MATH3560 Mid-Session Test
Due Date: Midnight Friday 19th March 2021
Instructions
• Use only those methods available to Mesopotamian scribes. You are per-
mitted to use variables x, y, z and modern mathematical terms to describe
your working.
• You are required to show your working, except for the multiplication steps.
• Leading or trailing zeros may be used to indicate magnitude.
Questions
1. A rectangular throne room has base 40 ninda, width 10 ninda and height
8 ninda. What is the distance between opposite points in the room (i.e.
from one corner of the floor to the opposite corner on the ceiling)?
2. You borrow 1 gur of grain from the National Assyrian Bank and pay 20
percent interest, compounded annually. If you make annual repayments
of 13 = 0 : 20 gur, how long (to the nearest month) will it take for you to
owe 23 = 0 : 40 gur of grain?
3. Find a two-digit approximation for
√
3.
4. A canal (Figure 1). The length 30, upper width 25, lower width 18 and
height 14 (all units are ninda). The canal is made deeper and now has a
total capacity of 3 : 20 : 00 cubic ninda. How much deeper is the canal?
What is the new lower width?
5. Explain the following question and answer.
Question: A little canal (Figure 2). The length 20. On the first day a
depth of 4 was dug and the lower width was 6. On the second day,
more was dug and the lower width became 12 (all units are ninda).
A total of 1 : 06 : 00 cubic ninda was extracted. What is the upper
width w of the little canal?
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Figure 1: Cross section of the canal.
Answer: You, with your doing. 3, the reciprocal of 20, to volume raise.
The surface 3 : 18 you see. 15, the reciprocal of 4, to 3 : 18 raise.
49 : 30 you see. 49 : 30 repeat, 1 : 39 you see. The two widths, add.
18 you see. To 1 : 39 raise, 29 : 42 you see. May your head hold.
49 : 30 square, 40 : 50 : 15 you see. From 40 : 50 : 15 tear out the
29 : 42 which your head was holding. 11 : 08 : 15 you see. 12 square,
2 : 44 you see. To 11 : 08 : 15 add, 13 : 32 : 15 you see. What is the
square root of 13 : 32 : 15? 28 : 30 is the square root. From 49 : 30
tear out 28 : 30. 21 you see. 21, the upper width. The procedure.
Figure 2: Cross section of the little canal.
2
1 Answers
1. The distance d between opposite points on the floor satisfies d2 = 402 +
102 = 28 : 20. The distance between opposite points on the floor and
ceiling is then √
82 + d2 =
√
29 : 24 = 42.
2. At the end of the first year, you owe (1 − 20) × 1 : 12 = 48. At the end
of the second year you owe (48− 20)× 1 : 12 = 33 : 36. So this will reach
2
3 = 0 : 40 gur somewhere between the first and second year. By linear
interpolation this will be about half way between the first and second years
since:
48− 40
48− 33 : 36 =
8
14 : 24
= 33 : 20.
Which is one year plus 33 : 20× 12 = 6 : 40 months.
3. Use an initial approximation a = 1 : 40 since it is regular and a2 = 2 : 46 :
40 ≈ 3. Then √
3 ≈ 1
2
(1 : 40 + 3÷ 1 : 40) = 1 : 44.
Since
√
3 = 1 : 43 : 55 : 23 · · · we see this is a good approximation.
4. Divide the volume by the length to find the area of the cross section 6 : 40.
So
6 : 40 =
25 + w
2
× (14 + x).
Or
13 : 20 = (25 + w)× (14 + x).
The reciprocal-slope is
run
rise
=
25− 18
14
= 30 =
25− w
14 + x
.
So
30× 13 : 20 = 252 − w2.
In other words, w2 = 3 : 45 so w = 15. This can be used to deduce
x = 25−w30 − 14 = 6.
5. The method is similar to that of BM 85194. Divide the volume by the
length of the canal to find the area of the cross section 1 : 06 : 00÷ 20 =
3 : 18. Divide the vertical axis by 4. The area is now 3 : 18÷ 4 = 49 : 30.
Multiplication by a factor of w− 18 = w−121+h/4 makes the right triangle into
an isosceles right triangle. Double the area into a gnomon with sides w,
then complete the square by adding 122
w2 = 122 + (w − 18)× 2× 49 : 30.
3
Or, equivalently
w2 − 1 : 39w = −29 : 42 + 122.
This quadratic equation is solved in the standard way, by completing the
square
(w − 49 : 30)2 = 49 : 302 − 29 : 42 + 122 = 13 : 32 : 15.
So w = 49 : 30 ± 28 : 30 = 21 or 1 : 18 and the scribe has chosen the
former.
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