A can of soup can feed 3 adults or 5 children. If there are 5 cans of soup and 15 children are fed, then how many adults would the remaining soup feed?
Show answer
Answer: B — 6 adults.
Show hints
Hint 1 of 2
First find how many cans the 15 children use.
Still stuck? Show hint 2 →
Hint 2 of 2
The leftover cans each feed 3 adults.
Show solution
Approach: use cans for children, then convert the rest
15 children need 15 ÷ 5 = 3 cans, leaving 5 − 3 = 2 cans.
To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains 60 pills, then the supply would last approximately
Show answer
Answer: D — 8 months.
Show hints
Hint 1 of 2
How many doses are in 60 pills if each dose is half a pill?
Still stuck? Show hint 2 →
Hint 2 of 2
Each dose covers two days.
Show solution
Approach: doses, then days
60 pills ÷ ½ per dose = 120 doses, and each dose lasts 2 days, so 240 days.
If each of the three operation signs +, −, × is used exactly once in one of the blanks in the expression 5 __ 4 __ 6 __ 3, then the value of the result could equal
Show answer
Answer: E — 19.
Show hints
Hint 1 of 2
Remember multiplication happens before addition and subtraction.
Still stuck? Show hint 2 →
Hint 2 of 2
Try placing × between 6 and 3.
Show solution
Approach: try placements, respecting order of operations
Take 5 − 4 + 6 × 3: the multiplication gives 18 first.
Subtract the two right triangles cut off from the rectangle to leave ABDF.
Still stuck? Show hint 2 →
Hint 2 of 2
△BCD and △FED are right triangles with one leg = a midpoint half-length.
Show solution
Approach: rectangle minus two corner right triangles
Rectangle ACDE has area 32 × 20 = 640. The two right triangles cut off to leave ABDF are △BCD (legs BC = 16, CD = 20, area 160) and △FED (legs FE = 10, ED = 32, area 160).
Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he number the pages of his scrapbook with these digits?
Show answer
Answer: D — 119.
Show hints
Hint 1 of 2
The only limit is the supply of 2's — count how many 2's the page numbers use.
Still stuck? Show hint 2 →
Hint 2 of 2
Pages 1–99 use 20 twos; then keep going until the 2's run out.
Show solution
Approach: track how the 2's get used up
Pages 1–99 use 20 twos (ten in the units place, ten in the tens place). That leaves 2 twos.
Pages 102 and 112 use one 2 each, exhausting the supply; pages 113–119 need no 2, but 120 would, so he can reach 119.
Five runners, P, Q, R, S, T, have a race. P beats Q, P beats R, Q beats S, and T finishes after P and before Q. Who could NOT have finished third in the race?
Show answer
Answer: C — P and S.
Show hints
Hint 1 of 2
Combine the clues into chains: P is ahead of Q, R, and T, and S is behind Q.
Still stuck? Show hint 2 →
Hint 2 of 2
Count how many runners must be ahead of each person.
Show solution
Approach: see who is forced too high or too low
P beats Q, R, and T, and P < T < Q < S, so P is always first — it can't be third.
S comes after Q, which comes after both P and T, so at least three runners beat S, putting it 4th or later — also never third. So the answer is P and S.
The figure below shows a triangular ‘staircase’ array of numbers. The first row has 1 number, the second row has 3, the third row has 5, and so on (the kth row has 2k−1 numbers, in order).
1
2 3 4
5 6 7 8 9
10 11 12 13 14 15 16
…
What number is directly above 142 in this array of numbers?
Show answer
Answer: C — 120.
Show hints
Hint 1 of 2
Row k ends at the perfect square k², and holds 2k − 1 numbers.
Still stuck? Show hint 2 →
Hint 2 of 2
Find which row holds 142, then the number sitting one row up and aligned with it.
Show solution
Approach: use the row structure of the triangle
Rows end at 1, 4, 9, 16, …, k², so 142 is in row 12 (122–144), as its 21st of 23 entries.
Row 11 (101–121) sits centered above, so directly above the 21st entry is the 20th entry of row 11: 101 + 19 = 120.
A checkerboard consists of one-inch squares. A square card, 1.5 inches on a side, is placed on the board so that it covers part or all of the area of each of n squares. The maximum possible value of n is
Show answer
Answer: E — 12 or more.
Show hints
Hint 1 of 2
Don't keep the card lined up with the grid — tilt it.
Still stuck? Show hint 2 →
Hint 2 of 2
A tilted card pokes its corners into many extra squares.
Show solution
Approach: tilt the card to cross more grid lines
Lined up, the 1.5-inch card touches only up to a 3 × 3 block (9 squares).
But tilting it lets its corners reach into still more squares, so it can cover 12 or more.
