MATHEMATICAL PROBLEMS.

Art. 305.-PROB. I. The sum and difference of two numbers given, to find those numbers.

RULE.

Subtract the difference from the sum, and divide the remainder by 2. The quotient will be the smaller number. Then add the given difference to the smaller number, and this sum will be the larger number.

EXAMPLE.

An assembly of 344 persons is convened in two rooms, one of which has 142 persons more in it than the other. How many are in each ?

Operation.

344-142-202; then 202-2=101 persons, in one room; then 101+142=243, in the other.

Art. 306.-PROB. II. The sum of two numbers, and the difference of their squares given, to find those numbers.

RULE.

Divide the difference of their squares by the sum of the numbers, and the quotient will be their difference. We then have their sum and difference, to find each number, by Prob. I.

EXAMPLE.

A. and B. played at marbles, having at first 14 each; but after playing several games, B. having lost some of his, would not play any longer, and it was found that the difference of the squares of what each then had, was 336. How many did B. lose?

Thus 336-14+14=12 difference; 14-half sum, and 12÷2=6 half difference. Then 14+6=20, A. retired with; and 14—6—8, B. retired with: then 14-8-6, B. lost.

Art. 307.-PROB. III. The difference of two numbers, and the difference of their squares given, to find those numbers.

RULE.

Divide the difference of their squares by the difference of their numbers, and the quotient will be their sum; then proceed by Prob. I.

EXAMPLE.

Said William to John, Father gave me $12 more than he gave Charles; and the difference of the squares of our separate parcels is 288. How much did he give each? Thus, 288÷12=24, the sum: then 24-12-2=6; then 12+6= $18, William's share; and 12-6=$6, Charles had given him. Art. 308.-PROB. IV. The sum of two numbers and their quotient given, to find those numbers.

RULE.

Add 1 to the quotient, and by this sum divide the sum of the two numbers: this will give the less number. Subtract the less number from the sum, and you will obtain the greater number.

EXAMPLES.

1. Divide 100 into two such parts, that if the greater be divided by the less, the quotient will be just 30.

Operation.

Thus 100÷30+1=31, the less part; then 100-3319634, greater.

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2. The sum of A. and B.'s ages is 45, and if you divide A.'s by B.'s the quotient will be 4. What is the age of each?

Ans. A.'s 36 years; B.'s 9 years.

Art. 309.-PROB. V. The difference of two numbers, and the quotient given, to find those numbers.

RULE.

The difference of the two numbers divided by the quotient less 1, will be the less number. Add the less number to the difference, and you will have the greater number.

A greyhound, in pursuit of a hare, ran three times as fast as the hare, and when he overtook the hare he had run 30 rods more than she. How many rods did each run?

Operation.

30-3-1-15 rods, the hare ran; then 15+30=45 rods, the greyhound ran.

Art. 310.-PROB. VI. To find the true weight of any quantity when weighed in each scale of a balance, whose beam is unequally divided.

RULE.

Take the square root of the product of the different weights, for the true weight.

A parcel of sugar weighs in one scale 25 lbs.; in the other 30 lbs. What was its true weight?

√25 X 30=27.856.

Art. 311.-PROB. VII. The base and perpendicular given, to find the hypotenuse.

RULE.

The square root of the sum of the squares of the base and perpendicular will be the hypotenuse.

This rule is illustrated by the following figure.

If the base of a right-angled triangle be 9 feet, and the perpendicular 12, what is the hypotenuse?

Art. 312.-PROB. VIII.

Given the base and sum of the

perpendicular and hypotenuse of a right-angled triangle, to

find the perpendicular.

RULE.

From the square of the sum subtract the square of the base, and divide the remainder by twice the sum, and the quotient will be the perpendicular.

A tree, 100 feet in height, is broken off-the top of the tree reaches the ground 30 feet from the bottom, while the part broken off rests on the stump. How high from the ground was it broken off? Ans. 45 feet.

Art. 313.—PROB. IX. Given the base and the difference of the hypotenuse and perpendicular of a right-angled triangle, to find the perpendicular.

RULE.

From the square of the base subtract the square of the given difference, and divide the remainder by twice the difference.

EXAMPLE.

If the base of a right-angled triangle be 30 feet, and the difference of the other two sides 6 feet, what is the length of the perpendicular? Ans. 72 feet.

Art. 314.-PROB. X. To find the diameter of the earth, from the known height of a distant mountain, whose summit is just visible in the horizon.

RULE.

From the square of the distance, divided by the height, subtract the height.

The highest point of the Andes is about 4 miles above the bed of the ocean. If a straight line from this touch the surface of the water at the distance of 178 miles, what is the diameter of the earth? Ans. 7940.

Art. 315.-PROB. XI. To find the greatest distance at which a given object can be seen on the surface of the earth.

RULE.

To the product of the height of the object into the diameter of the earth, add the square of the height; and extract the square root of the sum.

1. If the diameter of the earth be 7940 miles, and Mount Etna 2 miles high, how far can it be seen at sea?

Ans. 126+ miles.

√7940×2+22=126.

OBS.-The actual distance at which an object can be seen is increased

by the refraction of the air.

2. A man standing on a level with the ocean, has his eye raised 5 feet above the water. To what distance can he see the surface? Ans. 27 miles.

Art. 316.-PROB. XII. To find the height of an object at sea, or on the surface of the earth, having only the distance given.

RULE.

From the given distance, take the distance which the elevation of the eye above the surface will give, found by the last problem; then divide the square of the remainder by the diameter of the earth, and the quotient will be the height required.

Art. 317.—PROB. XIII. To find the contents of squared timber.

RULE.

Multiply the mean breadth by the mean thickness: the product, multiplied by the length, will give the contents.

Required the contents of a log, the length 24 feet 6 inches, mean breadth 1 foot 1 inch, and mean thickness 1 foot 1 inch. Ans. 28 feet 9 inches 6".

Art. 318.-PROB. XIV. To find the contents of round timber.

COMMON RULE.

Take one fourth of the mean girt, and square it, and multiply it by the length, for the contents.

OBS.-1. Tapering timber should be divided into pieces of eight or ten feet long, and these parts should be computed separately, and added.

2. In order to reduce the tree to such a circumference as it would have without its bark, a deduction is generally made of or of an inch for every foot of quarter-girt for young oak, ash, beech, etc.; but 1, or even 1 inch, must be allowed for old oak, for every foot of quarter-girt.

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3. The common rule gives the contents too small, by 3 feet on every feet of contents; yet it is universally used in practice, being originally introduced in order to compensate the purchaser of round timber for the waste occasioned by squaring it.

RULE II.-Take one fifth of the girt, and square it, and multiply by twice the length, for the contents.

1. Required the contents of a tree 24 feet long, and its girts at the ends 14 and 2 feet?

Ans. 96 feet, by the common rule; the true content is 122.88 feet.