class="homepageheader"1961 Old Farmer's Almanac
Cover  TOC
Problems  Answers

Farmers must be a lot smarter than we tend to give them credit for being. These math
and logic puzzles that appeared in the 1961 Old Farmer's Almanac are not a
duffer's
task to complete. Be careful to consider units of measure based on the venues (farming,
shipping, etc.).
Puzzle I is a relatively simple trigonometry problem, although the wording of the
problem statement is very confusing; it took some head scratching to figure out what
was meant. My answer is off from theirs  probably due to working it out with an electronic
calculator rather than an slide rule (i.e., rounding errors).
Puzzle II is somewhat shocking in its constitution, but considering the era, not so
surprising.
Puzzle III required me to opt for a graphical solution since I could not come up with
enough independent equations for the number of unknowns. If you look at the OFA page
scan, you will see typesetter's errors that add to the difficulty. The provided solution
is no help either, at least for me.
Puzzle IV seems utterly nonsensical, but it is actually legitimate); in fact, I was
convinced it was also a typesetter's error until I looked at the answer and figured out
what was expected. Maybe those types of challenges were common in 1961 and didn't need
to be explained to farmers.
Problem V was solved graphically rather than doing the hard geometry. After setting
Visio's page scale to display the proper line dimensions, I created three lines of the
specified lengths with 120° between them at the vertex, then drew an equilateral triangle
that intersected them all at the sides. Visio provided the triangle side length measurements,
so I needed only to calculate the area of the resulting triangle. My answer, however,
is half of the given answer, but I suspect the OFA answer is wrong. It appears they multiplied
the full length of the equilateral triangle base by the height rather than half the base
times the height. I'll be glad to be proven wrong.
Number VI was easy enough after being certain to use exactly the information provided
in the problem statement (still, it was somewhat ambiguous).
Puzzle VII is left as an exercise for the reader ;)
Puzzle VIII was a
lead pipe cinch.
My solution to some of the problems are presented below along with the OFA's answers.
The others are left as an exercise for the reader :)
Math & Logic Puzzles from the 1961 Old Farmer's Almanac
I
If a pole standing perpendicular, 40 feet in length, when the sun is bearing south,
cast a shadow 88 feet (on level ground) what is the breadth of a river running due N.
E. and S. W. within 22 feet, nearest place on the north side of another pole 230 feet
high which at the same time casts the extremity of its shadow 12 feet beyond the river.
II
There is a story that a ship's crew of 30, half whites and half blacks, were short
of provisions, and it became necessary that half of them should be thrown overboard.
It was agreed that they should be placed in a ring on deck, by the captain, and that
as he counted round and round, every tenth man should be thrown over, until the crew
should be reduced one half. He so placed them that all the blacks were taken. What was
the order of arrangement ? Can you so place them?
III
Two farmers, A and B, purchased a tract of land containing three hundred acres, for
which they paid $600. They wish to divide the tract into farms, so that A will pay 25
cents per acre more than B, and each one to pay an equal portion of the purchase money
($300 apiece). How many acres will each man have, and what will it cost him per acre?
IV
Wood
Mr. being at the . of King of terrors, they perfume for his Quakers and who, which
and what, and they penny for Dr. Hound tis IMAGE HERE Dr. Haypreservers and little
devil behold scarlet his assistance, but before he arrived the not legally good
taker
changed color and the was ten
mills for.
V
William owned a tract of land whose three sides were all equal in length, and the
centre of the house was in that particular spot, that the nearest distance to the three
sides was 150, 200 and 230 perches respectively. Required the area of the equilateral
triangle tract in acres and perches?
VI
On a mountain's high summit a person does stand. Whose height is five feet and
fourninths of a span. As he gazes far out upon the broad deep, A light from the
ocean his vision does meet; But this light was reflected from a beacon near by,
Which stood above water two hundred feet high. If admitting the distance from
the seat of his vision To be just one mile to the point of reflection, Then the
space from this point to the base of the tower, Is quite one hundred and fifty feet
more. Now the thing wished to be known is the perpendicular height From the base
of the mountain to the point at his feet.
VII
There is a house 66 feet long by 72 wide, and from the bottom or foundation of the
building to the peak is 43 feet. Required the distance from either lower corner of the
house to the centre of the peak or ridgepole?
VIII
If one pin was dropped into the hold of the "Great Eastern" the first week, two pins
the second, four the third, and so on, doubled each week for a year, (52 weeks) how many
pins would there be, what would be the weight of them, allowing 200 pins to the ounce,
and how many "Great Easterns" would be required to carry them, her tonnage being 22,500
tons?
Answers to Math & Logic Puzzles (as provided in the OFA)
In this case, a picture is needed to explain the Puzzle I
solution to this problem because writing it out would take
1,000 words (or more).
I. Breadth of River, 327.464 feet.
Note: There was a comma in the decimal point place on the printed page  which is
OK for EU and Canadian folks, but confusing to Americans  so I changed it here).
My solution: See graphic to the right.
II. There is a sort of rhyme that used to be our rule for this,
which runs thus:
Two before One,
Three before Five,
Here two, there two;
Save Four alive;
Here one, there one;
Three that are cast,
Now one, twice two;
Whip Jack at last.
My solution: I didn't work this one.
The given method for a solution to problem #3 is a bit fuzzy, so I plotted a graphical
solution based on writing the equation shown from the information provided in the problem
statement.
1961 Old Farmer's Almanac Puzzle III Graphical Solution
III. $2.13,† A's farm per acre. $1.88†† B's farm
per acre.
Rule.  Square the average price of one acre. Square the difference of price
between two acres. Add the two squares together. Extract the square root of the sum.
Add the average price of one acre to the result, and you will have the real price of
two acres.
Please to demonstrate the rule.
Solution.
 √(2.0577821² + 0.12½²) + $2.00 = $4.0155643†,
which is the real price of two acres (one of A's and
one of B's).
half sum ± 1/2 difference
Then $2.0577821 + 12½ cents = $2.13278221 †
A's farm per acre;
and $2.0577821  12½ cents = $1.88278221
†† B's farm per acre.
My solution: See graphic to the right.
IV. Mr. Underwood being at the point (.) of death (King of
terrors) they sent (perfume) for his friends (Quakers) and relatives (who, which and
what) and they sent (penny) for Dr. Curtis (Houndtis) who (inclosed) a few lines (=
=) to Dr. Barnes (hay preservers) and implored (little devil, behold, scarlet) his
assistance, but before he arrived the invalid (not legally good) died (changed color)
and the (under)taker was cent (ten mills) for.
1961 Old Farmer's Almanac Puzzle V Solution
V. 2,427 acres, 121 1/2 rods. (probably wrong)
My solution: See graphic to the right and below.
Note: My answer is half of the given answer, but I suspect the OFA answer is wrong.
It appears they multiplied the full length of the equilateral triangle base by the height
rather than half the base times the height.
1961 Old Farmer's Almanac Puzzle VI Solution
VI. 4,218 ft., 8 inches.
My solution: See graphic to the right.
VII. 66.4 ft.
My solution: I didn't work this one.
1961 Old Farmer's Almanac Puzzle VIII Solution
VIII. To Curious Question  Number of pins, 4,503,599,627,370,495;
weight, 628,292,358 tons; number of "Great Easterns" required to carry them, 27,924;
she carries one ton of pins to every ton measurement.
My solution: See graphic to the right and below.
Using Algebra
Calculator:
# of Pins = 2^52 = 4,503,599,627,370,495
Weight of Pins = 4.503599627370495E15 pins x (1 ounce / 200 pins)
= 22517998136852 ounces
= 1407374883553 pounds
= 31275 long tons
(used in shipping)
# Great Easterns (a ship) = Weight of Pins / 22,500 = 27,924 Ships
Posted June 12, 2018
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