演習例
GAC 数学 トピック 1 13:05 英語ナレーション・英語＋中文字幕 burning-in
章
Transcript
Here is the map for this practice pack. Four questions, each tied to a sheet you already know: Q1 and Q2 belong to sheet one point one, Q3 to sheet one point two, and Q4 to sheet one point four. Do not treat these as a preview of your centre's assessment; they are original practice. Try each question before you open its answer, and remember that a clear, stated method shows your reasoning better than a long report. The next slide shows the answering method every worked example will follow.
Every answer in this pack follows the same four steps, and they make your reasoning easy to check. First, write what is known, with units. Second, say why the rule applies to this question. Third, write the rule in symbols before any numbers appear. Fourth, put in the numbers, one calculation stage per line. The symbols matter: a line that jumps straight to numbers hides the reasoning. And the check at the end is part of the answer, not extra. Keep this structure in view while you try each question.
Here is the first question. A price starts at two hundred units. It rises by ten percent, then falls by ten percent. Your task has two parts: find the final price, and explain why it is not two hundred units. The two percentages match, so many people expect the starting price to return. Before the next slide, write the known numbers and your method. Say what each percentage is calculated on. The following slides work through the rise and the fall, one stage per line.
Start with the rise. We know the starting price, two hundred units, and the rate, ten percent. The reason for the method: each percentage is calculated on the current price, so we apply a multiplier at each stage. The increase equation in symbols comes first: the new price equals the old price times one plus r over one hundred. Then substitute: two hundred times one point one zero gives two hundred and twenty units. Symbols first, then numbers, on separate lines.
Now the fall, and it uses the price after the rise, not the starting price. The decrease equation in symbols: the new price equals the old price times one minus r over one hundred. Substitute: two hundred and twenty times zero point nine zero gives one hundred and ninety-eight. So the final price is one hundred and ninety-eight units. The diagram shows the journey: two hundred, then two hundred and twenty, then one hundred and ninety-eight, with each multiplier acting on the price at that stage.
Why does the price not return to two hundred? The two multipliers act together as one combined multiplier. In symbols, the final price over the starting price equals one plus r over one hundred, times one minus r over one hundred. Substituting gives one point one zero times zero point nine zero, which is zero point nine nine. The final price keeps ninety-nine percent of the start. The overall loss is one percent, not zero. Equal percentages up and down do not cancel, because they multiply, not add.
Here is the second reason, in actual amounts. The rise added twenty units, which is ten percent of two hundred. The fall removed twenty-two units, which is ten percent of two hundred and twenty. Ten percent of two hundred and twenty is larger than ten percent of two hundred, so the fall removes more than the rise added. This is the same one percent gap, seen from the other direction. When you explain your answer, quote both amounts and their different bases.
Every answer in this pack ends with a check, and here are two ways to check this one. The combined multiplier was zero point nine nine, so the loss fraction is one minus zero point nine nine, which is zero point zero one. One percent of the starting price: two hundred times zero point zero one gives two units. The direct difference: two hundred minus one hundred and ninety-eight also gives two units. Both routes agree, so the answer checks. Now compare your own method.
Question two reverses the direction of the change. After a twenty percent increase, a figure is eighty-four. Your task: find the original figure. The number you want is the base the increase was calculated on, and that base is unknown. Be careful: the increase was not calculated on eighty-four. Before the next slide, write what is known and pick your operation. Ask yourself what the increase multiplied the original by, and how to undo that multiplication.
Set up the increase equation before touching any numbers. In symbols, the new figure N equals the original P zero, times one plus r over one hundred. That names the relationship the question gives you. Now rearrange it so the original stands alone: divide both sides by the multiplier, giving P zero equals N over one plus r over one hundred. Notice the operation: dividing by the multiplier, never subtracting the same percentage from the new figure. The next slide substitutes the numbers.
Now substitute in stages. First, the original equals eighty-four over one plus twenty over one hundred. Next, work out the multiplier: one plus twenty over one hundred is one point two. Then divide: eighty-four over one point two gives seventy. So the original figure is seventy. Keep each stage on its own line, with the symbolic form repeated before the numbers. That repetition is what lets a reader check your method, and it is what makes your reasoning easy to follow.
Check by running the increase forward again. Seventy times one point two returns eighty-four, exactly the stated figure, so the reversal is correct. Now look at the trap. Subtracting twenty percent of eighty-four gives sixty-seven point two, which is wrong, because the increase was calculated on seventy, not on eighty-four. This is the most common error in reverse percentage questions. To undo a percentage change, divide by the multiplier. Subtracting the same percentage from the new figure answers a different question.
Question three moves to algebra. Solve the statement nine minus two x is less than or equal to three, and show your solution on a number line. Solve it like an equation, but watch the last operation carefully. As you work, notice which operations preserve the inequality direction and which one might change it. Write each stage on its own line. When you think you have the solution, think about how to show the boundary value, and which direction the shading goes.
Start like an equation. The known statement is nine minus two x is less than or equal to three. Subtract nine from both sides: minus two x is less than or equal to three minus nine, which is minus six. One rule to hold in mind for the next step: dividing both sides by a negative number reverses the inequality sign. The subtraction stage needed no reversal, because subtracting preserves the direction. The next slide does the division and shows the result.
Now divide both sides by minus two, and reverse the sign: x is greater than or equal to three. The diagram shows this solution on the number line. The shading starts at three and points right, because every value greater than three satisfies the statement. Notice the filled circle at three. That is not decoration: it says the boundary itself is included. An open circle would exclude three, and that would be a different solution. The next slide checks the boundary with numbers.
Check the boundary directly. Substitute x equals three: nine minus two times three is nine minus six, which is three. Both sides are equal, so three satisfies the statement and belongs to the solution. That is why the circle is closed. And why does the shading run right? Values greater than three satisfy the statement, while values below three fail it. The next slide tests one value on each side, so you can see both outcomes with real numbers.
Test one value from each side of the boundary. Inside the solution, take x equals four: nine minus two times four is one, and one is less than or equal to three, so four satisfies the statement. Outside, take x equals two: nine minus two times two is five, and five is greater than three, so two fails. Both tests agree with the solution x greater than or equal to three. If your own answer fails a test like this, recheck the division stage first.
Question four moves to geometry. Two similar models have a length ratio of three to five. The smaller model has a surface area of thirty-six square centimetres. Find the larger surface area. The key hint: areas do not scale by the same factor as lengths. Before the next slide, decide what one length of the larger model is, and then think about what happens to an area, which is a product of two lengths. Write the known ratio and the known area first.
Set up the scaling language. Let k be the scale factor from a smaller length to the matching larger length. Because each area is a product of two lengths, areas scale as the square of k. The general relation in symbols: the larger area over the smaller area equals k squared. The length ratio three to five gives the scale factor: the larger length over the smaller length is five thirds. Hold that thought: the next slide squares it for the area ratio.
Now the area ratio: the larger area over the smaller area equals k squared, which is five thirds squared, giving twenty-five ninths. Rearrange the relation for the larger area: the larger area equals the smaller area times k squared. Substitute: thirty-six times twenty-five ninths gives one hundred. So the larger surface area is one hundred square centimetres. Notice the order: relation in symbols, rearrangement, then numbers, one stage per line. The next slide checks the ratio and states the limit.
Check by comparing the two areas directly. One hundred over thirty-six simplifies to twenty-five ninths, and k squared, five thirds squared, also gives twenty-five ninths. The area ratio twenty-five to nine is exactly the square of the length ratio five to three, so the answer checks. Then the limit: only areas scale by twenty-five ninths. If a question asked for a length of the larger model, you would multiply by five thirds instead. Match the scaling factor to the quantity.
Here are the four traps this pack has guarded against, in one table. First, treating matched percentages as fixed amounts; apply a multiplier to the current price instead. Second, subtracting twenty percent of the new figure; divide by the increase multiplier instead. Third, keeping the sign when dividing by a negative; reverse it. Fourth, scaling areas by the length factor; square the scale factor for areas. Use this table when you review your own answers.
Check yourself before you close this pack. Can you write each rule in symbols before the numbers appear? Can you run the reverse check and the boundary check yourself on a fresh copy? A check you have not performed is only a claim. Then keep practising: try the matching exercises on sheets one point one, one point two and one point four, record the step where your method differed, and try a new question. A stated method with a check shows your reasoning clearly; a bare number does not.