5. Alfred, Edward, and Herbert come each with his pail to a well; when a question arises about the quantity of water in the well: but none of them knowing how much his pail will hold they cannot settle the dispute. Luckily Mary comes up with a pint measure, by aid of which they discover that Alfred's pail holds half a gallon more than Edward's and a gallon more than Herbert's: but before the precise content of any pail is found out an accident happens and Mary's measure is broken. They are now however in a position to ascertain the quantity of water in the well: for they find that it fills each pail an exact number of times; and that the number of times it fills Edward's is greater by eight than the number of times it fills Alfred's, and less by forty than the number of times it fills Herbert's. How much water was there in the well?

Ans. 15 gallons.

6. Seven-ninths of the stronger of two glasses of wine and water of equal size is mixed with two-ninths of the weaker, and the remainder of the weaker with the remainder of the stronger. The stronger of these two new glasses is a certain number of times stronger than the weaker; and the stronger of the two original glasses was twice the same number of times stronger than the weaker. Compare the strengths of the two original glasses: the strength of a glass of wine and water being defined to be the ratio of the quantity of wine to the quantity of the whole mixture in the glass. Ans. 4: 1.

7. Two Tyrolese Jäger agree to shoot at a mark on the following terms: each is to shoot an even number of times fixed for each beforehand, and for every time that either hits he is to receive from the other a number of kreuzers equal to the whole number of times that he misses. They have two matches without varying the conditions. In the first match, the second Jager misses as often as the first hits in the second match, and the first Jager misses twice as often as the second hits in the second match; and the second Jager has to pay to the first a balance of 4 kreuzers. In the second match, each hits exactly the number of times most favourable to him, and the second has to pay to the first a balance of 36 kreuzers. How many times did each hit and miss in each match?

Ans. In the first match, the first Jager hit 4 times and missed 16; and the second Jager hit 6 times and missed 10.

In the second match, the first Jager hit and missed 10 times; and the second Jager hit and missed 8 times.

(x+y)(x+2)=a*,)

(y+z)(y+x)=b3,

(z+x)(z+y)=c3.)

Ans. x= abc (-1+1+1), and similarly y and z.

11. x3+y3+z3= 3xyz,

2

3a-x+8=3b-y+x=3c−x+y.)

Ans. x-a-b, y=b-c, z=c-a.

12. Obtain in the simplest form the value of x which satisfies the equation

(x+a−b-c−d)(x−a+b-c−d)(x−a−b+c−d)(x−a−b−c+d)
=(x−2a)(x−2b)(x−2c)(x−2d),

and investigate three equations independent of x which include all those
relations betwen a, b, c, d, for any one of which the equation becomes an
identity.
(2) a+b=c+d,]

Ans. (1) x=(a+b+c+d).

a+c=b+d, a+d=b+c.)

13. Two numbers are as 3 : 5, and their G.C.M. is 555: what are the numbers? Ans. 1665 and 2775.

14. Each of two points moves uniformly in the circumference of a circle one goes round 5 times while the other goes round twice; and at the end of 21590 days they return for the first time to the place from which they started. How long does each take to go round?

Ans. 4318 days, and 10795 days. 15. Four chronometers, which gain 6, 15, 27, 35 seconds a day respectively, shew true time. After how many days will this happen again? Ans. 43200 days.

16. Two points revolve in the circumferences of two circles. At starting they are in a common diameter of the circles; and before arriving again simultaneously in their initial positions, they have been in common diameters n times or m times, according as they have revolved in

the same or in opposite directions. Prove that m+n and m-n must have a common divisor 4, and no higher common divisor, and that the ratio of the periods of revolution of the points is m+n: m-n.

17. Three particles start at the same point to move uniformly in the circumference of a circle, and the period of the first: that of the second: that of the third :: a prime number: a second prime number: a third prime number. The first and second arrive together at the starting point 6 min. before the first and third, and 28 min. before the second and third arrive there together; also when the first and second arrive there together, they have been together as many times as the second and third when they arrive there together; and all three arrive there together 1 h. 55 min. 30 sec. after starting. Determine the periods of revolution. Ans. 1 min., 34min., 51min., respectively.

18. If the G.c.M. of m+n and m−n be 4: prove that the G.C.M. of m and n will be either 2 or 4; and that when it is 2, each of is odd,

and when it is 4, one of

4

m n
4

is odd, and the other even.

m n

"
2 2

19. From the longer of two rods a piece equal to the shorter is cut off: with the shorter and the remainder of the longer a similar operation is performed; and so on, until the rods are of equal length. Prove that if this last length be taken for the unit of measurement, the lengths of the original rods will be represented by numbers prime to each other.

Explain that there are many relative lengths of the original rods for which the cut rods will never become equal, however long the operation be continued.

20. Define commensurable magnitudes; and prove that magnitudes which are commensurable with the same magnitude are commensurable with one another.

21. The adjacent sides of a rectangular parallelogram are respectively equal to the hypothenuses of two right-angled triangles whose sides are commensurable with the unit of linear measurement. Prove that its area will be commensurable or incommensurable with the corresponding unit of square measurement according as its sides are or are not commensurable with each other.

22. Shew how to find geometrically lines equal to the G.C.M. and the L.C.M. of two given commensurable straight lines.

23. Prove that

a*(b2-c3)+b*(c2—a2)+c' (a3—b3)
a3(b−c)+b2 (c—a)+c2(a−b)

=(a+b)(b+c)(c+a).

24. Obtain the continued product of

also of

a+b+c+d, a+b-c-d, a-b-c+d, a-b+c-d;

-a+b+c+d, a−b+c+d, a+b-c+d, a+b+c-d:

and shew that the sum of these products=16abcd.

25. Find the continued product of n such trinomials as
x2-ax+a2, x-a2x2+a', x®-a*x*+a®, x1-a"x"+a', &c.

16

26. Find the sum of n such fractions as

2x-a 4x3-2a1x 8x2-4a1x3 x2-ax+a3' x1-a2x2+a1' x3—a1x1+a® ›

89

&c.

Ans.

2m t

+a

2mx2-1+maxTM-1 2x+a
2m x2+ax+a2'

where m=2".

27. Prove that

(a2—b®)3+ (b3—c3)2+(c2—a3)3 =(a+b)(b+c)(c+a).
(a−b)3+(b−c)3+(c—a)3

28. For what unit of time will the durations 115743. and 360360". be represented by numbers prime to each other? Resolve the same question for 36 hours and 2.76 hours.

Ans. (1) 18 seconds. (2) 432 seconds.

29. Divide 1+2x2+1+x12+2 by 1+2x+x: writing down the (2n-m+1)th and (2n+m+1)th terms in the quotient; the (2n-m)a and (2n+m)th remainders; and the complete quotient.

Ans. (1) (-1).(2n−m+1).x2”—”.

...

...

...

...

(2) (-1).(2n-m+1).x2+.

(3) (−1)TM{(2n−m+1)x2-m+ (2n−m)x2n-m+1}+2x2n+1+x4n+8 ̧ (4) (-1){(2n-m+1)2+m+(2n-m+2)x2+m+1}+x4n+8.

(5) 1-2x+3x3-...+(2n+1)xa3”—2nx2n+1+...+x1”.