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question:13.298. Two athletes are running on the same closed track of a stadium. The speed of each is constant, but the first one takes 10 seconds less to run the entire track than the second one. If they start running from a common starting point in the same direction, they will meet again after 720 seconds. What fraction of the entire track length does each runner cover in one second?
answer:Solution.Let V_{1}, V_{2} be the speeds of the athletes, l be the length of the track. According to the problem, we have the system left{begin{array}{l}frac{l}{V_{2}}-frac{l}{V_{1}}=10, 720 V_{1}-720 V_{2}=l,end{array}right. from which frac{l}{V_{1}}=80, frac{l}{V_{2}}=90.Then the first athlete runs frac{V_{1}}{l}=frac{1}{80} part of the length of the entire track per second, and the second athlete runs frac{frac{l}{V_{2}}}{l}=frac{1}{90}.Answer: frac{1}{80} and frac{1}{90}.
question:3. In the diagram, B D is perpendicular to B C and to A D. If A B=52, B C=21, and A D=48, what is the length of D C ?
answer:3. Since triangle A D B and triangle C B D are right-angled, we can apply the Pythagorean Theorem in each triangle.Therefore, A B^{2}=A D^{2}+B D^{2} or 52^{2}=48^{2}+B D^{2} and so B D^{2}=52^{2}-48^{2}=2704-2304=400. Since B D>0, then B D=sqrt{400}=20.Furthermore, D C^{2}=B D^{2}+B C^{2}=20^{2}+21^{2}=400+441=841.Since D C>0, then D C=sqrt{841}=29.ANSWER: D C=29
question:Hicham runs 16 mathrm{~km} in 1.5 hours. He runs the first 10 mathrm{~km} at an average speed of 12 mathrm{~km} / mathrm{h}. What is his average speed for the last 6 mathrm{~km} ?(A) 8 mathrm{~km} / mathrm{h}(B) 9 mathrm{~km} / mathrm{h}(C) 10 mathrm{~km} / mathrm{h}(D) 6 mathrm{~km} / mathrm{h}(E) 12 mathrm{~km} / mathrm{h}
answer:When Hicham runs 10 mathrm{~km} at an average speed of 12 mathrm{~km} / mathrm{h}, he takes frac{10}{12}=frac{5}{6} hours to run this distance.Since Hicham runs for a total of 1.5 hours, then he runs the last 6 mathrm{~km} in frac{3}{2}-frac{5}{6}=frac{9}{6}-frac{5}{6}=frac{4}{6}=frac{2}{3} hours.Since he runs 6 mathrm{~km} in frac{2}{3} hours, his average speed for this segment is frac{6}{2 / 3}=9 mathrm{~km} / mathrm{h}.ANsWER: (B)
question:João's father has between 200 and 300 books in his library. One fifth of these books are in English, one seventh in French, one quarter in Italian, and the rest are in Spanish. What is the total number of Spanish books in this library?#
answer:SolutionThe number of books is a multiple of 5, 7, and 4. Since these numbers do not have common prime factors, it must necessarily be a multiple of (5 cdot 7 cdot 4 = 140). In the given range, there is a unique multiple of 140, namely the number 280. Therefore, the number of books that are not in Spanish isfrac{280}{5}+frac{280}{7}+frac{280}{4}=56+40+70=166Thus, there are (280-166=114) Spanish books.
question:Professor Célia, in a class about sequences, decides to play a guessing game with patterns:I) First, she chooses a Natural number.II) Cláudia must say the double of its successor.III) Marcelo must say the triple of the predecessor of the number said by Cláudia.IV) Finally, Ademar must say the quadruple of the successor of the number said by Marcelo.a) If Professor Célia chooses 3, what will be the sequence formed by the 4 numbers?b) Diani was in the bathroom and when she came back, she heard Ademar saying 184. What was the number chosen by Professor Célia?c) Create an expression to determine the number chosen by Professor Célia if Ademar said the result is x.
answer:Solutiona) If Professor Célia chooses 3, Cláudia should say 2 cdot 4=8, Marcelo should say 3cdot7 = 21, and Ademar should say 4 cdot 22=88.b) We need to analyze the problem "backwards," using inverse operations. If Ademar said 184, then Marcelo must have said frac{184}{4}-1=46-1=45, and consequently, Cláudia must have said frac{45}{3}+1=16. Finally, Professor Célia must have said frac{16}{2}-1=7.c) Using the reasoning from the previous item, if Ademar said x, then Marcelo must have said frac{x}{4}-1, Cláudia must have said frac{frac{x}{4}-1}{3}+1, and the number chosen by Professor Célia wasfrac{frac{frac{x}{4}-1}{3}+1}{2}-1=frac{x-16}{24}
question:17 cdot 154 Let P_{1}, P_{2}, P_{3} be the equilateral triangles constructed on the sides AB, BC, CA of the right triangle triangle ABC (with C as the right angle). Then the (quad) of P_{1} is the sum of the ( ) of P_{2} and P_{3}.(A) area, area.(B) perimeter, perimeter.(C) sum of interior angles, sum of interior angles.(D) altitude on side AB, altitudes on sides BC and CA.(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
answer:［Solution］By the Pythagorean theorem and the given conditions, we knowA B^{2}=A C^{2}+B C^{2} text {. }Also, quad S_{p_{1}}=frac{sqrt{3}}{4} A B^{2}, S_{p_{2}}=frac{sqrt{3}}{4} B C^{2},S_{p_{3}}=frac{sqrt{3}}{4} A C^{2} text {. }Therefore,S_{p_{1}}=S_{p_{2}}+S_{p_{3}} .Hence, the correct choice is (A).