I am working on a project that involves me basically having to relearn about three years of math into about four months. I’ve gone through books. I’ve done problems. Now, I need the practice.
I’m looking for alegebra and geometry problems. I need problems like “If Joe does it in four hours and Sally does it in five, how long will it take them to do it if they work together” and “So many students are taking English, so many are taking math, so many are taking English and Math, how many are taking neither” and “If Mary drives 4 hours at 55 miles per hour and it took her six hours to reach her destination 250 miles away, what was her average speed on the last leg of her trip?”
Go easy and slow with me. It’s been nearly 20 years since I stretched my brain in this way.
Oh, and if anyone else wants to play, you’re welcome.
Okay, I just realized the example I gave meant Mary drove 15 miles an hour on the last leg of her trip. Let’s hope she wasn’t on a highway, or enountered heavy road construction. :smack:
Mary is driving a two mile stretch of road, and she’s taken the first mile at 30 mph. At what speed will she have to drive the second mile so that her average speed is 60 mph?
(This is a hard problem at the level of algebra I, but definitely doable.)
Just dug through the recycling bag next to my desk for these – they’re all from the Mensa Puzzle a Day calendar. None are all that difficult.
Six teenagers can eat 12 pizzas in one hour. At the same rate, how many pizzas can four teenagers eat in two hours?
Hal rolls a die and remembers the number. Then he performs the following sequence of operations on that number, in this order:
Add 1.
Square the result.
Subtract 6.
Multiply the result by 8.
Add 4.
Hal ends up with the number 84. What number did he roll on the die?
Wally and Norm each own several T-shirts. If Wally were to give Norm six T-shirts, they would have an equal number of T-shirts. If instead Norm were to give Wally six T-shirts, then Wally would have twice as many T-shirts as Norm. How many does each have?
Find the six-digit number in which the second digit is two less than the first, the third is two less than the second, and the fourth is two less than the third. The fourth is the difference when the third is subtracted from the fifth. The sum of all the digits is 38.
Tracy is now three times as old as Bob was two years ago. In four years, Bob will be as old as Tracy is now. How old are they now?
Find the number that best completes the following sequence:
6 31 156 ___ 3906
I’d encourage you to show your work. Repeat the problem, show how you’d set it up in an equation and the steps you take to solve it. We can throw in some helpful hints to jar your memory as you go. For example, ultrafilter’s problem can be solved with the use of the multiplicative property of equality and the additive property of equality. These are handy rules to keep in your head as you go through algebra, you’ll be using them a lot.
When we’re working with our kids we find the worksheet generator on the website for the math curriculum we use to be very useful. It goes up to prealgebra, but some of the problems can be pretty challenging. It can also generate answer sheets which don’t just give the final answer, but have the whole problem worked out, so you can see the steps involved. It doesn’t really do word problems though. There are a LOT of downloadable pages in the downloads section, ranging from worksheets, to tests, to teacher materials. If you have good reference on how to do the work and just need problem sets, this is a good way to get them.
Enjoy,
Steven
ETA: And that’s what I get for taking so long to compose this post. Glad to see you got it, hope some of the references here are useful.
Six teenagers can eat 12 pizzas in one hour. At the same rate, how many pizzas can four teenagers eat in two hours? One teenager can eat two pizzas in an hour. Four can eat 16 in two hours.
Hal rolls a die and remembers the number. Then he performs the following sequence of operations on that number, in this order:
Add 1.
Square the result.
Subtract 6.
Multiply the result by 8.
Add 4.
Hal ends up with the number 84. What number did he roll on the die? Working backwards, I got two.
Wally and Norm each own several T-shirts. If Wally were to give Norm six T-shirts, they would have an equal number of T-shirts. If instead Norm were to give Wally six T-shirts, then Wally would have twice as many T-shirts as Norm. How many does each have? 6? I don’t think I did that right.
Okay, I know there’s a way to figure this out. The first number has to be either a nine or an eight, but I can’t figure out how to make them all add up to 38.
The key to Baker’s problem is the volume of a cake the mix makes. He said each cake mix makes 2 9" by 1 1/2" cakes. So you figure out the volume of that cake, multiply it by the three mixes, then solve for the unknown in your large cake pan size formula.
The general formula for area of a right cylinder is the area of the circle times the height of the cylinder. Volume = pi * radius[sup]2[/sup]* height. So since a circle’s area is pi times the radius squared, we just have to add in a factor for height. The radius of a 9" cake layer is 4.5". The height is 1.5" and pi is approximately 3.14159. So radius squared is ~20.25 * pi = 63.62 square inches for the area of the circle the cake takes up, and now we factor in the height with 63.62 * 1.5 to get 95.43 cubic inches per cake. Each cake mix makes two, so we have 190.86 cubic inches of cake from each mix. We have three mixes. 190.86 * 3 = 572.58. Now we know how much our big cake pan needs to hold, and we have a limit on how tall it can be(say we’re making a wedding cake and don’t want it to be too tall). So we have all but one of the terms for our volume equation for the final cake. This is where the algebra comes in. Remember, V = pi * r[sup]2[/sup] * h. We know V is 572.58 cubic inches. We know h is 2", and of course we also know pi. So we have 572.58 = pi * r[sup]2[/sup]* 2. We can take the two and divide it by both sides by the multiplicative property of equality. So 572.58 / 2 = pi * r[sup]2[/sup] or 286.29 = pi * r[sup]2[/sup]. Now the second step is the same, but using pi as our divisor. 286.29 / pi = r[sup]2[/sup] or 91.13 = r[sup]2[/sup]. Now we can take the square root of both sides, and we get sqrt(91.13) = r or 9.55 = r. Since the problem asked for diameter, we get a diameter of 19.1" for our large cake pan.
Checking Baker’s spoiler, I see our answers don’t match. I’ve double checked mine and I think it’s clean. If Baker posts his solution we can see where one of us went wrong.
You’re right – that’s wrong. Try again. Hint: you need to come up with an algebraic equation that has only one unknown, either the “W” or the “N.”
Yes.
Yes, that’s where you’re off. Take another look at which side needs to be “times three.”
You did one of these steps backwards. Reread and redo.
There’s a third possibility for the first digit. Figure out what you can starting from the three possible first digits before you worry too much about what “38” has to do with it.
Taking a closer look at ultrafilter’s question, I’d have to argue it’s not quite what the OP was looking for.
If Mary’s traveled one mile at 30 mph it means she’s taken two minutes to go one mile. Since she only has two miles to go, and an average of 60 MPH is her goal, how much time does she have left to make up that last mile? Well, let’s figure out how long it should take to go 2 miles at 60 mph, subtract the time she has already spent on the road, and then figure out how fast she has to go to cover that second mile in the time she would have left. 2 miles at 60 MPH is 2 minutes because 60 mph is a one to one miles to minutes ratio. At 30 mph, with one mile down, she’s already been on the road 2 minutes. So there is no possible way she can get a 60 MPH average unless she can cover the remaining mile in 0 minutes. If she goes 600 MPH over the next mile(taking 16.667 seconds) she’s still at 2:17 for the trip, which works out to ~ 53 MPH average.
If the OP is looking for problems to solve for drill, then questions which are essentially trick questions, aren’t really going to help.
A house is twice as old as its oven was when the house was as old as the oven is. The sum of their ages is 49 years. How old is the oven and how old is the house?
