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4. George divided 16 apples equally among 4 boys; how many did he give each?

5. Charles wishes to divide 24 oranges among 4 companions; how many can he give each?

6. A lady paid 24 shillings for 8 yards of satin; what did it cost her per yard?

7. A gentleman bought 4 yards of broadcloth for $20; what did he pay per yard ?

8. If 3 cents will buy a peach; how many can you buy for 9 cents ? how many for 12 cents ? how many for 15 cents ? how many for 18 cents ? how many for 21 cents ? how many for 24 cents ? how many for 27 cents? how many for 30 cents? how many for 33 cents? how many for 36 cents ?

9. If beef is 4 cents a pound; how many pounds can you buy for 8 cents ? how many for 12 cents ? how many for 16 cents? how many for 20 cents? how many for 24 cents ?

10. When stage fare is 5 cents a mile; how far can you be carried for 10 cents ? how far for 15 cents? how far för 20 cents ? how far for 25 Cents? how far for 30 cents ? bow far for 35 cents? how far for 40 cents ? 11. A teacher gave 48 cents to 8 boys; how many

did he 12, Twelve yards of broadcloth cost 60 dollars; what was the cloth a yard?

EXAMPLES-To be performed on the slate. 1. Divide 12 by 4

Dem. We first set down 4 Dividend,

our divisor, and at the right hand Divisor 4) 12 (3 Quotient.

write 12, our dividend, separated

from our divisor by a small ver12

tical line. We then inquire how 0

often 4 is contained in 12 ; and considering division to be exactly

the reverse of multiplication, we readily determine. In multiplication, we have two factors given to find their product and 4, our divisor, is one of those factors, and 12, our dividend, is the product arising from the multiplication of 4, by the other factor; consequently we must think of some figure by which we may multiply 4 to produce 12. The mind readily fixes itself on 3; as the other factor, for a quotient figure; because the divisor and the quotient. are the factors in multiplication, and the dividend the product. We then put 3 at the right of our dividend for a quotient, separated from the dividend by a small vertical line; we next multiply our divisor by 3, by saying 3 times 4. are 12; placing the product under

give each?

our dividend; we then subtract and find that we have no remainder,

PROOF.--According to the first method: Multiplicand, 4 DEM.—Here, we take the divisor and quoMultiplier; 3 tient, and multiply them together; which

clearly demonstrates the relation which this

rule bears to multiplication. In division, we Product, 12

only separate the factors and give them díffer

ent names; the 4 which we here have for a. multiplicand; we had before, for a divisor; and the 3, our multiplier, we had then for a quotient; they are niultiplied together in division the same as in multiplication, only under different names, producing the same product. When we divide by 4, our quotient is a fourth part of our dividend; and when we multiply 4 by 3, our multiplier is à fourth part of our product; consequently, our dividend and product must be alike; because 3 is the fourth part of either. Multiplying 4 by 3 produces 12; it is then plain, that 12 contains the 4, our multiplicand, 3 times; therefore when we divide by. 4, we must suppose the 12 to contain 4 three times: Proof.—According to the second method.

When we divide twelve by 4, we inquired how often 4, 4

was contained in 12, and we found it to be contained 3 times.

Consequently, if we put down the divisor three times and add, 4

our amount must equal the dividend; therefore, it is plain,

that division may be proved by adding the divisor as often as 12 the quotient expresses a unit; as our example shows by its pro

ducing the dividendo PROOF.--According to the third method.

Dem.-As division is a short way of performing subtrac4 tion, it is plain, that it may be proved by subtraction, or the

same work he performed by subtraction. In division we take 8. out the three i's from the dividend at one operation; conse4 quently the work is greatly abridged, especially, when our

quotient contains many figures, as our proof shows, which is 4

the manner of performing the work by subtraction, and illus174 trates the relation which this rule bears to subtraction. By 0

subtracting 12 in division, at once from our dividend, we liad no remainder; and by subtracting 4 three times, we have no

remainder, so that the same result is produced in both cases. But in performing this work by subtraction we would necessarily

lrave to count the number of subtractions, which is at once shown by ** our quotient in division. PROOF.--According to the fourth method. Dividend,

That division may be made to Divisor, 3) 12 (4 Quotient, prove itself is evident, because when

we divided 12 by 4, we found that 12

13 contained“4 three times; then it 0

follows, that 12 contains 3 four

times, as our proof shows; then it is E

12

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plain, if 4 can be subtracted from 12 three times, that 3 may be subtracted from 12 four tïies.

Here we offer an example where the dividend will have to be separated into different parts. 2. Divide 2460 by, 5.

Here we first assume two of the Dividend, Quo.,

left hand figures of our dividend, beDivisor, 5) 2460 (492

cause the left hand figure does not

contain the divisor; we then inquire 20

how often 5 is contained in 24, and 46

we suppose it contained 4 times, placing the 4 ät the right for a quotient;

we then say, 4 times 5 are 20, pla10

cing it under the assumed part of the 10

dividend; we next subtract 20 from

24; and to the right hand of the re0

myainder, we join 6, the next figure of our dividend, and then consider

how often can be subtracted from 46, and finding it to be 9 times, we then set 9 in the quotient, and multiply 5 by 9, which gives us 45, which we place under 46; we then subiract and find the remainder to be I, to which we join a cipher, the next figure of our dividend; and we again inguire how often 5iis contained in 10: we suppose it to be twice, setting 2 in the quotient; we then say 2 times 5 are ten, placing it under the remaining part of our dividend; then subtracting, we find it leaves nothing of our dividend; that is, we have no remainder. The work is now done, and the quotient shows how many times 5 may be subtracted from the dividend; for the operation is already performed in the shortest possible

wayi that is, by division.

Dem. Our divisor, 5, in our second example, expresses greater simple value than 2; the left hand figure of our dividend; consequently we cannot have in our quotient, units of the first order of our divi. dend; butówlien we assume the next figure of our dividend we have 24, which according to the laws of notation expresses 24 hundreds; we then can have units in our quotient of the second order of our dividend, which is hundreds; so our first quotient figure must express hundreds; because that part of the dividend producing the first figure of the quotient, is hundreds. We then place 4 in the quotient, without regard to the local part of the dividend producing it; because every figure at the right hand of that part of the dividend used, successively brings one in the quotient possessing the same local place of the part of the dividend producing it. By bringing down our next quotient figure to the remainder, we have 46 tens, according to the rules of notation, (for there is another figure to be brought down from our dividend,) in which the divisor is contained 9 times, which is nine tens; because the part of cur dividend producing it, is tens. When we join the next figure of our dividend to the remainder, we have ten units, consequently our next quotient figure, 2, must be 2 units; bcc cause the part of the dividend producing the 2, is units.

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NOTE.-By proving this example by multiplication the student will discover, that our remainders arise from the carriage in multiplication; considering our quotient and divisor as the two factors that have produced the dividend.

That no obscurity may shadow our work in this rule, we repeat the same example, with our dividend divided into its proper parts; and likewise separating our quotient, giving to each part its proper value.

Divisor, 5) 2460 Dividend.
The first part of our dividend is=2400

5X40Q=2000 400, the 1st quotient. 1st remainder= 400

add 60 The 2nd part of our dividend= 460 90 the 2nd quotient.

5X90= 450 2nd remainder- 10

add 0 The 3rd part of our dividend= 10 2, the 3rd quotient.

5X2= 10 492 Answer, or total Last remainder- 0

sum of the quo

tients. Here it is plain that our dividend can be resolved into 2400+6040. Although in our usual method of work, we would consider the first part of our dividend, 24 and our quotient, 4; yet our first quotient figure expresses 400, and the first part of our dividend 2400; and our remainder 460, as may be seen in the operation, and so with regard to the rest. 3. Divide 46346 by 6.

This is the shortest meDividend,

thod of proving this rule

The asteDivisor 6) 46346 (7724 Quotient. by addition.

risks shew the numbers 42*

which are to be added ; and the dotted lines, the

order in which they are to :42*

be added. .. 14

DEM.--This method of ...12*

proof is plain, when we

consider, that division is :.26

only a short way of per24

forming several subtrac

tions. Here, we add the 2*

remainder to the several Proof 46346

products or subtrahends,

which gives a sum equal to the dividend; which in subtraction, would be called the minuend; the remainder added to the last subtrahend equals that part of the divi

:43

dend directly above, and so of the rest, till the amount equals the divi dend. When we consider that these subtractions have exhausted the dividend; (except the remainder 2) it is then plain, if these subtra, hends be collected together with the remainder, the sum must equai the dividend.

Case I.— The work is usually performed by what is called short division, when the divisor does not exceed 12, and the quotient is then placed directly under the dividend.

When the divisor exceeds 12,' it is called long division ; and the quotient must then be placed at the right hand of the dividend as in the examples given above.

EXAMPLES
Daividend

We first say 2 in 4 1. Divisor 2) 4 6 7 8 3 2

twice, placing the 2 un

der; then 2 in 6, three 2 3 39 16 Ans.

times; then 2 in 7, three 2

times; and 1 over, 10 Proof 46 7 8 3 2

which we suppose the 8 to be joined, and say ?

in 18 nine times; then 2 in 3 once, and we have 1 over; to which we suppose 2, the next figure, to be joined, and say, 2 in 12, six times.

Dem.-To suppose the next figure of the dividend to be joined to the remainder, wherever we have a remainder, is considering the remainder to be so many tens, which must always be correct; for the remainder arises from the carriage in multiplication, and as it takes 10 of an inferiour column or place, to make 1 in the next superiour; it follows, that 1 in a superiour column or place, is equal to 10 in the next inferiour; therefore we may always consider the remainder to be tens as long as there are figures at the right in the dividend. 2. 3) 4 6 7 8 9

NOTE.- Where there is Ans. 1 5 5 9 6-1 Rem.

a remainder, in the proof of 3

the work, it must be added

to the product.
Proof. 46 7 8 9
3.
4.

5. 4) 63 784 4) 683 46 5) 7 6 3 7 8 4 6 3 6.

7.

8. 6) 1 2 367 34 7 ) 8 6 6 3 2 4 6 8 ) 8 3 4 6 7

10. 9) 67 68 36 10 ) 8 6 7 0 3 24 11 ) 3 4 6 6 7 8

12. 12) 676 1 26

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11.

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