Difference between revisions of "1951 AHSME Problems"
(→Problem 2) |
|||
Line 1: | Line 1: | ||
== Problem 1 == | == Problem 1 == | ||
− | The percent that <math>M</math> is greater than <math>N</math> | + | The percent that <math>M</math> is greater than <math>N</math> is: |
<math> \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} </math> | <math> \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} </math> | ||
Line 21: | Line 21: | ||
== Problem 4 == | == Problem 4 == | ||
− | A barn with a roof is rectangular in shape, <math>10</math> yd. wide, <math>13</math> yd. long and <math>5</math> 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: | + | A barn with a flat roof is rectangular in shape, <math>10</math> yd. wide, <math>13</math> yd. long and <math>5</math> 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: |
<math> \mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720 </math> | <math> \mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720 </math> | ||
Line 28: | Line 28: | ||
== Problem 5 == | == Problem 5 == | ||
− | Mr. | + | 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: |
− | <math> \mathrm{(A) \ A | + | <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> |
[[1951 AHSME Problems/Problem 5|Solution]] | [[1951 AHSME Problems/Problem 5|Solution]] | ||
Line 36: | Line 36: | ||
== Problem 6 == | == Problem 6 == | ||
− | [[ | + | 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> | ||
+ | |||
+ | [[1951 AHSME Problems/Problem 6|Solution]] | ||
== 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: | ||
+ | |||
+ | <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]] | ||
== 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: | ||
+ | |||
+ | <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]] | ||
== 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: | ||
+ | |||
+ | <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]] |
Revision as of 14:40, 30 May 2011
Contents
[hide]- 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 . 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
Problem 11
Problem 12
Problem 13
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: