Question 95.

4 add 14 equal 53 miles per hour, down the river; and 4 less 14 equal 24. miles per hour, up the river. Then 54 ald 24 equal 8 miles per hour, by the two boats Then say, as 8:1:: 20 to 24 hours. Then as the two boats will complete the 20 miles in 24 hrs. and the boat from Wheeling goes at the rate of 5 miles an hour, and the boat going to Wheeling, goes at the rate of 23 miles an hour, it will be thus: 54X132 miles by the boat from Wheeling, and 2×24=6} miles by the boat going to Wheeling. Ans.

PROOF. 13 add 64 equal 20 miles, the distance.

Question 96.

1728 inches in a solid foot; then 1728X4➡6912 inc. 12 = 576 feet. ,25X,25X4, or,2500. inch square will take 4 inches in length to

make a cubic inch.

Question 97.

Her age 4x4-16; then say, as 16 :: 1 to 36: then 36÷3-12: then say, as : 12 :: 1 to 42: then 42÷3— 14 years. Ans.

PROOF. Her age 14; then 14x8=42, and 2 of 42-12, and 12×3=36, and of 36=16, whose square root is 4.

Question 98.

RULE. To half the sum add half the difference, and that sum is the greatest number; and from half the sum take half the difference, and the remainder is the least number.

48 (sum)÷2=24, half sum; then 14 (diff.) ÷2=7, half the difference; Then 24+7-31, greater number; and 24-7—17, less number. Ans.

Question 99.

This question is to measure a quadrant, or the fourth part of the circumference of a circle. The height of the signpost is the radius, or semidiameter of the circle; therefore it will be 40 feet, equal to the semidiameter; and 40×2=80, the whole diameter. Now, to find the circumference, say,

As 7: 22:: 80 to 2514 feet, the cir. Then 2514, (the fourth part of the cir.) equal 624 feet, the curve line formed by the top-end of the sign. post, in falling. Now, as 1 : 1 :: 629 to 62 days 20 hrs. 344 min. Ans.

'Then say, as 4715 4715

4715

4713

10: 1988 to $4 21298 cts. for A. 10: 1858 to 3 942 290 cts. for B. 10 :: 869 to 1 841440 cts. for C. Proof $10 00

Question 101.

Suppose the distance 10 miles, from Washington to Cambridge; then Endley's horse will be 10 hours in hauling the load of salt, that is, a mile an hour. Endley's and Laurence's will haul the same load in 6 hours, that is, a mile and two-thirds per hour by the two horses. Endley's and Skinner's will haul the same in 4 hours, that is, 24 miles per hour by them two. Now, to find the rate per hour for Laurence's horse, and Skinner's.

Endley's in 10 hours, that is, 10-101 mile per hour. E's and L's in 6 hours, that is, 106=1 and two-thirds per hour. E's and S's in 4 hrs. that is, 104-24 per hour. Now Endley's and Laurence's went a mile and two-thirds an hour, and Endley's went alone a mile an hour; therefore Laurence's horse will go two-thirds of a mile an hour, and E's and S's travel 23 miles an hour. E's will go or travel alone one mile an hour; therefore S's horse will travel 14 miles an hour. Then the proportions are-Endley's one. mile, Laurence's two-thirds of a mile, Skinner's a mile and a half per hour. Then 1+1=3. Then say,

RULE. From the centre to the given height being 13 semidiameter, mul-tiply the square of 1 by the weight, and the product will be the answer. 1,5 X 1,5 X 400 900 lbs. Ans. Question 103.

180209, whose square root is 3, that is, 3 semidiameters from the earth's centre, which is equal to 4000 × 3 = 12000 miles from the earth's centre. Then 12000 less 4000, (the earth's semidiameter,) equal 8000 miles above the surface of the earth. Ans.

Question 104.

The proportion is as 1 is to 2, to lose half its weight; therefore it will be, As 1: 4000X4000 :: 2 to 32000000. Then 32000000=5656,85 miles from the centre of the earth. Then 5656,85 less 4000, (the earth's semidiameter,) equal 1656,85 miles above the surface, or in the air. Ans.

1 mile equals 320 rods; then 320 multiply 320-160-640 acres; and 320 multiply 4 multiply 10, equal 12800 rails: then 12800 multiply 12800÷640, equal 256000 rails, which will enclose 256000 acres, which is 20 miles square.

Z

Question 107.

Pepper 6 lbs. 13 lbs. ginger. Ginger 19 lbs. 44 lbs. cloves. Cloves
10 lbs. 63 lbs. sugar, at 5 cts. a pound. Then 13×4,75 × 53 X5 X100=
1945125. Then 6 multiply 19 multiply 10, equal 1140. Then 1945125÷
1140-1706,25 ets. which is 17 dolls. 64 cts. per 100 lbs. Ans.

Question 108.

As 13 is added to the number, it will be 45 + 13 = 58:
then say, as 45 80: 58 to 103 dollars 115 cents in the
same time; and in twice that time, the 58 persons will spend
103 dolls. 11 cts. (,)2=206 dolls. 223 cts. Then
say, as $206,223 cts.: 70 :: $252 to 85 dolls. 53 cts. 5+
mills per hhd. which x 2 = 171 dolls. 7 cts, nearly. Ans.

Question 109.

22×4 88 miles, the clerk is before the messenger; and
32-22-10 miles, the messenger gains on the clerk each
day. Then say, as 10: 1 :: 88 to 8 days. Then 8x
32-2813 miles, the messenger travelled, before he over-
took the clerk. The whole distance 400-2813=1183
miles this side of Philadelphia. Ans.

Question 110.

Three-eighths of 2 dolls. 20 cts. equal 824 ets. gained per yard, when sold
at 2 dolls. 20 cts. Then 2 dolls. 20 cts. less 82 cts. equal 137 cts. per yard,
first cost. Raised to 2 dolls. 70 cts. from which take 1 dol. 37 cts. first cost,
and the gain is 1 dol. 321⁄2 cts. Then say, as 137: 1323 :: 100 to 96 dolls.
36 cts. 3 mills per cent. Ans.

Question 111.

The son's share to the mother's, is as 2 to 1, and the
mother's to the daughter's, is as 2 to 1; therefore their 3
shares are 4, 2, 1, which added is 7. Then of course their
real shares are,, and . Now --; therefore say,
: 8000 :: to 7000 dolls. Ans.

8

AS T

4.

Question 112.

8

Suppose A's gain to be 2; then, by the question, B's is 3,
and C's will be 4, making 9, the sum.

As 9

:

936

Then say,

:: 2 to 208 A's gain. Ans.
: 936 :: S to 312 B's

9

Then say,

of 105

- 70 A's

3x5x7=105, common denominator.

As 3 : 2 :: 1 to; then

Now find her age by this proportion: As 3: 1 :: 45 to 15 years, her age, when married. Then 45 + 15 = 60, his age, when married 15 years. Then find her age by the following proportion:

As 2 : 1 :: 60 to 30 years, her age, when married 15 years,
Therefore, when married, he was 45, and his wife 15. Ans.

As 2 : 5 ::

As 4 : 7 ::

Question 114.

2000 to 5000 for B; then say,

5000 to 8750 for A.

Then A got $8750, B 5000, and C 2000, making the whole legacy 15750 dollars. Ans.

Question 115.

3X39, the square of A's distance; and 6×6=36, the square of B's distance. Then,

As 91: 36 to 4. A is 4 times as warm as B. Ans.

Question 116.

The velocity of the ram is 24 feet, and the weight of the ball 42 lbs. Compounded make a fraction of 24-4.

Then it will be 4×8000=45714 feet in a second. Ans.

Question 117.

As 190: 126 to 113 dollars 40 cts. Then the first
cost was $132-113,40 = 18 dollars 60 cents. Then say,
As 90 : 1 :: 18,60 to 20 gallons of water. Ans.

Question 118.

1st. Suppose the body was 40 inc. long, then the tail must be 20+8=28.
Length of tail 28 inches.

Head

2nd. Suppose the body was 60 inc. long, then the tail must be 30+8=38.-

64 inc. whole length of the fish. Ans. ·

PROOF. The body was 32 inches long, just as long as the head and tail.
The head 8 ine. and tail 24 inc. making 32 inc.

As 100

192 :: 550 to 1056 A's÷8=132 dolls.

= 108

As 100 192 :: 450 to 864 B's 8
A's was valued at $132, and B's at $108 per hhd. Ans.

Question 120.

180÷4-45, and 100-6=94; then 94: 100 :: 50 to
53 dolls. 1914 cts. Then 53,1914-45-8 dolls. 1934 cts.
Then say, as 53,1914: 8,1914 :: 100 to 15 dollars 39+
cents per cent. loss. Ans.

I omitted bringing in the fractions in the 1st and 2nd numbers.