Proportion / Ratio / Rate - SAT Math
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Vikki is able to complete 4 SAT reading questions in 6 minutes. At this rate, how many questions can she answer in 3 1/2 hours?
Vikki is able to complete 4 SAT reading questions in 6 minutes. At this rate, how many questions can she answer in 3 1/2 hours?
First, find how many minutes are in 3 1/2 hours: 3 * 60 + 30 = 210 minutes. Then divide 210 by 6 to find how many six-minute intervals are in 210 minutes: 210/6 = 35. Since Vikki can complete 4 questions every 6 minutes, and there are 35 six-minute intervals we can multiply 4 by 35 to determine the total number of questions that she can complete.
4 * 35 = 140 problems.
First, find how many minutes are in 3 1/2 hours: 3 * 60 + 30 = 210 minutes. Then divide 210 by 6 to find how many six-minute intervals are in 210 minutes: 210/6 = 35. Since Vikki can complete 4 questions every 6 minutes, and there are 35 six-minute intervals we can multiply 4 by 35 to determine the total number of questions that she can complete.
4 * 35 = 140 problems.
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If an airplane is flying 225mph about how long will it take the plane to go 600 miles?
If an airplane is flying 225mph about how long will it take the plane to go 600 miles?
Speed = distance /time; So by solving for time we get time = distance /speed. So the equation for the answer is (600 miles)/ (225 miles/hr)= 2.67 hours; Remember to round up when the last digit of concern is 5 or more.
Speed = distance /time; So by solving for time we get time = distance /speed. So the equation for the answer is (600 miles)/ (225 miles/hr)= 2.67 hours; Remember to round up when the last digit of concern is 5 or more.
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If the ratio of q to r is 3:5 and the ratio of r to s is 10:7, what is the ratio of q to s?
If the ratio of q to r is 3:5 and the ratio of r to s is 10:7, what is the ratio of q to s?
Multiply the ratios. (q/r)(r/s)= q/s. (3/5) * (10/7)= 6:7.
Multiply the ratios. (q/r)(r/s)= q/s. (3/5) * (10/7)= 6:7.
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The first term in a sequence is m. If every term thereafter is 5 greater than 1/10 of the preceding term, and m≠0, what is the ratio of the second term to the first term?
The first term in a sequence is m. If every term thereafter is 5 greater than 1/10 of the preceding term, and m≠0, what is the ratio of the second term to the first term?
The first term is m, so the second term is m/10+5 or (m+50)/10. When we take the ratio of the second term to the first term, we get (((m+50)/10))/m, which is ((m+50)/10)(1/m), or (m+50)/10m.
The first term is m, so the second term is m/10+5 or (m+50)/10. When we take the ratio of the second term to the first term, we get (((m+50)/10))/m, which is ((m+50)/10)(1/m), or (m+50)/10m.
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Two cars were traveling 630 miles. Car A traveled an average speed of 70 miles per hour. If car B traveled 90 miles an hour, how many miles had car A traveled when car B arrived at the destination?
Two cars were traveling 630 miles. Car A traveled an average speed of 70 miles per hour. If car B traveled 90 miles an hour, how many miles had car A traveled when car B arrived at the destination?
We first divide 630 miles by 90 miles per hour to get the amount of time it took car B to reach the destination, giving us 7 hours. We then multiply 7 hours by car A’s average speed and we get 490 miles.
We first divide 630 miles by 90 miles per hour to get the amount of time it took car B to reach the destination, giving us 7 hours. We then multiply 7 hours by car A’s average speed and we get 490 miles.
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If Jack ran at an average rate of 7 miles per hour for a 21 mile course, and Sam ran half as fast for the same distance, how much longer did it take for Sam to run the course than Jack?
If Jack ran at an average rate of 7 miles per hour for a 21 mile course, and Sam ran half as fast for the same distance, how much longer did it take for Sam to run the course than Jack?
Using the rate formula: Distance = Rate x Time,
Since Jack’s speed was 7 mph, Jack completed the course in 3 hours
21 = 7 x t
t = 3
Sam’s speed was half of Jack’s speed: 7/2 = 3.5
21 = 3.5 x t
t = 6
Therefore it took Sam 3 hours longer to run the course.
Using the rate formula: Distance = Rate x Time,
Since Jack’s speed was 7 mph, Jack completed the course in 3 hours
21 = 7 x t
t = 3
Sam’s speed was half of Jack’s speed: 7/2 = 3.5
21 = 3.5 x t
t = 6
Therefore it took Sam 3 hours longer to run the course.
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STUDENT ATHLETES WHO USE STEROIDS MEN WOMEN TOTAL BASKETBALL A B C SOCCER D E F TOTAL G H I
In the table above, each letter represents the number of students in each category. Which of the following must be equal to I?
| STUDENT ATHLETES WHO USE STEROIDS | |||
|---|---|---|---|
| MEN | WOMEN | TOTAL | |
| BASKETBALL | A | B | C |
| SOCCER | D | E | F |
| TOTAL | G | H | I |
In the table above, each letter represents the number of students in each category. Which of the following must be equal to I?
Since G is the total number of male athletes that use steroids and H is the total number of female athletes that use steroids, the sum of the two is equal to I, which is the total number of all students using steroids.
Since G is the total number of male athletes that use steroids and H is the total number of female athletes that use steroids, the sum of the two is equal to I, which is the total number of all students using steroids.
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A particular ball always bounces back to 2/5 of the height of its previous bounce after being dropped. After the first bounce it reaches a height of 175 inches. Approximately how high (in inches) will it reach after its fifth bounce?
A particular ball always bounces back to 2/5 of the height of its previous bounce after being dropped. After the first bounce it reaches a height of 175 inches. Approximately how high (in inches) will it reach after its fifth bounce?
The first bounce reaches a height of 175. The second bounce will equal 175 multiplied by 2/5 or 70. Repeat this process. You will get the data below. 4.48 is rounded to 4.5.
The first bounce reaches a height of 175. The second bounce will equal 175 multiplied by 2/5 or 70. Repeat this process. You will get the data below. 4.48 is rounded to 4.5.
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express 7/8 as a ratio
express 7/8 as a ratio
a ratio that comes from a fraction is the numerator: denominator
7/8 = 7:8
a ratio that comes from a fraction is the numerator: denominator
7/8 = 7:8
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1 meter contains 100 centimeters.
Find the ratio of 1 meter and 40 centimeters to 1 meter:
1 meter contains 100 centimeters.
Find the ratio of 1 meter and 40 centimeters to 1 meter:
1m 40cm = 140cm. 1m = 100cm. So the ratio is 140cm:100cm. This can be put as a fraction 140/100 and then reduced to 14/10 and further to 7/5. This, in turn, can be rewritten as a ratio as 7:5.
1m 40cm = 140cm. 1m = 100cm. So the ratio is 140cm:100cm. This can be put as a fraction 140/100 and then reduced to 14/10 and further to 7/5. This, in turn, can be rewritten as a ratio as 7:5.
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When television remotes are shipped from a certain factory, 1 out of every 200 is defective. What is the ratio of defective to nondefective remotes?
When television remotes are shipped from a certain factory, 1 out of every 200 is defective. What is the ratio of defective to nondefective remotes?
One remote is defective for every 199 non-defective remotes.
One remote is defective for every 199 non-defective remotes.
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A mile is 5280 feet
Susan is able to walk a fast pace of 4 miles per hour. How many feet will she walk in 40 minutes?
A mile is 5280 feet
Susan is able to walk a fast pace of 4 miles per hour. How many feet will she walk in 40 minutes?
Calculate the number of feet walked in an hour. Then calculate what fraction of an hour 40 minutes is.
5280 * 4 = 21120 feet walked in an hour
60 minutes in an hour, so 40 minutes = 2/3 hour (40/60)
21120(2/3) = 14080 feet walked in 40 minutes
Calculate the number of feet walked in an hour. Then calculate what fraction of an hour 40 minutes is.
5280 * 4 = 21120 feet walked in an hour
60 minutes in an hour, so 40 minutes = 2/3 hour (40/60)
21120(2/3) = 14080 feet walked in 40 minutes
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On a desk, there are 
papers for every 
paper clips and 
papers for every 
greeting card. What is the ratio of paper clips to total items on the desk?
On a desk, there are
Begin by making your life easier: presume that there are 
papers on the desk. Immediately, we know that there are 
paper clips. Now, if there are 
papers, you know that there also must be 
greeting cards. Technically you figure this out by using the ratio:
By cross-multiplying you get:
Solving for 
, you clearly get 
.
(Many students will likely see this fact without doing the algebra, however. The numbers are rather simple.)
Now, this means that our desk has on it:
papers
paper clips
greeting cards
Therefore, you have 
total items. Based on this, your ratio of paper clips to total items is:
, which is the same as 
.
Begin by making your life easier: presume that there are
By cross-multiplying you get:
Solving for
(Many students will likely see this fact without doing the algebra, however. The numbers are rather simple.)
Now, this means that our desk has on it:
Therefore, you have
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In a garden, there are 
pansies, 
lilies, 
roses, and 
petunias. What is the ratio of petunias to the total number of flowers in the garden?
In a garden, there are
To begin, you need to do a simple addition to find the total number of flowers in the garden:
Now, the ratio of petunias to the total number of flowers in the garden can be represented by a simple division of the number of petunias by 
. This is:
Next, reduce the fraction by dividing out the common 
from the numerator and the denominator:
This is the same as 
.
To begin, you need to do a simple addition to find the total number of flowers in the garden:
Now, the ratio of petunias to the total number of flowers in the garden can be represented by a simple division of the number of petunias by
Next, reduce the fraction by dividing out the common
This is the same as
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In a classroom of 
students, each student takes a language class (and only one—nobody studies two languages). 
take Latin, 
take Greek, 
take Anglo-Saxon, and the rest take Old Norse. What is the ratio of students taking Old Norse to students taking Greek?
In a classroom of
To begin, you need to calculate how many students are taking Old Norse. This is:
Now, the ratio of students taking Old Norse to students taking Greek is the same thing as the fraction of students taking Old Norse to students taking Greek, or:
Next, just reduce this fraction to its lowest terms by dividing the numerator and denominator by their common factor of 
:
This is the same as 
.
To begin, you need to calculate how many students are taking Old Norse. This is:
Now, the ratio of students taking Old Norse to students taking Greek is the same thing as the fraction of students taking Old Norse to students taking Greek, or:
Next, just reduce this fraction to its lowest terms by dividing the numerator and denominator by their common factor of
This is the same as
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In the reptile house at the zoo, the ratio of snakes to lizards is 3 to 5. After the zoo adds 15 more snakes to the exhibit, the ratio changes to 4 to 5. How many lizards are in the reptile house?
In the reptile house at the zoo, the ratio of snakes to lizards is 3 to 5. After the zoo adds 15 more snakes to the exhibit, the ratio changes to 4 to 5. How many lizards are in the reptile house?
In order to maintain a proportion, each value in the ratio must be multiplied by the same value:
Before and after the snakes arrive, the number of lizards stays constant.
Before new snakes — Snakes : Lizards = 3_x_ : 5_x_
After new snakes — Snakes : Lizards = 4_x_ : 5_x_
Before the new snakes arrive, there are 3_x_ snakes. After the 15 snakes are added, there are 4_x_ snakes. Therefore, 3_x_ + 15 = 4_x_. Solving for x gives x = 15.
There are 5x lizards, or 5(15) = 75 lizards.
In order to maintain a proportion, each value in the ratio must be multiplied by the same value:
Before and after the snakes arrive, the number of lizards stays constant.
Before new snakes — Snakes : Lizards = 3_x_ : 5_x_
After new snakes — Snakes : Lizards = 4_x_ : 5_x_
Before the new snakes arrive, there are 3_x_ snakes. After the 15 snakes are added, there are 4_x_ snakes. Therefore, 3_x_ + 15 = 4_x_. Solving for x gives x = 15.
There are 5x lizards, or 5(15) = 75 lizards.
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A cafeteria with 40 tables can sit 600 people. Some tables can sit 10 people and some can sit 20 people. What is the ratio of the number of 10-person tables to the number of 20-person tables?
A cafeteria with 40 tables can sit 600 people. Some tables can sit 10 people and some can sit 20 people. What is the ratio of the number of 10-person tables to the number of 20-person tables?
Let x be the number of 10-person tables, and y be the number of 20-person tables. Since there are 40 tables in the cafeteria, x + y = 40. 10_x_ represents the number of people sitting at 10-person tables, and 20_y_ represents the number of people sitting at 20-person tables. Since the cafeteria can seat 600 people, 10_x_ + 20_y_ = 600. Now we have 2 equations and 2 unknowns, and can solve the system. To do this, multiply the first equation by 10 and subtract it from the second equation. This yields 0_x_ + 10_y_ = 200; solving for y tells us there are 20 tables that seat 20 people. Since x + y = 40, x = 20, so there are 20 tables that seat 10 people. The ratio of x:y is therefore 1:1.
Let x be the number of 10-person tables, and y be the number of 20-person tables. Since there are 40 tables in the cafeteria, x + y = 40. 10_x_ represents the number of people sitting at 10-person tables, and 20_y_ represents the number of people sitting at 20-person tables. Since the cafeteria can seat 600 people, 10_x_ + 20_y_ = 600. Now we have 2 equations and 2 unknowns, and can solve the system. To do this, multiply the first equation by 10 and subtract it from the second equation. This yields 0_x_ + 10_y_ = 200; solving for y tells us there are 20 tables that seat 20 people. Since x + y = 40, x = 20, so there are 20 tables that seat 10 people. The ratio of x:y is therefore 1:1.
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A lawn can be mowed by 
people in 
hours. If 
people take the day off and do not help mow the grass, how many hours will it take to mow the lawn?
A lawn can be mowed by 
The number of hours required to mow the lawn remains constant and can be found by taking the original 
workers times the 
hours they worked, totaling 
hours. We then split the total required hours between the 
works that remain, and each of them have to work 
and 
hours: 
.
The number of hours required to mow the lawn remains constant and can be found by taking the original 
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The ratio of 
to 
is 
to 
, while the ratio of 
to 
is 
to 
.
What is the ratio of 
to 
?
The ratio of
What is the ratio of
Since the ratios are fixed, regardless of the actual values of 
, 
, or 
, we can let 
and 
In order to convert to a form where we can relate 
to 
, we must set the coefficient of 
of each ratio equal such that the ratio can be transferred. This is done most easily by finding a common multiple of 
and 
(the ratio of 
to 
and 
, respectively) which is 
Thus, we now have 
and 
.
Setting the 
values equal, we get 
, or a ratio of 
Since the ratios are fixed, regardless of the actual values of
In order to convert to a form where we can relate
Thus, we now have
Setting the
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The flow of water through a certain pipe is 20 cubic meters per minute. How many minutes would it take for 4 of such pipes to fill 2 tanks, if each tank is a cube with a side length of 20 m?
The flow of water through a certain pipe is 20 cubic meters per minute. How many minutes would it take for 4 of such pipes to fill 2 tanks, if each tank is a cube with a side length of 20 m?
The flow of water through one pipe is 20 m3 / minute.
Thus, the flow of water through 4 pipes is 80 m3 / minute.
Since each tank is a cube with a side length of 20m, the volume of each tank is:
Volume of one tank = (20 m)3 = 8000 m3.
The total volume of two tanks is 2 * 8000 m3 = 16,000 m3
Therefore, the total minutes for 4 pipes to fill 2 tanks is:
16,000 m3/(80 m3/min) = 200 minutes
80 m3/min
The flow of water through one pipe is 20 m3 / minute.
Thus, the flow of water through 4 pipes is 80 m3 / minute.
Since each tank is a cube with a side length of 20m, the volume of each tank is:
Volume of one tank = (20 m)3 = 8000 m3.
The total volume of two tanks is 2 * 8000 m3 = 16,000 m3
Therefore, the total minutes for 4 pipes to fill 2 tanks is:
16,000 m3/(80 m3/min) = 200 minutes
80 m3/min
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