ESSAY Problem Answer

Transcription

1 ESSAY Problem Answer PROBLEM 1 A store has a big sale. For every 2 shirts purchased at regular price, a third shirts can be bought for 10,000 rupiahs. Mr. Bendot has bought twelve shirts for 1,200,000 rupiahs. What is the regular price for one shirt? Solution: Price for 3 shirts is = rupiahs (score1) 4 Regular price for 2 shirts is = rupiahs (score1) Regular price for 1 shirts is = rupiahs (score1) 2 PROBLEM 2 Starting from point A, Ali runs 500 m eastward. Then he runs westward for another 750 m. Then he runs eastward again for 350 m, turns back and runs westward for 450 m. How far is he from point A at the end of the run? Solution: (score 0,5) (score 0,5) (score1) (score1) = The distance between Ali and his house is 350m. Hal 1 dari 7

2 PROBLEM 3 In the figure, IMSO is a square and T is a point on the side MS. The triangle IMT has area 23 cm 2 and the triangle TSO has an area 9 cm 2. What is the length of a side of the square? I M Area of IMSO = 2x (IMT + OST) (1 ½) = 2 x (23 + 9) = 64.(1) Then the length of a side of the square is 8..(1/2) 23 P O 9 T S PROBLEM 4 David gave Sita a half of his candies. Sita gave John a half of the candies that she received from David. John kept 8 of those candies and gave the remaining 10 to Sheela. How many candies did David have originally? Solution : John s candies = 18 (score1) Sita s candies = 36 (score1) David s candies = 72 (score1) PROBLEM 5 In each of the first six tests, Brad s mark is always 4 points higher than Jim s. For in each of the other four tests, Jim s mark is always 1 point higher than Brad s. What is the difference between their average marks of the ten test? (T1+ T6)=24+(A1+ A6) (1) (T7+ T10)+4=(A7+ +A10) (1) [(T1+ T10)- (A1+ +A10)]/10=20/10=2 (1) Another Solution: The total marks of the two differs by 6x4-4x1=20 (1.5) Average differs by 20/10=2 (1.5) Answer : 2 Hal 2 dari 7

3 PROBLEM 6 In the figure, every vertex of a smaller square is the middle of a side of the larger square. The total area of the shaded parts is 21 cm 2. What is the total area (in cm 2 ) of the unshaded portion? The shaded area is (1- ½ + ¼ - 1/8 + 1/16-1/32) = 21/32 of the largest square (1 ½ ) The unshaded area is 11/32 of the largest square. (1) The unshaded area is 11/32 : 21/32 x 21 cm 2 = 11 cm 2. (½ ) PROBLEM 7 The number of parrots in forest A decreases by 120 per year, while the number of parrots in forest B increases by 80 per year. There were 7,340 parrots in forest A in the year 2000 and 4,200 parrots in forest B in year In what year will the number of parrots in forest B start to exceed the number of parrots in forest A? Solution: In year (score 1) In year (score 1) Answer : year 2017 (score 1) Year Forest A Forest B ? ? ? Hal 3 dari 7

4 PROBLEM 8 The design given below is made of a number of semi circles. The horizontal diameter of the circle is cut into three equal lengths. If the area of the large circle is 6 m 2, what is the area (in m 2 ) of the shaded part? There are three circle which are the largest, the midle and the smallest. The largest area is 6 m 2, the midle area is 8/3 m 2 and the smallest area is 2/3 m 2. (1 ½ ) The shaded area = The midle area - the smallest area (1 ½ ) The shaded area = 8/3 2/3 = 2 m 2 PROBLEM 9 Rudy is given three positive whole numbers by his teacher and is told to add the first two and then multiply the result by the third. Instead, he multiplies the first two and adds the third to the result. Surprisingly, he still gets the correct answer which is 14. How many different possible values could the first number be? Solution Let the three positive integer be a, b, and c. From the given information (a + b) c = 14 and (a b) + c = 14. From the first equation we see that c must be a factor of 14, the only positive value of c are 1, 2, 7, and 14. (1) If c = 1, then a + b = 14 and ab = 13, so a = 1 and b = 13 or a = 13 and b = 1 The only possible value of a and b are 1 and 13. If c = 2, then a + b = 7 and ab = 14. For this a = 3 and b = 4 or a = 4 and b = 3(1) If c = 7, then a + b = 2 and ab = 7. There are no possibility for a and b here. If c = 14, then a + b = 1 and ab = 0. There are no possibility for a and b here. Therefore, the four possible values for a are 1, 13, 3, and 4. (1) (-0,5 point for missing one possibility) Hal 4 dari 7

5 PROBLEM 10 In the figure, ABC is a right triangle, AX = AD and CY = CD. What is the measure of XDY (in degrees)? X A D B Y C Since AX = AD and CY = CD then Let CDY = CYD = α. Then DCY = 180 2α and Since ΔAXD and ΔCDY are isosceles triangle. Δ AXD isosceles triangle, then AXD = ADX = Notice that ADX + XDY + YDC = 180 or (1/2) 135 α + XDY + α = 180 or XDY = 45 (1/2) XDY is 45 degrees BAC = 2α 90. (1) 180 ( 2α 90 ) = 135 α. (1) 2 PROBLEM 11 Given three numbers. If each number is added to the average of the other two numbers, the results are 65, 69, and 76. What is the average of the three numbers? Solution Each number appears twice in the three numbers 65, 69, and 76. (1) Sum of the three numbers is ( )/2=105. (1) Average is 105/3=35 (1) (Another Solution ) Let the numbers be a, b, and c. We obtain. b + c a + = 65 or 2a + b + c = 130. (score 0,5) 2 Analogy for the other two, a + 2b + c = 138 (score 0,5) a + b + 2c = 152 (score 0,5) Solve the equations 4a + 4b + 4c = 420 or a + b + c = 105 (score 1) a + b + c Or = 35. (score 0,5) 3 Average is 35. Hal 5 dari 7

6 PROBLEM 12 In the figure, AB is a diameter of the circle, BD=BE, and DAC=27 0. What is the measure of the angle ACD? Since BD = BE, ΔEDB isosceles, then EDB = BED (score 0,5) ABD = ACD, (score 0,5) BED + BDE = ABD (score 0,5) BED = BDE = ½ ABD BDA = 90 0, ΔACD (score 0,5) CAD + ACD + BDE + BDA = (score 0,5) ACD + ½ ACD = 180 0, ACD = 42 0 (score 0,5) PROBLEM 13 Ahmad and George take the same route of 7 km that start and ends at the same point. They start at the same time, take the route in opposite directions, and finish at the same time. Ahmad walks at the constant speed all along the route. George walks at constant speed for the first 3 km then increases his speed by 7 km/hr for the remaining distance. They meet in the middle of the walk only once, that is when Ahmad has covered 4,5 km of the distance. How much time do the two people need to finish the 7 km distance? Solution Ahmad = 4,5 km George = 2,5 km Speed A : speed G = 4,5 : 2,5 = 9 : 5(score 1) 9 George 0,5 km Ahmad x0,5 = 0, 9km 5 Remain Ahmad = 7 (4,5 +0,9) = 1,6 km Remain George = 7 (2,5 + 0,5) = 4 km(score 0,5) 18 Speed A = x 7 = 3,6km/ hr A = 5,4 km G = 3 km 35 Speed A : speed G = 1,6 : 4 = 2 : 5(score 0,5) 9:5 = 18:10 ; 2 : 5 = 18 :45 Speed G = initial : later = 10 : 45 (score 0,5) 10 difference 7 km/h, initial speed G = x 7 = 2km/ hr 35 Hal 6 dari 7

7 Speed A = 9 5 Time needed x 2 = 3,6km/ hr (score 0,5) 7 3, = = 1 hr (score 0,5) maximum score = 3 Note for Marking Scheme : 1. The correct result answer, without explanation was given 1 point. 2. The wrong calculating / wrong step of work with correct result answer awarded 1 point. 3. Miss calculating / mistype deducted ½ point. 4. Answer given in the drawing / sketch without explanation awarded full point. 5. Missing unit answer were not deducted, but false unit will be deducted ½ point. Problem 2: Opposite direction answer are allowed Problem 9: Counting factor from 14 awarded 1 point FIN Hal 7 dari 7