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> \ | + | <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> \ | + | <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 < | + | 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> |
[[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 < | + | 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> |
[[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 < | + | 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> |
[[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 < | + | 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> |
[[1951 AHSME Problems/Problem 16|Solution]] | [[1951 AHSME Problems/Problem 16|Solution]] |
Revision as of 18:00, 19 July 2011
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 See also
Problem 1
The percent that is greater than
is:
Problem 2
A rectangular field is half as wide as it is long and is completely enclosed by yards of fencing. The area in terms of
is:
Problem 3
If the length of a diagonal of a square is , then the area of the square is:
Problem 4
A barn with a flat roof is rectangular in shape, yd. wide,
yd. long and
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:
Problem 5
Mr. A owns a home worth dollars. He sells it to Mr. B at a
profit based on the worth of the house. Mr. B sells the house back to Mr. A at a
loss. Then:
Problem 6
The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:
Problem 7
An error of is made in the measurement of a line
long, while an error of only
is made in a measurement of a line
long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:
Problem 8
The price of an article is cut To restore it to its former value, the new price must be increased by:
Problem 9
An equilateral triangle is drawn with a side length of 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:
Problem 10
Of the following statements, the one that is incorrect is:
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 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)
Problem 12
At o'clock, the hour and minute hands of a clock form an angle of:
Problem 13
can do a piece of work in
days.
is
more efficient than
. The number of days it takes
to do the same piece of work is:
Problem 14
Problem 15
Problem 16
If in applying the quadratic formula to a quadratic equation
it happens that , then the graph of
will certainly: