History of Mathematics Project | New York State Regents Exams in Mathematics

History of
Mathematics

Project

1907

New York State Regents Exams in Mathematics

Exams from yesteryear

Standardized tests and examinations have long been an important tool allowing teachers and administrators to evaluate student knowledge and the effectiveness of educational practices. While modern standardized exams employ multiple-choice questions and automated grading, the exam sheets of yesteryear were answered and graded by hand. The National Museum of American History at the Smithsonian holds a fascinating collection of standard examinations in mathematics administered in the state of New York in the early decades of the 1900s.

New York State Regents Exams in Mathematics

In the 1860s, the Regents of the University of the State of New York began planning examinations to test the knowledge and abilities of students graduating from high school. Examples of these examinations in mathematics were collected by Brooklyn high school mathematics teacher L. Leland Locke between 1907 and 1919. This collection of problems, grouped into arithmetic, algebra, geometry, trigonometry and algebra, provides insight into the very high level of mathematical mastery expected of students graduating from high school in the early 20th century.

Artifact dimensions

1.5 cm × 12.9 cm × 20.4 cm

Artifact origin

New York, United States

Current artifact location

National Museum of American History, Washington

Catalog number

2011.0129.08

Timeline

Interactive Content

Computational Explanation

The 1907 mathematics examination of the Regents of the University of the State of New York consists of a collection of problems grouped by subject into arithmetic, algebra, geometry, trigonometry and algebra. While many of the problems are straightforward to solve with pencil and paper and memorized facts, the level of mathematical knowledge and sophistication expected of incoming college freshmen taking this exam is impressive.

The majority of test problems can be solved using the Wolfram Language in concert with the Wolfram Knowledgebase—two powerful tools that would not have been available to students at the turn of the previous century. A select number of problems and solutions picked from each section of the exam are highlighted below.

Arithmetic

Problem 5

This straightforward problem can be solved in its entirety beginning with a scan of the numbers. To do so, select the portion of the image corresponding to the column of numbers and rotate to the vertical:

In[]:=

image=ImageRotateImageTake

,{50,140},{90,750},Right

Out[]=

The text can now be extracted from the image using the optical character recognition built into the Wolfram Language:

In[]:=

p5=TextRecognize[image]

Out[]=

39384396849395369823396495829378398493788396489239874936934883959356392498293965

Split the resulting string containing numerals and newlines, and convert from strings of digits to a list of computable numbers:

In[]:=

numbers=FromDigits/@StringSplit[p5]

Out[]=

{3938,4396,8493,9536,9823,3964,9582,9378,3984,9378,8396,4892,3987,4936,9348,8395,9356,3924,9829,3965}

Finally, compute the sum:

In[]:=

Total[numbers]

Out[]=

139500

Problem 9

The first part of problem 9 asks for the length of the meter in inches. This can be found using the Wolfram Language's built-in unit conversion function:

In[]:=

UnitConvert["Meters","Inches"]

Out[]=

5000

127

While the conversion to (international) inches is exact, several digits of precision suffice for practical purposes:

In[]:=

N[UnitConvert["Meters","Inches"],4]

Out[]=

39.37

Parts (a) and (b) ask for the canonical units associated with two physical quantities. These too can be obtained directly in the Wolfram Language:

In[]:=

QuantityVariableCanonicalUnit["Weight"]

Out[]=

Newtons

In[]:=

QuantityVariableCanonicalUnit["Volume"]

Out[]=

Meters

The final part of problem 9 asks for the sum of 37 decimeters, 23 centimeters, 18 dekameters and 2 kilometers. It can be solved directly by interpreting the input as a computed quantity:

In[]:=

Interpreter["ComputedQuantity"]["37 dm + 23 cm + 18 Dm + 2 Km"]//N

Out[]=

2.18393

It is also straightforward to compute the answer using the Wolfram Language's built-in quantity arithmetic:

In[]:=

Quantity[37,"Decimeters"]+Quantity[23,"Centimeters"]+Quantity[18,"Dekameters"]+Quantity[2,"Kilometers"]

Out[]=

218393

If desired, the result can be converted to US customary units:

In[]:=

UnitConvert[

218393

,"Imperial"]//N

Out[]=

1.35703

Problem 12

Recognize the text using the optical character recognition built into the Wolfram Language:

In[]:=

p12=TextRecognize[

]

Out[]=

Find the square root of 170,569

The problem text can now be semantically interpreted:

In[]:=

SemanticInterpretation[p12,_,Hold]//InputForm

Out[]//InputForm=

Hold[Sqrt[170569]]

It can also be interpreted and immediately evaluated:

In[]:=

SemanticInterpretation[p12]

Out[]=

413

Alternatively, computing directly using the Wolfram Language immediately gives the solution:

In[]:=

Sqrt[170569]

Out[]=

413

Advanced Arithmetic

Problem 4

Problem 4 can easily be solved as a system of two simultaneous linear equations:

Solvex+y1300,

x

y,{x,y}

Out[]=

Problem 10

Problem 10 calls for the summation of the positive integers from 1 to 100. It can be solved directly using the built-in function Sum:

In[]:=

Quantity[Sum[k,{k,100}],"USCents"]

Out[]=

5050

Expressed in dollars and cents:

In[]:=

UnitConvert[%//N]

Out[]=

50.50

Without the use of computational software or knowledge of the closed form for a sum of this type, a solution can also be found by noting that by pairing numbers upward from 0 and downward from 99, i.e. 0 + 100  1 + 99  2 + 98  …  49 + 51, a sequence of 50 sums each equal to 100 is obtained. Adding to this the middle number 50 then gives the sum 50×100 + 50  5050. The solution of this problem is attributed to Carl Friedrich Gauss, who being assigned this problem as a schoolboy immediately wrote "5050" (and nothing else) on his slate, saying, "There it lies."

Generalizing to n tickets gives a sum given by a so-called triangular number:

In[]:=

Sum[k,{k,n}]

Out[]=

n(1+n)

… which can also be written in terms of a binomial coefficient as:

In[]:=

Binomial[n+1,2]

Out[]=

n(1+n)

To verify, plug in

n100

to obtain the same result as before:

In[]:=

Binomial[100+1,2]

Out[]=

5050

Elementary Algebra

Problem 2

Problem 2 is easily solved using the built-in Wolfram Language function Factor:

In[]:=

Factor/@{a^2-1+b^2+2ab,8-x^3,35x^2y^2-27b^2z^2+21b^2y^2-45x^2z^2,a^6+b^6,x^4+13x^2-68}//Column//TraditionalForm

Out[]//TraditionalForm=

(a+b-1)(a+b+1)

-(x-2)(

+2x+4)

)(7

-9

)

(

)(-

)

(x-2)(x+2)(

+17)

Problem 7

Similarly, the square root of the polynomial requested in problem 7 can be found using Factor:

In[]:=

Simplify[Factor[Sqrt[x^6-2x^5-x^4+3x^2+2x+1]],x∈Reals]

Out[]=

Abs[1+x+

]

Note the presence of an absolute value, which is needed to keep the resulting expression non-negative:

In[]:=

Plot[{1+x+

,Abs[1+x+

]},{x,-2,3},PlotStyle{AbsoluteThickness[4],Automatic},PlotLegends"Expressions"]

Out[]=

	1+x+ 2 x - 3 x
	1+x+ 2 x - 3 x 

… as it must be given that the original expression is always non-negative:

In[]:=

Reduce[x^6-2x^5-x^4+3x^2+2x+1≥0,x,Reals]

Out[]=

True