Logic & Word Problemswork-backwardcareful-counting
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
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Line the terms up so a changeover happens as early and as late in the 8 years as possible.
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Start mid-term and end mid-term to squeeze in extra principals.
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Let year 1 be the last year of one principal's term. Two full 3-year terms then fill years 2–7.
A fourth principal starts in year 8, for 4 principals.
Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
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Find the common starting height from Shea's 20% growth.
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Ara grew half as many inches as Shea — match inches, not percents.
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Shea grew 20% to reach 60, so the start was 60 ÷ 1.2 = 50 inches, meaning Shea grew 10 inches.
Ara grew half that, 5 inches, ending at 50 + 5 = 55 inches.
The number 64 has the property that it is divisible by its units digit. How many whole numbers between 10 and 50 have this property?
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Group the numbers by their units digit and test divisibility.
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Units digit 1, 2, or 5 always works; 0 never does (no division by 0).
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Units digit 1, 2, and 5 each give 4 working numbers (12 total). Then 33 (digit 3), 24 and 44 (digit 4), 36 (digit 6), and 48 (digit 8) work; digits 0, 7, 9 give none.
Each row is 100 ft long — find the cheapest way to fill one row.
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Staggering the joints forces the alternate rows to use a 1-ft block at each end.
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A row of all 2-ft blocks needs 100 ÷ 2 = 50 blocks. To stagger the joints, the in-between rows use 49 two-ft blocks plus a 1-ft block on each end = 51 blocks.
Four rows of 50 and three rows of 51: 200 + 153 = 353 blocks.
In order for Mateen to walk a kilometer (1000 m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
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The '25 times' clue gives the length; the '10 times' clue gives the perimeter.
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Then perimeter = 2(length + width) gives the width.
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Length = 1000 ÷ 25 = 40 m; perimeter = 1000 ÷ 10 = 100 m.
From 100 = 2(40 + W) we get W = 10, so area = 40 × 10 = 400 m².
Then compare the slanted sides to see which perimeter is larger.
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Both shapes have area 1 (I is a 1×1 parallelogram; II is two triangles of area ½).
They share two equal slant sides, but I's other two sides are unit length while II has a longer slant, so II's perimeter is bigger — meaning I's is less (choice E).
E
Same area, but quadrilateral I has the smaller perimeter.
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each type. How many dimes must you have?
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Start with one of each coin (41¢) and work out what the other 5 coins make.
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The leftover must end in a 1, which pins the number of pennies.
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One of each coin uses 1 + 5 + 10 + 25 = 41¢, leaving 61¢ in 5 more coins. To end in a 1, exactly one more is a penny, leaving 60¢ in 4 coins.
Those 4 must include quarters (4 dimes reach only 40¢); two quarters leave 10¢ = two nickels.
No extra dimes are needed, so there is just the original 1 dime.
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
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Keiko can only land on 0 or 1 head, so those are the only two cases you have to match.
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Find Ephraim's chance of 0 heads and of 1 head, then pair each with Keiko's.
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Keiko gets 1 head or 0 heads, each with probability ½. Ephraim's two tosses give 1 head with probability ½ (HT or TH) and 0 heads with probability ¼ (TT).
Both at 1 head: ½ · ½ = ¼. Both at 0 heads: ½ · ¼ = ⅛.
The little cube hides part of the big top, but its own top covers exactly that hidden patch — so what area is genuinely new?
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Only the four side walls of the small cube add surface; the top is a wash.
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The original cube's surface area is 6 · 2² = 24.
The small cube hides a 1 × 1 patch of the big top, but its own top sits directly above that patch, so the upward-facing area is unchanged. Only its 4 side walls are new, adding 4.
The increase is 4 / 24 = ⅙ ≈ 16.7%, closest to 17%.
There is a list of seven numbers. The average of the first four numbers is 5, and the average of the last four numbers is 8. If the average of all seven numbers is 647, then the number common to both sets of four numbers is
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Add the two group-sums together — every number is counted once, except the shared one, counted twice.
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So (sum of the two fours) minus (sum of all seven) leaves exactly the common number.
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The first four total 4 · 5 = 20 and the last four total 4 · 8 = 32. Adding gives 52, which counts every number once except the shared middle one, counted twice.
All seven total 7 · 6⁴⁄₇ = 46, counting each number once.
Subtracting strips one copy of everything, leaving the doubled number: 52 − 46 = 6.
Triangle AFG is isosceles with apex angle A — that pins both of its base angles.
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The base angle at F is the exterior angle of triangle BFD, so it already equals ∠B + ∠D.
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In triangle AFG the apex ∠A = 20° and the two base angles are equal, so each is (180° − 20°) / 2 = 80°.
Lines AD and BE cross at F, so ∠AFG is the exterior angle of triangle BFD at F. An exterior angle equals the sum of the two remote angles, which here are exactly ∠B and ∠D.
Therefore ∠B + ∠D = 80°.
D
80°.
Another way: supplement, then angle sum (as MAA presents it)
From the isosceles triangle, ∠AFG = 80°, so the angle inside triangle BFD at F is its supplement, 180° − 80° = 100°.
The angles of triangle BFD sum to 180°, so ∠B + ∠D = 180° − 100° = 80°.
The area of rectangle ABCD is 72. If point A and the midpoints of sides BC and CD are joined to form a triangle, the area of that triangle is
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Slice the triangle out by removing the three right triangles in the corners — each is a clean fraction of the whole rectangle.
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Two corners take a quarter each and the third takes an eighth; whatever's left is the answer.
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The triangle is what remains after cutting off the three corner right triangles. Measured against the whole rectangle, the ones at corners B and D are ¼ each (a full side and a half-side as legs), and the one at corner C is ⅛ (two half-sides).
Together the corners take ¼ + ¼ + ⅛ = ⅝ of the rectangle, leaving ⅜.
