SMML MEET 3 ROUND 1
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1 ROUND 1 1. How many different 3-digit numbers can be formed using the digits 0, 2, 3, 5 and 7 without repetition? 2. There are 120 students in the senior class at Jefferson High. 25 of these seniors participate in student government, but not in math team. 47 participate on math team, but not in student government. The number of seniors involved in both activities is 20% of the number of seniors involved in neither activity. If choosing a random senior from Jefferson High, what is the probability of choosing someone involved in both activities? 3. Adam, Barry, Carl and David are each asked to pick a random digit from 1 through 9. What is the probability that exactly 3 of them pick a multiple of 3?
2 ROUND 1 - Solutions 1. How many different 3-digit numbers can be formed using the digits 0, 2, 3, 5 and 7 without repetition? There are 5 digits, but only 4 of them can be in the first position (all but 0). This means there are 4 options for the first digit. There are then 4 options for the second digit (since any of the remaining can be used). This then leaves 3 options for the third digit. To calculate the total number of possibilities multiply: There are 120 students in the senior class at Jefferson High. 25 of these seniors participate in student government, but not in math team. 47 participate on math team, but not in student government. The number of seniors involved in both activities is 20% of the number of seniors involved in neither activity. If choosing a random senior from Jefferson High, what is the probability of choosing someone involved in both activities? Let G be the set of seniors involved in student government and M be the set of seniors involved in math team. We can use the Venn Diagram at the right to view this data. This implies: G x M P(choosing a random senior involved in both activities) x Adam, Barry, Carl and David are each asked to pick a random digit from 1 through 9. What is the probability that exactly 3 of them pick a multiple of 3? This is a binomial setting with 4 trials, 3 out of 9 desired outcomes and 6 out of 9 undesired outcomes. P(3 of 4 people picking a multiple of 3) =
3 ROUND 2 1. Simplify. 2. Write the following in fully simplified and factored form:. 3. Simplify.
4 ROUND 2 - Solutions 1. Simplify. 2. Write the following in fully simplified and factored form:. 3. Simplify. We can use the rational root theorem and synthetic division for both the numerator and denominator (or factor the denominator by grouping).
5 ROUND 3 1. In quadrilateral ABCD, the measure of = the measure of = 90. AB = 5, AD = 12, and BC = 7. Find CD in simplest radical form. 2. The figure below shows 2 sets of parallel lines that meet to form a rectangle with one corner at point X (as shown). These lines also divide the horizontal line shown into three congruent segments. Find the length of. X 4 6 Y 3. In the figure shown, and are both perpendicular to. CB = 9, XB = 4 and the area of AXY is the area of ABC. Find the length of. Y C A X B
6 ROUND 3 - Solutions 1. In quadrilateral ABCD, the measure of = the measure of = 90. AB = 5, AD = 12, and BC = 7. Find CD in simplest radical form. In the figure, divide the quadrilateral into two right triangles. Using the Pythagorean Theorem, we find: 7 C B 90 º 5 90 º A 12 D 2. The figure below shows 2 sets of parallel lines that meet to form a rectangle with one corner at point X (as shown). These lines also divide the horizontal line shown into three congruent segments. Find the length of. Since the parallel lines meet to form a rectangle, we know that they meet at right angles. This implies and are similar right triangles. Since the lines form 3 congruent segments on the horizontal, we know that DB = BA = AY = 6. Using proportions for the similar triangles, we know CB = 2. Likewise, and are similar right triangles. Using proportion, we know CX = 2. Using Pythagoras, X E we know. 4 C 2 D 6 B 6 A 3. In the figure shown, and are both perpendicular to. CB = 9, XB = 4 and the area 6 Y of AXY is the area of ABC. Find the length of. is similar to since they are both right triangles containing. Therefore, their sides are proportional. Let AX = b and YX = h. Y C Area of Area of h 9 Meanwhile, from similarity A b X 4 B This implies YC = 5.. Using the Pythagorean Theorem, AY = 10 and AC = 15. 5
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