A child's wading pool contains 200 gallons of water. If water evaporates at the rate of 0.5 gallons per day and no other water is added or removed, how many gallons of water will be in the pool after 30 days?
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Find the total amount lost over the 30 days first.
For a sale, a store owner reduces the price of a $10 scarf by 20%. Later the price is lowered again, this time by one-half of the reduced price. The price is now
Each of the letters W, X, Y, and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If WX − YZ = 1, then the sum of W and Y is
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For the difference to be exactly the whole number 1, each fraction should itself be a whole number.
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Try making one fraction 3/1 and the other 4/2.
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To get a difference of exactly 1, both fractions should be integers: 3/1 = 3 and 4/2 = 2 work, since 3 − 2 = 1.
At Annville Junior High School, 30% of the students in the Math Club are in the Science Club, and 80% of the students in the Science Club are in the Math Club. There are 15 students in the Science Club. How many students are in the Math Club?
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Find how many students are in both clubs, using the Science Club side.
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That same number is 30% of the Math Club.
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Both clubs share 80% of 15 = 12 students.
Those 12 make up 30% of the Math Club, so the Math Club has 12 ÷ 0.3 = 40 students.
Nisos Isles. In 1998 the islands have 200 people, and the population triples every 25 years. Estimate the year in which the population will be about 6000.
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6000 is about 30 times the starting 200.
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Each tripling multiplies by 3 — how many triples reach about 30×?
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6000 ÷ 200 = 30, and three triplings give ×27 ≈ 30.
Three 25-year periods is 75 years after 1998, about 2075.
Nisos Isles. In 1998 the islands have 200 people, and the population triples every 25 years. The total area is 24,900 square miles, and the Queen requires at least 1.5 square miles per person. In about how many years from 1998 will the population reach the maximum the islands can support?
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First find the largest population the land allows: area ÷ 1.5.
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Compare that to 200 and count how many triplings it takes.
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The land supports 24,900 ÷ 1.5 = 16,600 people, about 83 times today's 200.
Since 3⁴ = 81 ≈ 83, that's four triplings, or 4 × 25 = 100 years.
Each fold doubles the holes by reflecting across its crease.
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Unfold one step at a time, mirroring the punched hole each time.
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Undoing the left-to-right fold mirrors the hole across the vertical crease, and undoing the bottom-to-top fold mirrors both across the horizontal crease.
Tamika selects two different numbers at random from the set {8, 9, 10} and adds them. Carlos takes two different numbers at random from the set {3, 5, 6} and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?
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List Tamika's possible sums and Carlos's possible products.
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Compare each of the 3 × 3 equally likely pairings.
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Tamika's sums are 17, 18, 19; Carlos's products are 15, 18, 30 — each value equally likely.
Tamika beats 15 all 3 times, beats 18 once (with 19), and never beats 30: that's 3 + 1 + 0 = 4 of 9 cases.
Let PQRS be a square piece of paper. P is folded onto R, and then Q is folded onto S. The area of the resulting figure is 9 square inches. Find the perimeter of square PQRS.
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Each fold halves the area.
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After two folds the figure is one-fourth of the square's area.
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Folding P onto R halves the square, and folding Q onto S halves it again, leaving one-fourth of the area: s²/4 = 9, so s² = 36 and s = 6.
A 4 × 4 × 4 cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?
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Count the cubes that touch nothing — not the four walls, not the bottom — and subtract.
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Those untouched cubes form a small block held away from every wall and the floor.
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A cube avoids all four walls and the bottom only if it lies in the inner 2 × 2 columns and above the bottom layer: 2 × 2 × 3 = 12 cubes.
Terri builds a sequence of positive integers by these rules: if the integer is less than 10, multiply it by 9; if it is even and greater than 9, divide it by 2; if it is odd and greater than 9, subtract 5. Find the 98th term of the sequence that begins 98, 49, … .
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Generate terms until they start repeating.
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Once you spot the repeating block, use its length to jump ahead to the 98th term.
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The sequence runs 98, 49, 44, 22, 11, 6, 54, 27, 22, 11, … — from the 4th term on it cycles 22, 11, 6, 54, 27 with length 5.
From the 4th term, (98 − 4) = 94 steps and 94 ÷ 5 leaves remainder 4, landing on the 5th entry of the cycle: 27.
A board of 8 columns has squares numbered left to right, top to bottom (row one is 1–8, row two is 9–16, and so on). A student shades square 1, then skips one and shades square 3, skips two and shades square 6, skips three and shades square 10, and continues this way until every column has at least one shaded square. What is the number of the shaded square that first achieves this?
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The shaded squares are the triangular numbers 1, 3, 6, 10, 15, … .
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A square's column is set by its value mod 8 — you need all 8 remainders to appear.
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The shaded squares are triangular numbers, and a square's column depends on its value mod 8.
Every remainder shows up early except 0; the first triangular number divisible by 8 is 120, so the last column fills at square 120.
Three generous friends redistribute their money as follows: Amy gives Jan and Toy enough to double each of their amounts; then Jan gives Amy and Toy enough to double theirs; finally Toy gives Amy and Jan enough to double theirs. Toy had $36 at the beginning and $36 at the end. What is the total amount the three friends have?
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The total never changes — you only need to find it at one moment.
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Toy's money doubles in the first two rounds, so track it up to just before his own turn.
Show solution
Toy's $36 doubles in each of the first two rounds: 36 → 72 → 144 just before his turn.
He ends with $36, so he gave away 144 − 36 = 108, which exactly doubled Amy and Jan — meaning they held $108 then.
The total, unchanged throughout, is 144 + 108 = $252.
