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Difference between revisions of "1951 AHSME Problems"
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Mr. A owns a home worth <math>10,000</math> dollars.  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>10,000</math> dollars.  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 } &#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>
+
<math> \textrm{(A)}\ \text{A comes out even} \qquad\textrm{(B)}\ \text{A makes 1100 on the deal} \qquad\textrm{(C)}\ \text{A makes 1000 on the deal}</math>
 +
<math>\textrm{(D)}\ \text{A loses 900 on the deal} \qquad\textrm{(E)}\ \text{A loses 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 } &#36;</math>\text{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> \textrm{(A)}\ \text{the volume of the box} \qquad\textrm{(B)}\ \text{the square root of the volume} \qquad\textrm{(C)}\ \text{twice the volume}</math>
 +
<math> \textrm{(D)}\ \text{the square of the volume} \qquad\textrm{(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]]
Line 52: Line 54:
 
== 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]]
  
 
== Problem 10 ==
 
== Problem 10 ==
 +
Of the following statements, the one that is incorrect is:
 +
 +
<math> \textrm{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}</math>
 +
<math> \textrm{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}</math>
 +
<math> \textrm{(C)}\ \text{Doubling the radius of a given circle doubles the area.}</math>
 +
<math> \textrm{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}</math>
 +
<math> \textrm{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}</math>
  
 
[[1951 AHSME Problems/Problem 10|Solution]]
 
[[1951 AHSME Problems/Problem 10|Solution]]
  
 
== Problem 11 ==
 
== Problem 11 ==
 +
 +
The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1 \minus{} r}</math> where <math> a</math> denotes the first term and <math> \minus{} 1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is:
 +
 +
<math> \textrm{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textrm{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textrm{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textrm{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textrm{(E)}\ \text{none of these}</math>
  
 
[[1951 AHSME Problems/Problem 11|Solution]]
 
[[1951 AHSME Problems/Problem 11|Solution]]
  
 
== Problem 12 ==
 
== Problem 12 ==
 +
 +
At <math> 2: 15</math> o'clock, the hour and minute hands of a clock form an angle of:
 +
 +
<math> \textrm{(A)}\ 30^{\circ} \qquad\textrm{(B)}\ 5^{\circ} \qquad\textrm{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textrm{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textrm{(E)}\ 28^{\circ}</math>
  
 
[[1951 AHSME Problems/Problem 12|Solution]]
 
[[1951 AHSME Problems/Problem 12|Solution]]
  
 
== Problem 13 ==
 
== Problem 13 ==
 +
 +
<math> A</math> can do a piece of work in <math> 9</math> days. <math> B</math> is <math> 50\%</math> more efficient than <math> A</math>. The number of days it takes <math> B</math> to do the same piece of work is:
 +
 +
<math> \textrm{(A)}\ 13\frac {1}{2} \qquad\textrm{(B)}\ 4\frac {1}{2} \qquad\textrm{(C)}\ 6 \qquad\textrm{(D)}\ 3 \qquad\textrm{(E)}\ \text{none of these answers}</math>
  
 
[[1951 AHSME Problems/Problem 13|Solution]]
 
[[1951 AHSME Problems/Problem 13|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>\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>
  
 
[[1951 AHSME Problems/Problem 16|Solution]]
 
[[1951 AHSME Problems/Problem 16|Solution]]

Revision as of 17:00, 19 July 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 $10,000$ dollars. 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:

$\textrm{(A)}\ \text{A comes out even} \qquad\textrm{(B)}\ \text{A makes 1100 on the deal} \qquad\textrm{(C)}\ \text{A makes 1000 on the deal}$ $\textrm{(D)}\ \text{A loses 900 on the deal} \qquad\textrm{(E)}\ \text{A loses 1000 on the deal}$

Solution

Problem 6

The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:

$\textrm{(A)}\ \text{the volume of the box} \qquad\textrm{(B)}\ \text{the square root of the volume} \qquad\textrm{(C)}\ \text{twice the volume}$ $\textrm{(D)}\ \text{the square of the volume} \qquad\textrm{(E)}\ \text{the cube of the volume}$

Solution

Problem 7

An error of $.02"$ 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:

$\mathrm{(A) \ } \text{greater by }.18 \qquad\mathrm{(B) \ } \text{the same} \qquad \mathrm{(C) \ } \text{less} \qquad\mathrm{(D) \ } 10\text{ times as great} \qquad\mathrm{(E) \ } \text{correctly described by both}$

Solution

Problem 8

The price of an article is cut $10 \%.$ To restore it to its former value, the new price must be increased by:

$\mathrm{(A) \ } 10 \% \qquad\mathrm{(B) \ } 9 \% \qquad \mathrm{(C) \ } 11\frac{1}{9} \% \qquad\mathrm{(D) \ } 11 \% \qquad\mathrm{(E) \ } \text{none of these answers}$

Solution

Problem 9

An equilateral triangle is drawn with a side length of $a.$ 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:

$\mathrm{(A) \ } \text{Infinite} \qquad\mathrm{(B) \ } 5\frac{1}{4}a \qquad \mathrm{(C) \ } 2a \qquad\mathrm{(D) \ } 6a \qquad\mathrm{(E) \ } 4\frac{1}{2}a$

Solution

Problem 10

Of the following statements, the one that is incorrect is:

$\textrm{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}$ $\textrm{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$ $\textrm{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$ $\textrm{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}$ $\textrm{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}$

Solution

Problem 11

The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1 \minus{} r}$ (Error compiling LaTeX. Unknown error_msg) where $a$ denotes the first term and $\minus{} 1 < r < 1$ (Error compiling LaTeX. Unknown error_msg) denotes the common ratio. The limit of the sum of their squares is:

$\textrm{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textrm{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textrm{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textrm{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textrm{(E)}\ \text{none of these}$ (Error compiling LaTeX. Unknown error_msg)

Solution

Problem 12

At $2: 15$ o'clock, the hour and minute hands of a clock form an angle of:

$\textrm{(A)}\ 30^{\circ} \qquad\textrm{(B)}\ 5^{\circ} \qquad\textrm{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textrm{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textrm{(E)}\ 28^{\circ}$

Solution

Problem 13

$A$ can do a piece of work in $9$ days. $B$ is $50\%$ more efficient than $A$. The number of days it takes $B$ to do the same piece of work is:

$\textrm{(A)}\ 13\frac {1}{2} \qquad\textrm{(B)}\ 4\frac {1}{2} \qquad\textrm{(C)}\ 6 \qquad\textrm{(D)}\ 3 \qquad\textrm{(E)}\ \text{none of these answers}$

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

If in applying the quadratic formula to a quadratic equation

\[f(x) \equiv ax^2 + bx + c = 0,\]

it happens that $c = \frac{b^2}{4a}$, then the graph of $y = f(x)$ will certainly:

$\mathrm{(A) \ have\ a\ maximum } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad$ $\mathrm{(C) \ be\ tangent\ to\ the\ xaxis} \qquad$ $\mathrm{(D) \ be\ tangent\ to\ the\ yaxis} \qquad$ $\mathrm{(E) \ lie\ in\ one\ quadrant\ only}$

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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