18. Divide 60 into two parts proportional to 2 and 3. 19. Divide 90 into parts proportional to 2, 3 and 4.

20. Two men start in business with a capital of $7500. One of them furnishes $4000 and the other $3500. At the end of a year the profits are $3250. How much is each man's share?

21. A man starts in business with a capital of $5000 and in 3 months admits a partner with a capital of $4500. At the end of the year the profits amount to $3750. How much is each man's share?

22. A piece of work was to have been done by 10 men in 20 days, but at the end of two days 3 men left. How long did it take the remaining 7 men to complete the work?

23. If the interest on $325 is $72.50 in a given time, how much is the interest on $850 for the same time?

24. Two cog wheels work together; one has 36 cogs and the other 14. How many revolutions does the smaller one make while the larger one makes 28 revolutions?

METHOD OF ATTACK

230. In solving any arithmetical problem the student will find the following suggestions useful:

(1) The first essential is a thorough understanding of the proper relations between the conditions given. This requires some form of analysis leading to a complete statement of the conditions.

(2) The solution should involve no unnecessary work. Cancellation and other convenient short methods should be used if possible.

(3) All arithmetical work should be carefully checked. The student must realize that accuracy is of the highest importance and that to secure accuracy his work must always be checked. Any arithmetical work that has an error in it is valueless. The check also gives the student a means of knowing for himself whether he has a correct result or not. He has no need of answers to his problems.

Ex. 1. If the time of the beat of a pendulum varies as the square root of its length, and the length of a pendulum that beats seconds is 39.2 in., find the length of a pendulum that beats 50 times a minute.

Solution. The given pendulum beats 60 times per minute, the required pendulum beats 50 times per minute.

Since the longer the pendulum the more slowly it beats, the required pendulum is longer than the given one.

Therefore, the square root of the lengths of the pendulums are in the ratio 88, or g

60

309

Check either by changing the order of the factors and performing the multiplication again, or by casting out the nines.

Ex. 2. The greatest possible sphere is cut from a cube, one of whose edges is 3 ft. Find the portion of the cube cut away.

Solution. The volume of the cube is 38 cu. ft. The volume of the sphere is π × (3)3 cu. ft. Therefore the portion cut away is 33 cu. ft. TX (3) cu. ft. Without performing the operations indicated the student can by cancellation and combination of terms write the result thus,

Ex. 3. Find the area of a square field whose diagonal is 50 rods.

Solution. Let x = one side of the square field.

Ex. 4. Find the area of the circle which is equal in

area to two circles whose radii are 5 in. and 7 in,

Solution. Let r = the radius of the required circle.

Then its area in square inches = πr2 = π × 52 + π × 72 = π(52 +72) =π × 74, or 232.48 sq. in.

Check each step in the work.

Here, instead of multiplying π by 25 and then by 49 and adding the results, time is saved by adding 25 and 49 and multiplying π by the sum, 74.

231. The foot pound is used as a unit of work. This unit is defined as the amount of work required to overcome the resistance of one pound through a space of one foot. The rate of work is generally defined by using the term horse power. An engine of one horse power can do 33000 foot pounds of work in one minute, i.e. can overcome a resistance of 33000 pounds through a space of one foot in one minute.

Ex. 5. What horse power is an engine exerting that draws a train with a uniform speed of 40 miles an hour against a resistance of 1000 pounds?

Solution.

The amount of work done in one hour is 1000 × 40 × 5280 foot pounds.

The amount of work done in one minute is pounds.

1000 × 40 × 5280

foot

60

2 16

232. The student will notice that in each of the above exercises, first, the relations between the given conditions are carefully established; and second, a complete statement of these conditions is written out and the work shortened as much as possible by cancellation or otherwise, before the processes

of multiplication and division are used. Frequently students in solving such problems will perform the operations indicated at each step, thus doing a large amount of unnecessary work. By carefully studying these model solutions the student will see where the unnecessary work can be avoided.

As indicated in Art. 41, it is a good plan, whenever possible, to estimate the result mentally and to compare this rough estimate with the result found by solving the problem. This will prevent large errors and such errors as arise from misplacing the decimal point.

EXERCISE 51

1. Find the area bounded by 6 equal coins whose centers are at the vertices of a regular hexagon, the diameter of each coin being 2.38cm.

2. A crescent is bounded by a semi-circumference of a circle whose radius is 15 inches, and by the arc of another circumference whose center is on the first arc produced. Find the area and perimeter of the crescent.

3. A horse is tied with a 50 ft. rope to one corner of a barn 30 ft. by 40 ft. Find the area he can graze

over.

4. A well 30 ft. deep and 4 ft. in diameter is to be dug. If a cubic foot of earth weighs 12 lb., how much work is to be done?

5. A horse drawing a wagon along a level road at the rate of 2 mi. an hour does 29216 foot pounds of work in 3 min. What pull in pounds does he exert in drawing the wagon?