« AnteriorContinuar »
matical signs, the solution of a simple question in the lesson of the day, received this reply,—“I don't know how.” And when you have inquired, if it was not as easy to write the signs of the several operations to be performed as to perform them, have you not invariably been answered, " I can do the work and procure the answer given in the Book, but I don't know how to represent it on the Blackboard !" If you
have not had observation of this kind your scholars have been more fortunate in their previous train. ing than many of mine. So frequently and constantly has my attention been called to the deficiency of my scholars in this particular, that I have come to regard it as one of the prominent things to receive attention at every recitation in Arithmetic.
To explain in brief what I mean by WRITING THE SOLUTION OF PROBLEMS ON THE BLACKBOARD, and for the purpose of aiding the class of teachers referred to above, a little in this direction, let me suppose that the lesson of the day contains the following questions.
1. Alexander Green bought of John Fortune a box of sugar containing 475 lbs. for $30. He sold 1-3 of it for 8 cents per pound, and 2-3 of the remainder at 10 cents per pound. What is the ralue of what remains at 12 1-2 cents per pound, and what does Green make by the purchase and sale ?
The class having properly studied the lesson will be able, and should be required, to dispense with books at the recitation, and this for the two-fold purpose of disciplining them to habits of attention and relieving them of the aid of statements copied into their books from the work of others. The teacher reads the example distinctly twice, and three times if need be, and then requests one, two or a dozen of the class to go to the Black-board and indicate the solution of both questions. These correctly done will stand as follows: 175 x 23:1-312 1-2 = value of what remains.
475 X 1-3 X .08 + 475 X 2-3 X 2-3 X.10 + 475 X 2-3
x 1-3 x .12 1-2 30. Gain. 2. Bought 78 acres, 3 roods, and 30 rods, of land for $7,000, and having sold 10 House Lots, each 30 rods square, for $8.50 per square rod, I dispose of the remainder at 2 cents per square foot. How much do I gain by the bargain? 78 X 4+3 X 40 + 30 - 30 X 30 X 10 X 272 1.4 x .02
+ 30 X 30 X 10 X 8.50 7000. Gain.
3. How many yards of paper will it take to cover the walls of a room that is 15 1-2 feet long, 11 1-4 feet wide and 7 3-4 feet high, the paper being 30 inches wide ? and what would the plastering of the room cost at 10 cents per square yard? . 15 1-2 + 11 1-4 X 2 X 7 3-4 1 X 3 X 30-12
= No. yards in length. 15 1.2 + 11 1-4 X 2 X 7 3-4 + 15 1-2 X 11 1.4 1 X9
X.10 = Cost of plastering: 4. Two men in Boston hire a carriage to go to Concord, N. H., and back again for $25, the distance being 72 miles, with the privilege of taking in three more persons. Having gone 20 miles they take in C. At Concord they take in D, and when within 30 miles of Boston, on their return, they take in E. How much shall each
$25-144 x [20 X 1-2 + (72 - 20) X 1-3 + (72 -- 30) X
1-1 + 30 x 1-5] = A & B each pays. $25.144 x [(72 -- 20) x 1-3 + (72 -- 30) X 1-4 + 30 x
1-5) = C pays. $25-144 x [(72 — 30) X 1.4 + 30 x 1-5]
=D pays. $25.144 X 30 X 1-5 E
pays. 5. A. Atwood can hoe a certain field in 10 days. With the assistance of his son Jerry he can hoe it in 7 days, and with the assistance of his son Jacob he can hoc it in 6 days. In how many days con Jerry and Jacob hoe it together ? and in how many days will the three hoe it together? Work.
Work. Day. (1-7 1-10) + (1-6 --- 1.10):1:: 1: or 1
X1 Ans. to first. (1.7 - 1.10)+(1.6 1-10)
1-10) + 1-10 X1=Ans. to 2d.
Work. Day. (1-7 — 1-10) + (1-6 — 1-10) + 1-10 :1:: 1 : or 1
. (1-7 — 1-10)+(1-6 — 1-10) + 1.10
The foregoing will, doubtless, be sufficient to show what I mean by writing solutions on the Blackboard, and I proceed to a second point of importance in teaching Arithmetic,-second not in importance, but only in the order of recitation, and that is, TRAINING SCHOLARS TO EXPLAIN IN EXACT MATHEMATICAL
AND GRAMMATICAL LANGUAGE THESE WRITTEN SOLUTIONS.
In doing this, every teacher will need to exercise much patience and perseverance. A majority of scholars not having been accustomed to do this, will at first make hard work of it, both for themselves and their teacher. But let neither teacher nor scholar spare effort or time to secure proficiency in this matter. A little, accurately, thor. oughly and understandingly done in arithmetic, is worth more to the scholar than going through the book a dozen times in the way that many do. And in this particular, especially, will the superiority of one teacher over another be manifest, for, as in teaching reading, so in mathematical reasoning, the teacher must exhibit, over and over again, a specimen for the observation and imi. tation of the learner.
To illustrate, briefly, in this, take the last of the foregoing Examples. In how many days will Jerry and Ja cob hoe the field together? Statement.
- 1-10) +(1.6 — 1-10) X1= No. days Explanation. It will take them as many days as that part of the work which they can together do in one day is contained times in unity, or the whole work to be done. By the first condition of the question, the Father and Jerry can do the work in 7 days; therefore, in one day, they can do 1-7 part of it; but the Father alone can do it in 10 days, therefore, in one day, he can do 1-10 part of it. Hence 1-7 of the work, what the Father and Jerry can do in one day, minus 1-10, what the Father alone can do in one day, will represent what Jerry can do in one day. By the second condition, the Father and Jacob can do the work in 6 days; therefore, in one day, they can do 1-6 part of it. Hence 1-6 of the work, what they together can do in one day, minus 1-10 of the work, what the Father can do in one day, will represent what Jacob can do in one day.
Hence 1-7 — 1-10 of the work, what Jerry will do in one day, plus 1.6 — 1-10 of the work, what Jacob will do in one day, will represent what they both will do in one day, and it will take as many times one day for them to do the work required, as the part of the work thus indicated as done by them in one day, is contained times in unity or the work to be done.
One example further, to indicate what seems to me to be the proper phraseology in these explanations, and that from Colburn's Mental Arithmetic, the best of all Arithmetics in the hands of a competent, and the poorest of all in the hands of an incompetent, teacher. " Six sevenths of forty-two are four ninths of how many fifths of forty ? Expression. 42
X 6 7
X 9 4 40-5
= 10 1-8. Explanation. One seventh of forty-two is six. If six is one seventh of Forty-two, six sevenths will be six times as many. Six times six are thirty-six. (Thirty-six is four ninths of how many fifths of forty ?) If thirty-six is four ninths of some number, one ninth of that number will be one fourth as muny. One fourth of thirty-six is nine. If nine is one nintio of the number, nine ninths, which make the number, will be nine times as many. Nine times nine are eighty-one. (Eighty-one is how many fifths of forty ?) One fifth of forty is eight, and eight is contained in eightyone, ten and one eighth times. Therefore, six sevenths offor. ty two are four ninths of ten and one eighth times one fifth of forty.
In this simple example I have marked the process of reasoning with italics, to distinguish it from its application, and this because within my own observation, many teachers entirely neglect this, and thereby despoil the entire process of mathematical beauty-the beauty of precision.
One thing more in reference to Colburn's Lessons. Though designed to be purely a Mental Arithmetic, teachers of it will find, if they make the experiment, that training their classes to write the solutions on the Plack
d, will greatly increase their interest in the exercise and at the same time more effectually qualify them for the study of written arithmetic.
There is much more that I would like to say on these points, but I designed to be brief, and will close by requesting some one of your readers to furnish for our common benefit, through your columns, a solution for the following, found in Greenleaf's Higher Arithmetic. (I quote from memory.)
Four men, A, B, C and D, bought a Grindstone whose diameter was forty inches. It they pay for it equally, it is required to find what number of inches, in equity, each shall grind off from the semi-diameter in the order of the letters above, allowing the diameter of the orifice for the axle to be four inches ?
In the solution requested, I would not only wish for the simple mathematical formulæ, but the accompanying diagram and verbal explanation, or in other words the entire discussion of the question, both to the eye and the ear, which the teacher would require in the school-room.
It is not what we earn, but what we save, that makes us rich; it is not what we eat, but what we digest, that makes us strong; it is not what we read, but what we remember, that makes us learned.