if (2y − 1) {x2+ 2 + √(x2 + 4x + 3)} = (2x − 1) {y2 + 2 + √(y2+ 4y + 3)}, or x2+2+√(x2 + 4x+3) y2 + 2 + √(y* + 4y + 3) 2x-1 = 2y-1 Now x + 4x + 3 = (x2 + 2x + 1) (x2 — 2x + 3) = uv u = x2 + 2x + 1 and v = x2 - 2x+3; Substituting the value y=x in (2), we have = x+4 2x-1' 4; .. x= 11 4x2-16x+16=27; .. 2x-4=± 3 √√3; .. x = 4 (4 ± 3 √3); y= prove that xy+yz + zx = x2 (b− y) = ay (y—n), y' (a-x)=bx (x – n). x2 + xy + y2 = c2, x2 + x2 + z y2 + yz + z2 = a3, 1 √ { } (2a°b′ + 20°c® + 2c°a3 — a* − ba— and shew how to solve the equations. 5. Determine c so that 5x + 2y = c may have ten positive integral solutions excluding zero values, and c may be as great as possible. 7. Shew that if n and N are very nearly equal, N n+ N 4n + N+ n and that the error is approximately very nearly, (N-n)* 8n (N+n)3° 8. A man's income consists partly of a salary of £200 a year, and partly of the interest at 3 per cent. on capital, to which he each year adds his savings; his annual expenditure is less by £95 than five-fourths of his income; shew that whatever be the original capital its accumulated value will approximate to £6000. If the original capital be £1000, shew that it will be doubled in about thirty years; having given 10. If x be any prime number, except 2, the integral part of (1+2), diminished by 2, is divisible by 4x. 11. If any number of integers taken at random be multiplied together, shew that the chance of the last figure of their product being 5 continually diminishes as the number of integers multiplied together increases. 12. Two purses contain sovereigns and shillings; shew that if either the total numbers of coins in the two purses are equal, or if the number of sovereigns is to the number of shillings in the same ratio in both, then the chance of drawing out a sovereign is the same when one purse is taken at random and a coin drawn out as it is when the coins are all put in one purse and a coin drawn out. If neither of these conditions holds, the chance is in favour of the purse taken at random whenever the purse with the greater number of coins has the smaller proportion of sovereigns. LV. MISCELLANEOUS PROBLEMS. 748. We have already given in previous Chapters collections of problems which lead to simple or quadratic equations; we add here a few examples of somewhat greater difficulty with their solutions. 1. Each of three cubical vessels A, B, C, whose capacities are as 1 8 27 respectively, is partially filled with water, the quantities of water in them being as 1 : 2 : 3 respectively. So much water is now poured from A into B and so much from B into C as to make the depth of water the same in each vessel. After this 1284 cubic feet of water is poured from C into B, and then so much from B into A as to leave the depth of water in A twice as great as the depth of water in B. The quantity of water in A is now less by 100 cubic feet than it was originally. How much water did each of the vessels originally contain? Let x number of cubic feet in A originally; Now when the depth of the fluid is the same in all, it is clear that the quantities vary as the areas of the bases of the vessels, that is, are as 1 : 49. .. (since 6x is the total quantity) the quantity in A= 12x 27x and the quantities in B and C are 7 7 3x 7 Again, when the depth in A is twice that in B, A contains half as much as B. Now A contains x-100; .. B contains 2(x-100), and C' 2. Three horses A, B, C start for a race on a course a mile and a half long. When B has gone half a mile, he is three times as far ahead of A as he is of C. The horses now going at uniform speeds till B is within a quarter of a mile of the winning post, C is at that time as much behind A as A is behind B, but 1 th the distance between A and B is only 11 of what it was after B 1 rd 53 of had gone the first half mile. C now increases his pace by what it was before, and passes B 176 yards from the winning post, the respective speeds of A and B remaining unaltered. What was the distance between A and C at the end of the race? Let 11x= distance (in yards) between B and C at end of first mile, 33x B and A = .... Hence, after C increases his pace, the speeds of A, B, C will be proportional to 1320 + 30x, 1320, and 54 53 (1320 + 5x) respectively. |