How to find decimal equivalent to a percentage - PSAT Math
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55 and 1/2% of 23 is about what?
55 and 1/2% of 23 is about what?
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55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Tap to see back →
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
If 
of 
is equal to 
of 
, and 
of 
is equal to 
of 
, then what percent of 
is 
?
If
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We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
Tap to see back →
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen: 
have signed.
Sophomores: 
have signed.
Juniors: 
have signed.
Seniors: 
have signed.
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen:
Sophomores:
Juniors:
Seniors:
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
55 and 1/2% of 23 is about what?
55 and 1/2% of 23 is about what?
Tap to see back →
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Tap to see back →
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
If of is equal to of , and of is equal to of , then what percent of is ?
If
Tap to see back →
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
Tap to see back →
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen: have signed.
Sophomores: have signed.
Juniors: have signed.
Seniors: have signed.
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen:
Sophomores:
Juniors:
Seniors:
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
55 and 1/2% of 23 is about what?
55 and 1/2% of 23 is about what?
Tap to see back →
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Tap to see back →
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
If of is equal to of , and of is equal to of , then what percent of is ?
If
Tap to see back →
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
Tap to see back →
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen: have signed.
Sophomores: have signed.
Juniors: have signed.
Seniors: have signed.
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen:
Sophomores:
Juniors:
Seniors:
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
55 and 1/2% of 23 is about what?
55 and 1/2% of 23 is about what?
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55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Tap to see back →
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
If of is equal to of , and of is equal to of , then what percent of is ?
If
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We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
Tap to see back →
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen: have signed.
Sophomores: have signed.
Juniors: have signed.
Seniors: have signed.
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen:
Sophomores:
Juniors:
Seniors:
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
55 and 1/2% of 23 is about what?
55 and 1/2% of 23 is about what?
Tap to see back →
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
55 and 1/2% can be written as a decimal: 0.555. To see what number is about 55.5% of 23, multiply 0.555 by 23. Answer: 12.765 or about 13.
Another route is to say that 55.5% is about half of 23. Half of 23 is 11.5. Since 55.5% is greater than 50%, 13 is the logical choice instead of 11.
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Let x and y be numbers such that x and y are both nonzero, and x > y. If half of x is equal to thirty percent of the positive difference between x and y, then what is the ratio of x to y?
Tap to see back →
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
We need to find expressions for fifty percent of x and for thirty percent of the positive difference between x and y. Then, we can set these two expressions equal to each other and determine the ratio of x to y.
Fifty percent of x is equal to one-half of x, which is the same as multiplying x by 0.50.
50% of x = 0.5_x_
Thirty percent of the positive difference between x and y means that we need to multiply the positive difference between x and y by thirty percent. Because x > y, the positive difference between x and y is equal to x – y. We then need to take thirty percent of the quantity x – y. Remember that to convert from a percent to a decimal, we move the decimal two spaces to the left. Therefore, 30% = 0.30. We can now multiply this by (x – y).
30% of x – y = 0.30(x – y)
Now, we set the two expressions equal to one another.
0.5_x_ = 0.30(x – y)
Distribute the right side.
0.5_x_ = 0.3_x_ – 0.3_y_
The ratio of x to y is represent by x/y. Thus, we want to group the x and y terms on opposite sides of the equations, and then divide both sides by y.
0.5_x_ = 0.3_x_ – 0.3_y_
Subtract 0.3_x_ from both sides.
0.2_x_ = –0.3_y_
Divide both sides by 0.2
x = (–0.3/0.2)y
Divide both sides by y to find x/y.
x/y = (–0.3/0.2) = –1.5.
Because the answers are in fractions, we want to rewrite –1.5 as a fraction. We can write –1.5 as –1.5/1 and then mutiply the top and bottom by 2.
(–1.5/1)(2/2) = –3/2
The answer is –3/2
If of is equal to of , and of is equal to of , then what percent of is ?
If
Tap to see back →
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
We are told that 50% of x is equal to 25% of y. We need to represent these two pieces of information as algebraic expressions. We can convert 50% and 25% to decimals by moving the decimals two places to the left. Thus, 50% = 0.50, and 25% = 0.25. To find 50% of x, we multiply x by 0.50. In other words, 50% of x = 0.50_x_. Likewise, 25% of y = 0.25_y_. We now set 0.50_x_ and 0.25_y_ equal to one another.
0.50_x_ = 0.25_y_
Let's divide both sides by 0.25 to get rid of decimals.
2_x_ = y
Next, we are told that 40% of y is equal to 60% of z. We will represent 40% and 60% as 0.40 and 0.60, respectively. Thus, we can write the following equation:
0.40_y_ = 0.60_z_
Ultimately, we are asked to find x as a percentage of z. This means we want to find an equation with x and z, but not y. If we solve for y in the second equation, and then substitute this value into the first, we can eliminate y.
Let's take the equation 0.40_y_ = 0.60_z_ and divide both sides by 0.40.
y = 1.5_z_
Now, we can take 1.5_z_ and substitute this for y in the first equation.
2_x_ = 1.5_z_
In order to find x as a percent of z, we must solve for x in terms of z. This means we must divide both sides of the equation by 2.
x = 0.75_z_
x is 0.75 times z. We can represent 0.75 as 75%, because in order to convert from a decimal to a percent, we need to move the decimal two spaces to the right. Therefore, if x = 0.75_z_, then x = 75% of z.
The answer is 75.
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
To get on the ballot for the student body president at Harding High School, a student must turn in a petition with the signatures of 8% of the students from each of the four classes - freshman, sophomore, junior, and senior. There are 342 freshmen, 312 sophomores, 270 juniors, and 268 seniors enrolled at McKinley.
Tom has a petition with the signatures of 25 students from each of the four classes. Can he get on the ballot with the signatures he has, and if not, why not?
Tap to see back →
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen: have signed.
Sophomores: have signed.
Juniors: have signed.
Seniors: have signed.
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.
In order to answer the question, we must find out the percent of each class that has signed Tom's petition, and compare it to 8%.
Freshmen:
Sophomores:
Juniors:
Seniors:
Tom has the necessary signatures from members of the top three classes, but he cannot get on the ballot yet because he has not gathered enough signatures from freshmen.