How to find a ratio - PSAT Math
Card 1 of 133
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?
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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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Alan is twice as old as Betty. He will be twice as old as Charlie in 10 years. If Charlie is 2 years old, how old is Betty?
Alan is twice as old as Betty. He will be twice as old as Charlie in 10 years. If Charlie is 2 years old, how old is Betty?
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If Charlie is 2 years old now; in 10 years he will be 12 years old. At that point, Alan will be twice as old as Charlie. Twice 12 is 24. This means that Alan is currently 10 years younger than 24, or 14. Since Alan is currently twice as old as Betty, she must be half of 14, or 7.
If Charlie is 2 years old now; in 10 years he will be 12 years old. At that point, Alan will be twice as old as Charlie. Twice 12 is 24. This means that Alan is currently 10 years younger than 24, or 14. Since Alan is currently twice as old as Betty, she must be half of 14, or 7.
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The ratio of 10 to 14 is closest to what value?
The ratio of 10 to 14 is closest to what value?
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Another way to express ratios is through division. 10 divided by 14 is approximate 0.71.
Another way to express ratios is through division. 10 divided by 14 is approximate 0.71.
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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?
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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?
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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?
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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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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?
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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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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?
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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?
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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 
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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 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?
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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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You are planning a party. The maximum number of people the reception hall can hold is 1 person for every 5 square feet of space. If the hall is 60 feet wide and 50 feet long, how many people can you invite?
You are planning a party. The maximum number of people the reception hall can hold is 1 person for every 5 square feet of space. If the hall is 60 feet wide and 50 feet long, how many people can you invite?
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Total area of hall = 60ft * 50ft = 3000 ft2
At 1 person per 5 square feet, 3000 ft2 / 5 ft2 per person = 600 people
Total area of hall = 60ft * 50ft = 3000 ft2
At 1 person per 5 square feet, 3000 ft2 / 5 ft2 per person = 600 people
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A gallon contains 8 pints. Each pint contains 2 cups. How many cups are in a 10-gallon jug of water?
A gallon contains 8 pints. Each pint contains 2 cups. How many cups are in a 10-gallon jug of water?
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Find the number of cups in a gallon, then calculate cups in 10 gallons.
If 8 pints = 1 gallon, then 16 cups = 1 gallon
16 cups * 10 gallons = 160 cups
Find the number of cups in a gallon, then calculate cups in 10 gallons.
If 8 pints = 1 gallon, then 16 cups = 1 gallon
16 cups * 10 gallons = 160 cups
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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?
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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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Two numbers have a ratio of 5 : 2. If they are positive and differ by 21, what is the value of the larger number?
Two numbers have a ratio of 5 : 2. If they are positive and differ by 21, what is the value of the larger number?
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We can rewrite this as two equations
5/2 = x/y
x – y = 21
Solve y for x in the second equation: x = 21 + y → y = x – 21
Substitute back into the first equation and solve:
5/2 = x/(x – 21)
5(x – 21)/2 = x → (5x – 105)/2 = x → 5x – 105 = 2x → 5x – 2x = 105 → 3x = 105; x = 105/3 = 35
Since we know the numerator must be larger (given the prompt), the answer is x, or 35.
We can rewrite this as two equations
5/2 = x/y
x – y = 21
Solve y for x in the second equation: x = 21 + y → y = x – 21
Substitute back into the first equation and solve:
5/2 = x/(x – 21)
5(x – 21)/2 = x → (5x – 105)/2 = x → 5x – 105 = 2x → 5x – 2x = 105 → 3x = 105; x = 105/3 = 35
Since we know the numerator must be larger (given the prompt), the answer is x, or 35.
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A bag contains 240 marbles that are either red, blue, or green. The ratio of red to blue to green marbles is 5 : 2 : 1. If one-third of the red marbles and two-thirds of the green marbles are removed, what fraction of the remaining marbles in the bag will be blue?
A bag contains 240 marbles that are either red, blue, or green. The ratio of red to blue to green marbles is 5 : 2 : 1. If one-third of the red marbles and two-thirds of the green marbles are removed, what fraction of the remaining marbles in the bag will be blue?
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First, we need to figure out how many red, blue, and green marbles are in the bag before any are removed. Let 5x represent the number of red marbles. Because the marbles are in a ratio of 5 : 2 : 1, then if there are 5x red marbles, there are 2x blue, and 1x green marbles. If we add up all of the marbles, we will get the total number of marbles, which is 240.
5x + 2x + 1x = 240
8x = 240
x = 30
Because the number of red marbles is 5x, there are 5(30), or 150 red marbles. There are 2(30), or 60 blue marbles, and there are 1(30), or 30 green marbles.
So, the bag originally contains 150 red, 60 blue, and 30 green marbles. We are then told that one-third of the red marbles is removed. Because one-third of 150 is 50, there would be 100 red marbles remaining. Next, two-thirds of the green marbles are removed. Because (2/3)(30) = 20, there would be 10 green marbles left after 20 are removed.
To summarize, after the marbles are removed, there are 100 red, 60 blue, and 10 green marbles. The question asks us for the fraction of blue marbles in the bag after the marbles are removed. This means there would be 60 blue marbles out of the 170 left in the bag. The fraction of blue marbles would therefore be 60/170, which simplifies to 6/17.
The answer is 6/17.
First, we need to figure out how many red, blue, and green marbles are in the bag before any are removed. Let 5x represent the number of red marbles. Because the marbles are in a ratio of 5 : 2 : 1, then if there are 5x red marbles, there are 2x blue, and 1x green marbles. If we add up all of the marbles, we will get the total number of marbles, which is 240.
5x + 2x + 1x = 240
8x = 240
x = 30
Because the number of red marbles is 5x, there are 5(30), or 150 red marbles. There are 2(30), or 60 blue marbles, and there are 1(30), or 30 green marbles.
So, the bag originally contains 150 red, 60 blue, and 30 green marbles. We are then told that one-third of the red marbles is removed. Because one-third of 150 is 50, there would be 100 red marbles remaining. Next, two-thirds of the green marbles are removed. Because (2/3)(30) = 20, there would be 10 green marbles left after 20 are removed.
To summarize, after the marbles are removed, there are 100 red, 60 blue, and 10 green marbles. The question asks us for the fraction of blue marbles in the bag after the marbles are removed. This means there would be 60 blue marbles out of the 170 left in the bag. The fraction of blue marbles would therefore be 60/170, which simplifies to 6/17.
The answer is 6/17.
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If the ratio of 
to 
is 
, what is the ratio of 
to 
?
If the ratio of
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You will get this problem wrong if you do not pay attention to what is being asked. The problem states that the ratio of m to n is 
.
Because the problem asks for the ratio of 3_n_ to m, we have to multiply 13 * 3 = 39 to get 3_n_ and 5 * 1 = 5 to get m (or 1_m_).
Then the requested ratio of 3_n_ to m is 39 to 5 or 
.
You will get this problem wrong if you do not pay attention to what is being asked. The problem states that the ratio of m to n is
Because the problem asks for the ratio of 3_n_ to m, we have to multiply 13 * 3 = 39 to get 3_n_ and 5 * 1 = 5 to get m (or 1_m_).
Then the requested ratio of 3_n_ to m is 39 to 5 or
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Tom wants to buy an aquarium. He has found that that he needs one filter for the aquarium for every 40 creatures or plants he puts in the tank. The fish he wants to get also requires special plants be present at 2 plants for every 5 fish. These plants need 4 cleaning fish for every 3 plants in order to flourish.
If Tom is going to buy 3 filters, how many of the original fish he wanted will his aquarium support?
Tom wants to buy an aquarium. He has found that that he needs one filter for the aquarium for every 40 creatures or plants he puts in the tank. The fish he wants to get also requires special plants be present at 2 plants for every 5 fish. These plants need 4 cleaning fish for every 3 plants in order to flourish.
If Tom is going to buy 3 filters, how many of the original fish he wanted will his aquarium support?
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3 filters can support a total of 120 creatures/plants. The fish he wants need 2 plants for every 5 fish. The plant needs 4 cleaning fish per 3 plants. Thus for every 15 of the fish he wants, he needs 6 plants and 8 cleaning fish.
This gives us a total of 29 creatures. We can complete this number 4 times, but then we are left with 4 spots open that the filters can support.
This is where the trick arises. We can actually add one more fish in. Since 1 plant supports up to 2.5 fish (2:5), and 2 cleaning fish support up to 1.5 plants, we can add 1 fish, 1 plant, and 2 cleaning fish to get a total of 120 creatures. If we attempt to add 2 fish, then we must also add the 1 plant, but then we don't have enough space left to add the 2 cleaning fish necessary to support the remaining plant.
Thus Tom can buy at most 61 of the fish he originally wanted to get.
3 filters can support a total of 120 creatures/plants. The fish he wants need 2 plants for every 5 fish. The plant needs 4 cleaning fish per 3 plants. Thus for every 15 of the fish he wants, he needs 6 plants and 8 cleaning fish.
This gives us a total of 29 creatures. We can complete this number 4 times, but then we are left with 4 spots open that the filters can support.
This is where the trick arises. We can actually add one more fish in. Since 1 plant supports up to 2.5 fish (2:5), and 2 cleaning fish support up to 1.5 plants, we can add 1 fish, 1 plant, and 2 cleaning fish to get a total of 120 creatures. If we attempt to add 2 fish, then we must also add the 1 plant, but then we don't have enough space left to add the 2 cleaning fish necessary to support the remaining plant.
Thus Tom can buy at most 61 of the fish he originally wanted to get.
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If 1 nib is equal to 6 nubs, and 2 nubs is equal to 3 nabs, how many nabs are in 5 nibs?
If 1 nib is equal to 6 nubs, and 2 nubs is equal to 3 nabs, how many nabs are in 5 nibs?
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First, find the ratio of nabs to nibs. We know that 1 nib is equal to 6 nubs. We also know that 2 nubs is equal to 3 nabs:
We can multiply both sides of the second equation by 3:
Now we can combine our two equations:
Our question asks for the number of nabs in 5 nibs. We know from the equation above that 1 nib is equal to 9 nabs. Multiply both sides by 5 to find how many nabs are equal to 5 nibs.
Our final answer is 45.
First, find the ratio of nabs to nibs. We know that 1 nib is equal to 6 nubs. We also know that 2 nubs is equal to 3 nabs:
We can multiply both sides of the second equation by 3:
Now we can combine our two equations:
Our question asks for the number of nabs in 5 nibs. We know from the equation above that 1 nib is equal to 9 nabs. Multiply both sides by 5 to find how many nabs are equal to 5 nibs.
Our final answer is 45.
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Alan is twice as old as Betty. He will be twice as old as Charlie in 10 years. If Charlie is 2 years old, how old is Betty?
Alan is twice as old as Betty. He will be twice as old as Charlie in 10 years. If Charlie is 2 years old, how old is Betty?
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If Charlie is 2 years old now; in 10 years he will be 12 years old. At that point, Alan will be twice as old as Charlie. Twice 12 is 24. This means that Alan is currently 10 years younger than 24, or 14. Since Alan is currently twice as old as Betty, she must be half of 14, or 7.
If Charlie is 2 years old now; in 10 years he will be 12 years old. At that point, Alan will be twice as old as Charlie. Twice 12 is 24. This means that Alan is currently 10 years younger than 24, or 14. Since Alan is currently twice as old as Betty, she must be half of 14, or 7.
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