QuanticaPrealgebraOpen in the courseSome problems arrive as a wall of words that is hard to hold in your head all at once. The strongest first move is often not a calculation. It is a sketch. A path becomes a map, a schedule becomes a timeline, and a set of clues becomes dots and arrows you can point at. Once the information is on paper, the answer is usually easy to read straight off it.
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- Draw the loaf and mark the cuts. Nine pieces need one fewer cut than pieces.
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The sketch shows \(8\) cut marks producing \(9\) pieces, since each cut adds exactly one piece to the count.
$$8 \times 30 = 240 \text{ seconds} = \boxed{4} \text{ minutes}$$
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- Count gaps first. The path holds \(180 \div 15 = 12\) gaps.
- Dots outnumber the gaps between them by exactly one.
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The path divides into \(180 \div 15 = 12\) gaps, and the drawing shows one more lamp than gaps, since both ends hold a lamp.
$$12 + 1 = \boxed{13}$$
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- Laid end to end they would span \(60\) cm, but the overlapped stretch was counted twice, once inside each board.
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End to end, the boards would cover \(30 + 30 = 60\) cm, but the drawing shows the \(8\) cm overlap belonging to both boards at once, so it was counted twice.
$$60 - 8 = \boxed{52}$$
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- Mark the pier, step \(40\) east to the ice cream stand, then step \(25\) back west for the arcade.
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On the sketch, the arcade lands \(40 - 25 = 15\) meters east of the pier.
$$\boxed{15}$$
Without the line, east and west blur together. With it, the subtraction is obvious.
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- Total the east-west motion and the north-south motion separately. The wandering collapses to one net east amount and one net north amount.
- Net \(6\) east and net \(8\) north are the legs of a right triangle, and the straight-line distance is its hypotenuse.
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East-west nets to \(9 - 3 = 6\) blocks east, and north-south nets to \(2 + 6 = 8\) blocks north, so takeoff and landing sit at two corners of a right triangle with legs \(6\) and \(8\).
$$d^2 = 6^2 + 8^2 = 100, \qquad d = \boxed{10}$$
Opposite directions cancel, so a wandering trip always collapses to one net east-west amount and one net north-south amount.
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- Track the peak height at each press. Press one peaks at \(7\), then each later press peaks \(4\) higher than the last, since \(7\) up and \(3\) back gain \(4\) per full cycle.
- The story ends the moment a peak reaches \(25\). Do not subtract the slip after the winning press.
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The peaks climb \(7, 11, 15, 19, 23, 27\), gaining \(4\) per full cycle. The fifth peak of \(23\) is still short of the wall, and the sixth peak of \(27\) clears it, so it takes \(\boxed{6}\) presses.
Dividing \(25\) by the net gain of \(4\) misses that the winning press never gives back its \(3\) cm.
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- Kai immediately after Lena glues them into a block, Lena then Kai. Try the block in positions \(2\)-\(3\) and \(3\)-\(4\), since Mira must fit before Lena.
- If Lena and Kai take \(2\) and \(3\), the only slot before Kai that is not first is taken by Lena. Push the block later and see what opens up.
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Kai right after Lena glues them into a block, Lena then Kai. If that block sits at \(2\) and \(3\), Mira takes \(1\) and Jo has no slot left that is before Kai and not first. So the block sits at \(3\) and \(4\), Mira takes \(1\), and Jo takes \(\boxed{2}\). The order Mira, Jo, Lena, Kai fits every clue.
Four drawn slots turn each clue into a physical constraint you can test in seconds, instead of a sentence you have to keep rereading.
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- Only Drift connects to Ebb, so any trip to Ebb must arrive through Drift.
- Find the fastest way from Coral to Drift, then add the final ride.
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Ebb connects only to Drift, and Drift connects only to Bell, so the trip has to finish Bell, then Drift, then Ebb. Coral reaches Bell directly, so Coral to Bell to Drift to Ebb takes \(\boxed{3}\) rides, and nothing shorter works.
Drawn as dots and lines, the dead-end branch out to Ebb is obvious in a way a written route list never is.
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- List the maximum counts so far, \(2, 4, 7\), and look at the jumps between them. Each new chord adds one more region than the chord before it added.
- The jumps run \(2, 3, 4, 5\), so the fourth chord brings the count to \(11\), and the fifth continues the pattern.
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The counts climb \(2, 4, 7\), with jumps of \(2\), then \(3\). A new chord that crosses \(k\) old chords passes through \(k + 1\) regions and splits each one, so the fourth chord adds \(4\) and the fifth adds \(5\).
$$2, \; 4, \; 7, \; 11, \; \boxed{16}$$
Drawing the small cases gave the numbers, and counting crossings explains why each jump is one bigger than the last.
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West \(12\) and east \(12\) cancel, leaving only the \(5\) blocks north.
$$\boxed{5}$$
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The trip nets to \(5\) km east and \(12\) km north, a right triangle with hypotenuse
$$\sqrt{5^2 + 12^2} = \sqrt{169} = \boxed{13}$$
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The round peaks run \(6, 10, 14, 18, 22\). The fifth peak touches \(22\) exactly, and the game ends there, no slide back.
$$\boxed{5}$$
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With the Nia-Pia block at positions \(1\) and \(2\), Owen must finish after Nia but not last, which forces Owen into \(3\) and Quinn into \(4\). With the block at \(2\) and \(3\), Owen would have to take \(4\), which the clue forbids. So the order is Nia, Pia, Owen, Quinn.
$$\boxed{3}$$
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Six sections come from \(5\) cuts, and \(5 \times 2 = \boxed{10}\) minutes.
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Between \(25\) trees sit \(24\) gaps of \(6\) meters each,
$$24 \times 6 = \boxed{144}$$
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Without overlap the boards would span \(90\) cm, and the shelf lost \(90 - 78 = 12\) cm to the doubled section.
$$\boxed{12}$$
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The drawing shows F hanging off station E alone. The quickest way to E from A is A to B, then the direct B to E line, and then E to F.
$$\boxed{3}$$
The B to E shortcut is nearly invisible in the written list and impossible to miss in the picture.
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The jumps climb \(2, 3, 4, 5, 6\), so six chords reach
$$16 + 6 = \boxed{22}$$
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The perimeter is \(2(50 + 30) = 160\) meters, which splits into \(160 \div 10 = 16\) gaps. The path is a loop, so the sixteenth gap ends exactly where the first flag already stands, and flags equal gaps at \(\boxed{16}\).
On a straight path flags would outnumber gaps by one, so line or loop is the whole question in fencepost problems.