Item #671 MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY
MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY
MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY
MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY
MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY
MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY
MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY

MATH CYPHER & LEDGER 1857--1915 YORK, PA - JOINED CIVIL WAR UNION ARMY

Item #671

This 6.25 x 7.5” hardcover book contains a gorgeous, faded marble cover, on which is written “Daniel Bailey”. The front cover and first few pages are detached. Inside the front cover and first page of the notebook contain this name again, along with “York County Academy” in York, Pennsylvania, and the dates 1855-57. The first part of the notebook contains Daniel’s handwritten notes on simple proportion and compound proportion, which are accompanied by his work solving word problems and mathematical formulas. The middle section of the notebook contains blank pages. If one goes to the back of the book and flips it upside down, there are ledgers of payments made from 1859 to around 1915. These first ledger pages describe the logged payments as: “Money paid by Matilda Weltz, Administrator of the estate of F. E. Bailey, Deceased, for Daniel D. Bailey during his College course at Gettysburg.”
This notebook belonged to Daniel Doudal Bailey, who left Pennsylvania College in Gettysburg when the Civil War started, to become a Corporal with Company G, Twelfth Regiment, Pennsylvania Reserves. He was wounded and captured during a battle, then later exchanged and sent to Camden Hospital. Unfortunately, he died from his wounds in 1862.[source][source -pg. 868, or 1030/1308]
Rules for Simple Proportion
I. Write that number for the third term which is of the same denomination as that in which the answer is required.
II. Then if from the nature of the question the answer ought to be greater than the third term place the greater the other two numbers for the second term and the less for the first but if the answer ought to be less than the third term place the less number for the second term and the greater for the first.
III. Multiply the third term by the second and divide by the first the quotient will be the answer in the same denomination as the third term.

Compound Proportion Rules
I. Write that number for the third term in the denomination as the answer is required.
II. Yoke any two numbers slide and arrange them in simple proportion [illegible] (2, 2) that is if reference to these numbers are greater than the third term with the greater number for the second and the less for the first but if reference of than numbers the answer ought to be less than the third term with the 2 and the [illegible] for [illegible].
III. Take any other 2 numbers of the same kind and arrange them in the same manner and so on [illegible] the numbers are employed.
IV. Multiply the third by the contrived product of the second and divide the result by the contrived product of the first first term the [illegible] will be the fourth term.

Price: $250.00

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