Grades 6, 7 and 8
SOLUTIONS
RULES
• Work the problems in any order.
• If you think you have finished one correctly, tell one of the
organizers. If it is really correct, he or she will certify that it
is correct.
• Don’t feel shy about asking for hints.
• Don’t feel shy about getting up, walking around, or talking
with anybody you want to talk.
• If you want to write on one of the whiteboards, we have
markers available.
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1. A beaker filled to the brim with water weighs 5 pounds, while the same beaker filled halfway weighs 3.25
pounds. How many pounds of water can the beaker hold?
Solution. The difference in weight between full and half full is 5 − 3.25 = 1.75 pounds. This tells us that
the weight of half the water needed to fill the beaker is 1.75 pounds, thus the beaker holds 2 × 1.75 = 3.5
pounds.
2. Which is greater 333,333 x 444,444 or 222,222 x 666,667? By how much? The use of a calculator to solve
this problem is strictly forbidden.
Solution.
One way is to see that all numbers but one are multiples of 111,111:
333, 333 × 444, 444
=
(3 × 111, 111) × (4 × 111, 111) = 12 × 111, 111 × 111, 111,
222, 222 × 666, 667
=
(2 × 111, 111) × [6 × (111, 111) + 1] = 12 × 111, 111 × 111, 111 + 2 × 111, 111.
The second number is larger, by 222, 222.
3. When Peter broke his piggy bank, it contained no more than 100 coins. He divided the coins into piles of 2
coins each, but was left with one extra coin. The same happened when Peter divided the coins into piles of 3
coins, piles of 4 coins, and piles of 5 coins. Each time he was left with one extra coin. How many coins were
in the piggy bank? (Assume there is more than one coin in the bank.)
Solution. The number of coins has to be odd, so that one is left over when you divide into piles of 2 coins.
To have one left over after dividing into piles of 5 it has to be a multiple of 5 plus 1; the only possible such
numbers under 100 are
1
11
21
31
41
51
61
71
81
91.
Of these, only 61 differs by 1 from a multiple of 3 and 4. So the number is 61.
4. Here is a series of figures.
The first consists of one square. How many squares are there in the 100th figure? How many squares are
there in the first 100 figures altogether?
Solution. One sees that the 100-th figure will have 199 squares. To see how many squares there are in
the first 100 figures, notice that we can take the first figure and attach it to the second figure so as to make
a square out of the second figure:
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The total number of little squares has not changed. Now we can take the square on the left, made up out of
four little squares, and attach it to the next figure to get a 3 × 3 square of little squares.
And so forth:
Continuing this way, we see that all the little squares of the first 100 figures can be put together to form a
100 × 100 square. The answer is 10,000 squares.
5. Angela has 7 potatoes, Minh has 5, and Greg has none. They combine their potatoes to make a bowl of
mashed potatoes and share the bowl of potatoes equally among them. In exchange for his share, Greg gives
Minh and Angela 12 pieces of chocolate. How should they divide the chocolate between them, if they are to
be fair?
Solution. Minh should get 3 bars, Angela 9. In fact, each one eats 4 potatoes, Minh contributes one
(a fourth) of Greg’s potatoes, Angela contributes three (3/4) of Greg’s potatoes. So Minh should get 1/4,
namely 3, of Greg’s chocolate, Angela the rest.
6. Ten gnomes are playing checkers. Each plays one game with each of the others. (a) How many games did
each gnome play? (b) How many games were played in total?
Solution. (a) Each gnome plays 9 games.
(b) One way of answering this is to order the gnomes so that we have a first gnome, a second gnome, etc.
The first gnome played 9 games, the second gnome played an additional 8 games, the third seven more, etc.
So the total number is 9 + 8 + · · · + 2 + 1 = 45. A more visual way is to draw a graph. Place the gnomes
in a circle, join each pair with a line representing the game that pair played. How many lines are there in
all? We can count all the lines, but that is sort of messy and not up to our high standards. There are 9 lines
emanating from each gnome, 90 lines. But since each line is shared by two gnomes, we have counted them
twice. Divide by 2 to get 45.
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7. Is it possible to cut several circles out of a square of side 10 cm, so that the sum of the diameters of the circles
would be 5 meters or more?
Solution.
Yes, the trick is to make them very small.
8. Given a triangle ABC with ∠A = 90◦ and AB = BC = 1, and a point M chosen at random on AC, is it
possible to tell what the sum of the distances from M to AB and from M to BC will be? Be sure to draw a
picture!
Solution.
The sum of distances is always 1.
9. In the figure below, ∠C = 90◦ , |AD| = |DB|, DE is perpendicular to AB, |AB| = 20 and |AC| = 12.
Determine the area of the quadrilateral ADEC.
Solution. The area of the quadrilateral equals the area of triangle ABC minus the area oftriangle DBE.
The area of ABC can be computed using side CB as the base and side AC as the height. We are given AC;
cB can be computed using the theorem of Pythagoras. It works out to
p
p
AB 2 − AC 2 = 202 − 122 = 16.
The area of triangle ABC is thus [ABC] = 21 · 12 · 16 = 96. To compute the area of triangle DBE we notice
that because both triangles ABC and EBD are right triangles and share the angle at B, they are similar.
Thus
DE
AC
AC
12
=
, hence DE =
× DB =
× 10 = 7.5.
DB
CB
CB
16
The area of DBE is now seen to be [DBE] =
1
2
· 7.5 · 10 = 37.5. Thus
[ADEC] = [ABC] − [DBE] = 96 − 37.5 = 58.5.
10. The front door of a mansion has an opening of the shape shown below, covered by a metal plate that can
slide back when Paunchy the butler needs to look out and identify a visitor. It slides into the door. Here is
how things look when the plate completely closes the opening. The shape of the left edge is made up of two
quarter circles and one half circle.
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Paunchy slides the lid back by 2 inches. What is the area of the opening that has been uncovered? (colored
green in the picture).
Solution. The area “gained” must equal the area “lost.” In other words the uncovered area must equal
the area that slid into the door. That last area is easily computed, it is 5 × 2 = 10 square inches.