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Difference between revisions of "1951 AHSME Problems"
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== Problem 5 ==
 
== Problem 5 ==
−
Mr. A owns a home worth <math>\</math><math>10,000</math>.  He sells it to Mr. B at a <math>10 \%</math> profit based on the worth of the house. Mr. B sells the house back to Mr. A at a <math>10 \%</math> loss.  Then:
+
Mr. A owns a home worth <math>&#36;</math>10,000<math>.  He sells it to Mr. B at a </math>10 \%<math> profit based on the worth of the house. Mr. B sells the house back to Mr. A at a </math>10 \%<math> loss.  Then:
  
−
<math> \mathrm{(A) \ } \text{A comes out even.} \qquad\mathrm{(B) \ } \text{A makes } \</math><math>\text{1100 on the deal} \qquad\mathrm{(C) \ } \text{A makes } \</math> <math>\text{1000 on the deal} \qquad \mathrm{(D) \ } \text{A loses } \</math> <math>\text{900 on the deal} \qquad\mathrm{(E) \ } \text{A loses } \</math> <math>\text{1000 on the deal} </math>
+
</math> \mathrm{(A) \ } \text{A comes out even.} \qquad\mathrm{(B) \ } \text{A makes } &#36;<math>\text{1100 on the deal} \qquad\mathrm{(C) \ } \text{A makes } &#36; </math>\text{1000 on the deal} \qquad \mathrm{(D) \ } \text{A loses } &#36; <math>\text{900 on the deal} \qquad\mathrm{(E) \ } \text{A loses } &#36; </math>\text{1000 on the deal} <math>
  
 
[[1951 AHSME Problems/Problem 5|Solution]]
 
[[1951 AHSME Problems/Problem 5|Solution]]
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The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:
 
The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:
  
−
<math> \mathrm{(A) \ } \text{the volume of the box} \qquad\mathrm{(B) \ } \text{the square root of the volume} \qquad \mathrm{(C) \ } \text{twice the volume} \qquad\mathrm{(D) \ } \text{the square of the volume} \qquad\mathrm{(E) \ } \text{the cube of the volume} </math>
+
</math> \mathrm{(A) \ } \text{the volume of the box} \qquad\mathrm{(B) \ } \text{the square root of the volume} \qquad \mathrm{(C) \ } \text{twice the volume} <math></math> \qquad\mathrm{(D) \ } \text{the square of the volume} \qquad\mathrm{(E) \ } \text{the cube of the volume} <math>
  
 
[[1951 AHSME Problems/Problem 6|Solution]]
 
[[1951 AHSME Problems/Problem 6|Solution]]
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== Problem 7 ==
 
== Problem 7 ==
  
−
An error of <math>.02"</math> is made in the measurement of a line <math>10"</math> long, while an error of only <math>.2"</math> is made in a measurement of a line <math>100"</math> long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:
+
An error of </math>.02"<math> is made in the measurement of a line </math>10"<math> long, while an error of only </math>.2"<math> is made in a measurement of a line </math>100"<math> long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:
  
−
<math> \mathrm{(A) \ } \text{greater by }.18 \qquad\mathrm{(B) \ } \text{the same} \qquad \mathrm{(C) \ } \text{less} \qquad\mathrm{(D) \ } 10\text{ times as great} \qquad\mathrm{(E) \ } \text{correctly described by both} </math>
+
</math> \mathrm{(A) \ } \text{greater by }.18 \qquad\mathrm{(B) \ } \text{the same} \qquad \mathrm{(C) \ } \text{less} \qquad\mathrm{(D) \ } 10\text{ times as great} \qquad\mathrm{(E) \ } \text{correctly described by both} <math>
  
 
[[1951 AHSME Problems/Problem 7|Solution]]
 
[[1951 AHSME Problems/Problem 7|Solution]]
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== Problem 8 ==
 
== Problem 8 ==
  
−
The price of an article is cut <math>10 \%.</math> To restore it to its former value, the new price must be increased by:
+
The price of an article is cut </math>10 \%.<math> To restore it to its former value, the new price must be increased by:
  
−
<math> \mathrm{(A) \ } 10 \% \qquad\mathrm{(B) \ } 9 \% \qquad \mathrm{(C) \ } 11\frac{1}{9} \% \qquad\mathrm{(D) \ } 11 \% \qquad\mathrm{(E) \ } \text{none of these answers} </math>
+
</math> \mathrm{(A) \ } 10 \% \qquad\mathrm{(B) \ } 9 \% \qquad \mathrm{(C) \ } 11\frac{1}{9} \% \qquad\mathrm{(D) \ } 11 \% \qquad\mathrm{(E) \ } \text{none of these answers} <math>
  
 
[[1951 AHSME Problems/Problem 8|Solution]]
 
[[1951 AHSME Problems/Problem 8|Solution]]
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== Problem 9 ==
 
== Problem 9 ==
  
−
An equilateral triangle is drawn with a side length of <math>a.</math> A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is:
+
An equilateral triangle is drawn with a side length of </math>a.<math> A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is:
  
−
<math> \mathrm{(A) \ } \text{Infinite} \qquad\mathrm{(B) \ } 5\frac{1}{4}a \qquad \mathrm{(C) \ } 2a \qquad\mathrm{(D) \ } 6a \qquad\mathrm{(E) \ } 4\frac{1}{2}a </math>
+
</math> \mathrm{(A) \ } \text{Infinite} \qquad\mathrm{(B) \ } 5\frac{1}{4}a \qquad \mathrm{(C) \ } 2a \qquad\mathrm{(D) \ } 6a \qquad\mathrm{(E) \ } 4\frac{1}{2}a <math>
  
 
[[1951 AHSME Problems/Problem 9|Solution]]
 
[[1951 AHSME Problems/Problem 9|Solution]]
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<cmath>f(x) \equiv ax^2 + bx + c = 0,</cmath>
 
<cmath>f(x) \equiv ax^2 + bx + c = 0,</cmath>
  
−
it happens that <math>c = \frac{b^2}{4a}</math>, then the graph of <math>y = f(x)</math> will certainly:
+
it happens that </math>c = \frac{b^2}{4a}<math>, then the graph of </math>y = f(x)<math> will certainly:
  
−
<math>\mathrm{(A) \ have\ a\ maximum  } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad</math> <math>\mathrm{(C) \ be\ tangent\ to\ the\ xaxis} \qquad</math> <math>\mathrm{(D) \ be\ tangent\ to\ the\ yaxis} \qquad</math> <math>\mathrm{(E) \ lie\ in\ one\ quadrant\ only}</math>
+
</math>\mathrm{(A) \ have\ a\ maximum  } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad<math> </math>\mathrm{(C) \ be\ tangent\ to\ the\ xaxis} \qquad<math> </math>\mathrm{(D) \ be\ tangent\ to\ the\ yaxis} \qquad<math> </math>\mathrm{(E) \ lie\ in\ one\ quadrant\ only}$
  
 
[[1951 AHSME Problems/Problem 16|Solution]]
 
[[1951 AHSME Problems/Problem 16|Solution]]

Revision as of 19:13, 19 June 2011

Problem 1

The percent that $M$ is greater than $N$ is:

$\mathrm{(A) \ } \frac {100(M - N)}{M} \qquad \mathrm{(B) \ } \frac {100(M - N)}{N} \qquad \mathrm{(C) \ } \frac {M - N}{N} \qquad \mathrm{(D) \ } \frac {M - N}{M} \qquad \mathrm{(E) \ } \frac {100(M + N)}{N}$

Solution

Problem 2

A rectangular field is half as wide as it is long and is completely enclosed by $x$ yards of fencing. The area in terms of $x$ is:

$(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}$

Solution

Problem 3

If the length of a diagonal of a square is $a + b$, then the area of the square is:

$\mathrm{(A) \ (a+b)^2 } \qquad \mathrm{(B) \ \frac{1}{2}(a+b)^2 } \qquad \mathrm{(C) \ a^2+b^2 } \qquad \mathrm{(D) \ \frac {1}{2}(a^2+b^2) } \qquad \mathrm{(E) \ \text{none of these} }$

Solution

Problem 4

A barn with a flat roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:

$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720$

Solution

Problem 5

Mr. A owns a home worth $&#36;$10,000$. He sells it to Mr. B at a$10 \%$profit based on the worth of the house. Mr. B sells the house back to Mr. A at a$10 \%$loss. Then:$ \mathrm{(A) \ } \text{A comes out even.} \qquad\mathrm{(B) \ } \text{A makes } $$\text{1100 on the deal} \qquad\mathrm{(C) \ } \text{A makes } &#36;$\text{1000 on the deal} \qquad \mathrm{(D) \ } \text{A loses } $ $\text{900 on the deal} \qquad\mathrm{(E) \ } \text{A loses } &#36;$\text{1000 on the deal} $[[1951 AHSME Problems/Problem 5|Solution]]

== Problem 6 ==

The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:$ (Error compiling LaTeX. Unknown error_msg) \mathrm{(A) \ } \text{the volume of the box} \qquad\mathrm{(B) \ } \text{the square root of the volume} \qquad \mathrm{(C) \ } \text{twice the volume} $$ (Error compiling LaTeX. Unknown error_msg) \qquad\mathrm{(D) \ } \text{the square of the volume} \qquad\mathrm{(E) \ } \text{the cube of the volume} $[[1951 AHSME Problems/Problem 6|Solution]]

== Problem 7 ==

An error of$ (Error compiling LaTeX. Unknown error_msg).02"$is made in the measurement of a line$10"$long, while an error of only$.2"$is made in a measurement of a line$100"$long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:$ \mathrm{(A) \ } \text{greater by }.18 \qquad\mathrm{(B) \ } \text{the same} \qquad \mathrm{(C) \ } \text{less} \qquad\mathrm{(D) \ } 10\text{ times as great} \qquad\mathrm{(E) \ } \text{correctly described by both} $[[1951 AHSME Problems/Problem 7|Solution]]

== Problem 8 ==

The price of an article is cut$ (Error compiling LaTeX. Unknown error_msg)10 \%.$To restore it to its former value, the new price must be increased by:$ \mathrm{(A) \ } 10 \% \qquad\mathrm{(B) \ } 9 \% \qquad \mathrm{(C) \ } 11\frac{1}{9} \% \qquad\mathrm{(D) \ } 11 \% \qquad\mathrm{(E) \ } \text{none of these answers} $[[1951 AHSME Problems/Problem 8|Solution]]

== Problem 9 ==

An equilateral triangle is drawn with a side length of$ (Error compiling LaTeX. Unknown error_msg)a.$A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is:$ \mathrm{(A) \ } \text{Infinite} \qquad\mathrm{(B) \ } 5\frac{1}{4}a \qquad \mathrm{(C) \ } 2a \qquad\mathrm{(D) \ } 6a \qquad\mathrm{(E) \ } 4\frac{1}{2}a $[[1951 AHSME Problems/Problem 9|Solution]]

== Problem 10 ==

[[1951 AHSME Problems/Problem 10|Solution]]

== Problem 11 ==

[[1951 AHSME Problems/Problem 11|Solution]]

== Problem 12 ==

[[1951 AHSME Problems/Problem 12|Solution]]

== Problem 13 ==

[[1951 AHSME Problems/Problem 13|Solution]]

== Problem 14 ==

[[1951 AHSME Problems/Problem 14|Solution]]

== Problem 15 ==

[[1951 AHSME Problems/Problem 15|Solution]]

== Problem 16 == If in applying the quadratic formula to a quadratic equation

<cmath>f(x) \equiv ax^2 + bx + c = 0,</cmath>

it happens that$ (Error compiling LaTeX. Unknown error_msg)c = \frac{b^2}{4a}$, then the graph of$y = f(x)$will certainly:$\mathrm{(A) \ have\ a\ maximum } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad$$ (Error compiling LaTeX. Unknown error_msg)\mathrm{(C) \ be\ tangent\ to\ the\ xaxis} \qquad$$ (Error compiling LaTeX. Unknown error_msg)\mathrm{(D) \ be\ tangent\ to\ the\ yaxis} \qquad$$ (Error compiling LaTeX. Unknown error_msg)\mathrm{(E) \ lie\ in\ one\ quadrant\ only}$

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

Problem 26

Solution

Problem 27

Solution

Problem 28

Solution

Problem 29

Solution

Problem 30

Solution

See also

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