(8) MONDAY, Jan. 4, 1875. 1 to 4. MR. FREEMAN. Arabic numbers. MR. COCKSHOT. Roman numbers. 1. FOR a house occupied by B, A pays a rent of £40 per annum by equal payments at the end of each quarter. ` B pays 4 by equal payments in advance at the beginning of each month. How much a month ought B to pay in order that at the end of the year, with simple interest reckoned at 3 per cent. per annum, A may have recovered the value of his own four payments with one-tenth additional? The answer is £3. 128. 11415. 2. Shew how to find the lowest common multiple of three algebraical expressions. If ,, are the lowest common multiples of B and C, of C and A, of A and B, respectively; if 91, 92, 9, are the highest common divisors of the same pairs; and if L, G are the lowest common multiple and highest common divisors of A, B, and C; prove that L G 2 3 = 919293' Let a, B, y be the powers of a simple factor P in A, B, C respectively. Suppose them to be in descending order of magnitude but without excluding the cases in which some of them are equal or zero. L and in is (B+ a + a-y-y—B) = a - y. + + (c+p) (c + q) (a + p) (a + q) (b+p) (b + q) (a−b)(a−c)(a+x) * (b−c)(b−a}(b+x) * (c−a)(c−b} (c+x) * (B) If the letters all denote positive quantities, prove that 4. Find in terms of the coefficients the sum and product of the roots of the equation ax2+2bx + c = 0. Find the condition that the roots of ax2+2bx + c = 0 may be formed from those of a'x2+ 26'x + c = 0 by adding the same quantity to each root. The difference of the roots in each equation being the same, therefore b2-ac b'2 - a'c' - (a) (c+a−2b) x2 + (a + b − 2c) x + (b + c − 2a) = 0. (B) ax+yz=ay + zx = az + xy = b2. C (a) x=1 is obviously one root, and therefore the other root is (B) b+c-2α c+a-2b therefore ax+yz = ay + zx = az + xy = b2; (y − z) (x − a) = 0; therefore x = y = z = a root of the quadratic with the roots obtained by a cyclical change of x, y, z. Of these six sets of roots three only are distinct, and the roots b2 - a2 are a, a, α 6. Investigate the formula for the number of combinations of n things taken r at a time, without assuming the formula for permutations. A selection of c things is to be made, part from a group of a things and the remainder from a group of b things. Prove that the number of ways in which such a selection may be made will never be greater than when the number of things taken from the group of a things is the integer next (a + 1) (c + 1) less than a+b+2 Let x, y be the numbers to be chosen respectively from the groups of a and b things. is to be a maximum for integral values of x and is to be a maximum for an integral value of x. y, Therefore the factor by which this expression is multiplied when we write x + 1 for x must be just less than 1. That is, that value of x must be taken which will first If there were n groups containing a,, a, a,... a, things (the a's being in descending order of magnitude), and if p things were to be chosen, the number of ways in which the selection could be made would never be greater than when the number taken from the group of a, things was the integer next less than vii. Shew that corresponding small increments of a number and its logarithm are proportional. Find n from the following data: log, 42563 = 4 6290322, log, 4256446290424, 10 Find to how many decimal places n can be determined by this method, given that log,e=43429. The result is n=425.6353 correct to four places of decimals, for generally μδ ·+... n μδ If 8 is so small that does not come into the tables, 8 n cannot be determined by the method of proportional parts; or the result is not to be relied upon beyond the fourth place of decimals. viii. Find a general expression in terms of a for all the angles whose cosecants are equal to coseca. Find all the solutions of sin 30- cos 0=0. 1 + sin2 0 = 3 sin cos 0? ix. Prove that cos (A-B)= cos A cos B+ sin A sin B for all values of A and B between 0 and 90°. Shew how the proof may be extended so as to include all values of A and B. |