25. Suppose a mixture of gold, brass, iron, and tin, to weigh 60lb., and the weight of gold and brass together 40lb., the gold and tin 451b., and the gold and iron, 3616.; how much was the weight of each? Ans, Gold 30lb., brass 911., iron 5416.,

and tin 141lb..

26. If 2 be 3, and 3 be 5, and 6 be 11, what is the half of 26, and the fourth of 27?

Ans. 55, and 2822.

27. If a stone of 40lb. weight was to break into four pieces, such that by help of them alone, a person might weigh 1, 2, 3, &c. lbs. to 40, what should the weight of each piece be?

Ans. 1, 3, 9, 27lbs. respectively.

28. The ages of A, B, and C are as follow: the age of A, together with half the sum of the other two, is 73; the age of B, together with one-third the sum of the other two, is 54; and the age of C, together with one-fourth of the sum of the other two, Their respective ages are required..

is 38.

Ans. A's 48, B's 32, C's 18 years.
29. A landed man two daughters had,
And both were very fair;

To each he gave a piece of land,
One round the other square.
At twenty shillings an acre just
Each piece its value had;

The shillings that did compass each
For it exactly paid.

If 'cross a shilling be an inch,

As it is very near,

Which was the greater fortune, she
Which had the round, or square?

Ans. Area of square=250905.6 acres.

Area of circle 197060.63125 acres. And of course each equal to as many pounds; which shew that she who had the square was much the better fortune. This question may be done by Position or by Algebra.

30. If a roll of butter weighs in one scale 2b., and being changed into the other, weighs 4lb.; what is the true weight of the butter?

Ans. 3lb.; for 2.25 x 4= 3; which is a true rule for all such cases, viz. the square root of the product of the respective weights.

31. I have seen a pack of cards, with 32 numbers on each, so contrived, that the numbers in their natural order, from 1 to 63, were distributed amongst the cards in such a manner that the same numbers are to be found on 1, 2, 3, 4, 5, and 6, of them, and others but on one or two of the cards. By help of these cards, one can tell another whatever number he may chance to think upon, if it does not exceed the highest, provided he be informed of the cards on which this number may be found. It is required to construct these cards, and explain how one may guess to such certainty.

On the following page you see, that 1, 2, 4, 8, 16, and 32 are in the 1st horizontal row, and no where else; but 1+2=3, shews that 3 is to be found in the first and second columns because it requires 1 and 2 to make 3. If it requires three or more of the top numbers to make one of the numbers in the table, as 1+2+4+8=15, the number 15 is found in all the columns in which 1, 2, &c. are found at top; and so it is with all the other numbers, which are generated in the same way. The horizontal rank of numbers, 1, 2, 4, &c. are in geometrical ratio, and the perpendicular rank on the left in arithmetical ratio; both together, the abscisae and ordinates of the logarithmic curve. The rank of geometricals at the head, viz. 1, 2, 4, &c. afford such combinations as make up all the other numbers. In whatever columns any number stands, the sum of the geometricals which stand over such number, is equal to it; and the reason why one can guess any number thought on, is, the sum of the numbers which stand over it, in the first horizontal row, makes it, and no other, exactly.

EXAMPLE.

Suppose I should fix upon the number 50, and ask you to tell which I thought on; you would ask me in which of the columns, and how many, does it stand in? I would say, in the second and two last columns. You would then put together 2+16+32

50, and tell me to a certainty; because 50 stands no where else but in the second and two last columns. Whatever number you guess upon, if only in the first, second, third, fourth, fifth, or sixth columns, it must be 1, 2, 4, 8, 16, or 32; because each of them are found but once; but where it takes two or more of the figures in the horizontal row of geometricals, to make the one you think on, the number is to be found in all the columns, exactly under the geometricals which compose it, and no where else. Hence it is easy both to make and understand them; and even to extend the cards to much higher numbers. If another column was added, its head number would be 64; from which a table could be generated, that would give all the numbers from 1 to 127; because 63+64-127.

32. Divide 40 into four such parts, that if to the first you add 4, from the second subtract 4, multiply the third by 4, and divide the fourth by 4, the sum, difference, product, and quotient will be all equal.

9.

Ans.,,, and .

33. Into what three whole parts can 117 be divid ed, so that if each of them be squared, their squares will be in arithmetical ratio? Ans. 9, 45, 63.

34. Divide 50 into four such parts, that when each of those parts are cubed, the sum of their cubes may be a cube.