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Let x represent a side of the greater, and y a side of the less gar den; then

x2 and y2 represent the respective areas of the two gardens. The Equations then are

x2+y2=1025;

x-y=5.

From these Equations, a will be found 25 rods; and y=20 rods.

22. Find two numbers such, that their sum, their product, and the difference of their squares shall all be equal to one another."

Let x+y represent the greater number, and x-y the less; then 2x represents their sum;

x2-y2 represents their product; and

4xy represents the difference of their squares.

By the conditions of the problem, the Equations are

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23. A merchant received $12 for a quantity of linen, and an equal sum, at 50 cents less per yard, for a quantity of calico, which exceeded the quantity of linen by 32 yards. What was the quantity of each?

Let x represent the number of yards of linen, and y the number of yards of calico; then

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12X

of a dollar, by the conditions of the

12 1

y=x+32.

From these Equations, x=16, and y=48 yards.

x

24. Find two numbers whose sum multiplied by the greater shall be equal to 192, and whose difference multiplied by the less shall be equal to 32.

Let x represent the greater, and xy the less number; then

x(x+xy)=192;
xy(x-xy)=32.

192y i+y

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Developing the first member of the second Equation, x2y-x2y2-32.

Substituting for 22 in this Equation, its value as found above,

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Clearing this Equation of fractions by multiplying it by the least common multiple of the denominators,

1

192y-192y2=32+32y; hence y=

3

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Since x2=

192 1+y'

we have x2=192÷÷÷14=144; hence x=12.

Since the less number is represented by xy, we have for its value, =4.

1

3

25. Three merchants gained $1444; of which their respective shares were such that B's added to the square root of A's, made $920; but if added to the square root of C's, it made $912. What was the share of each?

Let x represent A's share, and y B's share; then -x-y represents C's share.

1444.

We shall then have the following Equations:

y+√x=920;

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y+√1444-x-y=912.

Transposing y in the first Equation, and squaring, x= 846400-1840y+y2.

Proceeding in a similar manner with the second Equation, 1444-x-y-831744-1824y+y2.

Adding together the last two Equations,

1444-y=1678144-3664y+2y2.

for

This Equation will give y-$900; and substituting this value of y y in the third Equation, we find x=$400. Hence, C's share was 1444-900-400-$144.

26. The sum of three numbers in harmonical progression is 13, and the product of the two extremes is 18. What are the numbers?

Let x and y, respectively, represent the two extremes; then

2.xy

x+y

represents the mean term, (184).

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(x+y)2+2xy=13(x+y).

Substituting for xy in this Equation its value from the second Eq'n, (x+y)2+36=13(x+y).

This Equation is quadratic with reference to x+y, and will give x+y=9; hence x-9-y.

Substituting this value of x in the second Equation, we find y=3;
2×6×3
-4, is the value of the mean term.
6+3

then x 183=6; and

27. There is a rectangular field whose length is to its breadth as 4 to 3. A part of this field, which is equal to of the whole, being in meadow, there remain for ploughing 1296 square rods. What are the dimensions of the field?

Let a represent the length, and y the breadth; then xy represents the area of the field.

We shall then have for the first Equation

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And since is in meadow, there remains for ploughing; then

3xy

4

1296- =

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from which, and the first Equation, we shall find 48 rods, and y= 36 rods.

28. The sum of three numbers in geometrical progression is 21, and the sum of their squares 189. What are the numbers?

Let x and y represent the extremes; then
√xy represents the mean term, (189).
By the conditions of the problem,

x+√xy+y=21;
x2+xy+ y2=189.

Dividing the second Equation by the first,

x−√xy+y=9. Subtracting this Equation from the first,

36

2xy=12; hence √xy=6, and x=

y

Substituting these values of x and Vay in the first Equation,

36

+6+y=21.

y

36

From this Equation, y=12; hence =3, the value of x.

12

Then. 3x12=6, the mean term.

29. A and B set out from two places which are distant 110 miles, and traveled towards each other. A went 5 miles an hour; and the number of hours in which they met was greater, by four, than the number of miles C went per hour. What was B's rate of traveling?

Let x represent the number of miles B went per hour; then
x+4 represents the number of hours each was traveling;
5(x+4) represents the number of miles A traveled; and
x(x+4) represents the number of miles B traveled.
The Equation then is

5(x+4)+x(x+4)=110; whence x6 miles.

30. The arithmetical mean between two numbers exceeds the geometrical mean by 13, and the geometrical mean exceeds the harmonical mean by 12. What are the numbers?

Let x and y represent the numbers; then

x+y

represents the arithmetical mean, (179);

2

√xy represents the geometrical mean, (189); and
2xy
represents the harmonical mean, (184).

x+y

By the conditions of the problem,

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x+y

2

+12.

√xy=

Clearing this Equation of fractions, and squaring both sides,

4xy=(x+y)2-52(x+y)+676.

- 13.

Equating the two values of √y from the second and third Equations,

2xy x+y

+12=

x+y

2

-13.

Clearing this Equation of fractions, transposing, &c.,

4xy=(x+y)2—50(x+y).

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