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PROPOSITION XV.-If four quantities be in continued proportion, the first is to the fourth, as the cube of the first is to the cube of the second ; that is, in the triplicate ratio of the first and second. Let
C = bd;
(2) multiplying (1) by (2), ac = id; whence, by Prop. II, a:d=b':c;
PROBLEMS IN PROPORTION.
To show some of the applications of the preceding principles, we give the following problems :
1. Find two numbers, the greater of which shall be to the less as their sum to 42, and as their difference to 6.
Let x = the greater, and y = the less. By the conditions, s
(1) x : y = x+y : 42,
2 : y = x -y :6. Prop. V, x+y: 42= x-y:6,
(3) Prop. III, c+g : 2-9=42 : 6,
(4) Prop. VII,
2a : 2 = 48 : 36, Prop. IX, :y=4:3,
(6) From (1) and (6), Prop. V, 4:3= x+y: 42,
2. Divide the number 14 into two such parts that the quotient of the greater divided by the less, shall be to the quotient of the less divided by the greater, as 16 to 9.
Let x = the greater, then 14—* = the less.
14 By the conditions,
= 16 : 9. Multiplying terms, Prop. IX, ** :(14—~) = 16:9, extracting square root,
3:14- = 4:3, by composition, Prop. VI, 2:14 = 4:7, dividing consequents,
3:2 = 4:1,
x = 8
14- = 6
3. There are three numbers in continued proportion; their sum is 52, and the sum of the extremes is to the mean as 10 to 3. Required the numbers.
Three numbers in continued proportion may be represented by X, xy, xy’; for we observe that the product of the extremes will then be equal to the square of the mean. Hence,
x+xy +xy' = 52, (1) By the conditions,
sy' +x : xy = 10: 3. (2) From (2),
y' +1:y= 10:3, (3) or,
y+1:2y = 10:6, by Prop. VII, y+2y+1:y-2y+1 = 16 : 4, taking square root,
y+1:y--1= 4:2, by Prop. VII,
27 : 2 = 6:2,
y:1= 3:1, whence,
Ans. and from (1),
4. The product of two numbers is 112; and the difference of their cubes is to the cube of their difference, as 31 to 3. What are the numbers ?
(1) By the conditions,
xy = 112, jo : (x-Y) = 31: 3.
(2) From (2), Prop. IX, x'+ry+y:x-2.ry+y = 31:3, (3) by Prop. VI,
3.xy: (x-y) = 28 : 3, (4) by substitution,
336 : (x-y)' = 28 : 3, (5) whence,
(x−y)' = 36, (6) or,
(7) From (1) and (7), we obtain x = 14, y = 8.
X-Y = 6.
5. What two numbers are those whose difference is to their sum as 2 to 9, and whose sum is to their product as 18 to 77 ?
Let x and y represent the numbers. By the conditions, sx-y: x+y = 2: 9,
(1) x+y :
XY = 18 : 77. From (1), Prop. VII,
23 : 2y = 11:7,
18y lly From (2), by substitution,
= 18 : 77,
(6) 7 7 by Prop. IX, 18y: 11y = 18 : 77,
(6) or, 18 : ily = 18 : 77,
(7) or, 1 : y = 1:7,
6. Two numbers have such a relation to each other, that if 4 be added to each, they will be in proportion as 3 to 4; and if 4 be subtracted from each, they will be to each other as 1 to 4. What are the numbers ?
Ans. 5 and 8. 7. Divide the number 27 into two such parts, that their product shall be to the sum of their squares as 20 to 41.
Ans. 12 and 15. 8. In a mixture of rum and brandy, the difference between the quantities of each is to the quantity of brandy, as 100 is to the number of gallons of rum; and the same difference is to the quantity of rum, as 4 to the number of gallons of brandy. How many gallons are there of each ? Ans. 25 of rum, and 5 of brandy.
9. There are two numbers whose product is 320; and the difference of their cubes is to the cube of their difference as 61 to 1. What are the numbers ?
Ans. 20 and 16. NOTE.-In the last example, put x+y= the greater, and c-y=the less.
10. Divide 60 into two such parts, that their product shall be to the sum of their squares as 2 to 5.
Ans. 40 and 20. 11. There are two numbers which are to each other as 3 to 2. If 6 be added to the greater and subtracted from the less, the sum
and the remainder will be to each other as 3 to 1. What are the numbers ?
Ans. 24 and 16.
12. There are two numbers which are to each other as 16 to 9, and 24 is a mean proportional between them. What are the numbers ?
Ans. 32 and 18.
13. The sum of two numbers is to their difference as 4 to 1, and the sum of their squares is to the greater as 102 to 5. What are the numbers ?
Ans. 15 and 9.
14. The number 20 is divided into two parts, which are to each other in the duplicate ratio of 3 to 1. Find the mean proportional between these parts.
Ans. 6. 15. There are two numbers in the proportion of 3 to 2; and if 6 be added to the greater, and subtracted from the less, the results will be as 9 to 4. What are the numbers ?
Ans. 39 and 26. 16. There are three numbers in continued proportion. The product of the first and second is to the product of the second and third, as the first is to twice the second; and the sum of the first and third is 300. What are the numbers ?
Ans. 60, 120 and 240. 17. The sum of the cubes of two numbers is to the difference of their cubes, as 559 to 127; and the square of the first, multiplied by the second, is equal to 294. What are the numbers ?
Ans. 7 and 6. 18. The cube of the first of two numbers is to the
of the second as 3 to 1, and the cube of the second is to the square of the first as 96 to 1. What are the numbers ?
Ans. 12 and 24. 19. Given the proportion (x-+1)*:(2-1)=2(x+1)': (x-1)', to find the value of x.
20. Prove that a:b=c:d, when (a+b+c+d) (a—6-c+d)=(a−6+c-d)(a+baad).
PERMUTATIONS AND COMBINATIONS.
335. The Permutations of things are the different results obtained by placing the things in every possible order. In forming permutations, all of the given elements, or a part only, may be taken at a time; but in any proposed system, the different results must contain the same number of things.
Thus, the permutations of the letters, a, b, c, taken two at a time,
ab, ba, ac, ca, bc, cb. The permutations of the same letters taken all at a time, are
cab, acb, abc, cba, bca, bac. NOTE.— The results obtained by permuting things, where less than all are taken at a time, are sometimes called variations or arrangements ; the word permutations would then be restricted to the case in which all the things are taken at a time.
336. The Combinations of things are the different collections that can be formed out of them, without regarding the order in which the things are placed, the same number of elements entering into all the results.
Thus, the combinations of the letters, a, b, c, taken two at a time,
ab, ac, bc. It will be observed that if the letters be regarded as factors, the combinations which may be formed by taking n at a time will constitute all the different products of the nth degree, of which the letters are capable.
337. To find the number of permutations of n things taken r at a time.
Suppose the things to be n letters, a, b, c, d.....
First :-If we take each of the n letters by itself, there will be in every case n -1 other letters, or n-1 reserved letters.
Now if to each of the n letters we annex each of the reserved letters successively, we shall form all the permutations with two