PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 149 Example 5. The middle digit of a number between 100 and 1000 is zero, and the sum of the other digits is 11. If the digits be reversed, the number so formed exceeds the original number by 495 find it. Let x be the digit in the units' place, then, since the digit in the tens' place is 0, the number will be represented by 100y+x. [Art. 76, Ex. 4.] And if the digits are reversed the number so formed will be represented by 100x+y. Again, since the sum of the digits is 11, and the middle one is 0, we have 10. x+y=11. From (1) and (2) we find x=8, y=3. Hence the number is 308. EXAMPLES XX. (2). 1. Find two numbers whose sum is 34, and whose difference is 2. The sum of two numbers is 73, and their difference is 37; find the numbers. 3. One third of the sum of two numbers is 14, and one half of their difference is 4; find the numbers. 4. One nineteenth of the sum of two numbers is 4, and their difference is 30; find the numbers. 5. Half the sum of two numbers is 20, and three times their difference is 18; find the numbers. 6. Six pounds of tea and eleven pounds of sugar cost $5.65, and eleven pounds of tea and six pounds of sugar cost $9.65. Find the cost of tea and sugar per pound. 7. Six horses and seven cows can be bought for $250, and thirteen cows and eleven horses can be bought for $461. What is the value of each animal ? 8. A, B, C, D have $290 between them; A has twice as much as C, and B has three times as much as D; also C and D together have $50 less than A. Find how much each has. 9. A, B, C, D have $270 between them; A has three times as much as C, and B five times as much as D; also A and B together have $50 less than eight times what C has. Find how much each has. 10. Four times B's age exceeds A's age by twenty years, and one third of A's age is less than B's age by two years: find their ages. 11. One eleventh of A's age is greater by two years than one seventh of B's, and twice B's age is equal to what A's age was thirteen years ago: find their ages. 12. In eight hours A walks twelve miles more than B does in seven hours; and in thirteen hours B walks seven miles more than A does in nine hours. How many miles does each walk per hour? 13. In eleven hours C walks 12 miles less than D does in twelve hours; and in five hours D walks 31 miles less than C does in seven hours. How many miles does each walk per hour? 14. Find a fraction such that if 1 be added to its denominator it reduces to, and reduces to on adding 2 to its numerator. 15. Find a fraction which becomes on subtracting 1 from the numerator and adding 2 to the denominator, and reduces to on subtracting 7 from the numerator and 2 from the denominator. 16. If 1 be added to the numerator of a fraction it reduces to ; if 1 be taken from the denominator it reduces to : required the fraction. 17. If be added to the numerator of a certain fraction the fraction will be increased by, and if be taken from its denominator the fraction becomes : find it. 18. The sum of a number of two digits and of the number formed by reversing the digits is 110, and the difference of the digits is 6: find the numbers. 19. The sum of the digits of a number is 13, and the difference between the number and that formed by reversing the digits is 27: find the numbers. 20. A certain number of two digits is three times the sum of its digits, and if 45 be added to it the digits will be reversed: find the number. 21. A certain number between 10 and 100 is eight times the sum of its digits, and if 45 be subtracted from it the digits will be reversed: find the number. 22. A man has a number of silver dollars and dimes, and he observes that if the dollars were turned into dimes and the dimes into dollars he would gain $2.70; but if the dollars were turned into half-dollars and the dimes into quarters he would lose $ 1.30. How many of each had he? 23. In a bag containing black and white balls, half the number of white is equal to a third of the number of black; and twice the whole number of balls exceeds three times the number of black balls by four. How many balls did the bag contain ? zero. PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 151 24. A number consists of three digits, the right hand one being If the left hand and middle digits be interchanged the number is diminished by 180; if the left hand digit be halved and the middle and right hand digits be interchanged the number is diminished by 454. Find the number. 25. The wages of 10 men and 8 boys amount to $22.30; if 4 men together receive $3.40 more than 6 boys, what are the wages of each man and boy? 26. A grocer wishes to mix sugar at 8 cents a pound with another sort at 5 cents a pound to make 60 pounds to be sold at 6 cents a pound. What quantity of each must he take? 27. A traveller walks a certain distance; had he gone half a mile an hour faster, he would have walked it in four-fifths of the time: had he gone half a mile an hour slower, he would have been 2 hours longer on the road. Find the distance. 28. A man walks 35 miles partly at the rate of 4 miles an hour, and partly at 5; if he had walked at 5 miles an hour when he walked at 4, and vice versâ, he would have covered two miles more in the same time. Find the time he was walking. 29. Two persons, 27 miles apart, setting out at the same time are together in 9 hours if they walk in the same direction, but in 3 hours if they walk in opposite directions: find their rates of walking. 30. When a certain number of two digits is doubled, and increased by 10, the result is the same as if the number had been reversed, and doubled, and then diminished by 8; also the number itself exceeds 3 times the sum of its digits by 18: find the number. 31. If I lend a sum of money at 6 per cent., the interest for a certain time exceeds the loan by $100; but if I lend it at 3 per cent., for a fourth of the time, the loan exceeds its interest by $425. How much do I lend? 32. A takes 3 hours longer than B to walk 30 miles; but if he doubles his pace he takes 2 hours less time than B: find their rates of walking. CHAPTER XXI. INDETERMINATE AND IMPOSSIBLE PROBLEMS. NEGATIVE RESULTS. INDETERMINATE AND IMPOSSIBLE PROBLEMS. 165. By reference to Art. 153 it will be seen that a single equation involving two unknown quantities is satisfied by an indefinitely great number of sets of values of the unknowns involved, and that it is essential to have as many equations expressing different, or independent conditions, as there are unknown quantities to be determined. If the conditions of a problem furnish a less number of independent equations than quantities to be determined, the problem is said to be indeterminate. If, however, the conditions give us a greater number of independent equations than there are unknown quantities involved, the problem is impossible. Suppose the problem furnishes 3x+y=10, From (1) and (2) we obtain x=5 and y=-5. From (2) and (3) we obtain x=2 and y=1. It is evident that these values cannot all be true at the same time, hence the problem is impossible. NEGATIVE RESULTS. 166. A is 40 years old, and B's age is three-fifths of A's. When will A be five times as old as B? Let x equal the number of years that will elapse. Then by the conditions or Therefore 40+x=5(24+x). 40+x=120+5, x= -20. According to this analysis, A will be five times as old as B in -20 years. The meaning of this result ought to be at once evident to a thoughtful student. Were the result any positive number of years we should simply have to count that number forward from the present time (represented by the word "is" in the problem); manifestly then the 20 years refers to past time. Hence the problem should read "A is 40 years old, and B's age is three-fifths of A's. When was A five times as old as B?" Suppose the problem read A is 40 years old, and B's age is three-fifths of A's. Find the time at which A's age is five times that of B. Let us assume that x years will elapse. Then by the conditions 40+x=5(24+x); .. x=-20. Interpreting this result as above, we see that we should have assumed that x years had elapsed. The student will notice that the word "will" in the first statement suggested that we should assume x as the number of years that would elapse, and that the negative result showed a fault in the enunciation of the problem; but that the problem as given in the next discussion permitted us to make one of two possible suppositions as to the nature of the unknown quantity, so that the negative result indicates simply a wrong choice. Hence in the solution of problems involving equations of the first degree negative results indicate (1) A fault in the enunciation of the problem, or (2) A wrong choice between two possible suppositions, as to the nature of the unknown quantity, allowed by the problem. Generally it will be easy for the student to make such changes as will give an analogous possible problem. 167. If we were to write the largest possible number, thus: 792481127 putting down any numerals, at random or not, and continuing the writing forever before placing a decimal point, we should have a number called infinity, and represented by the character |