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Math-tanong CEER 2012 - Set 1 Solutions

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 1What is the smallest three-digit number thatleaves 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 2The pie chart above shows the distribution of DVDrentals from Dibidi Doo Bee shop for a single night. If250 DVDs were rented that night, how many moreaction movies were rented than horror movies?(a) 10(b) 20(c) 22(d) 25 math-tanong CEER Supplement set 1
  5. 5. Question 2 SolutionTIP: For this type of problem, just subtract the ratesbefore applying the percentage – no need to multiplyfirst before subtracting.Those who rented actionmovies are 22% — 12% = 10%more than those who rentedhorror movies. Thus, there are 250(10%) = 250(0.1) = 25more action movies rented thanhorror movies. math-tanong CEER Supplement set 1
  6. 6. Question 3Suppose that n and p are integers greater than 1. If 5n isa perfect square and 75np is a perfect cube, what is thesmallest value of n + p? (a) 5 (c) 8 (b) 6 (d) 15 math-tanong CEER Supplement set 1
  7. 7. Question 3 SolutionIf 5n is a perfect square, the smallest possible value of nwould be 5. Now, 75np = 52 i 3 i npIf 75np is a perfect cube and n = 5 is the smallestpossible value of n, the smallest possible value for p is32 = 9 since 75np = 52 i 3 i 5 i 32 = 53 i 33Hence, the smallest possible value for n + p is 5 + 9 = 14 math-tanong CEER Supplement set 1
  8. 8. Question 4For real numbers a and b, define the operation a ∗ b = 3b − 2 aIf 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 SolutionBy 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 SolutionOf course, you can do substitution. By the definition, 2 ∗ 3 = 3 ( 3 ) − 2 (2 ) = 5 and 3 ∗ X = 3 ( X ) − 2 ( 3 ) = 3X − 6Q: Which of the choices will give a value of 5 whensubstituted to X in 3X − 6? Choice (c). math-tanong CEER Supplement set 1
  11. 11. Question 5A test is composed of 25 questions. The score of anexaminee is obtained by giving him 4 points for eachcorrect answer and deducting 1 point for each wronganswer. If an examinee answered all the questions andobtained a score of 70, how many questions did heanswer 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 ) = 70Correct x 4 4x 4 x − 25 + x = 70 1(25 − x) 5 x = 95Wrong 25 − x 1 = 25 − x x = 19 math-tanong CEER Supplement set 1
  13. 13. Question 6In the figure below, 1 2 . What is thevalue of x?(a) 40(b) `50(c) 60(d) 70 math-tanong CEER Supplement set 1
  14. 14. Question 6 SolutionDraw an extra line passing through the vertex ofangle x and parallel to 1 and 2 m° n°math-tanong CEER Supplement set 1
  15. 15. Question 6 SolutionIf 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 7If 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 SolutionWarning: DO NOT ATTEMPT to solvefor 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 9Erin calculated the average of 5 numbers to be 38. Thenshe found out that she had made an error and hadwritten 40 for one of the numbers when she should havewritten 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 SolutionLet S be the sum of the first 4 numbers that Erin has.If the average of 5 numbers is 38, including theerroneous 40, then S + 40 average = = 38 ⇒ S + 40 = 190 5But 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 5Then 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 10In the diagram, FDCB is a rectangle. Segment ED is 6units long, segment AB is 10 units long, and the measureof 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 SolutionFast 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 SolutionHence, we have the following:1. By the properties of a 30-60-90 triangle:  6  12 3EC = 2   = i =4 3  3 3 3 30° x 3 =6 2 1DC = EC = 2 3 x 2 x 2 math-tanong CEER Supplement set 1
  24. 24. Question 10 Solution2. Since BCDF is a rectangle, DC = BF, so DC = BF = 2 33. 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 Solution4. 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 SolutionExpert fast solution! Let x = EF and z = AE Based on theestablished 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

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