Institute of Chartered Accountants in England and.

The Accountants' manual, Volume 10 online

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Pearson, S. H. (R. Mellors). Nottingham
Pitts, J. W. (S. S. Dawson), Liverpool
Porter, R. R. (W. Porter). Blackburn

Rees. p. M. (E. E. Price), London

Robertson. F. (R. H. Holmes), Newcastle-upon-Tyne

Robertson, L. G. (P. D. Hannay). London



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164 December 1904 — Successful Candidates.

PINAL BXAl^iNATlOfi—contmved.

(In alphabetical order)

Russell, H. H. (A. J. Williams), Birmingham

Saunders, B. L. (F. Whinney), London
Slattery, F. W. (J. H. Stephens), London
Sloman, F. S. (J. J. Deller), London
SouTHWORTH, H. (R. E. Smalley) Preston
Swift, G. H. J. (J. H. Stephens), London
SwiNsoN, J. H. (E. Hayes), London

Tyler, R. B. (J. S. Mallam), London

Walker, J. S. (H. M. Branford), London
Walters, A. H. (H. Walters), London
Watson, H. (R. G. M. Creasey), London
Whitehouse, C. (H. Ashford), Birmingham
Wickenden, p. (R. G. M. Creasey), London
WoosTER, M. C. (P. H. Le Sueur), London
Wright, W. (J. A. S. Hassal), Liverpool

63 Candidates failed to satisfy the Committee.



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NRT II. VOL I.

The Bss(D)ennitaiinil§
Mamnuiaflo

Being the Answers to the
Examination Questions of the
Institute of Chartered Accoun-
tants.



Preliminary,
Intermediate, and Final.



JUNE 1905.



The Answers to the Questions set at the Preliminary Examination are by
the Rev, G. T. P. Strceter, of the Rectory, Orcheston St. Mary,
S.0„ Wilis,



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PRELIMINARY EXAMINATION, JUNE 1905.



The ANSWERS to the QUESTIONS set at the Preliminary Examination are
by the Rev. G. T. P. STREETER, of the Rectory, Orcheston St, Mary, S.O.,
Wilts.



ALGEBRA.

Q. 1. — How can you write down the product of two binomial
factors like x + a,x + bhy inspection ? Illustrate your answer by
writing down the products of

(x + 6) X (x + 7),
{x - b) X {x + 13),
(x + 0-2) X {x- 10).

A. 1. — ^Pat down for first term a^; for teoond x with a coefficieQt=the sum
of a and b ; for third, the product ab

(x+6) (a?+ 7) = a;«+18a;+42
(«-6) (a?+13) = a?>+ 8a;-66
(x+'2) (x-10) = a;«-9-8x-2

Q. 2.— Divide a'^-b^+c^+Sabc by a-b+c.

Having found the quotient, can you write down by inspection
the quotient of a* — 6* — c* — Sabc by a — 6 — c ? If so, do it,
explaining your method.

A. 2.—

a— 6+c



"^ = (fi+ab+b^
a—b



a»-6»+8a6c+c» I a«+a6+6*-(a-6) c+c«



{a^-2ab+b^) c-{a-b) c«



[a-6)c2+c»



Quotient =a»+a6+62 _ ac+ -fcc+c'-
The 2nd quotient is obtained by putting - c for c in the first, as the^
dividend and divisor of the second oouple of expressions is obtained
from the first by so doing, and=a'+a6+6*+ac-6c+c«

U 1



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158 PREUMINARy EXAMINATION.

Q. 3. — Beduce to a common denominator and simplify the
following expressions, checking your answer in each case by
putting x=2, 2/=l : —

(i.) 3 2 5|^.



,.j , _ 1 1



ic^-\-bxy-\-^y^ x^+7xy+12y*



A. 3.-






x2-y«
If x=2 and y=l

_ - + -=-, which is correct.
3 18 3



(ii.)



x^-^bxy+ey^ a;*H-7xy+12y«
1 1



(x+2y) (x+8y) (a;+3y) (a;+4y)

ar-|-4y -j- x-\-2y
(x+2i/) (X4 3y) (x4 42(r)

(X + 2y) (ar + 4^) (X + 31/) (x + 2y) (x + iy)
Substituting

_ 1 4. _J ^ == ^ ^_l_
4+10+6 4 + 14 + 12 4x6 12

i.e., -^ + J[^ = ?±-^ = J, = i. which is correct.
20 30 60 60 12



Q. 4. — A sum of £25 is to be divided in gratuities between three
clerks, so that the head clerk is to receive £5 more than the second,
and the second £5 more than the third. What sums will they each
receive ?



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ALGEBRA. 159



A. 4.—

Let £x => receipts of 3rd clerk

3« = 10

X =i?^= £3 63. 8d.
3

.*. 3rd clerk has £3 6 8 1
2nd ,, „ £8 6 8r Ana.
Head,, ,, £13 6 8)

£25



Q. 6. — Find either algebraically (without quoting formulae) or
graphically, using the squared paper provided, approximate values
of the roots of the equation

ir«-2x+0-5=0.

Explain your work if you use the graphic method, and state he
scales you have used.



A.


5.—












Algebraically












JC* -


2x + i


=








sfi -


2x + 1


= i










a: - 1 =




■■ ±




V2 =


= 1-414


X =


2 ± v'a

a










=




•586
2








=


1-707 or


293



v'2



Graphically .—Takhig a scale « i and drawing XOX^ and YOY^ at right
angles to each other, we find the curve represented to be a parabola
MNAP, oattmg OY in M, so that OM » i, OK in N and P, so that
ON s 'QQS and OP = 1*707, and turning at point A, whose co-
ordinates are (1, - }).



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160



PRELIMINARY EXAMINATION.



Q. 6. — A yard can be paved with 900 square tiles of a certain
size, or with 400 larger tiles measuring 3 inches more each way.
What are (i.) the sizes of the tiles, (ii.) the number of square feet in
the yard ?



A. 6.-



Let X ins. - length of small tile
.-. 900{ca = 400 (x + 3)« = Area in square inches.
600ic8 = 400 (6x + 9)
6a;» - 24aj = 36
x«-2_4 I2p^l44 36

= 32£

25

18

5
12 ± 18






^ reject - sign



30 «
= _ - = 6



/. a; + 3 = 9



Answers (1.) Size of tiles — 36 or 81 sq. ins.
(ii.)Area = 92°Al^,.ft.



144
900



= 226 sq. ft.



144
180



324

1



18



224



Q. 7. — Using the squared paper provided, find by the graphic
method the approximate roots of the equations

0-7a;+My = l-64, l-7a?-l-3i/=2-21.

Mark clearly on your diagram the axes of x and y and the scales
you have used.

A. 7.—

The equations represent each a straight line which out one another at
the point x = 1*594 and y = *385 nearly, .*. taking scale » } the
point of Qpntact (O) will be in the first quadrant. If AG be the first
line mentioned, it wiU cat YOY^ above the aads of ar, and BO the
second will oat it below.



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ALGEBRA. 161



A. 7 — {continued) »

To find these values, we have

1-lx - l'3y = 2-21

7x + I'ly =1-54

Subtract

Multiply by •?

Equation (2)

Subtract 2-78y = 1071



. X - 2'iy
Ix - l-68y
•7a: + l-ly


E


•67
•469
1-54



(2)



y = S^ = -386 nearly



.-. X - 2-4 x-385 = -67

a; = -67 + '924
= 1*594 nearly

Q. 8. — Solve the equations : —

(i.) I0y^x=l2 \

a?2/=32-5 J '
(ii.) x»-2/'=91 I
a:— y=l )
A. 8.—

(i.) lOy - X = 12 I

xy = 32-5 = 32J = ^ |
.-. lOj/8 - X2/ = 12y
lOj/* - 12y = xy ^^

2^ - ^ ;^a ^ 65^ ^ 13 J_

^ 5^ Si 20 4 ^ 25

= 325 + 36 _ 861

100 100

3 ^ 19

^6 10

y . l±i? = ?5 or - 13-
^ 10 . 10 10

2 10

x=g.xi=^-x^ = i3



y 2' 6
18



orx«-^X-^= -25



Ans. 13 and 2\ or - 25 and - IL



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162 PRELIMINARY EXAMINATION.



A. S— {continued).

(ii.) a;" - y^ = 91}

X - y = If

Divide a« + xy -\- y^ =^ 91
x'^ - 2xy -\- 1/* = 1



Subtract Sxy =90 /. oiy ^ SO

.-. ar2 + 2xy + 2/» = 121 .*. a; + y = ± 11

g - y = 1
2ic = 12 /. a; = 6l
2|r = 10 y = 5f

If a; + y = — 1 we have

Anb. 6 or - 5 and 5 or - 6



X = - 5)



Q. 9. — An examiner, marklDg different batches of examination
papers, finds he has taken the following times in marking : —

100 Algebra, 50 Geometry, 30 Higher Math., 32 hrs.

60 „ 100 „ 40 „ „ 34 hrs., 20 m.

30 „ 40 „ 100 „ „ 35 hrs.

Find the number of minutes occupied, on the average, in marking a
single paper in each subject.

A. 9—

Let X » number of minutes lor the Algebra paper
y = ,, „ „ Geom. ,,

a = M ,, „ Higher ,,

Then arranging and dividing by 10

lOa; + 5y + 8a = 192 \ (1)
5a; + lOy + 4a = 206 [ (2)
3a; + 4y + 10a = 210 ) (3) ^

Subtract (1) from (2) - 5a; -f 6y + a = 14 (4)
„ (2) „ (3) - 2a; - 6y + 6a = 4

- X - 3y H- 3a = 2
Multiply (4) by 3 - 15a; + 15y + 3a = 42
Subtract - 14a; + 13|^ ==40

- 7a; + 9y =20
and 3a = 2 4- a; -I- 3y



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ALGEBRA. 163



A. 9 — (continued).

Sub. in (1) /. lOx + 5y -f 2 4- ar + 3?/ = 192

11a; + 8y = 190

- 14a; 4- 18y = 40

Subtract 25a; - lOy = 150

5x - 2y = 30

20x - 8y = 120

11a; + 8y = 190

20a: - By = 120

Add 3ix -= 310 .-. a; = 10

also



and



6x


-2y


= 30






50


-2y


= 30








2y


= 50 -


30 =


20


Sz ^


2 + x


+ By






=


2 + 10 + 30


= 42




a =


14









/. y = 10



Ak8. 10 minutes each for Algebra and Geometry Papers.
14 yt ,, Higher Mathematics ,,

Q. 10. — Find the Highest Common Factor of

Are there any steps in the process which differ from the
arithmetical work of finding a Greatest Common Measure ? If so,
which, and why are they introduced ?

A. 10.—

a;» - lla;* + 88a; - 40

tx^ - 13ara + 54a; - 72

2a;» - 16a; + 32 = 2 (a;« - 8a; + 16) = 2 (a; - 4)*

Two steps, differing from the arithmetical work of finding a G.C.M.
are here introduced :—

(1) 2 as a factor of the H.C.F. may be rejected, as it cannot divide the

coefficient 11 in one expression.

(2) One factor (x - 4) is also rejected, as 16, the last term of

a;* - 8x + 16 cannot go into 40 in same expression.



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164 PRELIMINARY EXAMINATION.

A. 10^{continued).

They are introdaoed to simplify the solution, this would be more
difficult and less needful in arithmetic,
a; _ 4 I x« - llx« -f 38x - 40 I x2 - 7x + 10



X8 -


llx« -f 38x -
4x«

7a;» + 88x"
7a:« + 28a;


- 40




lOr -
lOar -


- 40

- 40



Aus. ic - 4.



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ARITHMETIC.



165



ABITHMETIC.

Q. 1. — The following were the percentages of Reserves on
Liabilities for the Bank of England in each week of 1903 : —



Date


Percentage


Date




Percentage


Date


Percentage


January 7


387


May


6


522


September 2


493


• 14


441




13


37-4


9


5I-I


• 21


460


«


20


51-5


16


544


28


474


m


27


507


23


539


February 4


486


June


3


496


30


439


11


492


»


10


530


October 7


429


18


480


m


17


53*1


» 14


440


25


477


m


24


51-6


m 21


44-9


March 4


45*4


July


I


420


28


496


II


470





8


51-5


November 4


47*4
48-5


18


487


m


15


519


II


. -, 25


47*2





22


520


18


499


April I


433


1


29


495


« • . ^5


49*5


8


48-1


August


5


499


December 2


437


15


481


«


13


49-6


9


450


• 22


516





19


471


16


437


29


490


1


26


48-2


23
30


369
329



Find the average percentage for the year to the second place of
decimals.

On the squared paper supplied draw a diagram exhibiting the
fluctuations of the percentage during the year, and insert a line
showing the average. Choose your own scales.

A. 1.—

let column added = 798-1

2nd „ „ = 840-8

3rd „ „ = 834:8

4 I 2473-7



13 I 618-426

47-571 Ans.
The diagram will be an oblique line ascending for the first six weeks, then
declining for three, then ascending not so high as before for two, &o., &c.

Q. 2. — Find the values of the following products by the shortest
methods you can : —

327-2 X 126, 7674 x 3-16, 6832 x 19-98.



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166 PRELIMINARY EXAMINATION.

Divide 3764-986 by 54-87, stating the answer to two places of
decimals.



A. 2.—



8272

5

16360

5
8180

5

40900 Ans.





816 = 3^^

7674
19
68066
7674


= m


= 3J =


= V




6


144806 . .












2413433 Ans.










19-98 = 20 -
6832
20


•02

multiplication by
Ans.






13664000

136-64 =
136503-36


02



5^-^X I 3764-986 | 68 64 Ans.

! 3292 2 !

I 472-7

I 438 4 I

34-4

32 4

20

20



Q. 3. — A second-class season ticket from a oertain station to
London costs £4 6s. 3d. for three months, while the second return
fare is 3s. 7^d. If I go up to town and back on 273 days in the
year, what do I save by taking out a season ticket ?



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ARITHMETIC.



167



A. 8.—



d.

7i = 3-6268.

ahil.

3-625

273



10 876
258 76
725

989 625 = £49 98. 7Jd.
£4 68. 3d. X 4 = 17 58. Od.
£82 48. lid.
Saving per year = £32 4s. 7Jd.



Ans.



Q. 4. — If a bicycle wheel is 28 inches diameter over all, how many
revolutions will it make in a 20 mile ride ?

(Take the ratio of the circumference of a circle to the diameter as
3*14159. •., but state your answer only to the nearest revolution.)

A. 4.—

Diameter = 28 in8.

Circumference = 28 x 314159 in8.
It will make one revolution in going its own circumference.

. _ 20 X 1760 X 36



176


A^b. —


28 X 314159


18


=


5 X 1760 X 86


1408

176

8168

814159

7


7 X 314159


_ 180 X 1760 X 100000
7 X 314159
31.680,000,000


2199113


2199113




219^U)^


31680-00 14406
2199113

968887

879644




89243
87964




1279
1096

184




Ans. =


■• 14405 revolutions.



Q. 5. — Find, to the nearest penny, the total interest on £1,000
invested in 2^ per cent. Consols at 89ff , allowing brokerage.



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168



PRELIMINARY EXAMINATION.



A. 6.—



Suppose the brokerage = i %
Price of stock = 89JJ
= 90A



+ i



AN8. = "^.-^



1000 j^

90iV



16
90

1440



1441



£ s d
27 15 1



1000 X 5 X 16
2 X 1441

40000 „
1441

40000
2882
11180
10087
1093

20

218(50

1441

7450

72a5

245

_12

1740

1441

299



Ans.



Q. 6. — If, in makiDg a road, one load of ballast is required for
every 3 sq. yds., how many loads will be required for a road 540 ft.
long and 36 ft. 6 in. wide ?
A. 6.—



540x36i

AN8. - ^ ^ g

540 x 73



54



= 730 loads.



Q. 7. — According to ** Whitaker's Almanack " the Japanese 20-
yen gold coin is worth £2 Os. llfd., while the Russian gold piece of
15 roubles is worth £1 lis. 9d. If Russia had to pay Japan in gold
a war indemnity of 150,000,000 roubles, what would be the
approximate equivalent in yen ?



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ARITHMETIC. 169

A. 7.—

By ChaiQ Rule.

Answer in yen == 150,000«000 roubles.
15 roubles = Slf x 12 pence.
491} pence <= 20 yen.

_ 150,000.0 00 X 127 X 3 X 20 X 4
^^®- " 16 X 1967

150,000,000 X 127 x 16 16

1967 _i6

804800.000,000 _„ ,^

= — w— ^*- ^

1967 I 804800 I 154 127

Li?5Z — ! 1[680

10810 480



9885



240_

80480



9760
Ans. = 154,000,000 yen approximately.

Q. 8. — What is meant by compound interest ? What will be the
Compoand Interest on £1,000 for two years at 2J per cent, payable
half-yearly ?



.. 8.—




Compound Interest is Interest on the Interest.


The question is the same as Compound Interest for 4 years at IJ %.


Operating fraction =
10 1 1000-


li 5 1
100 " 400 ~ 80




8| 100-




125




10 1 1012 5




8 1 101-26




12-66626




10 1 1026- 15625




8| 102-5166




12-8145




10 1037 9707




8 103797




12-974




1060*944 »
1000


Amount in 4 half-years.



60-944 s= Compound Interest.
Anb^. = £5 18s. lOid.



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170 PRELIMINARY EXAMINATION.

Q. 9. — If you had two long pieces of string, of unknown lengths,
and no measuring tape or rule, how could you nevertheless find out
the greatest common measure of the two lengths, supposing this to
have a considerable magnitude ? You may assume if you like, for
the sake of illustration, that the two strings are actually 51 and
39 ft. long.

A. 9.—

Let the two striogs bang together ; cut off the piece by which the
longer exceeds ; the greatest common measure is contained in it.

Measure this piece along the shorter, and, if any piece be over, cat
off that.

If nothing be over, the first piece cut is the greatest common measure.

If something, treat the two cut off pieces as tbe two strings ; measure
one into the other as often as it will go. If nothing be over, the second
piece is the greatest common measure, i.e., it will measure an exact number
of times in each string.

If anything be over, cut it ofi and measure it into the second piece, and
so on, till you get a piece which will exactly go into the one cut off before
it. The last will be the greatest common measure.

ft.
51
39

12 I 39 I 3
|_36j
3 I 12 I 4
I 12 I

The greatest common measure is the second piooe out off, viz., 8.
It is immaterial whether the pieces of siring be long or short.

Q. 10. — What is the difference between true discount and banker^;'
discount ?

What would be the true discount on £826 9s. 4d. due three years
and nine months hence, reckoning 4 per cent. Simple Interest ?



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AlUTHMETIC.



171



10.—

True discount = the interest on the present value.
Banker's discount = interest on the full sum.
Take £20 or any convenient sum.
Interest or Bankers' Discount on £20



^ 20 X 3 f X 4



100"
20 X 15
100



£



£ = £3



£3 = true discount on £23

£ £

23 . 3 : :

23



£


3 d




826 9 4




3


£


^4479 8


107


23




179




161




18




20 B.




368


16




28






188






138







Ans. £107 168. Od.



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172 PRELIMINARY EXAMINATION.



ENGLISH COMPOSITION.

Write an Essay on one, but not more than one, of the following
subjects : —

(1) The Profit and Pleasure to be derived from Reading.

(2) Modern Methods of Locomotion : their Advantages and

Disadvantages.

(3) The History of the Steam Engine.



The Profit and Pleasure to be derived from Beading.

Bacon says : " Beading maketh a full man " ; and it is
a necessary habit for anyone who aspires even to pass muster as
a well-informed man in the present day. But it may be too
multifarious, and many men read more than they can hold and
assimilate. Observation and conversation are no mean aids to
the acquirement of knowledge, and reflection and thought are
necessary to make the best use of what is acquired.

We all read nursery books, then school books, then newspapers,
then special works which treat of our calling or hobby. Then, for
mental training, first the Bible as a storehouse of religious truths
and, in our own version, a literary masterpiece.

Delight in stories is as old as the human race, and we
doubtless all read novels. But here a caveat is necessary, for
in them pleasure is the first thing, advancement of learning the
second, and sometimes a very bad second. A mind trained on
novels is like a body fed on sweetmeats ; we read them without
the trouble of thinking, a good thing often for a mind which
has already done its thinking and is jaded— but weakening for
one that has not. The mind that thinks not will rust; that



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ENGLISH COMPOSITION. 173

which is fed by impure garbage will cormpt. Beading with
many means only reading novels, and this eventually beoomes
a waste of time, or a turning of time to a bad use.

A course of History, or Science, or Art should be taken up
as primary, and more general knowledge as secondary ; novels
after these. Poetry is not generally cared for in youth ; it may
be studied in after life with a higher appreciation. Shakespeare
should be read through ; he yields both profit and pleasure. His
development of character, analysis of motive, eloquence of
description, and healthy tone, give his works an educational value
of a very high character. Milton, Spencer, Chaucer, Campbell,
Collins, Gray, Tennyson, Browning, Longfellow, and others, will
elevate as well as amuse.

It may be asked. Should one study an Encyclopaedia ? I fear
that in these days we must, as a work of reference, but it
tends to make our reading desultory. Desultory reading is not
without its value, but it is far below systematic. Still, many
men have been eminent — e.g., Samuel Johnson — on desultory
reading.

A systematic course of Historical reading will be of enormous
use. Firstly, our own country in Hume, Hallam, and Macaulay,
&c, ; secondly, French history, almost more interesting than our
own ; then the histories of the German empire, Italy (of marvellous
interest), Spain, Russia, Switzerland, Scandinavia, India,
Palestine, Turkey, the Saracenic Empire, Egypt, China, Japan,
the United States, Canada, Cape Colony, Australia— all will give us
an insight into life in olden times, and ability to take just views of
the present, in which our lot is cast for weal or woe. May it be
for the former in the highest spiritual sense I



N 2



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174 PRELIMINARY EXAMINATION.



GEOMETRY.

Q. 1. — How can you test on your drawing-board the angle of a
set-square which is supposed to be a right angle, basing your test on
Euclid's definition ?

A. 1. — Place one of the sides containing the right angle, the
base, on the lower side of the board, near the middle, then draw
a pencil line along the other, the perpendicular.

Now turn the set square over on the perpendicular, the angular
point remaining the same.

Draw a second pencil line along the perpendicular. If these
two pencil lines coincide the set square is correct.

Q. 2. — State and prove any construction for bisecting a given
straight line.

A. 2. — See EucUd, Book I., prop. 10.

Q. 3. — Are any of the following triangles impossible ? If so, state
why : —

(i.).a triangle with sides 3*7 inches, 4*4 inches, and 8-2

inches long ;
(li.) a triangle with angles 90°, 27°, 50° ;
(iii.) a triangle with two angles 67°, 114° ;
(iv.) a right-angled triangle with sides 3 inches and 4 inches
and base 5 inches long.

A. 3.—
(i) Impossible, as two sides 3-7 -f 4-4- are less than the

third 8-2.
(ii.) Impossible, as 90° + 27° + 50° = 167°, which is less

than 180°.
(iii.) Impossible, as 67° + 114° = 181°, which is greater

than 180°.
(iv.) Possible, as 5* = 3* -f 4^



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GEOMETRY. 175

Q. 4.<^Proye that, if a side of any triasgle be produced, the
exterior angle is equal to the two interior and opposite angles.

A. 4. — See Euclid, Book I., prop. 82.

Q. 5. — Find the area of a triangle with sides 13, 14, and 15 inches
long respectively.
A 5

Let » = li+-l* "t^^ = 1^= 21
2 2



Area = V'* (s - 13) (s - 14) (s - 16)
= v'21 X 8 X 7 X 6
= 7 V"3 X 2 X 4 X 6
= 7x6x2 = 84 square inches. Ans.

Q. 6. — State and prove a construction for dividing a straight line
into two parts so that the rectangle contained by the whole and one
part may be equal to the square on the other part.

How may this problem be expressed in algebraic terms ?

A. 6. — See Euclid, Book II., prop. 11.

Let a = whole line and x the one part



a — X and x are the two parts


/. (a - xy = ax


a* — 2ax + x^ = ax


x^^3ax + ^^


■ - „.+»»•


5a«


4


3a_ , a^5

2 "^ 2


^ 3a±(XA/5

X ^- -


=


"'-/^



as 2; < a the + sign must be dropped.



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176 PRELIMINARY EXAMINATION.

Q. 7. — State and prove a construction for finding the length of
the side of a square the area of which shall be equal to that of a
given rectangle.

Indicate the dimension of the lines on your figure for the case in
which the sides of the rectangle are 9 inches and 4 inches
respectively.

A. 7.— See Euclid, Book II., prop. 14.

As 9 X 4 = 36 = 6' the dimensions of the sides of the
square figure will be 6 inches.

Q. 8. — Show that the angle at the centre of a circle is double the
angle at the circumference on the same base.

A. 8.— See Euclid, Book III., prop. 20.

Q. 9. — Show that, if AB, AG are tangents to a circle and AD
bisects the angle BAG, AD passes through the centre of the circle.

Describe and sketch an instrument based on this property for
enabling you to find the centre of a circle or circular disc, or the
centre of the cross section of a circular cylinder.

A. 9. — If the line AD do not pass through the centre let the
centre be 0, and let it be, if possible, a point outside AD.

Join OA, OB, 00.

Now in triangles OAB, OAO

'.' OB and 00 are radii and AB and AO tangents at points
B and C.

/. each of the angles OB A, OCA is a right angle.

also OB = 00 and OA is common.

/. they are two right-angled triangles which have two sides of
the one equal respectively to two sides of the other.

.'. they are equal in every respect.

.-. angle OAB = angle OAO.

/. angle OAB = \ angle BAO.



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GEOMETRY. 177

But angle DAB = J angle BAG.

/. angle DAB = angle OAB the less to the greater, which is
impossible. .*. point O cannot be outside DA. /. DA passes
through the centre. Q.A.D.

(ii.) A pair of compasses with the arms fixed, and a rod fixed
bisecting the angle between the legs, just above the plane of the
legs, would answer the purpose for a circular disc.

Place the compasses so as to touch the circle in two points ;



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