Question 95.

4 add 14 equal 53 miles per hour, down the river; and 4 less 1 equal 24 miles per hour, up the river. Then 54 add 24 equal 8 miles per hour, by the two boats. Then say, as 8: 120 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 24 miles an hour, it will be thus: 5X13 miles by the boat from Wheeling, and 2×24➡6) miles by the boat going to Wheeling. Ans

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

Question 96.

1728 inches in a solid foot; then 1728×4➡6912 inc. ÷ 12➡ 576 feet. ,25,25X4, or,2500. Finch square will take 4 inches in length to make a cubic inch.

Question 97.

Her age 4×4-16; then say, as : 16 :: 1 to 36: then 36÷3=12: then say, as : 12 :: 1 to 42: then 42÷314 years. Ans.

PROOF. Her age 14; then 14×3=42, and 2 of 42=12, and 12×3-36, and 4 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 40X2=80, the whole diameter. Now, to find the circumference, say,

As 7: 22: 80 to 251 feet, the cir. Then 2514, (the fourth part of the cir.) equal 629 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. Aps.

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, 10÷÷÷6=1 and two-thirds per hour. E's and S's in 4 hrs. that is, 10-4-2 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 24 miles an hour. E's will go or travel alone one mile an hour; therefore S's horse will travel 1 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

Then say,

S.

RULE. From the centre to the given height being 1 semidiameter, mul. tiply the square of 13 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 X 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.

As 13 : .1

Question 105.

1 to 3. As 8 : 1 :: 1 to 1. Then -1; then as 18: 1 :: 1 to 20 days by B.

T04

5

Question 106.

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. 4 lbs. cloves. Cloves 10 lbs. 63 lbs. sugar, at 5 cts. a pound. Then 13×4,75 X63 X5 X100= 1945125. Then 6 multiply 19 multiply 10, equal 1140. Then 1945125÷÷ 1140-1706,25 cts. 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 11 cents in the same time; and in twice that time, the 58 persons will spend 103 dolls. 114 cts. (.) × 2206 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 overtook the clerk. The whole distance 400-2813 = 118 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 I dol. 37 cts. first cost, and the gain is 1 dol. 324 cts. Then say, as 137: 132 :: 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 peal shares are 4, 4, and 4. Now --; therefore say, As : 8000 :: to 7000 dolls. Ans.

Question 112.

8

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

Then say,

3×5×7=105, common denominator. Then say,

Then say, as 193: 15440 :: 70 to $5600 A's stock.

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 88750, B 5000; and C 2000, making the whole legacy 15750 dollars. Ans..

Question 115..

SX3-9, the square of A's distance; and 6×6=36, the square of B's distance. Then,

As 9 1 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 244.

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

Question 117.

As 1 90: 126 to 113 dollars 40 cts.

Then the first
Then say,

cost was $132-113,40= 18 dollars 60 cents.

As 90 : 1 :: 18,60 to 20 gallons of water. Ans.

Question 118.

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

Head

8

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

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

As 100

450

550

Question 119.

A's gain 550 dolls,
B's gain 450

100 diff.

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
58 dolls. 19 cts. Then 53,1914-45-8 dolls. 19 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.