 3/2 miles + 3/2 miles = 6/2 miles = 3 miles.
Since Mary traveled 3 times as quickly coming from school as she did going to school (6 miles per hour compared to 2 miles per hour), we know that Mary spent only a third of the time coming from school as she did going. If x represents the number of hours it took to get to school, then x/3 represents the number of hours it took her to return.
Knowing that the total trip took 1 hour, we have:
x + x/3 = 1
3x/3 + 1x/3 = 1
4_x_/3 = 1
x = 3/4
So we know it took Mary 3/4 of an hour to travel to school (and the remaining 1/4 of an hour to get back).
Remembering that distance = rate * time, the distance Mary traveled on her way to school was (2 miles per hour) * (3/4 of an hour) = 3/2 miles. Furthermore, since she took the same route coming back, she must have traveled 3/2 of a mile to return as well.
Therefore, the the total number of miles in Mary's round trip is 3/2 miles + 3/2 miles = 6/2 miles = 3 miles.
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According the pie chart, the degree measure of the sector representing the number of workers spending 5 to 9 years in the same role is how much greater in the construction industry chart than in the financial industry chart?
According the pie chart, the degree measure of the sector representing the number of workers spending 5 to 9 years in the same role is how much greater in the construction industry chart than in the financial industry chart?
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Since the values in the pie charts are currently in terms of percentages (/100), we must convert them to degrees (/360, since within a circle) to solve the question. The "5 to 9 years" portion for the financial and construction industries are 18 and 25 percent, respectively. As such, we can cross-multiply both:
18/100 = x/360
x = 65 degrees
25/100 = y/360
y = 90 degrees
Subtract: 90 – 65 = 25 degrees
Alternatively, we could first subtract the percentages (25 – 18 = 7), then convert the 7% to degree form via the same method of cross-multiplication.
Since the values in the pie charts are currently in terms of percentages (/100), we must convert them to degrees (/360, since within a circle) to solve the question. The "5 to 9 years" portion for the financial and construction industries are 18 and 25 percent, respectively. As such, we can cross-multiply both:
18/100 = x/360
x = 65 degrees
25/100 = y/360
y = 90 degrees
Subtract: 90 – 65 = 25 degrees
Alternatively, we could first subtract the percentages (25 – 18 = 7), then convert the 7% to degree form via the same method of cross-multiplication.
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6 contestants have an equal chance of winning a game. One contestant is disqualified, so now the 5 remaining contestants again have an equal chance of winning. How much more likely is a contestant to win after the disqualification?
6 contestants have an equal chance of winning a game. One contestant is disqualified, so now the 5 remaining contestants again have an equal chance of winning. How much more likely is a contestant to win after the disqualification?
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When there are 6 people playing, each contestant has a 1/6 chance of winning. After the disqualification, the remaining contestants have a 1/5 chance of winning.
1/5 – 1/6 = 6/30 – 5/30 = 1/30.
When there are 6 people playing, each contestant has a 1/6 chance of winning. After the disqualification, the remaining contestants have a 1/5 chance of winning.
1/5 – 1/6 = 6/30 – 5/30 = 1/30.
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If
then which of the following is equal to
?
If then which of the following is equal to
?
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To raise
to the exponent
, square
and then take the cube root.
![w^\frac{2}{3}=\sqrt[3]{w^2}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/177545/gif.latex)
![(\frac{1}{8})^\frac{2}{3}=\sqrt[3]{(\frac{1}{8})^2}=\sqrt[3]{\frac{1}{64}}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/177546/gif.latex)
![\sqrt[3]{\frac{1}{64}}=\frac{1}{4}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/177547/gif.latex)

To raise to the exponent
, square
and then take the cube root.
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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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For this question, the following trigonometric identities apply:
,


Simplify: 
For this question, the following trigonometric identities apply:
,
Simplify:
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To begin a problem like this, you must first convert everything to
and
alone. This way, you can begin to cancel and combine to its most simplified form.
Since
and
, we insert those identities into the equation as follows.

From here we combine the numerator and denominators of each fraction together to easily see what we can combine and cancel.

Since there is a
in the numerator and the denominator, we can cancel them as they divide to equal 1. All we have left is
, the answer.
To begin a problem like this, you must first convert everything to and
alone. This way, you can begin to cancel and combine to its most simplified form.
Since and
, we insert those identities into the equation as follows.
From here we combine the numerator and denominators of each fraction together to easily see what we can combine and cancel.
Since there is a in the numerator and the denominator, we can cancel them as they divide to equal 1. All we have left is
, the answer.
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If
pizzas cost
dollars and
sodas cost
dollars, what is the cost of
pizzas and
sodas in terms of
and
?
If pizzas cost
dollars and
sodas cost
dollars, what is the cost of
pizzas and
sodas in terms of
and
?
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If 10 pizzas cost x dollars, then each pizza costs x/10. Similarly, each soda costs y/6. We can add the pizzas and sodas together by finding a common denominator:

If 10 pizzas cost x dollars, then each pizza costs x/10. Similarly, each soda costs y/6. We can add the pizzas and sodas together by finding a common denominator:
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Factor out 7 from the numerator: 
This simplifies to 7.
Factor out 7 from the numerator:
This simplifies to 7.
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If
and
, then which of the following is equal to
?
If and
, then which of the following is equal to
?
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In order to solve
, first substitute the values of
and
provided in the problem:

Find the Least Common Multiple (LCM) of the fractional terms in the denominator and find the equivalent fractions with the same common denominator:



Finally, in order to divide by a fraction, we must multiply by the reciprocal of the fraction:


In order to solve , first substitute the values of
and
provided in the problem:
Find the Least Common Multiple (LCM) of the fractional terms in the denominator and find the equivalent fractions with the same common denominator:
Finally, in order to divide by a fraction, we must multiply by the reciprocal of the fraction:
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Simplify:

Simplify:
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Begin by simplifying the numerator and the denominator.
Numerator
has a common denominator of
. Therefore, we have:

Denominator
has a common denominator of
. Therefore, we have:

Now, reconstructing our fraction, we have:

To make this division work, you multiply the numerator by the reciprocal of the denominator:

Begin by simplifying the numerator and the denominator.
Numerator
has a common denominator of
. Therefore, we have:
Denominator
has a common denominator of
. Therefore, we have:
Now, reconstructing our fraction, we have:
To make this division work, you multiply the numerator by the reciprocal of the denominator:
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If
,
, and
, find the value of
.
If ,
, and
, find the value of
.
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In order to solve
, we must first find the values of
,
, and
using the initial equations provided. Starting with
:

Then:



Finally:



With the values of
,
, and
in hand, we can solve the final equation:





In order to solve , we must first find the values of
,
, and
using the initial equations provided. Starting with
:
Then:
Finally:
With the values of ,
, and
in hand, we can solve the final equation:
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Find the value of
if
and
.
Find the value of if
and
.
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In order to solve for
, first substitute
into the equation for
:


Then, find the Least Common Multiple (LCM) of the two fractions and generate equivalent fractions with the same denominator:

Finally, simplify the equation:


In order to solve for , first substitute
into the equation for
:
Then, find the Least Common Multiple (LCM) of the two fractions and generate equivalent fractions with the same denominator:
Finally, simplify the equation:
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Evaluate the following equation when
and round your answer to the nearest hundredth.

Evaluate the following equation when and round your answer to the nearest hundredth.
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1. Plug in
wherever there is an
in the above equation.

2. Perform the above operations.



1. Plug in wherever there is an
in the above equation.
2. Perform the above operations.
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Simplify:

Simplify:
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Begin by simplifying the numerator.
has a common denominator of
. Therefore, we can rewrite it as:

Now, in our original problem this is really is:

When you divide by a fraction, you really multiply by the reciprocal:

Begin by simplifying the numerator.
has a common denominator of
. Therefore, we can rewrite it as:
Now, in our original problem this is really is:
When you divide by a fraction, you really multiply by the reciprocal:
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A school's tornado shelter has enough food to last 20 children for 6 days. If 24 children ended up taking shelter together, for how many fewer days will the food last?
A school's tornado shelter has enough food to last 20 children for 6 days. If 24 children ended up taking shelter together, for how many fewer days will the food last?
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Because the number of days goes down as the number of children goes up, this problem type is inverse variation. We can solve this problem by the following steps:
20*6=24*x
120=24x
x=120/24
x=5
In this equation, x represents the total number of days that can be weathered by 24 students. This is down from the 6 days that 20 students could take shelter together. So the difference is 1 day less.
Because the number of days goes down as the number of children goes up, this problem type is inverse variation. We can solve this problem by the following steps:
20*6=24*x
120=24x
x=120/24
x=5
In this equation, x represents the total number of days that can be weathered by 24 students. This is down from the 6 days that 20 students could take shelter together. So the difference is 1 day less.
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