JUNIOR DIVISION 1998
1. Fred makes a cup full of coffee with no milk, drinks one third of it and
then tops it up with milk to make the cup full again. Then he drinks
one sixth of the mixture and again tops it up with milk to make a full
cup. After drinking another half cup, he tops it up with milk a third
time and then drinks the whole cup. How much milk did he drink?
Solution.
Initially there is no milk in the cup. Then Fred adds 13 cup of milk, then
1
cup of milk, then 12 cup of milk. He drinks everything, therefore his
6
total milk consumption is 13 + 16 + 12 = 1 cup of milk. (The alternative
solution, which involves adding up the amounts of milk drunk at each
stage, assuming the milk to be mixed in homogeneously, also leads to
the correct answer, but with rather more work.)
2. Two trains, both 200 metres long, travel on parallel tracks through a
tunnel. One travels at 100 Km/h, the other at 200 Km/h. The front
of the fast train enters the tunnel just as it passes the back of the slow
train. The back of the fast train leaves the tunnel just as it passes the
front of the slow train. How long is the tunnel?
Solution.
If the tunnel is of length l metres, then in the time interval of the
question, the fast train travels (l + 200) metres, while the slow train
travels (l − 200) metres. As the fast train travels at 200 km/h and the
slow train travels at 100 km/h, we have l+200
= l−200
which can be
200
100
rearranged to give l = 600 metres.
3. From the time Amanda gets up until her bus leaves at 8 a.m., she
uses one seventh of the time having a shower, then one sixth of the
remaining time getting dressed, and then one quarter of the remaining
time having breakfast, which she finishes at exactly 7:15 a.m. At what
time does she get up?
Solution.
We give two solutions, the first avoiding algebra.
Working backwards, eating breakfast took one quarter of the remaining
time and left 45 minutes, so from starting breakfast until 8 am was
45 ÷ (1 − 14 ) = 45 ÷ 34 = 60 minutes. Similarly, from starting dressing
until 8 am was 60 ÷ (1 − 1/6) = 72 minutes, and from getting up until
8 am was 72 ÷ (1 − 1/7) = 84 minutes. So she gets up at 6:36 am.
Alternatively, assume Amanda gets up x mins. before 8 am.
Then x7 minutes are spent showering, leaving 6x
mins.
7
Then 16 × 6x
= x7 mins. are spent dressing, leaving 5x
mins.
7
7
1
5x
15x
Then 4 × 7 = 5x
mins.
are
spent
breakfasting,
leaving
mins. and
28
28
it is now 7:15 am. which leaves 45 mins. till 8 am.
Therefore 15x
= 45, or x = 84 minutes. Hence Amanda gets up at 6:36
28
am. (A number of students tried to guess the answer by assuming that
x is a common multiple of 7, 6 and 4. While this is true, the answer
remains a guess.)
4. The postcode on an envelope is a four-digit number. The sum of the
digits is even. The first two digits form a two-digit number divisible
by 9. The first and third form the square of a whole number, and the
second and fourth form a multiple of 11. What is the postcode?
Solution.
Let the digits, from left to right, be a, b, c, d where each is a digit in the
range [0:9], apart from a which cannot be 0. Then a + b + c + d = 2n.
ab is divisible by 9, which means that the two digit number ab must be
one of 18, 27, 36, 45, 54, 63, 72, 81, 90 or 99.
Given that bd is divisible by 11, the two digits are identical, therefore
their sum is even. Since the sum of the four digits is even, the sum of
the digits a and c must also be even. But the two-digit number ac is
a perfect square, so it is one of 16, 25, 36, 49, 64, 81. Of these, only
the digits of 64 add to an even number. Hence a = 6 and c = 4, so
b = d = 3, and the required postcode is 6343. (Most of the credit was
given for careful elimination of possibilities, not simply for the correct
answer.)
5. Eight identical cardboard squares, are placed on a table one by one,
(they may overlap one another) and the result looks as follows:
f
g
h
c
d
e
a
b
Which square was the first to be placed?
Solution.
First one must deduce the placement of the squares (ignoring the order
of placement). From the figure, one can immediately deduce that a,
b, h and f occupy the four corners, and d the centre. Now c must be
placed centrally on the left-hand edge of the large square - otherwise it
would be entirely under, or on top of, either a or f . Similarly, g must
be placed midway along the top edge, and e must be placed midway
along the right-hand edge. Having established this, we can now work
out relative orderings. By inspection, we see that a > c, d, (where
a > b means here that a is placed after b.). Also, d > c, b, e, h, g, f,
and c > f, g > h, f > g, e > b, and h > e. Combining these, we get
the full ordering a > d > c > f > g > h > e > b. Hence b is placed
first. (Establishing the initial ordering - or something equivalent - is
essential for a complete answer. It is that aspect that was missing from
most solutions.)
6. A boy in a class of 9 students tells the teacher that the school bus
broke down. A girl in the class says that the boy is lying, then another
boy says the girl is telling the truth, another girl says the second boy
is lying and so on, each girl saying the previous boy is lying and each
boy saying the previous girl is telling the truth, up to the fifth boy who
says the fourth girl is telling the truth. Exactly four of the students
are lying. Did the school bus break down or not?
Solution.
We must check two assumptions. One that the bus broke down, and
the other that it didn’t. Assume the bus broke down. Let B1 denote
the first boy, and G1 denote the first girl, etc.
Under the stated assumption, B1 is telling the truth.
G1 says B1 is lying, so G1 is lying.
B2 says G1 is truthful, so B2 is lying.
G2 says B2 is lying, so G2 is truthful.
B3 says G2 is truthful, so B3 is truthful.
G3 says B3 is lying, so G3 is lying.
B4 says G3 is truthful, so B4 is lying.
G4 says B4 is lying, so G4 is truthful.
B5 says G4 is truthful, so B5 is truthful.
Thus four students are lying, and five are truthful. This is not sufficient
to prove that the bus broke down. We must repeat the above exercise
under the assumption that the bus didn’t break down. This reverses
all truth values, so that five students are lying and four are truthful.
Thus the only solution is that the bus broke down.
7. Given a square ABCD (corners labelled clockwise) with all sides of
length 10 cm, let E be the midpoint of AB and F the midpoint of BC.
Let X be the point of intersection of AF and EC. Find the area of the
quadrilateral AXCD.
Solution.
There are several ways of doing this. Here one, based on a solution
given by Bobbi Ramchen: By symmetry in the line BD, X lies on
BD and |BEX| = |BF X|, where | · · · | denotes the area enclosed.
As triangles with equal base length and height have the same area,
|AEX| = |BEX|, (bases AE = BE = 5cm.) and |BF X| = |CF X|
(bases BF = CF = 5cm.) Thus |AEX| = |BEX| = |BF X| = |CF X|.
Let this common area be A. Then
3A = |BEX| + |BF X| + |CF X| = |EBC| = 5 × 10/2 = 25cm2
Hence A = 25/3cm2 and |AXCD| = |ABCD| − 4A = 66 23 cm2 .
Here is one of the easier ways if you know some Cartesian geometry.
First, drop a perpendicular from the point X to the line CD, labelling
the point of intersection G. (Then the length XG is the height of the
triangle CXD). A horizontal line through X cuts AD at the point H,
where by symmetry HX = XG. We now proceed to find the lenght
HX. First, we erect a Cartesian coordinate system with the origin at
point D. We see that AF passes through (0, 10) and has gradient − 12 ,
so its equation is y = − 12 x + 10. Similarly, EC passes through (10,0)
and has gradient −2, hence its equation is y = −2x+20. They intersect
when the y (and of course x) coordinates are equal. Equating the two
equations for y gives
1
− x + 10 = −2x + 20,
2
which has solution x = 6 23 .
The area of the quadrilateral AXCD is twice the area of the triangle
DXC, which has base 10 and height 6 23 . Hence the area is 12 × 2 × 10 ×
6 23 = 66 23 sq. cm.
(A number of students used a scaled diagram to obtain an approximate
answer, an approach that attracted little credit.)