The Discussion on Mathematics National Examination Items for Junior High School Academic Year : 2009/2010

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1 The Discussion on Mathematics National Examination Items for Junior High School Academic Year : 009/00 Code : A. The result of 6 + (6:) (( 3) 3) is. A. 0 B. 3 C. 6 D. 9 This problem examines the student s ability in addition, subtraction, multiplication, and division of integers. Multiplication and division have higher hierarchies compared to addition and subtraction. This problem can be easily solved as follows: 6 + (6:) ((3) 3) = 6 +3 ( 9) = = 6 (C). Mother buys 40 kg of sugar. The sugar will be plastic packaged and sold in retail. The weight of each pack is 4 kg. So, she needs plastics to package. A. 0 packs B. 80 packs C. 0 packs D. 60 packs This problem examines the students to solve problem related to fraction. Forty (40) kg of sugar will be divided into small packs that each of them contains kg of sugar. So kg of sugar needs 4 packs of kg. Therefore, 40 kg of sugar 4 4 needs 40 x 4 =60 plastic packs. So, the number of plastic packs mother needs is 60. (D) 3. Finishing a particular work in 7 days needs 4 workers. After it has been done for 30 days, the work is stopped for 6 days. Assume that the worker s ability is the same each other. In order to finish the work as scheduled, the additional workers needed are..

2 A. 8 workers B. 6 workers C. 4 workers D. workers This problem examines the student s ability to solve problem related to the scale and comparison. To simplify the understanding about the problem, pay attention to the diagram below. Time: 7 days Workers: 4 people work Time: 30 days Workers: 4 people Result: work Stopped for 6 days Rest time: days Workers: 4 + n Result:..the rest work to finish Suppose the work must be finished is one work. Then in day, the 4 workers should finish of the work. 7 In one day, a worker should finish the work. 7 4 The work is done for 30 days by 4 people. So, the finished work is of the work The unfinished work is of the work. The work has been stopped for 6 days. So the rest time to finish the work as scheduled is = 36 days. If the additional workers necessary is n, then the present workers are 4 + n. They must be able to finish the 7 rest work in the rest time of 36 days. Consequently, 7 (4 n) n 36 4 n 8 n 4 So, the additional workers needed to finish the work on time are 4 workers. (C)

3 4. Andi buys 0 pairs of shoes at the price of Rp400, Seven of them are sold with the price of Rp50, each pairs. The other pairs are sold at the price of Rp40, each pairs, and the rest is donated. The benefit percentage obtained by Andi is A. 7 % B. 5 % C. % D. 30 % This problem examines the student s ability to solve the problem related to buying and selling. The price of 0 pairs of shoes = 400,000 Total of selling = (7 50,000) + ( 40,000) = 430,000 The profit = 430, ,000 = 30,000 profit The profit percentage = 00% buyingprice (430, ,000) = 00% 400,000 = 7 % So the profit percentage obtained by Andi is 7 % (A) 5. Someone borrows Rp4,000, on the installment plan in 0 months with the interest of.5 % per month. The amount of the installment per month is A. Rp44, B. Rp460, C. Rp47, D. Rp600, This problem examines the student s ability to solve the problem related to banking and cooperative. Suppose M is the borrowed money. M = 4,000, n is the number of installment, n = 0 i is the interest per month, i =.5 %

4 The amount of installment = installment per month + interest per month M = im n 4,000,000.5 = 4,000, = 400, ,000 = 460,000 So the amount of monthly installment is Rp460, (B) 6. Look at the pattern below carefully. 3 4 The number of the circles in the 0 th pattern is.. A. 380 B. 40 C. 46 D. 506 These circles have sequence pattern as follows:, 3, 3 4, 4 5, Then the 0 th term = 0 = 40 So the total circle at the 0 th term is 40. (B) 7. The next two terms of the sequence of 50, 45, 39, 3, are. A. 4, 5 B. 4, 6 C. 5, 7 D. 5, 8 This problem tests the ability to solve the problem related to the number sequence. The pattern of sequence: The above sequence pattern can be found by determining the difference between two sequential numbers. To find the nd number, subtract the first number by 5 so it will be 45, To find the 3 rd number, subtract the nd number by 6 so it will be 39, To find the 4 th number, subtract the 3 rd number by 7 so it will be 3,

5 From the pattern above, the fifth number and the sixth number can be determined, To find the 5 th number, subtract the 4 th number by 8 so it will be 4, To find the 6 th number, subtract the 5 th number by 9 so it will be 5, To find the 3 rd number, subtract the nd number by 6 so it will be 39, Therefore, the next two numbers of the number sequence of 50, 45, 39, 3,.. So, the next two terms of the sequence are 4 and 5. (A) 8. The result of (x )(x + 5) is. A. x x 0 B. x + x 0 C. x + 8x 0 D. x 8x 0 This problem examines the student s ability in multiplying algebraic expressions. The alternative solutions: Alternative : (x ) ( x + 5) = x(x + 5) (based on the distributive property) = x + 0x x 0 = x + 8x 0 (C) Alternative : Through geometric interpretation of the second number multiplication x x x x 5 0x 0 (x ) (x + 5) = x + 0x x 0 = x + 8x 0 (C) 9. The result of (4 x A. 3x 7 B. 3x 7 C. 3x 3 D. 3x 3 5) 5 x + 7 is.

6 This problem examines the ability in addition, subtraction, multiplication, division or quadratic of algebraic expressions. (4x 5) 5x 7 8x 0 5x 7 (Distributive property of multiplication toward subtraction) 8x 0 5x 7 8x 5x 0 7 3x 3 (C) 0. The simple form of A. x 3 x 3 B. x 3 x 3 C. x 3 x 3 D. x 3 x 3 x 3x 9 is. 4x 9 x 3x 9 (x 3)( x 3) x 3 4x 9 (x 3)(x 3) x 3 (C). If x + 7 = 5x then x + 3 is. A. 4 B. 4 C. 9 D. 4 This problem examines the student s ability in addition, subtraction, multiplication, division or quadratic in algebra. Alternative : x 7 5x x 7 5x 5x 5x (both segments are subtracted by 5 x ) 3x (both segments are subtracted by 7) 3x 8

7 3x x = 6 (both segments are divided by 3) So the value of x + 3 = = 9 (C) Alternative : x 7 5x x 6 5x 5 6 (show ( x + 3) at both segments) x 3 5 x 3 (both segments are subtracted by ) 6 3 5x x (both segments are subtracted by ) x (both segments are subtracted by 5(x + 3)) x 3 ( x + 3) = 9 3 (both segments are divided by 3) So the value of x 3 is 9 (C). If K = x 5 x 9}, x is natural number and L = x 7 x 3, x is whole number, so K L =.. A. 5, 6, 7, 8, 9, 0,,, 3 B. 5, 6, 7, 8, 9, 0,, C. 6, 7, 8, 9, 0 D. 7, 8, 9, 0 This problem examines the student s ability in determining the intersection or union Solving problem related intersection or union of two sets. K = 5, 6, 7, 8, 9 and L = 7, 8, 9, 0,, K L = 5, 6, 7, 8, 9, 0,, (B) 3. There are 69 applicants taking written test and interview test as a selection to get a job in a company. Finally, 3 applicants pass the interview test, 48 applicants pass the written test, and 6 applicants do not show up for the tests. The number of applicants accepted as employees is A. 3 applicants B. 7 applicants C. 5 applicants D. applicants

8 This problem examines the student s ability in determining the intersection or union of two sets and finishing problem related intersection or union of two sets. This problem can be solved first by drawing Venn diagram then making mathematics model from the existing information. From the first sentence in that problem, it can be concluded that to be accepted as an employee, an applicant must pass both written test and interview test. In this case, the data of applicants who pass both written and interview test has not been known yet. If the applicants who pass both of those tests is x, so the applicants who pass the written test is 48 x, and the applicants who pass the interview test is 3 x. Meanwhile, from the total of 69 applicants, there are 6 applicants who do not take both tests. Next, make Venn diagram as follows. S Written Tes tertulis Test Tes Interview wawancara test 48 - x x 3 - x 6 Then, the equation based on the description above is: (48 x ) + x + (3 x ) + 6 = x = 69 x = 7 x is the number of applicants who pass both written test and interview test. So the number of applicants accepted as employee is 7. (B) 4. A function is defined by the formula of f( x ) = 3 5 x. The value of f (4) is.. A. 3 B. 7 C. 7 D. 3 This problem examines the student s ability in solving the problem related to relations and function.

9 f( x ) = 3 5 x f(-4) = 3 5(4) = = 3 So the value of f (4) is 3. (D) 5. Gradient of a line with the equation of x 6y 9 = 0 is A. 3 B. 3 C. 3 D. 3 Alternative : Line equation with m gradient has general form of Therefore, the equation must be changed into x 6y 9 = 0 6y = x +9 x 9 y 6 y x 3 3 y mx n y mx n So the gradient of the line is 3. (C) Alternative : The gradient of the line passes through two points ( x, y ) and ( x, y ) can be y y determined by m x x Consequently, the gradient with that equation can be determined by choosing the two different points passed by the line. Suppose x = 0, then substitute it to the line equation.0 6y 9 = 0 3 y So, the line passes point 3 0,

10 Suppose y = 0, then substitute it to the line equation. x 6.0 9=0 9 x So, that line passes point 9, 0 From those two points, we can find the gradient, that is m y x y x (C) 6. Look at the picture carefully! The equation of line m is. A. 4y 3 x = 0 B. 4y 3 x + = 0 C. 4 x 3y = 0 D. 4 x 3y + = 0 Y m X This problem examines the student s ability in determining the gradient, line equation, and graphic. The line equation passes through x, ) and y, ) is ( x ( y So the line equation passes through (0, 3) and (4, 0) is y ( 3) x 0 0 ( 3) 4 0 y y y y x x x x y 3 x 3 4 4( y 3) 3x 4y 3x 0. So, the equation of m line is 4y 3x 0 (B)

11 7. Graphic of the line with equation 4x 3y is. A. Y C. Y 0 3 X X B. Y D. 0 X 4 Y X This problem examines the student s ability in determining gradient, line equation, and graphic. If y = 0, determine the intersection with x axis, 4x 3y 4x 3 0 4x 4x 4 4 x 3 The intersection with x axis, (3,0) Determine the axis point with y axis, if x=0

12 4x 3y y 3y 3y 3 3 y 4 The intersection with y axis (0, 4) So the graphic is.. (A) 4x y 3 8. It is known that 3x 5y The value of x y is.. A. B. 0 C. D. This problem is to examine the student s ability in solving system of linear equation of two variables. Equation () 4x + y = 3 can be changed into y = 3 4x then distributed to equation (), results: 3x + 5 (3 4x) = 3x + 5 0x = 7x = 5 7x = 7 7 x = 7 x = (3) Equation (3) is substituted to equation (): y = 3 4() = 3 4 = The value of x y = ( ) = + = (D)

13 9. The parking area for motorcycle and car can accommodate 30 vehicles. Once, the total wheels of vehicles are 90. If the number of motorcycle is x and the number of car is y, the linear equation system of two variables from the statement above is.. x y 30 A. x 4y 90 x y B. 4x y x y C. x 4y x y D. 4x y This problem examines the student s ability in solving linear equation system of two variables. The alternative solutions: It has two main procedures: understanding problem by reading and understanding the sentences carefully, and making mathematics model from this problem such as linear equation system. From the problem above, we know that the number of motorcycle is stated by x and the number of car is stated by y. In the first sentence, we can interpret that the total number of cars can be accommodated in that parking lot is 30 vehicles, consisting of car and motorcycle. Therefore we can make an equation x y 30..(i) Furthermore, from the second sentence, it is known that the total number of wheels is 90. Although it is not explicitly expressed in the sentence, we know that a motorcycle has two wheels and a car has four wheels (ideally). Therefore the equation is x 4y 90..(ii) From equation (i) and (ii), the linear equation system can be obtained is x y 30 x 4y 90 Answer: x y 30 x 4y 90 (A)

14 0. The length of AC is.. A. 4 cm B. 8 cm C. 30 cm D. 3 cm C A cm 35 cm B This problem examines the student s ability in solving the linear equation system of two variables. Based on Pythagorean Theorem, in the right angle above applies: AB + AC = BC AC = BC AB = 35 = 5 44 = 784 AC = = 8 So the length of AC is 8 cm. (B). Pay attention to the picture! PQRS is parallelogram with the length of TR = cm, PQ = 7 cm, and QR = 5 cm. So the length of PT is. A. 0 cm B. cm C. 4 cm D. 5 cm P T S R Q This problem examines the student s ability in solving problem related Pythagorean Theorem. T 5 S 7 R TS = TR SR = 7 = 5 By using Pythagorean Theorem, PT can be defined. PT = = P 7 Q 5 5

15 = = = 0 So the length of PT is 0 cm. (A). Look at the picture carefully! The shadowed area is the sketch of grass field. The width of the grass field is.. A.,400 m B.,900 m C.,400 m D.,00 m 40 m 0 m 5 m Jalan raya m m 75 m This problem examines the student s ability in calculating the width of the plane. IC = = = = = 30 AD = BC-IC = = 45 A 40 m B F G E 0 m 5 m H 75 m D m m I C The width of shadowed area= the width of ABCD-the width of EFGH = = = 900 So the width of the grass field is 900 m. (B) 3. Look at this plane carefully! The circumference of this plane is. A. 7 cm B. 9 cm C. 7 cm D. 4 cm 4 cm cm,5 cm

16 This problem examines the student s ability in calculating the circumference of the plane and the use of circumference concept in daily life. C D A B E F,5 cm L J K 4 cm H cm I G K = AB + BC + CD + EF + FG + GH + HI + IJ + JK + KL + LA = = 9 So the circumference of the plane is 9 cm. (B) 4. Father will build a circle-shaped park with the radius of 35 m. Surround the park will be planted with pine trees with interval of m. If one tree costs Rp5,000.00, the total cost of the pine trees planting is.. A. Rp5,900, B. Rp5,700, C. Rp5,500, D. Rp5,00, This problem examines the student s ability in calculating the circumference of the plane and the use of the concept of circumference in daily life. The circumference of the circle= πr = 7 35 = 0 The number of pine trees around the park = 0 : x = 0 Total cost of Pine tree planting = 0 x 5,000=5, So the total cost of Pine trees planting is Rp5,500, (C) Note: In this problem, one meter means one meter length of circular arc but it is necessary to know the concept of real distance is the shortest distance between two points.

17 So the distance of meter in the problem above should be one meter long of bowstring. 5. Pay attention to the rhombus of ABCD. A : B = :. The angle of C is A. 60 o D B. 90 o C. 0 o D. 50 o A C B This problem examines the student s ability in calculating the angle in the plane. The understanding of rhombus properties, two angles are supplementary each other and the comparison is needed to finish this problem. In rhombus, the opposite sides are parallel. It makes two angles, which are side by side, are parallel each other. So in that figure A is parallel with B, therefore A + B = 80 o.() From the comparison, it is known that A : B = :. So B = A.(ii). By substituting the equation (ii) to the equation (i), it will be: A + A = 80 o 3A = 80 o A = 60 o In rhombus, the two opposite angles have equal size. So A = C. Therefore C = 60 o (A) 6. Pay attention to the picture! l 4 6 m 5 3 The size of the angle no is 95 o and the size of the angle no is 0 o. The size of the angle no 3 is A. 5 o B. 5 o C. 5 o D. 35 o

18 This problem examines the student s ability in calculating the size of newly formed angle if two lines intersect each other. The size of angle no is 0 o, then the size of angle no 6 is 70 o (the size of two supplementary angles is 80 o ) The size of angle no is 95 o, then the size of angle no 5 is 95 o (the alternateinterior angles have similar size) The size of angle no 3 is added by the size of angle no 5 added by the size of angle no 6 is 80 o (total size of the angles in a triangle) So the size of angle no 3 is 80 o (95 o + 70 o ) = 5 o (B) 7. Look at the picture carefully! Suppose O is the center point of the circle. The size of angle AOB is. A. 5 o B. 30 o C. 45 o D. 60 o A O 30 B C This problem examines the student s ability in calculating the size of center angle and circumference angle in a circle. Remember the properties of circumference angle and center angle in a circle opposing to the same arc that the size of the center angle is twice of the circumference angle. Pay attention to O as the center of the circle, C is a point in a circle and AOB and ACB are opposing AB arc. Therefore AOB = ACB = 30 o = 60 o So, the size of AOB is 60 o. (B) 8. Pay attention to the picture! D cm P Q C A 6 cm B

19 P and Q are the center points of BD and AC diagonal. The length of PQ is.. A. 5 cm B. 4 cm C. 3 cm D. cm This problem examines the student s ability in solving problem using the concept of congruity. Alternative : AQ = QC AQ : AC = : BP = PD The first step is drawing the additional line by lengthening the segment of PQ line to the right and the left like in the picture. In ABS and CQS: CQS is congruent with CAB CQ CA CS CA CQ QS CA AB CQ QS CQ 6 QS AB D R A cm P 6 cm Q B S C QS 6 QS 3 In BCD and BSP BC BS BD BP CD SP BC CD BS SP

20 BS BS 3 PQ 6 PQ PQ 3 Alternative : DP = PB; CQ =QA DCT is congruent with BTA and PTQ With this congruity, if TB = x then DT = x so DB =3x 3 P is in the center of DB = x Therefore PT = x x PT PQ so PQ, so PQ =3 TB AB x 6 So the length of PQ is 3 cm. (C) 9. A photograph has the length of 30 cm and width of 0 cm ticked on a carton. The rest carton on the left, right, above the photograph is cm. If the photograph and carton are congruent, the rest carton under your photograph is. A. 5 cm B. 4 cm C. 3 cm D. cm This problem examines the student s ability in solving problem using the concept of congruity. The photograph and carton are congruent. AB AD PQ PS x x 5 3 x x x P S A D x Q B C R

21 5 x =0 5x 0 5x x 4 So the rest carton under the photograph is 4 cm. (B) 30. Look at the following two congruent triangles! C F + A + B D E The pairs of the lines in the same length are A. AB dan DE B. AC dan DE C. BC dan DE D. AB dan FE This problem examines the student s ability in solving problem using the concept of congruity. C E A + B F + D The lengths of the lines which have the same length are BC dan DE (C) 3. Look at the cube of ABCD.EFGH! The number of diagonal space is. E H F G A. B. 4 C. 6 D. D C A B

22 This problem examines the student s ability in determining the elements of threediagonal space of flat side. Make a list of space diagonal in cube of ABCD.EFGH, those are AG, BH, CE, DF. Therefore, there are 4 diagonal spaces in the cube of ABCD.EFGH. (B) 3. Pay attention to the picture below! (I) (II) (III) (IV) The pictures which are the nets of the cuboids are... A. I dan IV B. I dan III C. II dan III D. II dan IV This problem examines the student s ability in determining the nets of threedimensional shape. The nets of cuboids are picture I and III (B) 33. A cuboid-shaped pond has the length of 5 m, width of 3 m, and depth of m. The maximal volume of the pond is.. A. 6 m 3 B. 40 m 3 C. 30 m 3 D. 5 m 3 This problem examines the student s ability in calculating volume of three dimension shaped of flat-side and curved-side

23 The maximum volume it takes up is the same as the volume of the cuboids-shaped pond. The pond s volume = length width height = 5 3 = 30 So the maximum water it can take up is 30 m 3 (C) 34. A tube-shaped drum which has the radius of 70 cm and the height of 00 cm is full of oil. The oil will be poured into small tubes with the radius length of 35 cm and the height of 50 cm. The number of small tubes needed are A. tubes B. 4 tubes C. 6 tubes D. 8 tubes This problem examines the student s ability in calculating the volume of 3D shape of flat-side and curved-side. The volume of the tube = r r t Drum Volume Totalsmall tubes needed Small Tubes Volume So the amount of small tubes needed are 8 tubes. (D) 35. This picture is a prism with the trapezoid as the base of isosceles trapesium. The length of AB = 6 cm, BC = AD = 5 cm, CD = 4 cm, and AE = 5 cm. The width of prism surface is H G A. 450 cm B. 480 cm C. 500 cm D. 50 cm D E F C A B

24 This problem examines the student s ability in calculating the width of the surface of 3D shape of flat side and curved side. The width of surface area of the prism is the width of the whole volume of the prism. To determine the width of the whole prism s edge we can count the width of ABCD first, the width of EFGH, the width of ABFE, the width of BCGF, the width of CDHG, and the width of ADHE. The width of ABCD = (AB + CD) t (to find out the height of trapezoid we can use the Phytagorean theorem) = (6 + 4) 3 = 30 t The width of EFGH = the width of ABCD = 30 The width of BCGF = BC BF = 5 5 = 75 The width of ADHE = the width of BCGF = 75 The width of CDHG = CD DH = 4 5 = 0 The width of ABFE = AB AE = 6 5 = 90 The width of surface area of the prism = the width of ABCD + the width of EFGH + the width of ABFE + the width of BCGF + the width of CDHG + the width of ADHE = the width of ABCD + the width of BCGF + the width of CDHG + the width of ADHE = = 50 So the width of the prism surface is 50 cm (D) 36. Ali makes 5 large parachutes made from half-ball plastic. If the length of the diameter is 4 m and π = 3,4, the minimum width of the plastic needed is A. 88,4 m B. 376,8 m C. 66 m D. 753, 6 m

25 This problem examines the student s ability in calculating the width of the surface area of 3D shape with flat side and curved side Diameter = d = 4 m Radius = r = m The width of the ball surface = 4 π r The width of half ball = the width of the ball surface = 4 3,4 = 5, The minimum width of plastic needed = 5 the width of half ball = 5 5, = 376,8 So the minimum width of plastic needed is 376,8 m (B) 37. Look at the table below carefully! Score Frequency The median of the data in the table above A. 6.0 B. 6.5 C. 7.0 D. 7.5 This problem examines the student s ability in determining a measure of central tendency and using it to solve daily problems. To solve the problem no 37 needs the understanding of median concept.

26 Median of sets of data is a datum value located in the middle of datum value sorted from the smallest to the highest so it is divided into two parts. So there is 50 % of the datum which has higher score or the same as the median and 50 % of datum number which the scores are less than or the same as the median. There are two alternatives to determine the median. First alternative:. Sort the datum value from the smallest to the highest. Determine the value of the median that is by seeking the datum value located in the middle by crossing out the datum value located in the left side and the right side so we get datum value which is the middle. Second alternative:. Sort the datum value from the smallest to the highest.. Determine the position of the median n, n = the number of datum. 3. Determine the median value. From the problem above it can be determined that the datum amount is the number of the all frequencies that is = 4. By using the first alternative:. Sort the datum value from the smallest to the highest Find the datum value which is in the middle by crossing the datum value which is on the left side and on the right side so we can find the datum value which is in the middle Median Value So the median is 7 8 = 7,5. By using the second alternative:. Sort the datum value from the smallest to the highest

27 Datum ke= Datum ke=3. Determine the position of the median n, n = the number of datum The position of the median = n 4 5 = = =.5 So the median is lied between the th datum and the 3 th. The th datum value is 7 and the 3 th datum value is 8. So the median of the data th the datum value the3 is So the median of the data is 7.5 th datum value = 7 8 = 7.5 (D) 38. The average of mathematics score is 7, while the average score of male students is 69 and the average score of female students is 74. If there are 40 students in the class, the male students are.. A. 4 students B. students C. 8 students D. 6 students This problem examines the students ability in determining the measure of central tendency and using it in solving daily problems. To solve the problem no 38 it needs an understanding about average and linear equation system as well as uses it in problem solving. the number of data score From the problem, it is known that 7 = 40 The amount of data value = 7 40 = 880 If the male students is p, while female students is w And it is known that 69 = The totalscore of male students p and

28 the totalscore of female students 74 = w So the total score of the male students = 69 p = 69 p Total score of female students = 74 w = 74 w Total data score = Total score of male students + total score of female students 880 = 69 p + 74 w Equation 40 = p + w Equation w = 40 p Then substitute w = 40 p, to equation 880 = 69 p + 74 w 880 = 69 p + 74 (40 p) 880 = 69 p p = 5p 80 = 5 p 80 p = 5 = 6 So the number of male students is 6. (D) 39. Pay attention to this diagram! Frequency Student s score If 6 is a completeness score, the number of students who cannot achieve the completeness score is...

29 A. 7 students B. 0 students C. 4 students D. 8 students This problem examines the student s ability in presenting and interpreting the data To solve the problem no 39, it needs an understanding about bar chart and bar chart reading so it can be determined that the number of students who do not pass the completeness score is the number of frequency of the students who have score less than 6. The total number of students who do not pass the completeness score = = 0. So the total number of students who do not pass the completeness score is 0 (B) 40. The difference of the number of students who get score of 6 dan 9 in the diagram is... A. 9 students B. 6 students C. 5 students D. 4 students This problem examines the student s ability in presenting and interpreting data To solve the problem no 40, it needs an understanding about line diagram and line diagram reading so the difference between the total number of students who get 6 and 9 can be determined.

30 The difference of total number of students who get score of 6 and 9 is equal to the frequency of students who get score of 9 subtracted by students who get score of 6. The difference between the total number of students who get 6 and 9 are 9 5 = 4. So the difference between the students who get score of 6 and 9 is 4 students (D) Arranged by:. Wiworo. Untung Trisna Suwaji 3. Yudom Rudianto 4. Sri Purnama Surya 5. Nur Amini Mustajab 6. Choirul Listiani Reviewer: Th. Widyantini