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# Math-tanong CEER 2012 - Set 1 Solutions

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# Math-tanong CEER 2012 - Set 1 Solutions

The solutions to Practice Set 1

The solutions to Practice Set 1

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### Math-tanong CEER 2012 - Set 1 Solutions

1. 1. Math- Math-tanong PRACTICE SET 1 with Solutions
2. 2. Question 1 What is the smallest three-digit number that leaves a remainder of 1 when divided by 2, 3, and 5? (a) 121 (b) 151 (c) 181 (d) 211 math-tanong CEER Supplement set 1
3. 3. Question 1 Solution • Determine first the least common multiple (LCM) of 2, 3, and 5 – which is 30. • Now, find the smallest three-digit multiple of the LCM – that’s 120. • The smallest 3-digit number that leaves a remainder of 1 when divided by 2, 3, and 5 is the smallest 3-digit number that leaves a remainder of 1 when divided by 30, the LCM – that’s 121 ☺ math-tanong CEER Supplement set 1
4. 4. Question 2 The pie chart above shows the distribution of DVD rentals from Dibidi Doo Bee shop for a single night. If 250 DVDs were rented that night, how many more action movies were rented than horror movies? (a) 10 (b) 20 (c) 22 (d) 25 math-tanong CEER Supplement set 1
5. 5. Question 2 Solution TIP: For this type of problem, just subtract the rates before applying the percentage – no need to multiply first before subtracting. Those who rented action movies are 22% — 12% = 10% more than those who rented horror movies. Thus, there are 250(10%) = 250(0.1) = 25 more action movies rented than horror movies. math-tanong CEER Supplement set 1
6. 6. Question 3 Suppose that n and p are integers greater than 1. If 5n is a perfect square and 75np is a perfect cube, what is the smallest value of n + p? (a) 5 (c) 8 (b) 6 (d) 15 math-tanong CEER Supplement set 1
7. 7. Question 3 Solution If 5n is a perfect square, the smallest possible value of n would be 5. Now, 75np = 52 i 3 i np If 75np is a perfect cube and n = 5 is the smallest possible value of n, the smallest possible value for p is 32 = 9 since 75np = 52 i 3 i 5 i 32 = 53 i 33 Hence, the smallest possible value for n + p is 5 + 9 = 14 math-tanong CEER Supplement set 1
8. 8. Question 4 For real numbers a and b, define the operation a ∗ b = 3b − 2 a If 2 ∗ 3 = 3 ∗ X , what is the value of X? ? 11 ( a) 2 ( c) 3 9 ( b) ( d) 4 2 math-tanong CEER Supplement set 1
9. 9. Question 4 Solution By definition of the operation "∗ ": 2 ∗ 3 = 3 ∗ X ⇒ 3 ( 3 ) − 2 ( 2 ) = 3X − 2 ( 3 ) 5 = 3x − 6 3 x = 11 11 x= 3 math-tanong CEER Supplement set 1
10. 10. Question 4 Solution Of course, you can do substitution. By the definition, 2 ∗ 3 = 3 ( 3 ) − 2 (2 ) = 5 and 3 ∗ X = 3 ( X ) − 2 ( 3 ) = 3X − 6 Q: Which of the choices will give a value of 5 when substituted to X in 3X − 6? Choice (c). math-tanong CEER Supplement set 1
11. 11. Question 5 A test is composed of 25 questions. The score of an examinee is obtained by giving him 4 points for each correct answer and deducting 1 point for each wrong answer. If an examinee answered all the questions and obtained a score of 70, how many questions did he answer correctly? (a) 17 (c) 19 (b) 18 (d) 20 math-tanong CEER Supplement set 1
12. 12. Question 5 Solution Let x = the number of correct answers 25 – x = the number of wrong answers (since there are 25 questions in all) No. of Points Equation : Equatio n questions per Total points answered question 4 x − ( 25 − x ) = 70 Correct x 4 4x 4 x − 25 + x = 70 1(25 − x) 5 x = 95 Wrong 25 − x 1 = 25 − x x = 19 math-tanong CEER Supplement set 1
13. 13. Question 6 In the figure below, 1 2 . What is the value of x? (a) 40 (b) `50 (c) 60 (d) 70 math-tanong CEER Supplement set 1
14. 14. Question 6 Solution Draw an extra line passing through the vertex of angle x and parallel to 1 and 2 m° n° math-tanong CEER Supplement set 1
15. 15. Question 6 Solution If m and n are the angles formed as in the figure, then, using the ideas of angles formed by a transversal: m = 20 ( corresponding angles ) n = 180 − 150 = 30 (alternate interior 150, m° thenadjacent) n° ∴ x = m + n = 50 math-tanong CEER Supplement set 1
16. 16. Question 7 If 6 − x = 5.43 , what is the value of 6 + x ? (a) 6.43 (b) 6.57 (c) 17.43 (d) 17.51 math-tanong CEER Supplement set 1
17. 17. Question 7 Solution Warning: DO NOT ATTEMPT to solve for x! Sayang ang effort! 6 − x = 5.43 ⇒ − x = 5.43 − 6 = −0.57 x = 0.57 ∴ 6 + x = 6 + 0.57 = 6.57 math-tanong CEER Supplement set 1
18. 18. Question 9 Erin calculated the average of 5 numbers to be 38. Then she found out that she had made an error and had written 40 for one of the numbers when she should have written 30. What is the average of the correct 5 numbers? (a) 32 (b) 34 (c) 36 (d) 38 math-tanong CEER Supplement set 1
19. 19. Question 9 Solution Let S be the sum of the first 4 numbers that Erin has. If the average of 5 numbers is 38, including the erroneous 40, then S + 40 average = = 38 ⇒ S + 40 = 190 5 But the last number should be 30 instead of 40, so S + 40 − 10 = 190 − 10 ⇒ S + 30 = 180 The average, then, should have been S + 30 180 = = 36 5 5 math-tanong CEER Supplement set 1
20. 20. Question 9 Solution “Expert” shortcut solution: Start with S + 40 original average = = 38 ⇒ S + 40 = 190 5 Then the correct average is S + 30 ( S + 40 ) − 10 190 − 10 180 = = = = 36 5 5 5 5 math-tanong CEER Supplement set 1
21. 21. Question 10 In the diagram, FDCB is a rectangle. Segment ED is 6 units long, segment AB is 10 units long, and the measure of angle ECD is 60°.What is the length of segment AE? 3 ( a) 20 ( c) 20 − 2 3 ( b) ( d) 20 −4 3 2 math-tanong CEER Supplement set 1
22. 22. Question 10 Solution Fast facts about the figure: 1. Triangle CDE is a 30-60-90 triangle with angle DEC = 30 degrees. 2. Since DF is parallel to CB, and AE acts y y as a transversal, then 2 x 3 30° angle AEF is 30 2 degrees. Hence, x x triangle EAF is also a 2 x 2 30-60-90 triangle. math-tanong CEER Supplement set 1
23. 23. Question 10 Solution Hence, we have the following: 1. By the properties of a 30-60-90 triangle:  6  12 3 EC = 2   = i =4 3  3 3 3 30° x 3 =6 2 1 DC = EC = 2 3 x 2 x 2 math-tanong CEER Supplement set 1
24. 24. Question 10 Solution 2. Since BCDF is a rectangle, DC = BF, so DC = BF = 2 3 3. Since AB = 10, then AF = 10 − BF 30° = 10 − 2 3 2 3 2 3 math-tanong CEER Supplement set 1
25. 25. Question 10 Solution 4. Since EAF is a 30-60-90 triangle with angle AEF = 30 degrees, then AE = 2AF ( y = 2 10 − 2 3 ) ( = 2 10 − 2 3 ) 10 − 2 3 30° = 20 − 4 3 2 3 math-tanong CEER Supplement set 1
26. 26. Question 10 Solution Expert fast solution! Let x = EF and z = AE Based on the established facts earlier, you have the figure shown. x 3 12 3 =6⇒x = i =4 3 2 3 3 x 4 3 z = =2 3 z 2 2 y = 2 x 3 x 30° =6 y = 10 − = 10 − 2 3 2 2 x x z 2 ( y = = 10 − 2 3 ⇒ z = 2 10 − 2 3 ) 2 x 2 = 20 − 4 3 math-tanong CEER Supplement set 1
27. 27. END OF PRACTICE SET