All PSAT Math Resources
Example Questions
Example Question #1 : Proportion / Ratio / Rate
Two numbers have a ratio of 5 : 2. If they are positive and differ by 21, what is the value of the larger number?
14
70
35
None of the other answers
28
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.
Example Question #11 : How To Find A Ratio
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?
1/2
6/13
7/18
1/3
6/17
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.
Example Question #11 : How To Find A Ratio
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?
17
2
7
24
28
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.
Example Question #13 : Proportion / Ratio / Rate
The ratio of 10 to 14 is closest to what value?
0.57
0.71
1.40
0.24
0.04
0.71
Another way to express ratios is through division. 10 divided by 14 is approximate 0.71.
Example Question #12 : How To Find A Ratio
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?
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.
Example Question #11 : How To Find A Ratio
If the ratio of to is , what is the ratio of to ?
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 3n to m, we have to multiply 13 * 3 = 39 to get 3n and 5 * 1 = 5 to get m (or 1m).
Then the requested ratio of 3n to m is 39 to 5 or .
Example Question #11 : How To Find A Ratio
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?
135
75
120
50
90
75
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 = 3x : 5x
After new snakes — Snakes : Lizards = 4x : 5x
Before the new snakes arrive, there are 3x snakes. After the 15 snakes are added, there are 4x snakes. Therefore, 3x + 15 = 4x. Solving for x gives x = 15.
There are 5x lizards, or 5(15) = 75 lizards.
Example Question #13 : How To Find A Ratio
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?
3:7
7:3
1:7
6:7
7:5
6:7
Multiply the ratios. (q/r)(r/s)= q/s. (3/5) * (10/7)= 6:7.
Example Question #11 : Fractions
If 1 nib is equal to 6 nubs, and 2 nubs is equal to 3 nabs, how many nabs are in 5 nibs?
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.
Example Question #12 : Proportion / Ratio / Rate
Bob can build a house in 3 days. Gary can build a house in 5 days. How long does it take them to build a house together?
15/8 days
4 days
2 days
5/4 days
3/2 days
15/8 days
DO NOT pick 4 days, which would be the middle number between Bob and Gary's rates of 3 and 5 days respectively. The middle rate is the answer that students always want to pick, so the SAT will provide it as an answer to trick you!
Let's think about this intuitively before we actually solve it, so hopefully you won't be tempted to pick a trick answer ever again! Bob can build the house in 3 days if he works by himself, so with someone else helping him, it has to take less than 3 days to build the house! This will always be true. Never pick the middle rate on a combined rates problem like this!
Now let's look at the problem computationally. Bob can build a house in 3 days, so he builds 1/3 of a house in 1 day. Similarly, Gary can build a house in 5 days, so he builds 1/5 of a house in 1 day. Then together they build 1/3 + 1/5 = 5/15 + 3/15 = 8/15 of the house in 1 day.
Now, just as we did to see how much house Gary and Bob can build separately in one day, we can take the reciprocal of 8/15 to see how many days it takes them to build a house together. (When we took the reciprocal for Bob, 3 days/1 house = 1/3 house per day.) The reciprocal of 8/15 is 15/8, so they took 15/8 days to build the house together. 15/8 days is almost 2 days, which seems like a reasonable answer. Make sure your answer choices make sense when you are solving word problems!
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