Examples of putting Questions into Equations. and the distance which he goes is obtained from the pro portion 5:x+8=311: distance gone by Ist courier; whence, by art. 88, distance gone by 1st courier = 용 (x + 8). The distance gone by the second courier is obtained from the proportion whence 3:x 221: distance gone by 2d courier ; distance gone by 2d courier = x. But as both couriers go the same distance, the required equation is (x+8)=x. 4. A courier went from this place, n days ago, at the rate of a miles a day. Another has just started, in pursuit of him, at the rate of 6 miles a day. In how many days will the second courier overtake the first? Ans. If x = the required number of days, the required equation is 5. A regiment marches from the place A, on the road to B, at the rate of 7 leagues every 2 days; 8 days after, another regiment marches from B, on the road to A, at the rate of 31 leagues every 6 days. If the distance between A and B is 80 leagues, in how many days after the departure of the first regiment will the two regiments meet? Ans. If x = the required number of days, the required equation is +x+(x8) = 80. Examples of putting Questions into Equations. 6. A hostile corps has set out two days ago from a certain place, and goes 27 miles daily. Another corps wishes to march in pursuit of it from the same place, and so quickly that it may reach the other in 6 days. How many miles must it march daily to accomplish it? Ans. If x = the required number of miles, the required equation is 6 x 216. 7. From two different sized orifices of a reservoir, the water runs with unequal velocities. We know that the orifices are in size as 5: 13, and the velocities of the fluid are as 8:7; we know farther, that in a certain time there issued from the one 561 cubic feet more than there did from the other. How much water, then, did each orifice discharge in this space of time? Solution. Let x = the quantity discharged by the first orifice.. As the size of the second orifice is ths of that of the first, the water discharged from the second orifice, if it flowed at the same rate, would be But as the water flows from the second orifice with a velocity ths of that which it should have to discharge in the given time, its actual discharge must be 8. A dog pursues a hare. When the dog started, the hare had made 50 paces before him. The hare takes 6 paces to the dog's five; and 9 of the hare's paces are equal to 7 of the dog's. How many paces can the hare take before the dog catches her? Examples of putting Questions into Equations. Ans. If x = the required number of paces, the required equation is 9. A work is to be printed, so that each page may contain a certain number of lines, and each line a certain number of letters. If we wished each page to contain 3 lines more, and each line 4 letters more, then there would be 224 letters more on each page; but if we wished to have 2 lines less in a page, and 3 letters less in each line, then each page would contain 145 letters less. How many lines are there in each page? and how many letters in each line? Solution. Let x = the number of lines in a page, and we shall have z y = the number of letters in a page. But if there were 3 lines more in a page, and 4 letters more in a line, the number of letters in a page would be (x+3) (y+4)=xy+4x+3y+12, which exceeds the required number of letters in a page by 4x+3y+12 ; whence we have for one of the required equations 4x+3y+12 = 224 ; and, in the same way, the condition, that 2 lines less in a page and 3 letters less in a line make 145 letters less in a page, gives the equation or xy- (x-2) (y - 3) = 145 ; 3x+2y-6= 145. 10. Three soldiers, in a battle, make $96 booty, which they wish to share equally. In order to do this, A, who made most, gives B and C as much as they already had; in Examples of putting Questions into Equations. the same manner, B next divided with A and C, and after this, C with A and B. If, then, by these means, the intended equal division is effected, how much booty did each soldier make? the required equations are x+y+z=96 11. A certain number consists of three digits, of which the digit occupying the place of tens is half the sum of the other two. If this number be divided by the sum of its digits, the quotient is 48; but if 198 be subtracted from it, then we obtain for the remainder a number consisting of the same digits, but in an inverted order. What number is this? Ans. If x = the digit which is in the place of units, y = that in the place of tens, z that in the place of hundreds. The number is = 100 z + 10 y + x, and the required equations are 100 z+10 y+x-198=100 x + 10 y + z. 12. A person goes to a tavern with a certain sum of money in his pocket, where he spends 2 shillings; he then borrows as much money as he had left, and going to another tavern, he there spends 2 shillings also; then borrowing again as much money as was left, he went to a third tavern, Examples of putting Questions into Equations. where likewise he spent 2 shillings, and borrowed as much as he had left; and again spending 2 shillings at a fourth tavern, he then had nothing remaining. What had he at first? Ans. If x = the shillings he had at first, the required equation is 8x- 30 0. 13. A person possessed a certain capital, which he placed out at a certain interest. Another person, who possessed $10 000 more than the first, and who put out his capital 1 per cent. more advantageously than the first did, had an income greater by $ 800. A third person, who possessed $ 15 000 more than the first, and who put out his capital 2 per cent. more advantageously than the first, had an income greater by $ 1500. Required the capitals of the three persons, and the three rates of interest, Ans. If x = the capital of the first, y = his rate of interest per cent. the required equations are 10 000 y + x + 10 000 15 000 y + 2 x + 30 000 = 800, 1500. 14. A person has three kinds of goods, which together cost $2304. The pound of each article costs as many twenty-fourths of a dollar as there are pounds of that article; but he has one third more of the second kind than he has of the first, and 34 times as much of the third as he has of the second. How many pounds has he of each article? Ans. If x = the number of pounds of the first, the required equation is 24x2+2x2+x2=230 24 |