(b) "There is a rectangular field whose length exceeds its breadth by 16 yards, and it contains 960 yards; find its dimensions." (a) Squaring the first we get x2-2xy + y2 = 16; subtracting the second 2 y = 4 12 and x = 6 4 and y = 2 (b) Let x = breadth, whence x + 16 = length. The plus sign being taken, 24 is the breadth, and 241640, the length. 3. "A fast passenger train starts at 20 minutes past 12; it overtakes a luggage train, which travels 15 miles an hour, and after having gone 15 miles further overtakes a slow passenger train which travels 20 miles an hour. Another fast train, which travels at the same rate as the first, starts from the same station at 2 o'clock of that day, overtakes the luggage train, and, after having gone 65 miles further overtakes the slow train also, and finds that it has then travelled 120 miles. Required the rate at which the fast trains travel." Let x-rate per hour of the fast trains: now the former starts one hour and two-thirds before the second; therefore that time will elapse before the second comes to the point at which the first overtook the luggage train. We shall deduce our data from the consideration of this as a starting point. Let us first find, in terms of x, how far the slow passenger train was in advance when the luggage train was overtaken. Let y that distance in miles. The space travelled over by the first train after passing the luggage, and before overtaking the passenger train, was 15 miles: therefore y miles, plus the miles travelled from the time the fast train passed the luggage to the time of coming up with the passenger train, are equal to 15 miles. Now the first train travelled x miles per hour, and the slow 20 miles. Hence, the sum of the following series equals 15 miles. 20 х S = y + y X +yx + &c. ad infinitum (20)+ Now, the luggage train travels 15 miles per hour. Hence, by the time the second fast train arrives at the point at which the first overtook the luggage train, this the distance of the fast train's point of junction with these two trains from the termination of two respective series similar to the preceding: But from the question the difference of the distance represented by these expressions equals 65 miles. Solving this quadratic and taking the positive value of the root, we find, x = 30 N.B.-The length of the whole journey, 120 miles, does not enter into the solution. It will now merely serve to verify the result, and to indicate that the first fast train travelled 5 miles before it came up with the luggage train. SECTION II. 1. "Prove the formula for finding the sum of an arithmetical series. (a) "The first term of an arithmetical progression is 1; the common difference, 1; the sum of the series, 36. Required the number of terms." (b) (a) Let a = first term and d = common difference. Now the 2nd term = 1st term 3rd term = 2nd term وو + d = a + d + d = a + 2d 4th term = 3rd term + d = a + 3d -1.d nth term = (n−1)th term + d = a + n— Now S, the sum of the series, is equal to all these terms put together; or, S = a + (a + d) + (a + 2d) + (a + 3 d) +.. .... +(a+n-1. d) This series may be written in an inverse order; thus, S=(a+n-1. d) + .... + (a + 3d) + (a + 2d) n−1. +(a + (a + d) + a. And now the addition of any two terms that are vertically opposite in the series, will give (2a + n−1.d). For the sum of the first terms in each line this is sufficiently evident; the second term in the second line is d less than the preceding term; that is, it is (a + n- -2.d). This added to (a + d) the second term of the first line = (2a + n —1. d). The third term of the second line is (a + n−3. d); the third term of the first line is (a + 2d); the sum of these is (2 a + n−1. d), as before. The quantity (2 a+n-1. d) is repeated as many times as there are terms in the progression, that is n times in the present case. Hence, by adding the corresponding term of each line, we get 2S 2S (2a+n―1. d) + (2a + n−1. d)+ &c. to n terms (2a+n-1.d) n = n 2 = (a + 1); where 7 is the last term of the series, 2 and is equal to the first term, a, plus the common difference multiplied by the number of its place in the series less by one. (b) In the example given S = 36; a = 1; d and n is to be found. 1; Since is not given, formula (1) 2 (a) "When are quantities said to be proportional? (b) and when is one quantity said to vary as another? (c) Prove that if abcd; then a: a bc: ༩༤ d." (a) When the ratio between two quantities a, b, is the same as between two other quantities c, d; or when abcd the four quantities form a proportion which is written abc d, in which a is the same multiple, part, or parts of b, that c is of d. -(b). One quantity x, varies as another quantity y, when the magnitude of x depends on that of y. Thus, if x be the length and y the breadth of a parallelogram, and it be required to construct a similar parallelogram, having a given ratio to the first, x varies as y, or the length and breadth increase or diminish simultaneously. If a parallelogram be required equal in area but of different length from another parallelogram; then, as the length increases, the breadth diminishes, and the converse. In this case x varies inversely as y. |