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11. *Given x+y+z= 13, x +y tu 17, u t-% tu - 18, and y +z+u ==21, to find x, y, z, and u.

Ans. X

2, y = 5,2 6, and u= 10.

=

=

MISCELLANEOUS QUESTIONS,

PRODUCING SIMPLE EQUATIONS.

ODUCING The usual method of resolving algebraical questions, is first to denote the quantities, that are to be found, by x, y, or some of the other final letters of the alphabet; then, having properly examined the state of the question, perform with These letters, and the known quantities, by means of the common signs, the same operations and reasonings, that it would be necessary to make if the quantities were known, and it was required to verify them, and the conclusion will give the result sought.

Or, it is generality best, when it can be done, to denote only one of the unknown quantities by x or y, and then to determine the expression for the others from the nature of the question ; after which the same method of reasoning may be followed, as above. And, in some cases, the substituting for the sums and differences of quantities, or availing ourselves of any other mode, than a proper consideration of the question may suggest, will greatly facilitate the solution.

1. What number is that whose third part exceeds its fourth part by 16 ?

Let x = the number required.
1
1

1
Then its 5 part will be

part will be 5*, and its part ja. 3

4 1 1 And therefore

3a - 1

x == 16, by the question,

3 That is, a a * = 48, or 4x

= 192, 4 Hence x = 192, the number required. 2. It is required to find two numbers such, that their sum shall be 40 and their difference 16. Let x denote the least of the two numbers required,

Then will x + 16 = the greater number,
And x + x + 16 =40, by the question,

1

4

3x =

* This can be resolved by proceeding after the same manner as equations involving three unknown quantities; but the resolution of it may be greatly facilitated; by introducing into the calculation, beside the principal unknown quantities, a new unknown quantity arbitrarily assumed, such as, for example, the sum of all the rest; and when a little practised in such calculations, they become easy.

Let x =

24 That is, 2x = 40 - 16, or x = 12 = leasť number.

2 And x + 16 =12 +16=28= the greater number required.

3. Divide 10001. between A, B, and c, so that a shall have 721. more than B, and c 1001. more than A.

B's share of the given sum,
Then will : + 72 A's share,

And x + 172 = c's share,
Hence their sum is x + 2 + 72 + + 172,
Or 3x + 244 = 1000, by the question,
' That is, 3x = 1000 244 756,

756
Or x =

= 2521, = B.'s share,

3
Hence x + 72 = 3241. A's share,
And x + 172 = 4241.= c's share,
Also, as above, 2521. B's share.

Sum of all = 10001, the proof. 4. It is required to divide 1000l. between two persons, so that their shares of it shall be in the proportion of 7 to 9

Let x = the first person's share,
Then will 1000 x = second person's share,
And 2 : 1000 x ::7:9, by the question,
That is, 9x = (1000 -- x) X 7= 7000 - 73,

7000 Or 9x +7=7000, or x = = 4371. 10s. = 1st share,

16 and 1000 20 = 1000

. 4371. 10s. 5621. 10s. = 2d share. 5. The paving of a square court with stones, at 2s. a yard, will cost as much as the enclosing it with pallisades, at 5s. a yard; required the side of the square.

length of the side of the square sought,
Then 4x number of yards of enclosure,
And 22 number of yards of pavement,
Hence 4x X 5 = 20x = price of enclosing it,
And 22 x 2 2x3 = the price of the paving,

Therefore 2x2 = 20x, by the question,
Or, 2x 20, and x = 10, the length of the side required.

6. Out of a cask of wine, which had leaked away a third part, 21 gallons were afterwards drawn, and the cask being then gauged, appeared to be half full; how much did it hold?

Let x = the number of gallons the cask is supposed to have held.

1 Then it would have leaked away 5 gallons.

*

Let a =

Whence there had been taken out of it, altogether,

1
21 + 5 gallons,

3x = x by the question,

1 1 And therefore 21 + x =

3 2

3 That is, 63 + x = X, or 126 + 2x = 3.x,

Consequently, 3x – 2x = 126, or x = 126, the number of gallons required. 7. What fraction is that, to the numerator of which if 1 be

1 added, its value will be but if 1 be added to the denomi

3)

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y+1

Let the fraction required be represented

y' +1 1

1 Then

and

by the question.

4 39 Y Hence 3x +3=y, and 4x = y + 1, or x =

4
+
Therefore 3

+3=y, or 3y + 3 + 12 = 4y.
y +1

15 +1 16 That is, y = 15, and x

4, 4

4

4

4 Whence the fraction that was to be found is

15 8. A market-woman bought in a certain number of eggs at 2 a penny, and as many others at 3 a penny, and having sold them out again, altogether, at the rate 5 for 2d., found she had lost 4d. ; how many eggs had she ? Let x = the number of eggs of each sort,

1 Then will the price of the first sort,

2 * =

1 And 3 the price of the second sort,

40 But 5:2:: 2x (the whole number of eggs) :

5' 400 Whence the price of both sorts, when mixed together

5 at the rate of 5 for 2d

=

1 1 4x And consequently -x + 52 = 4, by the question.

2 3 5 That is, 15x + 10x 24x = 120, or x 120, the number of eggs of each sort, as required.

9. If a can perform a piece of work in 10 days, and B in 13; in what time will they finish it, if they are both set about it together ?

Let the time sought be denoted by X. Then the part done by A in one day,

10

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Consequently + = 1 (the whole work),

10 13 That is, 13x + 10x = 130, or 23x = 130.

130 15 Whence a

5 days, the time required.

23 23 10. If one agent A, alone, can produce an effect e, in the time a, and another agent B, alone in the time b; in whai time will both of them together produce the same effect ?

Let the time sought be denoted by X,
Then a:e::*: -= part of the effect produced by A.

exc

a

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Let 3

11. How much rye, at 4s. 6d. a bushel, must be mixed with 50 bushels of wheat, at 6s. a bushel, so that the mixture may be worth 58. a bushel ?

the number of bushels required.
Then 9x is the price of the rye in sixpences,
And 600 the prices of the wheat in ditto,
Also (50 + 2). X 10 the price of the wheat in ditto,
Whence 9x + 600 == 500 + 10X, by the question,

Or, by transposition, 10x - 9x = 600 - 500. Consequently x= 100, the number of bushels required.

12. A labourer engaged to serve for 40 days, on condition that for every day he worked he should receive 20d., but for every day he was absent he should forfeit 8d.: now, at the end of the time he had to receive 1l. lls. 8d.: how many days did he work, and how many was he idle ?

Let the number of days that he worked be denoted by a. Then will 40 a be the number of days he was idle, Also 20x the sum earned, and (40 — 2) X 8,

Or 320 – 8x the sum forfeited, Whence 20x — (320

(320 - 8x) == 380d. (=ll. 11s. 8d.), by the question, That is 20x - 320 + 8x

320 + 8x = 380, Or 28x = 380 + 320 = 700,

700 Consequently, x = = 25, the number of days he work

28 ed, and 40 x = 40 - 25 = 15, the number of days he was idle.

QUESTIONS FOR PRACTICE.

1. It is required to divide a line, of 15 inches in length, into two such parts, that one may be three fourths of the other.

Ans. 84 and 6%. 2. My purse and money together are worth 20s., and the money is worth 7 times as much as the purse, how much is there in it?

Ans. 173. 6d. 3. A shepherd being asked how many sheep he had in his flock, said, if I had as many more, half as many more, and 7 sheep and a half, I should have just 500; how many had he?

Ans. 197. 4. A post is one fourth of its length in the mud, one third in the water, and 10 feet above the water, what is its whole length?

Ans. 24 feet. 1

1 5. After paying away

of my money, and then of the remainder, I had 72 guineas left; what had I at first?

Ans. 120 guineas. 6. It is required to divide 3001. between A, B, and c, so that A may have twice as much as B, and c as much as A and B together.

Ans. A 1001., B 501., c 1501. 7. A person, at the time he was married, was 3 times as old as his wife: but after they had lived together 15 years, he was only twice as old; what were their ages on the weddingday?

Ans. Bride's age 15, bridegroom, 45.

4

5

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