Now, we'll talk about another math strategy that may be new to you. This is called backsolving. So in some algebra problems, in some word problems, all five answer choices are numbered. So, all five answer choices are numerical. Now of course, the way that you're supposed to do this, the way that the algebra teacher would want you to do this would probably be use algebra to get the answer. Read full transcript
Start from the problem. Work through the algebra and get the answer. And certainly, if you know how to do that, that's a perfectly valid way to approach it. But often, another solution method available would be to work backwards from the answer choices and this is called backsolving.
And it turns out that as you practice backsolving and get good at it, sometimes you'll find that it's actually much quicker than using algebra. So many times, the answers will be listed in either forward or backward numerical orders. So, smallest to the biggest or biggest to smallest. Not always, but often that's the case.
In backsolving, it always makes sense to start with the middle value. If the answers are listed from smallest to biggest or biggest to smallest, we would start with answer choice C right in the middle. One-fifth of the time, we would be lucky and get the answer on our first try. Even if that doesn't happen though, we get valuable information. If it turns out that C is too big, well, then D and E are also gonna be too big.
And so, we can eliminate three answers all at once. Conversely, if C is too small, then A and B are also too small and we can eliminate three answers all at once. Thus, if C is not the answer, we often should be able to eliminate three choices on the basis of one try. After that, we just have one more backsolve that we have to do from the two remaining answer choices.
Let's say, C was too low. So A, B and C are all out. Well, next, I can try either D or E. Let's just say, I try D. Either D works, so D is the answer or D doesn't work which means that E has to be the answer.
And so really with only to back solves, I can narrow it down to what the answer is. This is very valuable. Here's a practice problem. Pause the video and then we'll talk about this. So you may know how to do this with algebra and that's great, if you know how to do this with algebra.
I'm gonna demonstrate the back solving approach here. Backsolving approach certainly would be useful, if you don't understand the algebra. But even if you understand the algebra, it's good to appreciate another method of solution. And overtime, as you practice you may find that backsolving is even faster and more efficient than using algebra.
So as always with backsolving, we'll start with C, the middle value, x=6. Plug this into the left side of the equation, we get 6 squared is 36. So divide that 36 by 3, we get 12. 2 times 12 is 24. 24+1 is 25. All that is under the radical.
And so we get 5 plus the square root of 25, which is 5. 3 plus square root of 25, which is 3+5 and 3+5 is 8. And so, this actually equals what it should equal. We actually got the right answer on the first try. So, C is the answer. In that problem, we happened to get lucky.
Because we started with C and C happened to be the answer. About 20% of the time, we will get that lucky with that solving. Let's look at another question. So, here's a geometry question. Pause the video and then we'll talk about this. So, this is an interesting question.
There's really no advantage to approaching it with algebra. It's just a mess, if you try to approach it with algebra. And really, the most efficient solution with this is backsolving. So, we'll start with C. The shorter leg is 25. That's the numerical value we're given in C.
The longer leg is going to be two more than three times that. So, 3 times that is 75 plus 2 is 77. So, the long leg is 77. Well, so now, the area is gonna be one-half odd number times odd number. So, that's not gonna be an integer. So, we know right away C is not the answer.
Let's do a little approximation just so we can get a sense of is this too big or too small. So, let's approximate. We'll keep the short answer 25. That's a nice round number. Let's just round up the 77 to 80.
Suppose we had a short leg of 25, a long leg of 80. The area would be this. And so that happens to equal, do a little doubling and halving there. That happens to equal exactly 1,000. That's way too big. That's way too big.
So we know that C doesn't work and we also know that we can eliminate D, and E. And so now, we just have A and B. The answer has to be either A or B. Let's choose B. Short leg is 20, long leg is 62. We can figure out the area.
Half of 20 is 10 and 62 times 10 is 620. That's also to big, so that area is bigger than 400. And so we can eliminate B also. So now, we can eliminated four answer choices. We know that A must be the answer choice. And if you were pressed for time, if you were in the middle of the ACT math test and your pressed for time, you could stop right here and say, okay, A must be the answer choice.
I'll pick A and move on. But if you are simply doing practice or you feel like you have some extra time in the test, it's always a good idea if you have the time to check the final answer to make sure that it works. So, that's what we're gonna do right here just to verify that A actually does work. So, the value in A is 16.
Short leg is 16. The longer leg, 3(16) is 48 plus 2 is 50. So now, the area is gonna be one-half 16(50). That's (8)(50), doubling and halving and that we get 400 and that is the exact answer we were supposed to get for the area. And so, that works.
So, we go back to our answer choices and we select 16. On the two previous problems, it was relatively straightforward whether a bigger number answer choice would make the target number bigger or smaller. On some harder problems on the ACT math test, this might not be so clear. Don't automatically assume that making the answer choice bigger will make the target bigger.
You always have to think about the logic of the situation, the mathematical logic. What's actually going on in this situation? Here's a practice problem. Pause the video and then we'll go through this slowly. After taxes, Sally takes home a salary of J=$5,000 each month. She pays P percent of this to her rent and all her fixed bills each month, leaving her with K left.
She spends half of K on groceries, leaving her with L left. She spends one-third of L on gifts and puts two-thirds of L into her savings, and this would leave her with $200 for miscellaneous expenses. What is the value of P? So this is a very, very hard problem. This problem would be, again, almost impossible to do with straight forward algebra.
And so, backsolving is a much more efficient way to solve this problem. So, we're gonna start by picking C. And we're gonna get a value of P equals 50%. Well, then half of that is 2,500. So half goes to our fixed expenses and the other half, 2,500. That's K.
So half of that goes to groceries, so that leaves us with L=1,250. One-third of 1,250. Well, that's not gonna come out to a nice round number, but it's gonna be about $416. Two-fifths of L, that's gonna be $500.
When we take 1,250 and we subtract 416 and 500. Of course, what this is gonna be 416+500 is less than 1,000, we can see that. And so if we subtract, so it's gonna be less than, that value is actually going to be greater than if we did 2,500-1,000. Because we'd be subtracting more. We subtracted 1,000.
And then of course, 2,500-1,000 is 250. And so, turns out that this leaves us with more money. We're left with more leftover than 250. We should only be left with 250 leftover, but the choice of 50% left us with more leftover. So, C has a leftover amount that is too big.
Clearly, choice C is not the answer. It's not obvious though, in which direction we should eliminate. We have to think about this. As P goes up, this would make K smaller which makes L smaller which makes the leftover at the end smaller. Thus, if the leftover is too high, we need a larger value of P.
We can eliminate A, B and C and we can try either D or E. We will try E here. Let P be 70%. If 70% goes to taxes and fixed expenses, then 30% Is left as K. So, 30% of 5,000 is 1,500. Half of it goes to groceries.
So L, what's left after groceries? That's 750. We take a third of 750, that's 250. We take two-fifths of 750, that's 300. And if we subtract 750-250-300, we get 200 and that is the value that we need.
This is the answer. So we see that the answer here has to be E, 70. In summary, if all the answers to an algebraic or word problem are numerical, One possible solution method is backsolving. So we start with the number of middle value, the median of the five answer choices.
Often, this is gonna be answer choice C if they're listed in increasing or decreasing order. But even if they're not, you always start with the middle value. And then we note the one we start with is too big or too small, then you can use this information to eliminate other choices. You should be able to eliminate three answer choices after your first backsolving.
Sometimes we have to be careful in thinking about this, because making one thing bigger could make another thing smaller. We have to be very careful that we're eliminating in the right direction. And finally, I will reiterate, of course, as with all mental math. If this is the first time you're learning about backsolving, it's gonna seem unfamiliar.
It's gonna seem strange. It's gonna take you time to recognize it and do it. The more you practice it, the more efficient you will become at it. And when you become very efficient at it, you will find that in some cases, it's much more efficient than algebra.