BRITISH COLUMBIA SECONDARY SCHOOL
MATHEMATICS CONTEST, 2007
Junior Preliminary Problems & Solutions
1. Mr. Smith has three times as many girls as boys in his class. Ms. Perry has twice as many boys as girls in her class. Mr. Smith has 60 students in his class and Ms. Perry has 45 students. If the classes are combined into one class, the ratio of boys to girls is:
(A) 3 : 4 (B) 4 : 3 (C) 5 : 4 (D) 4 : 5 (E) 3 : 2
Solution
Since there are 60 students in Mr. Smith’s class, and three-quarters of them are girls, there must be 45 girls and 15 boys. Similarly, there must be 15 girls and 30 boys in Ms. Perry’s class. This gives a total of 45 boys and 60 girls in the combined class for a boys to girls ratio of 3 : 4.
Answer is (A).
2. In the diagram, each of the arcs is a semicircle. Of the total area inside the largest semicircle, the fraction that is shaded is:
(A) x
4 (B)
2
9 (C)
1 3
(D) 1
π (E)
4
9 x x x
Solution
Using the familiar formula for the area of a circle, the fraction that is shaded is
π
2x 2
− π2 x2
2
π
2 3x
2 2
= 1
3
Answer is (C).
3. A rhombus is a parallelogram with all sides equal. The rhombus shown has diagonals of lengths 2 units and 6 units. The perimeter of the rhombus is:
(A) 40 (B) 2√10 (C) 4√10
(D) 41+√3
(E) 8√10
Solution
By elementary reasoning on isosceles triangles we deduce that the two diagonals intersect at right angles. It follows from Pythagoras’ theorem that the perimeter of the rhombus is 4×√12+32=4√10.
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BCSSMC 2007 Junior Preliminary Problems & Solutions
4. The sum of the digits in the smallest positive integer that is divisible by 2, 4, 6, 10, 12, and 14 is:
(A) 3 (B) 6 (C) 9 (D) 15 (E) 18
Solution
We need to compute the least common multiple of the given numbers. To do this we calculate the least common multiple using the product of the primes to the maximal powers over all primes in the prime factorizations of the numbers. We have, 2 = 2, 4 = 22, 6 =21×31, 10 = 21×51, 12 = 22×31, and
14=21×71, so the least common multiple is 22×31×51×71=420, and the sum of its digits is 6.
Answer is (B).
5. Alan has thrown 24 passes and completed 25% of them. Over the rest of the season Alan completes all of his passes and he ends the season with an 80% pass completion record. The total number of passes Alan threw over the season was:
(A) 42 (B) 50 (C) 72 (D) 80 (E) 90
Solution
Letxdenote the number of passes Alan throws in the remainder of the season. Then x+6 x+24=
4 5, from which we findx=66. So the total number of passes Alan throws over the season is 66+24=90. Alternate solution: If Alan completes all of the rest of his passes, the total number of passes he did not complete is 75% of 24, which is 18. If Alan completes 80% of his passes over the whole season, he misses only 20%, or 1/5, of them. But this is 18 passes, so the total number of passes Alan throws is 5×18=90.
Answer is (E).
6. Let yn
ybe the largest prime number less thannand x n
x
be the smallest prime number greater than n. The expression
41+y35
7. A nine-digit integer has each of the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 appearing exactly once (in some order). The probability that the integer is divisible by 9 is:
(A) 1
Recall, an integer is divisible by 9 if and only if the sum of its digits is divisible by 9. Since the sum of the digits of the integer formed in the way described in the problem is 1+2+3+...+9=45 which is
divisible by 9, it follows that the probability that the integer is divisible by 9 is 1.
BCSSMC 2007 Junior Preliminary
Problems & Solutions Page 3
8. A two-digit integer is divided by the sum of its digits. The largest remainder that can occur is:
(A) 9 (B) 13 (C) 15 (D) 16 (E) 17
Solution
Ifn is a two digit number thenn = 10a+bwhere 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9. Then the sum of the digits isa+b, and 1≤a+b≤18, so division ofnbya+bleaves a remainder of at most 17. Working backwards from possible remainders of 17 we eliminate 99, 98, 89. Then we find that 16 into 79 leaves 4 with a remainder of 15 which must be the largest possible.
Answer is (C).
9. Three rectangular pieces are removed from the corners of a square piece of cardboard. The perimeter of the remaining portion is 40 cm and the total area of the three rectangles removed is 20 cm2. The area, in cm2, of the remaining piece of cardboard is:
(A) 20 (B) 60 (C) 80
(D) 380 (E) 1580
Solution
Evidently, removing the pieces does not change the perimeter so the side length must be 10 cm, and the area 100 cm2. The remaining piece must have area 80 cm2.
Answer is (C).
10. In the diagram, the angle atAis 60◦and the radius of the larger circle is 6. The radius of the smaller circle is:
(A) 2 (B) 3 (C) 3
2
(D) 4 (E) √6
A
C
B 6
Solution
LetCbe the centre of the larger circle and letBbe the base of the (necessarily) perpendicular line from Cto the lower of the two tangents to the two circles. The line segment ACbisects the angle at A, so that triangle ABCis a 30◦-60◦-90◦triangle with the 30◦angle atA. SinceBCis a radius of the larger circle and it is the side opposite the 30◦angle, the hypotenuseACis 12. If the smaller circle has centre Eand radiusr, the same argument shows thatAE=2r. ThenAC=2r+r+6=3r+6=12 sor=2.
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BCSSMC 2007 Junior Preliminary Problems & Solutions
11. In the diagram,ABCDis a rectangle,Fis the midpoint of sideAB, and Xis on the extension of sideBC. Further,AB= 33
5 andBC = 14
3. The length of the segmentBXfor which the area of triangleAFXis5
8 of the area of the rectangleABCDis:
(A) 35
= 774. Then, since trianglesAFX andFBXhave equal base and height, the area of
triangleABXis twice the area of triangleAFXand so is 277 4
12. King Henry took twenty-four of his knights on a hunting expedition. They stayed in one of Henry’s hunting lodges which had nine rooms, three on each side and one central room where Henry slept, as shown. The knights were assigned three to a room, but they were allowed to move among the rooms leaving more or less than three knights to a room, so long as there were always exactly nine knights on each side of the lodge. One night four friends of the knights came to the lodge disguised as knights. That night Henry made the rounds of the lodge and found that there appeared to be nine knights on each side of the lodge. The total number of knights and, possibly, disguised friends in the corner rooms was:
King
(A) 0 (B) 4 (C) 6 (D) 8 (E) 12
Solution
Let Xdenote the total number of persons in the corner rooms, and letYdenote the total number of persons in the interior side rooms. Then X+Y = 28 and, since each corner room is counted twice
when counting the total number of people on the sides of the lodge, 2X+Y = 4×9 =36, soX= 8
andY=20.
