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Prob. 11. There are two numbers whose product is 120, if 2 be added to the lesser, and 3 subtracted from the greater, the product of the sum and remainder will also be 120. What are the numbers ?
Ans. 15, and 8. Prob. 12. A person bought a certain number of sheep for 1201. If there had have been 8 more, each would have cost him ten shilliogs less. How may sheep were there?
Aps. 40. Prob. 13. A Merchant sold a quantity of brandy for 391. and gained as much per cent as the brandy cost him. Wbat was the price of the brandy ?
Ans. 301, Prob. 14. Two partners, A and B, gained 181. by trade. A's money was in trade 12 months, and he received for his principal and gain 26l. Also, B’s money, wbich was 301. was in trade 16 months. What money did A put into trade ?
Ans. 201. Prob. 15. A and B set out from two towns which were at the distance of 247 miles, and travelled the direct road till they met. A went 9 miles a day; and the number of days; at the end of which they met, was greater by 3 than the number of miles which B went in a day. How many miles did each go?
Ans. A 117, and B 130 miles. Prob 16. A man playing at hazard won at the first throw, as much money as he had in his pocket ; at the second throw, he won 5 shillings more than the square root of what he then had ; at the third throw, he won the square of all he then had ; and then he had 1121. 16s. What had be at first ?
Ans. 18 shillings. Prob. 17. If the square of a certain number be taken from 40, aud ihe square root of this difference be increased by 10, and the sum multiplied by 2, and the product divided by the number itself, the quotient will be 4. Required the number,
Aps. 6. Prob. 18. There is a field in the form of a rectangular parallelogram, whose length exceeds 'the breadth by 18 yards ; and it contains 960 square yards. Required the lengtb and breadth.
Ans. 40 and 24 yards. Prob. 19. A person being asked his age, answered, if you add the square root of it to half of it, and subtract 12, there will remain nothing. Required his age.
Ans. 16. Prob. 20. To find a number, from the cube of which, if 19 be subtracted, and the remainder multiplied by that cube, the product shall be 216.
Ans. 3, or - 2
Prob. 21. To find a number from the double of which if you subtract 12, the square of the remainder, minus 1, will be 9 times the number sought.
Ans. U, or 3 Prob. 22. It is required to divide 20 into two such parts, that the product of the whole and one of the parts, shall be equal to the square of the other.
Ans. 105-10, and 30—10/5. Prob. 23. A labourer dug two trenches, one of which was 6 yards longer than the other, for 171. 16s., and the digging of each of them cost as many shillings per yard as there were yards in its length. What was the leogth of each ?
Ans. 10, and 16 yards. Prob. 24. A company at a tavern had 81. lös. to pay, but before the bill was paid, two of them sneaked off, when those who remained had each 10s. more to pay. were there in the company at first ?
Ans. 7. Prob. 25. There are two square buildings, that are paved with stones, a foot square each. The side of one building exceeds that of the other by 12 feet, and both their pavements taken together contain 2120 stones. What are the lengths of them separately ?
Ans. 26, and 33 feet. Prob. 26. In a parcel which contains 24 coins of silver and copper,
each silver coin is worth as many pence as there are copper coins, and each copper coin is worth as many pence as there are silver coins, and the whole is worth 18 shillings. How many are there of each ?
Ans. 6 of one, and 18 of the other. Prob. 27. Two messengers, A and B, were despatched at the same time to a place 90 miles distant ; the former of whom riding one mile an hour more than the other, arrived at the end of his journey an hour before him. At what rate did each travel per hour ?
Ans. A went 10, and B 9. miles Prob. 28. A man travelled 105 miles, and then found that if he had not travelled so fast by 2 miles an hour, he should have been 6 hours longer in performing the journey. How many miles did he go per hour.
Ans. 7 miles. Prob. 29. Bought two flocks of sheep for 65l. 13s., one containing 5 more than the other. Each sheep cost as many shillings as there were sheep in the flock. Required the number in each flock.
Ans. 23, and 28. Prob. 30. A regiment of soldiers, consisting of 1066 men, is formed into two squares, one of which has 4 men more in a side than the other. What oumber of men are in a side of nach of the squares ?
Ans. 21, and 25.
Prob. 31. What oumber is that, to which if 24 be added, and the square root of the sum extracted, this root shall be less than the original quantity by 18 ?
Ans. 25. Prob. 32. A Poulterer going to market to buy turkeys, mét with four flocks. In the second were 6 more than three times the square root of double the number in the first. The third contained three times as many as the first and second ; and the fourth contained 6 more than the square of one-third the number in the third ; and the whole number was 1938. How many were there in each flock ?
Ans. the numbers were, 18, 24, 120, and 1770, respectively.
Prob. 33. The sum of two numbers is 6, and the sum of their 5th-powers is 1056. What are the numbers.
Ans. 4, and 2. § II SOLUTION OF PROBLEMS PRODUCING QUADRATIC EQUATIONS,
INVOLVING MORE THAN ONE UNKNOWN QUANTITY.
430. It is very proper to observe, that the solution of a problem, producing quadratic equations, involving two unknown quantities, will somelimes be very much facilitated by assuming x equal to their half 'sum, and y equal to their hair difference; thea (Art. 102), x+y will denote the greater, and the lesser. The solution, according to this method of notation, will in general, be more simple than that which would have been found, if the two unknown quantities were represented by x and y respectively.
Problem. 1. Required two numbers, such, that their sum, their product, and the difference of their squares, may be all equal.
Let x ty=the greater ; and x-y= the lesser ;
::: 2x=(x+y).(+4y=x-ya, and 2x=<+y) - (x-7)=4xy, S
by the problem. From the 20 equation, y=;:.y=* : Now by substituting this value of y, in the first, we leave 2r=r;..x-2x=1, aud z=1+ V 5.
431. The preceding problem leads also to the solution of the following
Prob. 2. To find two numbers, such, that their sum, their product, and the sum their squares, may be all equal?
Let, in the last problem, x+y=the greater, and my the lesser ; thien, by the problem, 2x=y", and 2x=(x+y)+(x−y)=22+-2y*;
..=+y; but 2r=;
... by addition, 3.c =2x”, and x=;
..x+y=-1+1-3, and r--y=7V - 3.
432. It is sometimes more expedient to represent one of the unknown quantities by x, and the other by ry, (Art. 425) The utility of this method of notation for eliminating one of the uoknown quantities, will appear evident, from the solution of the following problem.
Prob. 3. What two numbers are those, whose sum multiplied by the greater is 77; and whose difference, multiplied by the lesser, is equal to 12 ?
Letey=the greater, and r=the lesser; then by the problem, *?y2 +2y=77, and aʻy-=12;
12 77 ...2% and 2:2=
12 lem ; Let y=* ; then x= -16; . x=+4, and cy=
y--1 +7. Hence the numbers, by taking the positive values, are 4 and 7. Let also y=; then rr; ..x=+iv?, and xy=*x+iV2=F4V2. Hence the irrational numbers, IV and 2, will also answer the conditions of the problem.
433. When a problem expresses more than two distinct conditions, which require to be translated into as many equations ; the solution cannot be obtained by means of quadratics, unless that some of the equations are of the first degree ; for the final equation resulting from the elimination of the unknown quantities will, in general, be of a higher degree than the second. There are, however, some particular cases in which the unknown quantities may be eliminated by certain artifices, (which are best learned by experience), so as to have the final equation of a quadratic form.
Prob. 4. It is required to find three numbers, such, that the product of the first and second, added to the sum of their squares, shall be equal to 37 ; the product of the first and third added to the sum of their squares, shall be equal to 49 ; and the product of the second and third added to the sum of their squares, shall be equal to 61. Let x=the first number, y=the second, and z=the third. Then, 29+y'+xy=37 ;)
2+z+uz=49; by the problem. and yo+z2+yz=61 ;) By subtracting the first equation from the second, z-yota
12 (z~-y)x=12; .. z+y+r=
2-५ By subtracting the second equation from the third, y. -2°+
12 (y - 3)2=12 ; :; y+x+2= 12 12
and y-=z; 2y=x+z. 2-Y yox By substituting 2y for +2, in equations (a) and (b), we 12
12 and 3y=;
; .: zy-y=4, and yo-yx=4; y2 + 4 y-4
; .. ? y
y Now, by substituting these values of x and in the first of the original equations, it becomes
ky. -4 +ya+y.
by reduction, Y
y 49 ya - =-16 ; and, by completing the square, 49 49 2401–192
49 47 ym 592+
36 and, by taking the positive sign, y=+4;
16-4 ... by taking y=4, x =
y Hence the three numbers sought are 3, 4, and 5, which are in arithmetical progression. This relation appears also evident from the result 2y=x+2, found in the beginning of the solution.
-37 ; .'.
Prob. 3. There are three numbers, the difference of whose differen: