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数学策略：反解
Transcript
现在，我们将讨论另一个对你来说可能是新的数学策略。这叫做反解。
So in some algebra problems, in some word problems, all five answer choices are numbered.
所以，所有五个答案都是数字。当然，你应该这样做，
代数老师希望你这样做的方法可能是用代数来得到答案。
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.
事实证明，当你练习反解并擅长它时，有时你会发现它实际上比用代数快得多。 所以很多时候，答案会以向前或向后的数字顺序列出。 所以，从最小到最大或者从最大到最小。不总是这样，但经常是这样。
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. 即使不发生这种情况，我们也会得到宝贵的信息。如果事实证明，C太大，那么D和E也会太大。
所以，我们可以同时消除三个答案。相反地，如果C太小，那么A和B也太小 we can eliminate three answers all at once. Thus, if C is not the answer, 我们通常应该能够在一次尝试的基础上消除三个选择。在那之后，我们只需要从这两个中再做一个反解 剩下的答案选择。
比如说，C太低了。所以A，B和C都出来了。 好吧，接下来，我可以尝试D或E.让我们只是说，我试试D. 要么D起作用，所以D是答案，要么D不起作用，这意味着E必须是答案。
And so really with only to back solves, I can narrow it down to what the answer is. This is very valuable. 这里有个练习题。暂停视频，然后我们再谈这个。 所以你可能知道如何用代数来做这件事，如果你知道如何用代数来做这件事，那就太棒了。
I'm gonna demonstrate the back solving approach here. Backsolving approach certainly would be useful, 如果你不懂代数。但即使你懂代数， 欣赏另一种解决方法是很好的。而随着时间的推移，当你练习的时候，你可能会发现反解的速度更快，也更快 比使用代数更有效率。
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等于25。所有这一切都在激进之下。
所以我们得到5加上25的平方根，也就是5。3加25的平方根，即3+5，3+5等于8。 And so, this actually equals what it should equal. We actually got the right answer on the first try. 所以，C是答案。在那个问题上，我们碰巧走运。
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. 让我们看另一个问题。所以，这里有一个几何问题。 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. 实际上，最有效的解决方法是反解。我们从C开始。 较短的腿是25。这是用C表示的数值。
长腿的长度将是这一数字的两倍多。3乘以75加2等于77。 所以，长腿是77。好吧，现在，面积是奇数的一半乘以奇数。 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.
假设我们的短腿是25，长腿是80。区域应该是这个。 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.
我会选一个，然后继续。但是如果你只是在练习或者 您觉得您在测试中有一些额外的时间，如果您有时间检查最终答案以确保它有效，这总是一个好主意。 所以，这就是我们现在要做的，只是为了验证A是否真的起作用。所以，A的值是16。
Short leg is 16. The longer leg, 3(16) is 48 plus 2 is 50. 所以现在，面积是16（50）的一半。那是（8）（50），加倍和减半，我们得到400和 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 一个更大的数量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.
你总是要考虑情况的逻辑，数学逻辑。在这种情况下到底发生了什么？ 这里有个练习题。暂停视频，然后我们慢慢看。 After taxes, Sally takes home a salary of J=$5,000 each month. She pays P percent of this to her rent and 她每个月都有固定的账单，剩下的只有K。
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, 这会给她留下200美元的杂费。P的值是多少？ 所以这是一个非常非常困难的问题。问题是， almost impossible to do with straight forward algebra.
因此，BackSolving是一种更有效的解决这个问题的方法。所以，我们要开始挑选C. And we're gonna get a value of P equals 50%. Well, then half of that is 2,500. 所以半一半走到我们的固定费用，另一半，2,500。那是k。
一半的钱花在杂货上，剩下的是L=1250。 1250的三分之一。好吧，这不是一个好的整数，但是 it's gonna be about $416. Two-fifths of L, that's gonna be $500.
当我们取1250减去416和500。当然，416+500小于1000，我们可以看到。 因此，如果我们减去，所以它会少于，那个价值实际上将大于我们2,500-1,000岁。 因为我们会减去更多。我们减去1000。
当然，2500-1000等于250。所以，结果是，这给我们留下了更多的钱。 We're left with more leftover than 250. We should only be left with 250 leftover, but 50%的选择留给我们更多的剩余。所以，C的剩余量太大了。
显然，选择C不是答案。不过，我们应该从哪个方向消除，这并不明显。 我们得好好想想。当P上升时，这会使K变小，这会使L变小 这使得最后剩下的东西变小了。因此，如果剩余量太高，我们需要一个更大的P值。
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. 因此，30％的5,000占1,500。一半的是杂货。
那么我，杂货店之后还剩下什么？750美元。 我们取750的三分之一，也就是250。我们取750的五分之二，也就是300。 如果我们减去750-250-300，我们得到200，这就是我们需要的值。
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. 所以我们从中间值开始，五个答案的中间值。
通常，这是答案选择C，如果他们在增加或减少的顺序列出。 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, 然后您可以使用此信息来排除其他选择。你应该能够消除三个答案选择后 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. 如果这是你第一次学习反解的话，你会觉得很陌生。
会很奇怪的。你得花点时间去认识它，去做它。 你练习得越多，你的效率就越高。当你变得非常有效率的时候，你会发现在某些情况下， it's much more efficient than algebra.
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.
事实证明，当你练习反解并擅长它时，有时你会发现它实际上比用代数快得多。 所以很多时候，答案会以向前或向后的数字顺序列出。 所以，从最小到最大或者从最大到最小。不总是这样，但经常是这样。
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. 即使不发生这种情况，我们也会得到宝贵的信息。如果事实证明，C太大，那么D和E也会太大。
所以，我们可以同时消除三个答案。相反地，如果C太小，那么A和B也太小 we can eliminate three answers all at once. Thus, if C is not the answer, 我们通常应该能够在一次尝试的基础上消除三个选择。在那之后，我们只需要从这两个中再做一个反解 剩下的答案选择。
比如说，C太低了。所以A，B和C都出来了。 好吧，接下来，我可以尝试D或E.让我们只是说，我试试D. 要么D起作用，所以D是答案，要么D不起作用，这意味着E必须是答案。
And so really with only to back solves, I can narrow it down to what the answer is. This is very valuable. 这里有个练习题。暂停视频，然后我们再谈这个。 所以你可能知道如何用代数来做这件事，如果你知道如何用代数来做这件事，那就太棒了。
I'm gonna demonstrate the back solving approach here. Backsolving approach certainly would be useful, 如果你不懂代数。但即使你懂代数， 欣赏另一种解决方法是很好的。而随着时间的推移，当你练习的时候，你可能会发现反解的速度更快，也更快 比使用代数更有效率。
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等于25。所有这一切都在激进之下。
所以我们得到5加上25的平方根，也就是5。3加25的平方根，即3+5，3+5等于8。 And so, this actually equals what it should equal. We actually got the right answer on the first try. 所以，C是答案。在那个问题上，我们碰巧走运。
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. 让我们看另一个问题。所以，这里有一个几何问题。 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. 实际上，最有效的解决方法是反解。我们从C开始。 较短的腿是25。这是用C表示的数值。
长腿的长度将是这一数字的两倍多。3乘以75加2等于77。 所以，长腿是77。好吧，现在，面积是奇数的一半乘以奇数。 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.
假设我们的短腿是25，长腿是80。区域应该是这个。 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.
我会选一个，然后继续。但是如果你只是在练习或者 您觉得您在测试中有一些额外的时间，如果您有时间检查最终答案以确保它有效，这总是一个好主意。 所以，这就是我们现在要做的，只是为了验证A是否真的起作用。所以，A的值是16。
Short leg is 16. The longer leg, 3(16) is 48 plus 2 is 50. 所以现在，面积是16（50）的一半。那是（8）（50），加倍和减半，我们得到400和 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 一个更大的数量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.
你总是要考虑情况的逻辑，数学逻辑。在这种情况下到底发生了什么？ 这里有个练习题。暂停视频，然后我们慢慢看。 After taxes, Sally takes home a salary of J=$5,000 each month. She pays P percent of this to her rent and 她每个月都有固定的账单，剩下的只有K。
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, 这会给她留下200美元的杂费。P的值是多少？ 所以这是一个非常非常困难的问题。问题是， almost impossible to do with straight forward algebra.
因此，BackSolving是一种更有效的解决这个问题的方法。所以，我们要开始挑选C. And we're gonna get a value of P equals 50%. Well, then half of that is 2,500. 所以半一半走到我们的固定费用，另一半，2,500。那是k。
一半的钱花在杂货上，剩下的是L=1250。 1250的三分之一。好吧，这不是一个好的整数，但是 it's gonna be about $416. Two-fifths of L, that's gonna be $500.
当我们取1250减去416和500。当然，416+500小于1000，我们可以看到。 因此，如果我们减去，所以它会少于，那个价值实际上将大于我们2,500-1,000岁。 因为我们会减去更多。我们减去1000。
当然，2500-1000等于250。所以，结果是，这给我们留下了更多的钱。 We're left with more leftover than 250. We should only be left with 250 leftover, but 50%的选择留给我们更多的剩余。所以，C的剩余量太大了。
显然，选择C不是答案。不过，我们应该从哪个方向消除，这并不明显。 我们得好好想想。当P上升时，这会使K变小，这会使L变小 这使得最后剩下的东西变小了。因此，如果剩余量太高，我们需要一个更大的P值。
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. 因此，30％的5,000占1,500。一半的是杂货。
那么我，杂货店之后还剩下什么？750美元。 我们取750的三分之一，也就是250。我们取750的五分之二，也就是300。 如果我们减去750-250-300，我们得到200，这就是我们需要的值。
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. 所以我们从中间值开始，五个答案的中间值。
通常，这是答案选择C，如果他们在增加或减少的顺序列出。 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, 然后您可以使用此信息来排除其他选择。你应该能够消除三个答案选择后 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. 如果这是你第一次学习反解的话，你会觉得很陌生。
会很奇怪的。你得花点时间去认识它，去做它。 你练习得越多，你的效率就越高。当你变得非常有效率的时候，你会发现在某些情况下， it's much more efficient than algebra.
