Jamie counted the number of edges of a cube, Jimmy counted the corners, and Judy counted the faces. They then added the three numbers. What was the resulting sum?
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Answer: E — 26.
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Hint 1
A cube has 12 edges, 8 corners, and 6 faces — recall each, then add.
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?
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Answer: C — 5 tricycles.
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Hint 1
Imagine all 7 children on bicycles first, then see how many extra wheels you still need.
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Approach: assume all bicycles, then add the extra wheels
If all 7 rode bicycles, that's 7 × 2 = 14 wheels.
There are 19 − 14 = 5 extra wheels, and each tricycle adds exactly one extra wheel, so there are 5 tricycles.
Another way — solve the system:
With b bicycles and t tricycles: b + t = 7 and 2b + 3t = 19.
Subtract twice the first from the second: t = 19 − 14 = 5.
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Compared with Blake, Jenny scored 10 points higher on the first test, 10 points lower on the second, and 20 points higher on each of the third and fourth. By how much does Jenny's average exceed Blake's on these four tests?
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Answer: A — 10 points.
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Hint 1
You never need Blake's actual average — just track how far ahead or behind Jenny is on each test.
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Approach: average the point differences
Jenny's differences from Blake are +10, −10, +20, +20, which total +40.
Art, Roger, Paul, and Trisha bake cookies that are all the same thickness, in the shapes shown below (dimensions in inches). Each friend uses the same amount of dough, and Art's batch makes exactly 12 cookies.
Who makes the fewest cookies from one batch of dough?
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Answer: A — Art.
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Hint 1 of 2
With the same dough, whoever has the biggest cookie makes the fewest of them.
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Hint 2 of 2
Find each cookie's area; the largest area wins.
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Approach: same dough, so biggest cookie means fewest cookies
Everyone uses equal dough, so the person with the largest cookie makes the fewest.
Areas (square inches): Art ½(3 + 5)(3) = 12, Roger 2 × 4 = 8, Paul 3 × 2 = 6, Trisha ½(3)(4) = 6.
Art's cookie is the biggest, so Art makes the fewest.
Art, Roger, Paul, and Trisha bake cookies that are all the same thickness, in the shapes shown below (dimensions in inches). Each friend uses the same amount of dough, and Art's batch makes exactly 12 cookies.
Art's cookies sell for 60 cents each. To bring in the same total from one batch, how much should one of Roger's cookies cost, in cents?
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Answer: C — 40 cents.
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Hint 1 of 2
Same total money from one batch means each cookie's price scales with its size.
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Hint 2 of 2
Compare Roger's cookie area to Art's, then scale 60 cents by that ratio.
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Approach: price per cookie scales with cookie area
For a fixed batch and a fixed total price, each cookie's price is proportional to its area.
Roger's cookie is 8 in² and Art's is 12 in², a ratio of 8/12 = 2/3.
So Roger should charge 60 × 2/3 = 40 cents.
Another way — count the cookies:
Art: 12 cookies at 60¢ = 720¢ per batch, using 12 × 12 = 144 in² of dough.
Art, Roger, Paul, and Trisha bake cookies that are all the same thickness, in the shapes shown below (dimensions in inches). Each friend uses the same amount of dough, and Art's batch makes exactly 12 cookies.
How many cookies will be in one batch of Trisha's cookies?
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Answer: E — 24 cookies.
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Hint 1
A batch is the dough that makes 12 of Art's cookies. How does Trisha's cookie size compare to Art's?
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Approach: compare Trisha's cookie to Art's
A batch is the dough for 12 of Art's 12 in² cookies, i.e. 144 in².
Trisha's cookies are ½(3)(4) = 6 in², exactly half of Art's, so she makes twice as many: 24.
Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: "Ten percent off the listed price. Sale starts Monday." How much does a pair of shoes cost on Monday that cost $40 on Thursday?
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Answer: B — $39.60.
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Hint 1 of 2
A 10% increase followed by a 10% decrease is not a wash — the cut is taken off a bigger number.
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Hint 2 of 2
Multiply the two factors: ×1.1 then ×0.9.
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Approach: chain the two percent changes
Friday: 40 × 1.1 = 44. Monday: 44 × 0.9 = 39.60.
Equivalently 40 × 1.1 × 0.9 = 40 × 0.99 = 39.60 — a hair under the original.
When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?
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Answer: E — 1 (it always happens).
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Hint 1 of 2
6 = 2 × 3, so you just need a 2 and a 3 both showing somewhere.
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Hint 2 of 2
Only one face is hidden — check the worst case, where it's the 6.
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Approach: show it is certain
6 = 2 × 3, so the product is divisible by 6 as long as a 2 and a 3 both appear among the visible faces.
Only one face is hidden. If it isn't the 6, then the 6 is visible. If it is the 6, then 2 and 3 are both still visible.
Either way the product is divisible by 6, so the probability is 1.
In this addition problem, each letter stands for a different digit.
T W O
+ T W O
-------
F O U R
If T = 7 and the letter O represents an even number, what is the only possible value for W?
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Answer: D — W = 3.
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Hint 1 of 2
Two 3-digit numbers add to a 4-digit number, so F must be 1.
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Hint 2 of 2
Look at the hundreds column with T = 7: it forces O to be 4 or 5, and "O is even" picks one.
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Approach: work column by column
Two 3-digit numbers sum to the 4-digit number FOUR, so the leading carry makes F = 1.
Hundreds column: 7 + 7 = 14 (plus any carry from the tens) makes the hundreds digit O equal to 4 or 5; since O is even, O = 4, and nothing carried out of the tens.
Units: 4 + 4 = 8 gives R = 8 with no carry, so the tens column is simply W + W = U, again with no carry.
So 2W = U must stay below 10 and avoid the digits already used (7, 1, 4, 8): W = 3 gives U = 6, the only option. W = 3.
Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has four seats: one driver's seat, one front passenger seat, and two back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?
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Answer: D — 12 arrangements.
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Hint 1 of 2
Fill the most restricted seat first — only two people can take the driver's seat.
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Hint 2 of 2
Once the driver is chosen, the other three sit anywhere.
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Approach: seat the constrained person first
Only Bonnie or Carlo can drive: 2 choices for the driver's seat.
The other 3 people fill the remaining 3 seats in 3! = 6 ways.
The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?
Child
Eye Color
Hair Color
Benjamin
Blue
Black
Jim
Brown
Blond
Nadeen
Brown
Black
Austin
Blue
Blond
Tevyn
Blue
Black
Sue
Blue
Blond
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Answer: E — Austin and Sue.
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Hint 1 of 2
Write down Jim's two traits, then see who could even be in his family.
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Hint 2 of 2
Three of the other children share one trait among themselves — that locks in both families.
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Approach: group by a shared trait
Jim has brown eyes and blond hair, so a sibling must share one of those: only Nadeen (brown eyes), Austin (blond), or Sue (blond) qualify.
But Benjamin, Nadeen, and Tevyn all have black hair, so those three already form a valid family.
That forces Jim's family to be Jim, Austin, and Sue (all blond) — his siblings are Austin and Sue.
Drop perpendiculars from the short side down to the long side — the two slanted legs become familiar right triangles.
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Hint 2 of 2
The 10-8 and 17-8 legs hint at the 6-8-10 and 8-15-17 triples; that tells you how much the bottom overhangs the top.
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Approach: split into right triangles, then use the area
Drop perpendiculars from B and C to the long side AD. The legs give right triangles: AB = 10 with height 8 leaves a base of 6 (a 6-8-10 triangle), and CD = 17 with height 8 leaves 15 (an 8-15-17 triangle).
So AD = BC + 6 + 15 = BC + 21.
Area: ½(BC + AD)(8) = 164, so BC + AD = 41.
Then BC + (BC + 21) = 41, giving 2·BC = 20, so BC = 10.
