Difference between revisions of "1961 AHSME Problems"
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+ | {{AHSC 40 Problems | ||
+ | |year = 1961 | ||
+ | }} | ||
== Problem 1== | == Problem 1== | ||
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\textbf{(B)}\ -\frac{1}{25} \qquad | \textbf{(B)}\ -\frac{1}{25} \qquad | ||
\textbf{(C)}\ 25\qquad | \textbf{(C)}\ 25\qquad | ||
− | \textbf{(D)}\ -25 | + | \textbf{(D)}\ -25\qquad |
− | \textbf{(E)}\ 25\sqrt{-1 | + | \textbf{(E)}\ 25\sqrt{-1}</math> |
[[1961 AHSME Problems/Problem 1|Solution]] | [[1961 AHSME Problems/Problem 1|Solution]] | ||
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\textbf{(B)}\ \frac{30r}{a}\qquad | \textbf{(B)}\ \frac{30r}{a}\qquad | ||
\textbf{(C)}\ \frac{30a}{r}\qquad | \textbf{(C)}\ \frac{30a}{r}\qquad | ||
− | \textbf{(D)}\ \frac{10r}{a | + | \textbf{(D)}\ \frac{10r}{a}\qquad |
− | \textbf{(E)}\ \frac{10a}{r | + | \textbf{(E)}\ \frac{10a}{r} </math> |
[[1961 AHSME Problems/Problem 2|Solution]] | [[1961 AHSME Problems/Problem 2|Solution]] | ||
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\textbf{(B)}\ -\frac{2}{3}\qquad | \textbf{(B)}\ -\frac{2}{3}\qquad | ||
\textbf{(C)}\ -\frac{3}{2} \qquad | \textbf{(C)}\ -\frac{3}{2} \qquad | ||
− | \textbf{(D)}\ 6 | + | \textbf{(D)}\ 6\qquad |
− | \textbf{(E)}\ -6 | + | \textbf{(E)}\ -6 </math> |
[[1961 AHSME Problems/Problem 3|Solution]] | [[1961 AHSME Problems/Problem 3|Solution]] | ||
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we say that the set is closed under that operation. Then <math>u</math> is closed under: | we say that the set is closed under that operation. Then <math>u</math> is closed under: | ||
− | <math>\ | + | <math>\textbf{(A)}\ \text{Addition}\qquad |
− | \ | + | \textbf{(B)}\ \text{Multiplication} \qquad |
− | \ | + | \textbf{(C)}\ \text{Division} \qquad\\ |
− | \ | + | \textbf{(D)}\ \text{Extraction of a positive integral root}\qquad |
− | \ | + | \textbf{(E)}\ \text{None of these} </math> |
[[1961 AHSME Problems/Problem 4|Solution]] | [[1961 AHSME Problems/Problem 4|Solution]] | ||
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\textbf{(B)}\ (x-1)^4 \qquad | \textbf{(B)}\ (x-1)^4 \qquad | ||
\textbf{(C)}\ x^4 \qquad | \textbf{(C)}\ x^4 \qquad | ||
− | \textbf{(D)}\ (x+1)^4 | + | \textbf{(D)}\ (x+1)^4 \qquad |
− | \textbf{(E)}\ x^4+1 | + | \textbf{(E)}\ x^4+1 </math> |
[[1961 AHSME Problems/Problem 5|Solution]] | [[1961 AHSME Problems/Problem 5|Solution]] | ||
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\textbf{(B)}\ \log{2} \qquad | \textbf{(B)}\ \log{2} \qquad | ||
\textbf{(C)}\ 1 \qquad | \textbf{(C)}\ 1 \qquad | ||
− | \textbf{(D)}\ 0 | + | \textbf{(D)}\ 0\qquad |
− | \textbf{(E)}\ -1 | + | \textbf{(E)}\ -1 </math> |
[[1961 AHSME Problems/Problem 6|Solution]] | [[1961 AHSME Problems/Problem 6|Solution]] | ||
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\textbf{(B)}\ -\frac{15}{x}\qquad | \textbf{(B)}\ -\frac{15}{x}\qquad | ||
\textbf{(C)}\ -\frac{6x^2}{a^9} \qquad | \textbf{(C)}\ -\frac{6x^2}{a^9} \qquad | ||
− | \textbf{(D)}\ \frac{20}{a^3 | + | \textbf{(D)}\ \frac{20}{a^3}\qquad |
− | \textbf{(E)}\ -\frac{20}{a^3 | + | \textbf{(E)}\ -\frac{20}{a^3}</math> |
[[1961 AHSME Problems/Problem 7|Solution]] | [[1961 AHSME Problems/Problem 7|Solution]] | ||
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<math>\textbf{(A)}\ C_1+C_2=A+B \qquad | <math>\textbf{(A)}\ C_1+C_2=A+B \qquad | ||
\textbf{(B)}\ C_1-C_2=B-A \qquad\\ | \textbf{(B)}\ C_1-C_2=B-A \qquad\\ | ||
− | \textbf{(C)}\ C_1-C_2=A-B | + | \textbf{(C)}\ C_1-C_2=A-B \qquad |
− | \textbf{(D)}\ C_1+C_2=B-A | + | \textbf{(D)}\ C_1+C_2=B-A\qquad |
− | \textbf{(E)}\ C_1-C_2=A+B | + | \textbf{(E)}\ C_1-C_2=A+B</math> |
[[1961 AHSME Problems/Problem 8|Solution]] | [[1961 AHSME Problems/Problem 8|Solution]] | ||
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If <math>r</math> equals the product of <math>a^b</math> by <math>x^b</math>, then <math>x</math> equals: | If <math>r</math> equals the product of <math>a^b</math> by <math>x^b</math>, then <math>x</math> equals: | ||
− | <math>\textbf{(A)} a\qquad | + | <math>\textbf{(A)}\ a \qquad |
\textbf{(B)}\ 2a \qquad | \textbf{(B)}\ 2a \qquad | ||
\textbf{(C)}\ 4a \qquad | \textbf{(C)}\ 4a \qquad | ||
− | \textbf{(D)}\ 2 | + | \textbf{(D)}\ 2\qquad |
− | \textbf{(E)}\ 4 | + | \textbf{(E)}\ 4 </math> |
[[1961 AHSME Problems/Problem 9|Solution]] | [[1961 AHSME Problems/Problem 9|Solution]] | ||
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\textbf{(B)}\ \sqrt{28} \qquad | \textbf{(B)}\ \sqrt{28} \qquad | ||
\textbf{(C)}\ 6 \qquad | \textbf{(C)}\ 6 \qquad | ||
− | \textbf{(D)}\ \sqrt{63 | + | \textbf{(D)}\ \sqrt{63} \qquad |
− | \textbf{(E)}\ \sqrt{98 | + | \textbf{(E)}\ \sqrt{98}</math> |
[[1961 AHSME Problems/Problem 10|Solution]] | [[1961 AHSME Problems/Problem 10|Solution]] | ||
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\textbf{(B)}\ 40.5 \qquad | \textbf{(B)}\ 40.5 \qquad | ||
\textbf{(C)}\ 40\qquad | \textbf{(C)}\ 40\qquad | ||
− | \textbf{(D)}\ 39\frac{7}{8 | + | \textbf{(D)}\ 39\frac{7}{8} \qquad |
− | \textbf{(E)}\ \text{not determined by the given information | + | \textbf{(E)}\ \text{not determined by the given information}</math> |
[[1961 AHSME Problems/Problem 11|Solution]] | [[1961 AHSME Problems/Problem 11|Solution]] | ||
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\textbf{(B)}\ \sqrt[7]{2}\qquad | \textbf{(B)}\ \sqrt[7]{2}\qquad | ||
\textbf{(C)}\ \sqrt[8]{2}\qquad | \textbf{(C)}\ \sqrt[8]{2}\qquad | ||
− | \textbf{(D)}\ \sqrt[9]{2 | + | \textbf{(D)}\ \sqrt[9]{2}\qquad |
− | \textbf{(E)}\ \sqrt[10]{2 | + | \textbf{(E)}\ \sqrt[10]{2}</math> |
[[1961 AHSME Problems/Problem 12|Solution]] | [[1961 AHSME Problems/Problem 12|Solution]] | ||
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\textbf{(B)}\ t^2+t\qquad | \textbf{(B)}\ t^2+t\qquad | ||
\textbf{(C)}\ |t^2+t|\qquad | \textbf{(C)}\ |t^2+t|\qquad | ||
− | \textbf{(D)}\ t\sqrt{t^2+1 | + | \textbf{(D)}\ t\sqrt{t^2+1}\qquad |
− | \textbf{(E)}\ |t|\sqrt{1+t^2 | + | \textbf{(E)}\ |t|\sqrt{1+t^2}</math> |
[[1961 AHSME Problems/Problem 13|Solution]] | [[1961 AHSME Problems/Problem 13|Solution]] | ||
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\textbf{(B)}\ \frac{1}{2}\sqrt{2K}\qquad | \textbf{(B)}\ \frac{1}{2}\sqrt{2K}\qquad | ||
\textbf{(C)}\ \frac{1}{3}\sqrt{3K}\qquad | \textbf{(C)}\ \frac{1}{3}\sqrt{3K}\qquad | ||
− | \textbf{(D)}\ \frac{1}{4}\sqrt{4K | + | \textbf{(D)}\ \frac{1}{4}\sqrt{4K}\qquad |
− | \textbf{(E)}\ \text{None of these are correct | + | \textbf{(E)}\ \text{None of these are correct}</math> |
[[1961 AHSME Problems/Problem 14|Solution]] | [[1961 AHSME Problems/Problem 14|Solution]] | ||
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\textbf{(B)}\ \frac{y^3}{x^2}\qquad | \textbf{(B)}\ \frac{y^3}{x^2}\qquad | ||
\textbf{(C)}\ \frac{x^2}{y^3}\qquad | \textbf{(C)}\ \frac{x^2}{y^3}\qquad | ||
− | \textbf{(D)}\ \frac{y^2}{x^3 | + | \textbf{(D)}\ \frac{y^2}{x^3}\qquad |
− | \textbf{(E)}\ y | + | \textbf{(E)}\ y</math> |
[[1961 AHSME Problems/Problem 15|Solution]] | [[1961 AHSME Problems/Problem 15|Solution]] | ||
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\textbf{(B)}\ \frac{bh}{2h+2m}\qquad | \textbf{(B)}\ \frac{bh}{2h+2m}\qquad | ||
\textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad | \textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad | ||
− | \textbf{(D)}\ \frac{b(m+h)}{2m+h | + | \textbf{(D)}\ \frac{b(m+h)}{2m+h}\qquad |
− | \textbf{(E)}\ \frac{b(2m+h)}{2(h+m) | + | \textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}</math> |
[[1961 AHSME Problems/Problem 16|Solution]] | [[1961 AHSME Problems/Problem 16|Solution]] | ||
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In the base ten number system the number <math>526</math> means <math>5 \times 10^2+2 \times 10 + 6</math>. | In the base ten number system the number <math>526</math> means <math>5 \times 10^2+2 \times 10 + 6</math>. | ||
In the Land of Mathesis, however, numbers are written in the base <math>r</math>. | In the Land of Mathesis, however, numbers are written in the base <math>r</math>. | ||
− | + | Jones purchases an automobile there for <math>440</math> monetary units (abbreviated m.u). | |
He gives the salesman a <math>1000</math> m.u bill, and receives, in change, <math>340</math> m.u. The base <math>r</math> is: | He gives the salesman a <math>1000</math> m.u bill, and receives, in change, <math>340</math> m.u. The base <math>r</math> is: | ||
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\textbf{(B)}\ 5\qquad | \textbf{(B)}\ 5\qquad | ||
\textbf{(C)}\ 7\qquad | \textbf{(C)}\ 7\qquad | ||
− | \textbf{(D)}\ 8 | + | \textbf{(D)}\ 8\qquad |
− | \textbf{(E)}\ 12 | + | \textbf{(E)}\ 12 </math> |
[[1961 AHSME Problems/Problem 17|Solution]] | [[1961 AHSME Problems/Problem 17|Solution]] | ||
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\textbf{(B)}\ -1 \qquad | \textbf{(B)}\ -1 \qquad | ||
\textbf{(C)}\ 0 \qquad | \textbf{(C)}\ 0 \qquad | ||
− | \textbf{(D)}\ 1 | + | \textbf{(D)}\ 1\qquad |
− | \textbf{(E)}\ 12 | + | \textbf{(E)}\ 12 </math> |
[[1961 AHSME Problems/Problem 18|Solution]] | [[1961 AHSME Problems/Problem 18|Solution]] | ||
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Consider the graphs of <math>y=2\log{x}</math> and <math>y=\log{2x}</math>. We may say that: | Consider the graphs of <math>y=2\log{x}</math> and <math>y=\log{2x}</math>. We may say that: | ||
− | <math>\ | + | <math>\textbf{(A)}\ \text{They do not intersect}\qquad \\ |
− | \ | + | \textbf{(B)}\ \text{They intersect at 1 point only}\qquad \\ |
− | \ | + | \textbf{(C)}\ \text{They intersect at 2 points only} \qquad \\ |
− | \ | + | \textbf{(D)}\ \text{They intersect at a finite number of points but greater than 2} \qquad \\ |
− | \ | + | \textbf{(E)}\ \text{They coincide} </math> |
[[1961 AHSME Problems/Problem 19|Solution]] | [[1961 AHSME Problems/Problem 19|Solution]] | ||
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\textbf{(B)}\ \text{II and III}\qquad | \textbf{(B)}\ \text{II and III}\qquad | ||
\textbf{(C)}\ \text{I and III}\qquad | \textbf{(C)}\ \text{I and III}\qquad | ||
− | \textbf{(D)}\ \text{III and IV | + | \textbf{(D)}\ \text{III and IV}\qquad |
− | \textbf{(E)}\ \text{I and IV | + | \textbf{(E)}\ \text{I and IV}</math> |
[[1961 AHSME Problems/Problem 20|Solution]] | [[1961 AHSME Problems/Problem 20|Solution]] | ||
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\textbf{(B)}\ \frac{1}{8}\qquad | \textbf{(B)}\ \frac{1}{8}\qquad | ||
\textbf{(C)}\ \frac{1}{9}\qquad | \textbf{(C)}\ \frac{1}{9}\qquad | ||
− | \textbf{(D)}\ \frac{1}{12 | + | \textbf{(D)}\ \frac{1}{12}\qquad |
− | \textbf{(E)}\ \frac{1}{16 | + | \textbf{(E)}\ \frac{1}{16}</math> |
[[1961 AHSME Problems/Problem 21|Solution]] | [[1961 AHSME Problems/Problem 21|Solution]] | ||
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\textbf{(B)}\ 3x^2-4\qquad | \textbf{(B)}\ 3x^2-4\qquad | ||
\textbf{(C)}\ 3x^2+4\qquad | \textbf{(C)}\ 3x^2+4\qquad | ||
− | \textbf{(D)}\ 3x-4 | + | \textbf{(D)}\ 3x-4 \qquad |
− | \textbf{(E)}\ 3x+4 | + | \textbf{(E)}\ 3x+4 </math> |
[[1961 AHSME Problems/Problem 22|Solution]] | [[1961 AHSME Problems/Problem 22|Solution]] | ||
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\textbf{(B)}\ 70\qquad | \textbf{(B)}\ 70\qquad | ||
\textbf{(C)}\ 75\qquad | \textbf{(C)}\ 75\qquad | ||
− | \textbf{(D)}\ 80 | + | \textbf{(D)}\ 80\qquad |
− | \textbf{(E)}\ 85 | + | \textbf{(E)}\ 85 </math> |
[[1961 AHSME Problems/Problem 23|Solution]] | [[1961 AHSME Problems/Problem 23|Solution]] | ||
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<math>\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\ | <math>\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\ | ||
− | \textbf{(B)}\ \text{The middle book sells for \textdollar 36 }\qquad \\ | + | \textbf{(B)}\ \text{The middle book sells for \textdollar 36} \qquad \\ |
− | \textbf{(C)}\ \text{The cheapest book sells for \textdollar4 }\qquad \\ | + | \textbf{(C)}\ \text{The cheapest book sells for \textdollar4} \qquad \\ |
\textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\ | \textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\ | ||
\textbf{(E)}\ \text{None of these is correct } </math> | \textbf{(E)}\ \text{None of these is correct } </math> | ||
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\textbf{(B)}\ 26\frac{1}{3}\qquad | \textbf{(B)}\ 26\frac{1}{3}\qquad | ||
\textbf{(C)}\ 30\qquad | \textbf{(C)}\ 30\qquad | ||
− | \textbf{(D)}\ 40 | + | \textbf{(D)}\ 40\qquad |
− | \textbf{(E)}\ \text{Not determined by the information given | + | \textbf{(E)}\ \text{Not determined by the information given}</math> |
[[1961 AHSME Problems/Problem 25|Solution]] | [[1961 AHSME Problems/Problem 25|Solution]] | ||
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\textbf{(B)}\ -21.5 \qquad | \textbf{(B)}\ -21.5 \qquad | ||
\textbf{(C)}\ -20.5 \qquad | \textbf{(C)}\ -20.5 \qquad | ||
− | \textbf{(D)}\ 3 | + | \textbf{(D)}\ 3 \qquad |
− | \textbf{(E)}\ 3.5 | + | \textbf{(E)}\ 3.5 </math> |
[[1961 AHSME Problems/Problem 26|Solution]] | [[1961 AHSME Problems/Problem 26|Solution]] | ||
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== Problem 28== | == Problem 28== | ||
− | If <math>2137^{753}</math> is multiplied out, the units' digit | + | If <math>2137^{753}</math> is multiplied out, the units' digit in the final product is: |
<math>\textbf{(A)}\ 1\qquad | <math>\textbf{(A)}\ 1\qquad | ||
\textbf{(B)}\ 3\qquad | \textbf{(B)}\ 3\qquad | ||
\textbf{(C)}\ 5\qquad | \textbf{(C)}\ 5\qquad | ||
− | \textbf{(D)}\ 7 | + | \textbf{(D)}\ 7\qquad |
− | \textbf{(E)}\ 9 | + | \textbf{(E)}\ 9</math> |
[[1961 AHSME Problems/Problem 28|Solution]] | [[1961 AHSME Problems/Problem 28|Solution]] | ||
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\textbf{(B)}\ x^2-bx+ac=0 \qquad\\ | \textbf{(B)}\ x^2-bx+ac=0 \qquad\\ | ||
\textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad | \textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad | ||
− | \textbf{(D)}\ x^2+3bx-ca+2b^2=0 | + | \textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\ |
− | \textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0 | + | \textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0</math> |
[[1961 AHSME Problems/Problem 29|Solution]] | [[1961 AHSME Problems/Problem 29|Solution]] | ||
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\textbf{(B)}\ \frac{2a+b}{a+1}\qquad | \textbf{(B)}\ \frac{2a+b}{a+1}\qquad | ||
\textbf{(C)}\ \frac{a+2b}{1+a}\qquad | \textbf{(C)}\ \frac{a+2b}{1+a}\qquad | ||
− | \textbf{(D)}\ \frac{2a+b}{1-a | + | \textbf{(D)}\ \frac{2a+b}{1-a}\qquad |
− | \textbf{(E)}\ \frac{a+2b}{1-a | + | \textbf{(E)}\ \frac{a+2b}{1-a}</math> |
[[1961 AHSME Problems/Problem 30|Solution]] | [[1961 AHSME Problems/Problem 30|Solution]] | ||
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\textbf{(B)}\ 3:4 \qquad | \textbf{(B)}\ 3:4 \qquad | ||
\textbf{(C)}\ 4:3 \qquad | \textbf{(C)}\ 4:3 \qquad | ||
− | \textbf{(D)}\ 3:1 | + | \textbf{(D)}\ 3:1 \qquad |
− | \textbf{(E)}\ 7:1 | + | \textbf{(E)}\ 7:1 </math> |
[[1961 AHSME Problems/Problem 31|Solution]] | [[1961 AHSME Problems/Problem 31|Solution]] | ||
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\textbf{(B)}\ 10\qquad | \textbf{(B)}\ 10\qquad | ||
\textbf{(C)}\ 12\qquad | \textbf{(C)}\ 12\qquad | ||
− | \textbf{(D)}\ 15 | + | \textbf{(D)}\ 15\qquad |
− | \textbf{(E)}\ 18 | + | \textbf{(E)}\ 18 </math> |
[[1961 AHSME Problems/Problem 32|Solution]] | [[1961 AHSME Problems/Problem 32|Solution]] | ||
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\textbf{(B)}\ 1\qquad | \textbf{(B)}\ 1\qquad | ||
\textbf{(C)}\ 2\qquad | \textbf{(C)}\ 2\qquad | ||
− | \textbf{(D)}\ 3 | + | \textbf{(D)}\ 3\qquad |
− | \textbf{(E)}\ \text{More than three, but finite | + | \textbf{(E)}\ \text{More than three, but finite}</math> |
[[1961 AHSME Problems/Problem 33|Solution]] | [[1961 AHSME Problems/Problem 33|Solution]] | ||
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\textbf{(C)}\ \text{Both m and M are in S}\qquad\\ | \textbf{(C)}\ \text{Both m and M are in S}\qquad\\ | ||
\textbf{(D)}\ \text{Neither m nor M are in S}\qquad\\ | \textbf{(D)}\ \text{Neither m nor M are in S}\qquad\\ | ||
− | \textbf{(E)}\ \text{M does not exist either in or outside S} | + | \textbf{(E)}\ \text{M does not exist either in or outside S} </math> |
[[1961 AHSME Problems/Problem 34|Solution]] | [[1961 AHSME Problems/Problem 34|Solution]] | ||
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\textbf{(B)}\ 1\qquad | \textbf{(B)}\ 1\qquad | ||
\textbf{(C)}\ 2\qquad | \textbf{(C)}\ 2\qquad | ||
− | \textbf{(D)}\ 3 | + | \textbf{(D)}\ 3\qquad |
− | \textbf{(E)}\ 4 | + | \textbf{(E)}\ 4 </math> |
[[1961 AHSME Problems/Problem 35|Solution]] | [[1961 AHSME Problems/Problem 35|Solution]] | ||
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\textbf{(B)}\ \sqrt{17} \qquad | \textbf{(B)}\ \sqrt{17} \qquad | ||
\textbf{(C)}\ 4.25\qquad | \textbf{(C)}\ 4.25\qquad | ||
− | \textbf{(D)}\ 2\sqrt{5 | + | \textbf{(D)}\ 2\sqrt{5} \qquad |
− | \textbf{(E)}\ 4.5 | + | \textbf{(E)}\ 4.5 </math> |
[[1961 AHSME Problems/Problem 36|Solution]] | [[1961 AHSME Problems/Problem 36|Solution]] | ||
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\textbf{(B)}\ 58\qquad | \textbf{(B)}\ 58\qquad | ||
\textbf{(C)}\ 100\qquad | \textbf{(C)}\ 100\qquad | ||
− | \textbf{(D)}\ 116 | + | \textbf{(D)}\ 116\qquad |
− | \textbf{(E)}\ 120 | + | \textbf{(E)}\ 120</math> |
[[1961 AHSME Problems/Problem 37|Solution]] | [[1961 AHSME Problems/Problem 37|Solution]] | ||
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\textbf{(B)}\ s^2=8r^2 \qquad | \textbf{(B)}\ s^2=8r^2 \qquad | ||
\textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ | \textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ | ||
− | \textbf{(D)}\ s^2\le4r^2 | + | \textbf{(D)}\ s^2\le4r^2 \qquad |
− | \textbf{(E)}\ | + | \textbf{(E)}\ s^2=4r^2 </math> |
[[1961 AHSME Problems/Problem 38|Solution]] | [[1961 AHSME Problems/Problem 38|Solution]] | ||
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\textbf{(B)}\ \sqrt{2}/2\qquad | \textbf{(B)}\ \sqrt{2}/2\qquad | ||
\textbf{(C)}\ 2\sqrt{2}/3\qquad | \textbf{(C)}\ 2\sqrt{2}/3\qquad | ||
− | \textbf{(D)}\ 1 | + | \textbf{(D)}\ 1 \qquad |
− | \textbf{(E)}\ \sqrt{2 | + | \textbf{(E)}\ \sqrt{2}</math> |
[[1961 AHSME Problems/Problem 39|Solution]] | [[1961 AHSME Problems/Problem 39|Solution]] | ||
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\textbf{(B)}\ \frac{13}{5}\qquad | \textbf{(B)}\ \frac{13}{5}\qquad | ||
\textbf{(C)}\ \frac{13}{12}\qquad | \textbf{(C)}\ \frac{13}{12}\qquad | ||
− | \textbf{(D)}\ 1 | + | \textbf{(D)}\ 1\qquad |
− | \textbf{(E)}\ 0 | + | \textbf{(E)}\ 0 </math> |
[[1961 AHSME Problems/Problem 40|Solution]] | [[1961 AHSME Problems/Problem 40|Solution]] | ||
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* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
− | {{AHSME box|year=1961|before=[[1960 AHSME]]|after=[[1962 AHSME]]}} | + | {{AHSME 40p box|year=1961|before=[[1960 AHSME|1960 AHSC]]|after=[[1962 AHSME|1962 AHSC]]}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:33, 5 April 2023
1961 AHSC (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
| ||
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 |
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 Problem 31
- 32 Problem 32
- 33 Problem 33
- 34 Problem 34
- 35 Problem 35
- 36 Problem 36
- 37 Problem 37
- 38 Problem 38
- 39 Problem 39
- 40 Problem 40
- 41 See also
Problem 1
When simplified, becomes:
Problem 2
An automobile travels feet in
seconds. If this rate is maintained for
minutes, how many yards does it travel in
minutes?
Problem 3
If the graphs of and
are to meet at right angles, the value of
is:
Problem 4
Let the set consisting of the squares of the positive integers be called ; thus
is the set
.
If a certain operation on one or more members of the set always yields a member of the set,
we say that the set is closed under that operation. Then
is closed under:
Problem 5
Let . Then
equals:
Problem 6
When simplified, becomes:
Problem 7
When simplified, the third term in the expansion of is:
Problem 8
Let the two base angles of a triangle be and
, with
larger than
.
The altitude to the base divides the vertex angle
into two parts,
and
, with
adjacent to side
. Then:
Problem 9
Let be the result of doubling both the base and exponent of
, and
does not equal to
.
If
equals the product of
by
, then
equals:
Problem 10
Each side of is
units.
is the foot of the perpendicular dropped from
on
,
and
is the midpoint of
. The length of
, in the same unit, is:
Problem 11
Two tangents are drawn to a circle from an exterior point ; they touch the circle at points
and
respectively.
A third tangent intersects segment
in
and
in
, and touches the circle at
. If
, then the perimeter of
is
Problem 12
The first three terms of a geometric progression are . Find the fourth term.
Problem 13
The symbol means
is a positive number or zero, and
if
is a negative number.
For all real values of
the expression
is equal to?
Problem 14
A rhombus is given with one diagonal twice the length of the other diagonal.
Express the side of the rhombus is terms of , where
is the area of the rhombus in square inches.
Problem 15
If men working
hours a day for
days produce
articles, then the number of articles
(not necessarily an integer) produced by
men working
hours a day for
days is:
Problem 16
An altitude of a triangle is increased by a length
. How much must be taken from the corresponding base
so that the area of the new triangle is one-half that of the original triangle?
Problem 17
In the base ten number system the number means
.
In the Land of Mathesis, however, numbers are written in the base
.
Jones purchases an automobile there for
monetary units (abbreviated m.u).
He gives the salesman a
m.u bill, and receives, in change,
m.u. The base
is:
Problem 18
The yearly changes in the population census of a town for four consecutive years are, respectively, 25% increase, 25% increase, 25% decrease, 25% decrease. The net change over the four years, to the nearest percent, is:
Problem 19
Consider the graphs of and
. We may say that:
Problem 20
The set of points satisfying the pair of inequalities and
is contained entirely in quadrants:
Problem 21
Medians and
of
intersect in
. The midpoint of
is
.
Let the area of
be
times the area of
. Then
equals:
Problem 22
If is divisible by
, then it is also divisible by:
Problem 23
Points and
are both in the line segment
and on the same side of its midpoint.
divides
in the ratio
,
and
divides
in the ratio
. If
, then the length of
is:
Problem 24
Thirty-one books are arranged from left to right in order of increasing prices.
The price of each book differs by from that of each adjacent book.
For the price of the book at the extreme right a customer can buy the middle book and the adjacent one. Then:
Problem 25
is isosceles with base
. Points
and
are respectively in
and
and such that
.
The number of degrees in
is:
Problem 26
For a given arithmetic series the sum of the first terms is
, and the sum of the next
terms is
.
The first term in the series is:
Problem 27
Given two equiangular polygons and
with different numbers of sides;
each angle of
is
degrees and each angle of
is
degrees,
where
is an integer greater than
.
The number of possibilities for the pair
is:
Problem 28
If is multiplied out, the units' digit in the final product is:
Problem 29
Let the roots of be
and
. The equation with roots
and
is:
Problem 30
If and
, then
?
Problem 31
In the ratio
is
. The bisector of the exterior angle at
intersects
extended at
(
is between
and
). The ratio
is:
Problem 32
A regular polygon of sides is inscribed in a circle of radius
. The area of the polygon is
. Then
equals:
Problem 33
The number of solutions of , in which
and
are integers, is:
Problem 34
Let S be the set of values assumed by the fraction .
When
is any member of the interval
. If there exists a number
such that no number of the set
is greater than
,
then
is an upper bound of
. If there exists a number
such that such that no number of the set
is less than
,
then
is a lower bound of
. We may then say:
Problem 35
The number is to be written with a factorial base of numeration, that is,
where
are integers such that
and
means
. Find
Problem 36
In the median from
is given perpendicular to the median from
. If
and
, find the length of
.
Problem 37
In racing over a distance at uniform speed,
can beat
by
yards,
can beat
by
yards,
and
can beat
by
yards. Then
, in yards, equals:
Problem 38
is inscribed in a semicircle of radius
so that its base
coincides with diameter
.
Point
does not coincide with either
or
. Let
. Then, for all permissible positions of
:
Problem 39
Any five points are taken inside or on a square with side length . Let a be the smallest possible number with the
property that it is always possible to select one pair of points from these five such that the distance between them
is equal to or less than
. Then
is:
Problem 40
Find the minimum value of if
.
See also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by 1960 AHSC |
Followed by 1962 AHSC | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.