Page:The Rhind Mathematical Papyrus, Volume I.pdf/126

Problem 75
Another problem. 155 loaves of pefsu 20 are to be exelzanged for loaves of pefsu 30. How many of these will there be?

The amount of wedyet-flour in the 155 loaves of pefsu 20 is 7¹⁄₂¹⁄₄ hekat. Multiply this by 30; it makes 232¹⁄₂.

Do it thus: 155 loaves of pefsu 20, made from 7¹⁄₂¹⁄₄ hekat of wedyet-flour, can be exchanged for 232¹⁄₂ loaves of pefsu 30. It takes 7¹⁄₂¹⁄₄ hekat.

This problem is like Problems 72 and 73, and the solution given is like that for Problem 73.

Problem 76
Another problem. 1000 loaves of pefsu 10 are to be exchanged for a number of loaves of pefsu 20 and the same number of pefsu 30. How many of each kind will there be?

One loaf of each kind will take
¹⁄₂₀ and ¹⁄₃₀ of a hekat.

As parts of 30 these are
¹⁄1¹⁄₂ and 1, together 2¹⁄₂.

Multiply 2¹⁄₂ so as to get 30

	1	21⁄2
\	10	25
\	2	5
Total	12.

Therefore 2¹⁄₂ is biz of 30, so that ¹⁄₂₀¹⁄₃₀ equals ¹⁄₁₂. Two loaves, one of each kind, will take ¹⁄₁₂ of a hekat and 1 hekat will make 12 loaves of each kind.

The quantity of wedyet-flour in the 1000 loaves is 100 hekat. Multiply 100 by 12; the result is 1200, which is the number of loaves of each kind for the exchange. That is

1000

loaves

of

pefsu

10,

making

in

wedyet-flour

100

hekat can be exchanged for

1200

loaves

of

pefsu

20,

"

"

"

1⁄2

of

"

10

hekat

and

1200

"

"

"

30

"

"

"

1⁄4

of

"

15

"

In this problem, as in Problem 74, the author wishes to know how many loaves of two kinds can be made from a certain amount of zvedyet-flour, but in Problem 74 he uses half of the flour for one kind of loaf and half for the other. This time he wishes to