2008 Comp So

, !

TEAM ROUND

I.

2

.

Zip codes are of the form abcdefghji where each letter represents a digit. How many zip codes are possible where a is not O? There are 10 choices for each letter except a where there are 9 choices.

9 x 10 x 10 x 10 10 x 10 = 9 )( 108 Ani...

x

x -

I

Y

=

1.0625 =

10

1-. = 16

x

10 x 10

x

17 - . 16

~ and yare pqsitive and x < 50 and y <: 50.

17 34 -= -=- 51

16

32

4~

Only the first,two fractions work since 51 > 50> 17 + 16 + 34 + 32 = 33 + 66 = 99 Ans. 3.

A jumbo jet can travel its own length in 20 seconds. At this same speed,

(I)

the jet can taxi past a 710 foot long hangar in 70 seconds. So what is the length of the jumbo jet? Let x be the length of the jet. Let y be the speed of the jet in feet per second. Remember that the jet must completely pass the hangar! Then

x -

= 20

z First, let's find the radius of circle T. C = 2rrr 12rr = 2rrr r= 6 If we draw a line from T to Z we have made an equilateral triangle since both TX and TZ are radii of the circle making the measure of angle TZX 600 which means that the measure of angle XTZ is also 60°. Therefore XZ = XT = r = 6. Ans.

and

y 710+ x

70

=

y x = 20y 710 + x = 70y 710 + 20y = 70y 710 = 50y y=

TIO -=50

71

6. We are given a system of

5

20x 71 x = 20y = --5

inequalities: 2x> 3x- 3 =

71 x 4

=

3x- a>-6

284

How many positive integral values of a are there for which x = 2 is the only positive integer solution? Substitute 2 for x in the second equation. 3x2-a>-6 6-a>-6 12 - a> 0 a < 12 2x> 3x- 3 0>x-3 3>x x must be positive so x could be 1 or

4. The 10 letters in the name Cybil and Ronda are placed on identical cards so that each of the 10 cards contains one letter. Without replacement, two cards are selected at random. What is the probability that one letter is from each sister's name? The probability that a letter from Cybil's name is picked first is 5 5 1 5 5

2.

-x -= -x -=10

9

2

9

We must also look at what happens when we choose a value for a between 1 and 11. If x = 1, then 3-a>-6

18

Similary, the probability that a letter from Ronda's name is picked first is 5

-a>-9 a<9

18

5

5

10

5

So a must be 9,10 or 11.

18+ 18= 18= 9' ~ 5. Circle T has a circumference of 12rr inches and segment XY is a diameter. The measure of angle TXZ is 60°, So what is the length of segmentXZ?

3 Ans. 7.

We have two candles one of which is 20 cm taller than the other. They are both lit at the same time. 5 hours later they were both at the same height. The taller one bumed all of its wax in six hours while the shorter one burned all of its wax in 10 hours. So what is the ratio of the original height of the shorter candle

to the original candle?

Ifx

Let x be the height of the smaller candle. Then x + 20 is the heiglit the larger candle The taller one

of

x+ 20 burns at the rate of --hour.

The smaller

per

6

one bums

at the

X

rate of -

per hour.

10

After 5 hours

candle smaller

---

of the larger

6 5x

left and candle

x = -

10

2

left.

These

are

x

2

5

5.4

6

~J

+

y-

z 4 5

345 3

5

5.8

6 6

4

5

6.4

NO for for

9.

13 = ---

72

Ans.

In the figure, m
~

10

Since
I

---= -= longer 30 -3--Ans. Each of three six-sided

dice is rolled

once. So what is the probability that the three numbers rolled are the lengths of the sides of an obtuse triangle? In an obtuse triangle the length of the side opposite the obtuse angle, must be larger than then length of . the hypotenuse if the triangle was a right triangle .. ' So, if x, y and z are the sides of the triangle where z is opposite the obtuse angle and x <= y, then

z rel="nofollow"> .Jx2-;.-;z

and z < x + y.

So what values of the dice can give us an obtuse triangle? If x = y, then let's just make 3 columns for x,. y and z we can just list them

17+ y2

2

y 2

2.8

3

3

3

4.3

5

4

3.6 4.5

6x 6x 6

two

= 10 x + 20 = 30

4

vx'

3 4

39

x

x

Y

2 2

.-.--

6x = 2x + 40 4x = 40

8.

< y,

3 )( 3 + 5 )( 6 = 9 + 30 = 39 ways

20 ---6 2

shorter

NO

X

of the

values, i.e., the lengths of the candles after 5 hours of burning, equal.

x+

7.1

There are only 3 permutations the first set and 6 permutations the second set.

x+ 20 there was

5

5

height of the taller

5.6

Z

6

that

degree angle, the third angle must be 180-(90+26)= 180-116=64 If we now look at the triangle which contains the angle of yO, the angle to the upper right of that is also 64 by corresponding angles. That leaves the third angle in that triangle to be 180 - (64 + y) = 116 - y. By corresponding angles the rightmost angle in the five-sided polygon that contains angles A and B must also be 116 - y. Now five-sided polygons must have angles whose sum is (5 - 2) )( 180 = 3 )( 180 = 540 degrees. We already know the measures of angles A and B, the 900 angle (obviously!!!) and the one we got from corresponding angles (116 - y). So what about the one that has x as its exterior angle? Well that's just 360 - x. So: 540 = 28 + 360 - x + 74 + 90 + 116

-y 540 = 388 - x + 164 + 116 - Y 540 = 668 - x - y

x + y = 668 - 540 = 128 Ans.

10. A circular tabletop is divided into four congruent sectors by two diameters that are perpendicular to each other. Each sector will be painted with one of four colors. How many distinct ways can the table be painted? Since we don't have to use all 4 colors in painting the tabletop, let's start with just using one color. Clearly we have four ways.

How many combinations are there?

4x 3x 2 ---

2x 2

=

6 for each choice of

color that will be used twice so a total 6 x 4 = 24 for each type. Oh, but wait, red, blue, red, green will rotate to be the same as red, green, red, blue. Thus when the color chosen twice is in opposite sections, there are only 12 possible arrangements. Giving us 24 + 12 = 36 ways for 3 colors. 36 + 28 = 64 so far. Finally how about 4 colors? Here are all the possibilities with red in the upper left-hand comer.

Now let's look at two different colors. There can either be 2 sections of each or there can be 3 of one color and 1 of another. When there are 2 of each, we have two different ways we ~n ~o it. Either the same colors are together or they are opposite each other.

--

How many combinations are there?

4x 3 --

2

=

Are these all distinct? Yes. Butwe will gain no new combinations by putting another color in the top left comer since we can just rotate the . given circles around until we get the desired ~'or in the top left comer. Therefore, we have added only 6 more. 64 + 6 = 70 Ans.

6 for each of the two

circles above or a tp13l of 12 ways. That's: 12-'" 4 = 16 to far. Now let's lad< at having 3 One color and 1 of anottte-r:. tf we hMve 3 red, the oth.t secfion can be blue, green or yeltolN-thal'13" optl6ns. The set of thret!t same-<::okJt.td sections can be atfyone oftftj •• dOlors, thus there

d

anr
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