Improvements in the Setting of Arithmetical Tests for Elementary Schools.
Year: 1894

page_text: Date of Application, 1st Sept., 1894
Complete Specification Left, 27th May, 1895-Accepted, 29th June. 1395
PROVISIONAL SPECIFICATION.
Improvements in the Setting of Arithmetical Tests for Elementary
Schools.
I, CHARLES RHODES, of 11 Kennington Park Gardens, London, S.E.,
Schoolmaster, do hereby declare the nature of this invention to be as follows:-
My invention relates to the setting of arithmetical tests, or exercises in
arithmetic, for the various standards in elementary schools, by the use of which
tests the scholars in large classes can each be doing different examples in the
various rules, while the teachers can mark the work without the use of printed
answers much more rapidly than if other tests are used requiring printed answers.
Copying is absolutely prevented, and the rapidity of the marking enables the
teachers to pay attention to backward scholars without keeping the others
waiting. I do not have one method of testing running through all the rules in the
various standards, but various methods, each specially adapted for the particular
rule for which it is employed. The principal devices I employ are herein set
forth :-
:-
No. 1 This method I use for simple addition, subtraction, and multiplication,
and consists in the use of numbers which are multiples of 99, all the answers also being
multiples of 99, and any answer shown which is not a multiple of 99 is easily
detected and marked wrong by the teacher. To test any answer the teacher adds
the first, third, fifth &c. figures of that answer together, and then the second,
fourth, sixth, &c. If the one total be 9, or any multiple of 9, as 18, or 27, the
other total will be the same multiple of 9, but if the two totals are different, by
adding the two totals we shall get a multiple of 9. and their difference will be 11
or some multiple of 11. In other words, if one total be 9 the other must be 9 as
in 7425 or 46431. If one total be 18, the other must be 18, as in 7649631. I1
one total be 19 the other must be 8, as in 72468, for 19 +8=27 and 19 - 8 = 11.
If one total be 27, the other must be 27. If one total be 29 the other must be 7.
and so on. If each line of an addition sum of any number of lines, or of a
subtraction sum, or the top line of a multiplication sum be a multiple of 99, the
answer will be a multiple of 99. If a selection of multiples of 99 are printed on a
sheet, in figures large enough to be seen by all the class, the teacher will have no
difficulty in setting the class at work at addition, subtraction, or multiplication,
each line being numbered or lettered, and the children commencing at different
parts of the sheet.
No. 2 This method I use for the same rules as No. 1, and the testing is-more
rapid. It consists in making each line of the addition, each line of the subtraction,
and the top lines of the multiplication sums a multiple of 101, consequently the
answers will be multiples of 101. To test whether the answers shown are multiples
of 101, if the answer consists of 4 figures the first two figures are repeated,
as 8484.
If 5 or 6 figures occur in the answer, the right hand part of the answer
(2 figures) added to the left hand part of the answer (1 or 2 figures) will usually
give the middle part of the answer (two figures), but sometimes 101 more than the
middle part.
The following are instances

page_text: When there are more that 6 figures in the answer, as in 5374614, begin with
the two figures on the right, take that number from the next two, that remainder
from the next two, and so on, and we finally get the left-hand part of the answer.
In this particular instance 14 from 46 = 32, and 33 from 37 5. In some cases
it is necessary to add 101 before subtracting, but I usually select multipliers to
avoid that.
:-
-
No. 3 This method of setting sums may be used for the same three rules, but
it is especially useful for multiplication of five figures by three figures. According
to this method all the answers are multiples of 1001: Now 1001 = 143 × 7.
I make the multiplicands multiples of 143 and the multipliers multiples of 7,
consequently the answers are multiples of 1001. To test, add the number formed
by the three figures on the right to the millions, and that will either give the
thousands, or 1001 more than the thousands. If then on a large sheet are printed
60 multiples of 143 as multiplicands, and 60 multiples of 7 as multipliers, being
able to use any multiplicand on the sheet' with any multiplier; we have the
materials for 3600 easily tested sums. As an instance of the working of this rule
take 30459 (a multiple of 143) x 651 (a mttltiple of 7): The answer is 19,828,809,
in which 809+ 19828.
Other 'sums' testing in the same way may be made by using multiples of 77' with
multiples of 13; or multiples of 91 with multiples of 11, for 77 x 13'= 1001,
and 91 x 11 =1001:
=
No 4: This method is useful for multiplication of six figures by four figures, the
sums being set by using multiples of 137 as multiplicands; and multiples of 73'as
multipliers, or vice versa. Now 137 × 73 10,001, and consequently all the
answers will be multiples of 10. 01. They may be tested by adding the four figures
on the right to the left hand' part, when the result will be the intermediate
four figures or else 10001 more, eg. :-625131' (a multiple of 137) × 3504' (a
multiple of 73)=21 9045 9024 in which 9024 + 21 9045.
No. 5 This is useful for simple multiplication, especially of four figures by two
figures, and consists in using multiples of 334 with multiples of 3; or multiples
of 167 with multiples of 6, giving multiples of 1002 as products: In: each
answer therefore the left hand part will be half of the right hand part,
e.g. 3674 +-72
= 264528: Here 264 is half of 528.
No. 6-This is useful for multiplication of five figures by two. I use multiples
of 667 as multiplicands and multiples of 3 as multipliers, the answers being
multiples of 2: 01, and causing the left hand part to be double of the right hand
part,. e.g. 19346 (a multiple of 667) × 63 (a multiple of 3) 1 1218609, in
which 1218 is double of 609. As 2001 can also be divided by 23 and 29 the same
result may be attained by using these factors instead.
No. 7This I use for multiplication,, all the answers being inultiples of 20,001
or 10,002, the former being divisible by 59,3, or 113; and the latter by 3 and 3334.
By combining multiples of these factors to make multiplication sums; the four
right hand figures will either be half or double of the rest of the answer,
e.g. 393353 x 78 = 30681534. where the right hand part is half of the left hand'
part.
Take
No. 8-
-This I use for addition, subtraction, and multiplication of money, and
practice. All the answers are multiples of £2.0: 1 or a quarter of that amount,
viz. 10s. Od., i.e., in each answer there are twice as many pounds as pence.
£174 173 as an instance. In 7s. 3 d. there are 87+ pence, and £174: 103. =
£174 (double of 87). The difference between two such amounts will give
another amount testing in the same way. To set the multiplication sums I adopt the
same principle as in some of the simple rules. The factors for 481 (the number of
farthings in 10s. Otd.) are 13 and 37. I use a multiple of each, one being farthings
expressed in £ s. d., and the other being the multiplier. Take for instance 8s. 52d.
(a multiple of 37 farthings) and multiply by 221 (a multiple of 13). The answer
is £93 13 101, i.e. 461 pence and 93. pounds.

page_text: If more figures are required in the pounds to make longer sums, the additional
figures may be some definite and easily recognised multiple of the pounds obtained
as above by doubling the number of pence.
t
No. 9 This method I use for sums requiring pounds, shillings, and pence in
the answers, but no farthings. In each answer there are eleven times as many
pounds as pence. "Now £11:01=139 x 19 pence. By multiplying multiples of
one factor by multiples of the other we get such answers. e.g. £1497 (a
multiple of 139d.) x 57 (a multiple of 19)=£825 6 3, which gives 75 pence,
and 11 times as many pounds.
"All these methods of setting and testing sums depend upon one principle,
ie in each set of sums each answer is a multiple of a certain number or amount all
such multiples being easily recognised by certain tests, the working of which may
be proved algebraically. The principle of casting out nines and elevens, and
testing sums by remainders has been in use as a curiosity, but not the systematic
setting of sums so that the answers shall be multiples of 99. Each test was applied
separately, but I combine the two and have no remainders. All the other methods
of setting and testing I have discovered myself, and if the properties of such
numbers as 101 or 1001 have been noticed, before they have.never been used in the
setting of large numbers of sums. My only object in applying for a Patent is to
prevent others systematically setting and publishing other sets of sums testing
similarly to mine. Copyrighting would.only protect, the particular sums I publish,
and not the methods.
Dated this First day of Santambon 1201
Improvements in the Setting of Arithmetical Tests for Elementary
Schools.
I, CHARLES RHODES, of. 11 Kennington Park Gardens, London, S.E.,
Schoolmaster do hereby declare the nature of this invention and in what manner
the same is to be performed,, to be particularly, described and, ascertained, in and
by the following statement :-
My invention relates to the setting of arithmetical tests, or exercises..in
arithmetic, for the various standards.in elementary schools, by the use of which
tests the scholars in large classes can each be doing different examples in the
various rules, while the teachers can mark the work without the use of printed
answers much more rapidly than if other tests are used requiring printed answers.
Copying is absolutely prevented, and the rapidity of the marking enables the
teachers to pay attention to backward scholars without keeping the others waiting.
I do not have one method of testing running through all the rules in the various
standards, but various methods, each specially adapted for the particular rule for
which it is employed. The principal devices I employ are as follows:-
No. 1 This method I use for simple addition, subtraction, and multiplication,
and consists in the use of numbers which are multiples of 99, all the answers being
also multiples of 99, and any answer shown which is not a multiple of 99 is easily
detected and marked wrong by the teacher. To test any answer, the teacher. adds
the first, third, fifth, &c. figures of that answer together, and then the- second,
fourth, sixth &c. If one total be 9 or any multiple of 9, as 18. or 27, the other
total will be the same multiple of 9; but if the two totals are different, by adding
the two totals we shall get a multiple of 9, and their difference will be 11 or some
multiple of 11. In other words If one total be 9 the other must be 19, as
in 7425 or 46431. If one total be 18, the other must be 18, as in 7649631. If
one total be 19, the. other must be 8 as in 72468 for 19 +8=27, and 19-8 = 11.
If one total be 27; the other must be 27. If one total be 29, the other must be 7,

page_text: and so on. If each line of an addition sum of any number of lines, or of a
subtraction sum, or the top line of a multiplication sum be a multiple of 99, the
answer will be a multiple of 99. If a selection of multiples of 99 are printed on a
sheet, in figures large enough to be seen by all the class, the teachers will have no
difficulty in setting the class at work at addition, subtraction, or multiplication,
each line being numbered or lettered, and the children commencing at different
parts of the sheet.
No. 2: This method I use for the same rules as No. 1, and the testing is more
rapid. It consists in making each line of the addition, each line of the subtraction,
and the top lines of the multiplication sums a multiple of 101, consequently the
answers will be multiples of 101. To test whether the answers shown are multiples
of 101, if the answer consists of 4 figures, the first two figures are repeated,
as 8484. If 5 or 6 figures occur in the answer, the right hand part of the answer
(2 figures) added to the left hand part of the answer (1 or figures) will usually
give the middle part (2 figures) of the answer, but sometimes 101 more than the
When there are more than 6 figures in the answer, as in 5374614, begin with the
two figures on the right, take that number from the next two, that remainder from
the next two, and so on, and we finally get the left hand part of the answer.
In this particular instance 14 from 46 = 32, and 32 from 37
= : 5. In some cases
it is necessary to add 101 before subtracting, but I usually select multipliers to
avoid that.
No. 3 This method of setting sums may be used for the same three rules, but
it is especially useful for multiplication of five figures by three figures. According
to this method all the answers are multiples of 1001. To test, add the number
formed by the three figures on the right to the millions, and that will either give
the thousands, or 1001 more than the thousands. To set the sums I make the
multiplicands multiples of 143 and the multipliers multiples of 7, consequently the
answers are multiples of 1001, for 7 x 143 = 1001. It then on a large sheet are
printed 60 multiples of 143 as multiplicands, and 60 multiples of 7 as multipliers,
being able to use any multiplicand on the sheet with any multiplier we have the
materials for 3600 easily tested sums. As an instance of the working of this rule
take 30459 (a multiple of 143) × 651 (a multiple of 7). The answer 19,828,809,
in which 809+ 19 = 828.
Other sums testing in the same way may be made by using multiples of 77 with
multiples of 13, or multiples of 91 with multiples of 11, for 77 × 13 = 1001,
and 91 x 11 = 1001.
No. 4 This method is useful for multiplication of 6 figures by 4 figures, the
sums being set by using multiples of 137 as multiplicands, and multiples of 73 as
multipliers, or the versa. Now 137 x 73 = 10,001, and consequently all the
answers will be multiples of 10001. They may be tested by adding the 4 figures
on the right to the left hand part, when the result will be the intermediate
4 figures, or else 10001 more, e.g. 625131 (a multiple of 137) × 3504 (a multiple
of 73) 2190459024, in which 9024 +21
= : 9045.
-
No. 5 This is useful for simple multiplication, especially of 4 figures by
2 figures, and consists in using multiples of 334 with multiples of 3, or multiples
of 167 with multiples of 6, giving multiples of 1002 as products. In each auswer
therefore the left hand part will be half of the right hand part; e.g.,
3674 x 72 = 264528. Here 264 is half of 528.
No. 6: This is useful for multiplication of 5 figures by 2 figures. I use
multiples of 667 as multiplicands, and multiples of 3 as multipliers, the answere

page_text: =
being multiples of 2001, with the left hand part of the answer double of the right
hand part, e.g. 19346 (a multiple of 667) × 63 (a multiple of 3) 1218609, in
which 1218 is double of 609. As 2001 can also be divided by 23 and 29, the same
result may be attained by using these factors instead.
No. 7 This I use for multiplication, all the answers being multiples of 20,001
or 10,002, the former being divisible by 59, 3, or 113, and the latter by 3 and 3334.
By combining multiples of these factors to make multiplication sums, the four
right hand figures will either be half or double of the rest of the answer;
e.g., 393353 × 78 = 30681534, where the right hand part is halt of the left hand.
part.
-
No. 8-This I use for addition, subtraction, and multiplication of money; and.
practice. The whole or a part of each answer is a multiple of £2 0s. 1d., or of a
quarter of that amount, viz. 10. Otd., i.e., there are twice as many pounds as pence.
Take as an instance £174 178. 3td. In 7s. 3 d. there are 87 pence; and
£174 10s. £174 (double of 87). The difference between two such amounts
will give another amount testing in the same way. To set the multiplication sums,
I adopt the same principle as in some of the simple rules. The factors for 481
(the number of faithings in 10s. Otd.) are 13 and 37, I use a multiple of each, one
being farthings expressed in £ s. d., and the other being the multiplier. Take for
instance 8s. 5d. (a multiple of 37 farthings) and multiply by 221 (a multiple
of 13). The answer is £93 13 10, i.e., 46 pence and 93 pounds.
If more figures are required in the pounds to make longer answers, the additional
figures may be some definite and easily recognised multiple of the pounds obtained
as above by doubling the number of pence. As instances, I have in the answers
of one set of sums the left hand part of the pounds 4 times the right hand part, in
another 303 times, and in another 404 times. In the two last instances the left
hand part of the pounds is tested according to method No. 2. The answers for
the three lots of sums are formed from £8002 0s. 1d., £606,002 0s. 1d., and.
£808,002 0. 1d. respectively, factors being discovered for all three. The first is.
particularly useful for practice sums, giving a great variety.
No. 9: This method I use for sums requiring pounds, shillings, and pence in
the answers, but no farthings. In each answer there are 11 times as many pounds
as pence, the shillings being reduced to pence and reckoned with them.
Now £11 0s. 1d. 139 x 19 pence. By multiplying multiples of one factor by
multiples of the other we get such answers; e.g., £14 9s. 7d. (a multiple
of 139d.) x 57 (a multiple of 19) = £825 6s. 3d., which gives 75 pence and 11 times
as many pounds.
All the foregoing methods depend upon one principle, i.e., in each set of sums.
each answer is a multiple of a certain number or amount, all such multiples being
easily recognised by certain tests, the working of which may be proved.
I also employ the following methods of testing sums, either separately or in
combination. They do not depend like the preceding ones upon making each sẹt
of answers multiples of a common factor, so in order to set multiplication sums
the answers have to be examined separately for factors to be used as multipliers to
bring the required answers.
'
No. 10:This I use for sums requiring 5 figures or more in the answer, and I
adapt it both to simple and compound rules. In the simplest form of these answers,,
the first figure of the answer divided into the number formed by the next two
figures, will give the next figure of the answer as quotient, and the last figure as
remainder, e.g., 87492 would be a correct answer for 8 (the 1st figure) divided
into 74 (the next 2 figures) goes 9 times (the next figure) and 2 (the last figure):
over.
I also lengthen these answers by letting the additional figures depend upon some
of the five figures; e.g., in a certain set of division sums I let the remainder be the
product of the last 2 figures of the quotient. In some sets the answers are real
from left to right, and in others from right to left.
No. 11-This method I use for sums requiring answers in £ s. d. The answers

page_text: -
have two figures in the pounds, and if these two figures are reversed in position,
and the number so formed called pence, on reducing to shillings: the result will
be the shillings and pence of the answer, e.g., £89+8s. 2d. would be correct,
for 89 reversed in 98, and 98d. 88. 2d. If I wish to have farthings in the
answers, I usually let the number of farthings be the remainder when the--left hand
figure of the pounds is divided by 4. I. sometimes enlarge the number of: pounds
by having one of the other tests, such as the preceding one, for the left hand
figure of the pounds. Another variation I employ is to add on 10s. 1d. to the
shillings and pence obtained by reversing the pounds.
No. 12: This is a useful system for answers with 3 or 4 figures in the pounds,
and shillings, pence, and 'farthings. £762 13. 3 d. and £7624 13s.. 5 d. are
instances of answers testing in this way. The two. left hand figures of the pounds
multiplied together give the remaining figure or figures of the pounds. If there
are two more figures they form this product reversed; if one more figure, it is the
right hand figure of the product. In above. instances, 24 is 42 (e. product
of 7 and 6) reversed, and the 2 of 762 is the right hand figure of the product 42.
The number of shillings =sum of 2. left hand figures of the pounds. The number
of --pence
=1+units figure of pounds. The number of farthings:
when the left hand figure of the pounds is divided by 4.
= remainder
No. 13 Answers are in tons, cwts., qrs., 1lbs., ozs., drs. The sum of the
Dumbers in the tons, qrs., and ozs. the sum of the: numbers: in. the cwts., (lbs.,
and drs.; e.g., in 18 tons 10.cwts. 2 qrs. 16 lbs. 13 ozs. 7: drs., 18+ 2-+113= 33, and
7+16 +10 =*33.
No. 14 Answers are in mls.; fur.; pls., yds.; ft., in., and the sum of the mis., pls.,
and ft. is double of the sum of the fur., yds., and inches, e.g., in 12 mls. 7 fur.
18 pls., 1 yd., 2 ft., 8 in., 12 +18+2=32, ie., double of 7+1+8.
1
1
No. 15 Answers are in ac., rds., pls., yds., ft., in. The sum of the acres, roods,
and poles
= 4 times the number of sq. yds., while the sum of the sq. ft. and
sq. inches. also = 4 times the number of sq. yds., e.g.in. 30 ac. 2. rds. 86 pls.,
17 sq. yds., 8 sq. ft. 60 sq. inches, 30 +2+36 68, ie, 4 x 17, and
8+60 68.
.
No. 16 Answers are in qrs., bus., pks., gals., qts., pts. 2 added to the bushels
gives the left hand figure of the quarters, while the pks., gals., qts., and pts. added
together give the right hand figure of the quarters, 67 qrs., 4 bus., 2, pks., 1. gal.,
3 qts., 1 pt. forms an instance."
Take
No. 17 An adaptation of No. 10. applied to yds., ft., and inches.
8262 yds. 2 ft. 7 in. as an example. Adding the ft. and inches, we get 9, and that
divided into 26 (the next 2 figures) goes 2 (the next) and 8 (the last) over.
No. 18 This is used for addition, simple or compound rules. Each answer is
tested from one line of the sum, preferably the bottom line but one.
In simple
addition all the answer except the left hand figure is obtained by subtracting
1 from each figure in the testing line. The left hand figure I sometimes make
the same as some figure occupying a definite position in the sum, eg., the first
figure in the second line. In £ s. d. sums the answer is obtained by subtracting ‡d.
from the farthings in the testing line, 1d. from the pence, 1s. from the shillings,.
and 1 from each figure in the pounds. Should the figure "0" or a blank space
occur in the testing line, it will be represented in the. answer by the highest
quantity we could have-9 in simple addition, or in the pounds, 19 in the shillings,
Il in the pence, and d. in the farthings. For printing on large sheets to be hung
before the class I arrange it so that with whatever line we commence, a certain
number of consecutive lines shall have a self-testing answer.
Having now particularly described and ascertained the nature of my suid
invention, and in what manner the same is to be performed, I declare that what I
claim is:-
1. The setting and the publication of arithmetical tests, whether straight-

page_text: forward or in the form of problems, the answers to which test in any or all of
the various ways herein described.
2. The setting or publication of sums in which any of these methods form an
essential part of the test.
Dated this 27th day of Mar 1905
CHARLES RHODES.