Electronic library

 eBooksRead.com books search new books russian e-books
Edward Brooks.

Methods of teaching mental arithmetic, and key to the normal mental arithmetic ..

. (page 2 of 6)

can always be done, and the reason is as follows : We
require the pupils to multiply and divide until we reduce
to the number they begin with, we then (not knowing,
of course, what number they have) require them to add
some number to the number that they have, and then
require them to subtract the number they thought of,
which leaves the number we told them to add, and we
now know what number they have ; we can, therefore,
tell them to multiply or divide by anything we choose,
and knowing the number they have, we can readily name
the product or quotient. This little puzzle, even with
an advanced class, I have known to elicit a very high
degree of interest. When it is well understood, the
teacher may show the class that they need not reduce it
to once the number before adding the number which he
gives them, provided they subtract as many times the
number from the sum.

SECTION II.

NOTE. This section treats of fractions by using the
fractional iuord^ but not the expression consisting of two
figures with a line between them. This is regarded as a
valuable preparation to the more general treatment of the
following section.

f ^ OF THE - A

f UNIVERSITY )

18 KEY. [SECTION n

LESSON I.

7. SOL. If 1 yard of cloth cost 16 cents, 1 half of a
yard of cloth will cost 1 half of 16 cents, which is 8
cents.

25. SOL. If John had 21 cents and gave 2 thirds of
them to Susan, Susan received 2 thirds of 21 cents; 1
third of 21 cents is 7 cents, and 2 thirds of 21 cents are
2 times 7 cents, which are 14 cents.

LESSONS II. & III.
No Suggestions or Solutions necessary.

LESSON IV.

19. SUG. He sold 2 tenths -f 3 tenths -j- 4 tenths,
which are 9 tenths of his number, hence there remained
10 tenths 9 tenths, which is 1 tenth of his number,
that is, 1 tenth of 20 sheep, which is 2 sheep.

30. SUG. A gave B 20 tenths of a dollar, then 20
tenths + 5 tenths, which are 25 tenths, equals 1 half of
what I had, hence I then had 50 tenths of a dollar, and
at first 56 tenths of a dollar.

38. SUG. One yard cost 4 tenths of a dime, hence
for 12 tenths of a dime 3 yards may be bought.

LESSON V.

NOTE. The previous lessons in fractions were needed
to prepare the student for the solution of the problems

LESSON VII.] KEY. 19

in this lesson. Some authors introduce problems similar
to those in this lesson previous to any exercises in frac-
tions, but such an arrangement is evidently illogical.

47. SuG.The lemons were worth 24 cents, for which
you could buy 6 oranges at 4 cents each ; hence there
remained 9 6, or 3 oranges.

LESSON VI.

NOTE. The problems in this lesson are so simple that
no solutions need be given in the Key. The author
would remark that problems similar to the 29th and 30th
should frequently be given to the class. They afford an
excellent exercise for acquiring rapidity and accuracy of
mental computation. The teacher can readily extem-
porize them, making them as complicated as the advance-
in en t of the class may allow.

LESSON VII.

NOTE. The solution given in the Mental Arithmetic
for the 24th is the reverse of that given for the first pro-
blem of this lesson, and since the 24th problem is the
reverse of the first, the solution given seems most appro-
priate. The following solution to problems of this class
is, however, preferred by some teachers.

24. SOL. In one there are three thirds, hence one
third of the number of thirds equals the number of
ones ; 1 third of 6 is 2 ; therefore, in 6 thirds there are

20 KEY. [SECTION n.

LESSON VIII.

NOTE. The solution of the first problem is sometimes
given thus : If 3 is one half of some number, two halves
of that number is two times 3, or 6. The solution giveu
in the Arithmetic is perhaps rather more logical.

30. SOL. If $10 is 1 third of 6 times the cost of the dress, 3 thirds of 6 times the cost, or 6 times the cost of the dress is 3 times$10, which are $30 ; if 6 times the cost of the dress is$30, once the cost of the
dress is 1 sixth of $30, which is$5.

36. SOL. If 1 third of 1 half of some number is 8, 3
thirds of one half of that number is 3 times 8, or 24 ; if
1 half of some number is 24, 2 halves of the number
equals 2 times 24, which are 48.

LESSONS X. & XI.

NOTE. The method of solution in the previous part
of this section has been to pass from a collection of things
or parts to the single thing or part, and then pass from
the single thing or part to another collection, thus passing
from one collection to another. The process involves the
logic of comparison, and is very easy, because the rela-
tion between a unit and a collection, or the relation be-
tween a collection and the unit, is so evident that it is
readily apprehended. The pupils are now prepared to
discover relations existing between different collections
of units, that is, between different numbers, and they
should be led to apply these relations in the solutions of
problems. This is the object of Lessons X. and XI.,
and we commend them to the careful attention of the
teacher. After the pupil is familiar with the method of
running to the unit, it is a waste of time to require him
to do so every time he wishes to compare two numbers.

LESSON Xlf.] KEY. 2i

LESSON XII.

NOTE. This we deem a valuable lesson in preparing
for fractions, although not usually introduced into works
on Mental Arithmetic. The pupils should be drilled
upon it until they are entirely familiar with it. The
results will, of course, since the numbers >are small, be
determined by inspection. The pupils should be required
to frame definitions for themselves from the suggestive

SECTION III.

LESSON I.

1. SOL. In one there are f and in 3 there are 3 times
|, which are |, and | plus J- are 3. Therefore, in 3i
there are 2.

NOTE. One of the prominent features of this book is
the derivation of methods in fractions from the results of
analytic processes. Unless such methods are derived it
would be necessary to give the analysis in full for every
problem, which would become exceedingly tedious as the
problems became complicated. Our plan then is, first to
require the analytic solution, and after the pupils are
familiar with such solution, to derive a rule, or method
we prefer to call it, by which the results may be obtained
more briefly.

After the pupils have analyzed a few of the problems
in the first part of this lesson, lead them to see that they
can reduce mixed numbers to fractions by multiplying the
entire part by the denominator of the fractional part,
adding the numerator of the fractional part to the prod*

22 KEY. [SECTION III.

net, and using the result as the numerator of the
fraction.

A similar remark applies to the problems from the 18th
to the 26th, inclusive. The method is, Divide the nume-
rator l>y the denominator ; if there is a remainder place
it over the denominator, and unite this fraction to the
'quotient obtained l>y the division.

LESSON II.

15. SOL. It is evident that these fractions may be-'
reduced to 12ths. One equals |f, hence | equals J of
j|, which is y^, and f equals 2 times T 4 2 , or -%, &c.

NOTE. Having derived a method, we may now omit
the analysis, and proceed with this method as follows.

36. SOL. To reduce ^ to twentieths we multiply both
numerator and denominator by 10, and have i, equal to
ig } to reduce ^ to twentieths we multiply both nume-
rator and denominator by 5, and have |, equal to |g, &c,

46. SOL. If 60 cents was |- of i of what he then had,
\ of -J of what he then had was 1 of 60 cents, which is
12 cents, and | of ^ of what he then had was 4 times 12
cents, or 48 cents ; if 48 cents was i of what he then
had, | of what he then had equals 2 times 48 cents,
which are 96 cents; hence before he found 60 cents he
Lad 96 cents 60 cents, or 36 cents. Therefore, &c.

uts, at first.

LESSON III,

NOTE. We analyze the problems to the eleventh, and
then derive a method, which we apply in the solution of
the problems which follow.

LESSON IV.] KEY. 28

15. SOL. To reduce -fa to fifths we divide both nu-
merator and denominator by 2, and we have T 6 5 , equal to
J, &c.

23. SOL. To reduce | to its lowest terms we divide
both numerator and denominator by the greatest number
that will divide them both without a remainder, which
number is 4, and we have |, equal to , &c.

37. SOL. 4| melons = 2 ^ 4 melons. If 2 ^ 4 melons are
worth 6 lemons, ^ of a melon is worth ^V of 6 lemons,
which is 3 ^, or 1 of a lemon, and |- of a melon is worth
f> times i, or | of a lemon, and 7 melons are worth 7
\times f , which are 3 5 5 , or 8| lemons.

LESSON IV.

NOTE. This lesson treats of the addition and subtrac-
tion of fractions. We commence by adding and sub-
tracting fractions having a common denominator, then by
problems from 10 to 1.6 suggest the reduction to a com-
mon denominator, after which we reduce to common de-
nominator by the method derived in Lesson II., and then
add or subtract as the problem may require.

17. SOL. i is equal to |, and | is equal, to |; | plus
| are |. Therefore the sum of ^ and | is |.

47. SOL. If Thomas had | of a dollar and found \ of
a dollar, he then had J of a dollar plus | of a dollar;
J =-~h and = 7%> -h + T 6 5 *= \\- Therefore, &c.

50. SUG. | of the sum plus \ of the sum equals J
of the sum, which was $21 ; hence the sum was 24. 78. SOL. J of the number diminished by | of the number equals | of the number, which equals 36 ; if | of the number equals 36, &c. NOTE. The pupil may be led to see that when we wish to add two fractions having a unit for the nume- rator, we may take the sum of their denominators for the 3 24 KEY. [SECTION in. numerator of the sum, and the product of the deno- minator for the denominator of the sum ; also, that the difference of two fractions, whose numerators are a unit, is the difference of their denominators divided by their product. Thus, I + i = = &, and l i = LESSON V. 4. SOL. If a boy lost 4 marbles and found 10, he then had 10 4, which is 6 more than at first, which equals |, or i of what he had at first, &c. 14. SUG. We find$24 was what remained, which
equals | minus |, which is | of what he had at first, &c.

24. SUG. If he gave g of away, there remained the
difference between | of ^ and J of , which is | of .

28. SUG. | of f equals -|. If 60 feet is f of the
length of the shadow, diminished by 20 feet, 60 feet
plus 20 feet, -which are 80 feet, is f of the length of the

NOTE. In solving problems like the first in this lesson,
some say " let f equal what he had at first." This is
absurd. 4 of what he has will equal it whether you let
it or not.

LESSON VI.

35. ANS. By dividing the denominator of j by 2.

36. ANS. By dividing the denominator of | by 3.
39. SOL. Dividing the denominator by 4 we have 4

times | equals ->-, or 2^.

59. SOL. 2| =|; 3 times f miles equals 8 miles;
if twice the distance is 8 miles, &c.

WESSON VIII.] KEY. 25

. Hereafter, when a fraction is to be multiplied
by a number which will divide the denominator, insist
upon pupils dividing the denominator, instead of multi-
plying the numerator. This will require the continual
attention of the teacher, for pupils adhere to the method
of multiplying the numerator with wonderful tenacity.

LESSON VII.

42. SUG. After selling f of f of a barrel, there re-
mained i of f of a barrel, &c.

NOTE. Pupils will naturally solve this by obtaining |
of |, and then subtracting this from |. The solution
indicated is much more concise.

LESSON VIII.

NOTE. The problems in this lesson as far as to the
13th should be solved like the 1st, then by means of the
13th and 14th we derive a method, which we apply to
the solution of those which follow.

The first may also be solved thus : \ is equal to T 3 j, and
J of j 3 3 is JTJ ; hence A of { is J 2 . The solution given in
the Arithmetic is preferred, since it involves the principle
of obtaining a part of a fraction, which the other solution
does not.

15. SOL. By multiplying the numerators together for
the numerator of the result, and the denominators for the
denominator, we find | of J to equal |.

21. SOL. If a man, owning | of a farm, sold of it
to his neighbor, his neighbor received | of |, which is
A of it.

NOTE. The 30th to the 37th we solve like the 29th.
At the 37th we derive the method, and apply this method
to the solution of those which follow.

26 KEY. [SECTION in

38. SOL. By multiplying the numerators together foi
the numerator, and the denominators for the denominator
of the result, we find | of -| equals ^7, &c.

42. SOL. If Johnson, having | of a melon, gave | of
it to Martin, there remained | of | minus | of |, whict
is | of |, which equals i. Therefore, &c.

43. SOL. If I had | of a bushel of apples and gave |
of them away, there remained | of them minus | of them,
which is \ of them ; hence there remained \ of |, which
is r 3 g of a bushel.

50. SuG. If she shared them with her schoolmates,
they were divided equally between herself and 5 school-
mates, or 6 persons ; hence each received J of f = T V
of a pound.

54. SUG. It arose \ r of ^ = i, and then was ^ + g
= | of the first distance from the ground ; it fell \ of
i = 2 L, and then was f ^ = f of whole distance
from the ground.

LESSON IX.

12. ANS. Multiply the dividend by the fraction ob-
tained by inverting the terms of the divisor.

NOTE. Assign a few of the problems which have
been solved by the Analysis, and let the pupil apply the
Method.

13. SOL. If a yard of cloth cost f of a dollar, for $12 you can buy as many yards as | is contained times in 12, which are 12 X | S ? or 2 ^- NOTE. The 13th problem may also be solved by the following analytic process. Let the pupil understand both. 13. SOL. If$| will buy one yard, $i will buy \ of a yard, and$| will buy 5 times -*, which are f of a yard,
und $12 will buy 12 times |, which are 20 yards. LESSON XI.] KEY. 27 28. SOL. One is contained in | | of a time; and since one is contained in | | of a time, | is contained in | 3 times |, which are | of a time, and | is contained in | 5 of |, which is of a time. 49. ANS. Multiply the dividend by the fraction ob- tained by inverting the terms of the divisor. NOTE. Assign a few of the problems just analyzed, and require the class to solve them by the Method. 50. SOL. If a yard of muslin cost f of a dime, for | of a dime you can purchase as many yards as | is con- tained times in |, which is | X | = If > <> r lij- LESSON X. 16. SOL. i is 2 of f , and J, or one, is 3 times ^, which are f of f . It' one is ! of f , ^ is I of f , which is T 3 of |, and | is 4 times T 3 ff = jg, or f of f . Therefore, f is I <*f- NOTE. This problem may also be solved by reducing the fractions to a common denominator, thus : 16. SOL. = if* an d | = yf> an d T i s ^ e same part of 1$ that 12 is of 10; and 12 is {, or f of 10.
Therefore, | is f of f .

LESSON XI.

24. SOL. There were as many persons as f is con-
tained times in 8, which are 10, and since he shared
them, these included himself; hence there were 9 com-
panions.

26. SuG. SA = \ 5 . If ^ peaches are worth 5
apples, -i of a peach is worth -^ of 5 apples, or ^ of an
apple, and J of a peach is worth 3 times i, which are |
of an apple, and 10 peaches are worth 10 times J of an
apple, which are 6 apples.

28 KE?. [SECTION IV.

LESSON XII.

1. SOL. If $40 was | of what remained, -1 of what remained was ^ of 40, which is$10, and | of what re-
mained was 4 times 10, or $40. After spending | of his fortune, there remained | |, or | of his fortune, which equals$40, &c.

6. SUG. If he sold ^ of his sheep, ^ of his sheep
remained, which equals 15 ; hence he owned twice 15, 01
30. If -| of his cows remained, he sold | of them, which
equals 10 ; hence he had at first 15 cows.

12. SUG. He found 6 cents, hence he had 12 6,
or 6 cents less than he had at first ; but he had only | as
much as at first, hence 6 cents equals ^ of what he had
at first ; therefore he had at first 4 times 6 cents, or 24
cents.

hence she gave away 50 cents 10 cents, or 40 cents.

NOTE. Let the pupils be drilled upon the demon-
strations of the Propositions which are given at the close
of this lesson. With young pupils use special numbers
which they may give the general demonstration. For
valuable exercises of this character, see the "Normal
Primary Arithmetic/' pages 71, 72, 73, and 74.

SECTION IV.

NOTE. After the solution and remarks given for the
preceding sections, it will not be necessary to solve any
of the problems in this section. It consists principally
of the application of the previous principles to denominate
numbers. The teacher will find that the remarks upon

LESSON I.] ' KEY. 29

the tables may be made the means of eliciting much in-
terest in class.

In the solution of such problems as the 8th under Fed-
eral Money, the two numbers compared must of course be
reduced to the same denomination.

The 5th problem under Apothecaries' Weight, and the
9th under Avoirdupois, are intended to illustrate the
relation of the pound and ounce in the three different
weights. A pound of lead is absolutely heavier than a
pound of gold, since the pound Avoirdupois consists of
7000 grains, and the pound Troy of but 5760. An
ounce of silver, however, is heavier than an ounce of
feathers, since the Troy ounce is heavier than the Avoir-
dupois ounce, as may be seen by a very simple reduction,
that is, by ascertaining the number of grains in an ounce
of each weight. Thus, in Avoirdupois weight there are
16 oz. in a pound, hence 1 oz. = ^ of 7000 grs. = 437^ ;
but in a pound Troy there are 12 oz., hence 1 oz. = y 1 ^
of 5760 = 480 ; hence an ounce Troy is 42^ grs. heavier
than an ounce Avoirdupois.

SECTION V.

LESSON I.

IN the solution of the problems of this lesson let the
pupils observe the following suggestion. When we
reduce a collection to the unit, pass immediately from this
unit to the other collection of the same unit. Thus, if
we commence with a collection of horses, and reduce to
one horse, pass to the other collection of horses, instead
of first reducing some of the other numbers to the unit,
and then coming back to the horses.

Many of these problems can be solved without passing
to the unit, by employing the relations of the numbers,
as is illustrated in Lessons X. and XL, Section II.

30 KEY. [SECTION v.

9. SOL. If 10 oxen can eat 4 acres of grass in 6
days, one ox can eat 4 acres of grass in 10 times 6 days,
which are 60 days, and 30 oxen can eat 4 acres of grass
in -g^ of 60 days, which is 2 days; if 30 oxen eat 4 acres
of grass in 2 days, they will eat 1 acre in | of 2 days,
which is ^ of a day, and they will eat 8 acres in 8 times
J, which are |, or 4 days.

9. SOL. 2d. If 10 oxen eat 4 acres of grass in 6 days,
30 oxen, which are 3 times 10 oxen, can eat 4 acres of
grass in ^ of 6 days, which is 2 days ; and if they can
eat 4 acres of grass in 2 days, they will eat 8 acres,
which are 2 times 4 acres, in 2 times 2 days, which are
4 days.

10. SUG. Find how many men can perform the same
piece in 6 days, and then to perform a piece 3 times as
large will require 3 times as many days.

16. SOL. If 3 oranges are worth 9 cents, one orange
is worth ^ of 9 cents, which is 3 cents, and 8 oranges, or
4 melons, are worth 8 times 3 cents, which are 24 cents ;
if 4 melons are worth 24 cents, one melon is worth i of
24 cents, or 6 cents, and 10 melons are worth 10 times
6 cents, which are 60 cents.

NOTE. This problem can also be solved by com-
mencing with the melons, thus :

16. SOL. If 4 melons are worth 8 oranges, one melon
is worth | of 8 oranges, which is 2 oranges, and 10
melons are worth 10 times 2 oranges, which are 20
oranges ; if 3 oranges are worth 9 cents, one orange is
worth -1 of 9 cents, which is 3 cents, and 20 oranges are
worth 20 times 3 cents, which are 60 cents.

23. SOL. If A can do as much work in 2 days as B
can in 4 days, or C in 6 days, then B can do as much in
4 days as C can in 6 days, and B can do as much in 3
times 4 days, or 12 days, as C can in 18 days, which are
3 times 6 days.

25. SOL. If A can do 3 times as much in a day as B,
and B can do twice as much in a day as C, then A can

LESSON II.] KEY. 31

do 3 times twice as much as C, which arc 6 times as
much as C ; hence A can do as much in one day as C
does in 6 days, and A can do as much in of a day as C
can in one day, and A can do as much in |, or f of a day,
as C can in 4 days.

27. SUG. Find how long it will take 8 boys to do ^
of it, and then how long it will take 8 3, or 5 boys, to
do the other half.

28. SOL. If 9 men build 10 rods of wall in 8 days,
they can build 20 rods, which are 2 times 10 rods, in 2
times 8 days, which are 16 days, and they can build \ of
it in i of 16 days, which is 4 days, and | of it, what
remains, in 3 times 4 days, or 12 days; if 9 men can
build | of the wall in 12 days, ^ of the number, which
remains when | leave, can build it in 3 times 12 days,
which are 36 days, and the 20 rods will be built in 4 -f-
36, or 40 days.

29. SOL. If a measure of flour make 5 four-cent
loaves, it will make 5 times 4, or 20 one-cent loaves ; and
if it will make 20 one-cent loaves, it will make i of 20,
or 10 two-cent loaves.

LESSON II.

2. SUG. After finding the number of pupils to be
13, we find the number of questions by multiplying 2 by
13, and then adding 26, or by multiplying 4 by 13.

7. SOL. The difference between having 10 more and
30 more, is 20, and the difference between having 2 times
as, many as Robert and 4 times as many as Robert, is
twice as many as Robert ; hence twice Robert's number
equals 20, and Robert's number equals | of 20, or 10, &c.

7. SOL. 2d. By the first condition, twice Robert's
equals Morris' plus 10 ; by the second condition, 4 times
Robert's equals Morris' plus 30 : hence 4 times, minus 2
times, or 2 times Robert's equals 30 10, or 20, &c.

82 KEY. [SECTION v.

10. SUG. We find the cost of one orange is 4 cents,
and the cost of one apple is 2 cents, the difference between
the cosfc of each is 2 cents, the difference between the
cost of all is 18 cents ; hence there were as many of each
as 2 is contained times in 18, which are 9.

12. SOL. The difference between having 8 more and
12 less is 20, and the difference between having 6 times
i is many as Henry, and 2 times as many as Henry, is 4
times as many as Henry; hence 4 times Henry's number
equals 20, &c.

13. SOL. If one girl received 3 apples, and one boy
plus 4 apples, which are 7 apples; and since there were
the same number of each, and all received 28 apples,
there were as many of each as 7 is contained times in 28,
which are 4.

17. SOL. If one boy receives 2 cents, 3 boys will re-
ceive 3 times 2 cents, which are 6 cents, and one girl and 3
boys will receive 4 cents plus 6 cents, which are lO cents,
and they all received 60 cents ; hence there were as many
times one girl and 3 boys as 10 is contained times in 60,
which are 6 ; hence there were 6 girls, and 6 times 3, or
18 boys.

18. SUG. We find the difference between the cost
of 2 apples and one orange is 2 cents, and the difference
between the cost of all is 10 cents; hence there were as
many times one orange and 2 apples as 2 is contained
times in 10, which are 5; hence there were 5 oranges,
and 5 times 2, which are 10 apples.

LESSON III.

6. SOL. Three times a certain number equals | of
the number, which increased by | of the number equals
y of the number, which by a condition of the problem
equals 22, &c.

LESSON III.] KEY. 33

12. Suo. We find that | of the height, plus 10 feet,
equals | of the height ; then | of the height, minus | of
the height, which is | of the height, equals 10 feet, c.

16. SOL. If 2 times a number, plus 6, equals 3 times
the same number, plus 2, then 3 times the number, minus

2 times the number, which is once the number, equals 6
minus 2, which is 4.

19. SOL. If 4 times A's age, minus 10 years, equals

3 times A's age, plus 10 years, 4 times A's age equals 3
times A's age, plus 20 years ; and if 4 times A's age
equals 3 times A's age, plus 20 years, 4 times A's age,
minus 3 times A's age, which is once A's age, equals 20

 Using the text of ebook Methods of teaching mental arithmetic, and key to the normal mental arithmetic .. by Edward Brooks active link like:read the ebook Methods of teaching mental arithmetic, and key to the normal mental arithmetic .. is obligatory