A reflection across a vertical line swaps left and right but leaves up and down alone.
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Hint 2 of 2
Hold each choice up to a mirror placed on the dashed line — only one matches the original.
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Approach: mentally flip left↔right and keep up/down
Reflecting across the vertical dashed line swaps the left and right of every feature: the corner square moves to the opposite side of its box, and the slanted arms flip their lean.
Choice B is the only T-like shape whose corner square and arms are the mirror image of the original.
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?
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Answer: C — 149.
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Hint 1 of 2
To minimize the difference, make the 3-digit number as small as you can and the 2-digit number as large as you can.
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Hint 2 of 2
Pick the smallest hundreds digit first, then build each number from the digits you have left.
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Approach: smallest 3-digit minus largest 2-digit
Smallest 3-digit using three of {2, 4, 5, 6, 9}: lead with 2, then take the next two smallest ascending → 245.
Largest 2-digit from the remaining {6, 9} is 96. Difference: 245 − 96 = 149.
Many calculators have a reciprocal key 1/x that replaces the current number displayed with its reciprocal. For example, if the display is 00004 and the 1/x key is pressed, then the display becomes 000.25. If 00032 is currently displayed, what is the fewest positive number of times you must depress the 1/x key so the display again reads 00032?
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Answer: B — 2.
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Hint 1 of 2
Try the operation once, then again — does anything change after that?
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Hint 2 of 2
The reciprocal of the reciprocal is the original number.
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Approach: apply 1/x twice
Press once: 32 → 1⁄32. Press again: 1⁄32 → 32.
So the display reads 32 again after exactly 2 presses.
Pair up the faces that end up opposite each other after folding — opposite faces never share a corner.
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Hint 2 of 2
At every corner, the three meeting faces are one from each opposite pair, so pick the larger number from each pair.
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Approach: find opposite-face pairs from the net, then take the max from each pair
Fold the net with 2 as the front: 1 goes on top and 3 on the bottom (1↔3), 6 to the left and 4 to the right (6↔4), and 5 wraps around to the back opposite 2 (2↔5).
Each corner has one face from each pair, so the biggest corner sum is 5 + 3 + 6 = 14.
Jack had a bag of 128 apples. He sold 25% of them to Jill. Next he sold 25% of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?
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Answer: D — 71.
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Hint 1 of 2
Selling 25% means keeping 75% = 3⁄4.
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Hint 2 of 2
Multiply by 3⁄4 twice, then subtract 1 for the apple given to the teacher.
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Approach: keep ¾ twice, then subtract 1
After Jill: 128 × 3⁄4 = 96. After June: 96 × 3⁄4 = 72.
Each cube face is 1 m². Count only the faces that are visible — none of those touching the ground or another cube.
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Hint 2 of 2
Sum the exposed faces from the top, front, back, left, right; faces hidden by neighbors don't count.
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Approach: tally exposed faces by direction
Top faces (one per cube whose top isn't covered) total 10. Front-facing exposed faces total 6, back 6, and the two side walls together account for the remaining 11.
Pick a convenient side length for the square (say 4) so the pieces have whole-number sides.
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Hint 2 of 2
The cut parallel to the fold leaves the fold-side as one rectangle; the open side falls apart into two.
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Approach: pick side 4 and read the three rectangles off
Take the square 4 × 4. Folding halves the width to a 2 × 4 stack; cutting it in half parallel to the fold means the cut sits at 1 from the fold. The fold-side unfolds back into one 2 × 4 large rectangle; the open side becomes two separate 1 × 4 small rectangles.
Small perimeter 2(1 + 4) = 10, large perimeter 2(2 + 4) = 12, ratio 10⁄12 = 5⁄6.
