Download CD Solutions

Math Contest, Fall 2005
CD EXAM SOLUTIONS
1. In an election, there were two candidates. One of them received 65% of the votes and
got 1500 votes more than his competitor. How many people voted?
65
35
Solution. Let x be the number of people voted. Then
x =
x + 1500, thus
100
100
x = 5000.
2. Ms. Smith wrote two numbers on the board and asked her students to calculate the
product. The last digit of one of the numbers was eight, but it was not written clearly.
John mistook this digit for 6, and ended up with the product of 4740 as his answer.
Mary mistook this digit for 3, and ended up with the product of 4695 as her answer.
What should be the correct answer to the teacher’s problem?
Solution. The difference between John’s and Mary’s answers is 4740 − 4695 = 45,
which is 3 times the misunderstood second factor (since he multiplied by extra 3). So
the second number is 15, and the correct answer is 4740 + 2 × 15 = 4770 since the
correct last digit was 8, and not 6 that he used.
3. In the figure, AC is an arc of a circle with center O, BDA is a semicircle with center
M. If OA is perpendicular to BA and OA = BA = 14 cm, what is the total area of
the shaded part BCA?
Solution. The triangle OAB is a right isosceles triangle, thus 6 O = 45◦ . The area of
π. The area of the shaded region BCA is
the circular sector OCA is thus 18 π142 = 49
2
142
49
49
− 2 π = 98 − 2 π.
2
4. Suppose that a function f is such that f ( x1 ) − 3f (x) = x for every x 6= 0. Find f (2).
Solution. Consider x = 2. Then f ( 12 ) − 3f (2) = 2. Now consider x = 12 . Then
f (2) − 3f ( 12 ) = 21 . Solving the system f ( 12 ) − 3f (2) = 2, f (2) − 3f ( 12 ) = 21 we get
f (2) = − 13
.
16
5. Find the area of the triangle formed by the coordinate axes and the line through the
points (0, 4) and (4, 2).
Solution. An equation of this line is y = − 12 x + 4, the x-intercept is (8, 0), thus the
area is 12 (4 × 8) = 16.
6. The sum of the sixth and the ninth terms of an arithmetic sequence is 20, and their
product is 64. Find the tenth term if the first term is negative.
Solution. The system a6 + a9 = 20, a6 × a9 = 64 leads to the quadratic equation
z 2 − 20z + 64 = 0 where z = a6 . The solutions are z = a6 = 16 or z = a6 = 4. Then
a9 = 20 − a6 = 4 or a9 = 20 − a6 = 16. Thus for the difference of the sequence we
have d = −4 or d = 4. Here −4 does not work, so d = 4 and thus a10 = a9 + 4 = 20.
7. What are the values of n for which the system 3x + ny = 3, 2x − 4y = 1 has a solution
x < 0, y < 0?
and substitute this expression
Solution. Solve the second equation for x to get x = 1+4y
2
3
12+n
into the first equation. Then y = 2(6+n) , and so x = 2(6+n) . To make x < 0, we must
take −12 < n < −6, while y < 0 if n < −6. Thus, to have a solution x < 0, y < 0 we
must have −12 < n < −6.
8. The function f is given by the table
x
f(x)
1 2 3 4 5
4 3 5 1 2
If u0 = 3, and un+1 = f (un ) for n ≥ 0, what is the value of u2005 ?
Solution. Checking the pattern for the first few values we see that u0 = 3, u1 = f (3) =
5, u2 = f (5) = 2, u3 = 3, thus u3j+k = uk , u2005 = u1 = 5.
9. Find the sum of the first twenty odd numbers such that they leave a remainder of one
when divided by three.
Solution. Each such number can be written as 3n + 1, n = 0, 2, ..38, so 1 + 7 + 13 +
= 1160.
... + 115 = 116(20)
2
10. It is known that x and y are positive numbers and x + y = 5. What is the smallest
possible value of the expression x1 + y1 ?
5
5
Solution. x1 + y1 = x+y
= xy
= x(5−x)
. The smallest value of the fraction is achieved
xy
when the denominator is the largest, or when x = 25 (vertex of the parabola x(5 − x) =
5x − x2 ). Thus the smallest possible value of the expression x1 + y1 is 54 achieved when
x = 25 , y = 52 .
1
11. Compute N if N = 1 +
1
2+
2+
1
2 + ...
Solution. Rewrite as N − 1 =
1
1
2+
1
2 + ...
√
2
= N + 1, N − 1 = 1, N = 2.
2+
Thus
1
N −1
so
1
= 2+
N −1
1
1
2+
2+
= N + 1.
1
2 + ...
12. Find the ratio of the area of the triangle CED and the parallelogram ABCD (first
area to the second)
Solution. Denote the heights of triangles DEC and BEF by h1 and h2 respectively,
and the height of the parallelogram by h. Then the height of the parallelogram ABCD
h1
2
is equal to h = h1 + h2 . Since the triangles DEC and BEF are similar,
=
so
h2
3
2
h1 = h2 . The area of the parallelogram ABCD is equal to 3h = 3(h1 + h2 ) = 5h2 .
3
The area of the triangle DEC is 3h2 /2. Thus the ratio of the areas is 3/10.
13. Let s be an x−intercept of the function f (x) = x2 + 3x − 6. What is the value of the
product (s − 2)(s + 5)?
Solution. s2 + 3s − 10 = s2 + 3s − 6 − 4 = 0 − 4 = −4
14. Find the sum 502 − 492 + 482 − 472 + ... + 22 − 1
Solution. Using the formula a2 − b2 = (a − b)(a + b) we rewrite given expression as the
sum of arithmetic sequence and then compute the sum: 502 −492 +482 −472 +...+22 −1 =
99 + 95 + 91 + ... + 3 = (3+99)(25)
= 1275. (Here a = 3, d = 4, 99 = 3 + 4(n − 1), so
2
n = 25.)
15. Two positive integers are chosen at random. What is the probability that their product
is odd? Express your answer as a common fraction.
Solution. There are four possible combinations: (odd, odd), (even, odd),(odd, even),
(even, even), the chances of ”odd” and ”even” are 12 . Only the first combination
produces an odd product. Thus the probability is 41 = 0.25.
16. Find the volume of the solid created by rotating the triangle whose vertices are
(0, 3), (6, 3), and (6, 0) around the y−axis. Express your answer in terms of π.
Solution. The volume of the solid is the difference between the volume of the cylinder
and the volume of the cone: π(62 )(3) − 13 (π(62 )(3)) = 32 (π(62 )(3)) = 72π.
17. What is the last digit of the number 777778 ?
Solution. The last digits follow the same pattern as the last digits of the sequence
70 = 1, 71 = 7, 72 = 49, 73 = 343, 74 = 2401, .... Since 778/4 has the remainder of 2,
the last digit of the number 777778 is 9, the same as in 72 = 49.
18. An airplane was supposed to cover the distance of 2900 km. However, after covering
1700 km, it had to land and wait on the ground for 1h 30 min. After that it flew with
the average speed 50 km/h less than before. Find the original average speed of the
plane if it is known that the plane came to its destination 5 hours after departure.
1200
+ r−50
+ 1.5 = 5. This
Solution. Denote by r the original speed of the plane. Then 1700
r
2
equation becomes 0.035r − 30.75r + 850 = 0 which has two solutions 850, 28.5714285.
The second solution is too small for the original speed. Thus, the original speed was
850 km/h.