Difference between revisions of "1967 AHSME Problems"

Latest revision as of 13:14, 20 February 2020

1967 AHSC (Answer Key) Printable versions: Wiki • AoPS Resources • PDF
Instructions This is a 40-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive ? points for each correct answer, ? points for each problem left unanswered, and ? points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers. Figures are not necessarily drawn to scale. You will have ? minutes working time to complete the test.
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Problem 1

The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$ . If $5b9$ is divisible by 9, then $a+b$ equals

$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution

Problem 2

An equivalent of the expression

$\left(\frac{x^2+1}{x}\right)\left(\frac{y^2+1}{y}\right)+\left(\frac{x^2-1}{y}\right)\left(\frac{y^2-1}{x}\right)$ , $xy \not= 0$ ,

is:

$\text{(A)}\ 1\qquad\text{(B)}\ 2xy\qquad\text{(C)}\ 2x^2y^2+2\qquad\text{(D)}\ 2xy+\frac{2}{xy}\qquad\text{(E)}\ \frac{2x}{y}+\frac{2y}{x}$

Solution

Problem 3

The side of an equilateral triangle is $s$ . A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:

$\text{(A)}\ \frac{s^2}{24}\qquad\text{(B)}\ \frac{s^2}{6}\qquad\text{(C)}\ \frac{s^2\sqrt{2}}{6}\qquad\text{(D)}\ \frac{s^2\sqrt{3}}{6}\qquad\text{(E)}\ \frac{s^2}{3}$

Solution

Problem 4

Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$ , all logarithms to the same base and $x \not= 1$ . If $\frac{b^2}{ac}=x^y$ , then $y$ is:

$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$

Solution

Problem 5

A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:

$\text{(A)}\text{independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2}$

Solution

Problem 6

If $f(x)=4^x$ then $f(x+1)-f(x)$ equals:

$\text{(A)}\ 4\qquad\text{(B)}\ f(x)\qquad\text{(C)}\ 2f(x)\qquad\text{(D)}\ 3f(x)\qquad\text{(E)}\ 4f(x)$

Solution

Problem 7

If $\frac{a}{b}<-\frac{c}{d}$ where $a$ , $b$ , $c$ , and $d$ are real numbers and $bd \not= 0$ , then:

$\text{(A)}\ a \; \text{must be negative} \qquad \text{(B)}\ a \; \text{must be positive} \qquad$ $\text{(C)}\ a \; \text{must not be zero} \qquad \text{(D)}\ a \; \text{can be negative or zero, but not positive } \\ \text{(E)}\ a \; \text{can be positive, negative, or zero}$

Solution

Problem 8

To $m$ ounces of a $m\%$ solution of acid, $x$ ounces of water are added to yield a $(m-10)\%$ solution. If $m>25$ , then $x$ is

$\textbf{(A)}\ \frac{10m}{m-10} \qquad \textbf{(B)}\ \frac{5m}{m-10} \qquad \textbf{(C)}\ \frac{m}{m-10} \qquad \textbf{(D)}\ \frac{5m}{m-20} \\ \textbf{(E)}\ \text{not determined by the given information}$

Solution

Problem 9

Let $K$ , in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:

$\textbf{(A)}\ K \; \text{must be an integer} \qquad \textbf{(B)}\ K \; \text{must be a rational fraction} \\ \textbf{(C)}\ K \; \text{must be an irrational number} \qquad \textbf{(D)}\ K\; \text{must be an integer or a rational fraction} \qquad$ $\textbf{(E)}\ \text{taken alone neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is true}$

Solution

Problem 10

If $\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}$ is an identity for positive rational values of $x$ , then the value of $a-b$ is:

$\textbf{(A)}\ \frac{4}{3} \qquad \textbf{(B)}\ \frac{5}{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{11}{4} \qquad \textbf{(E)}\ 3$

Solution

Problem 11

If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$ , in inches, is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 12

If the (convex) area bounded by the x-axis and the lines $y=mx+4$ , $x=1$ , and $x=4$ is $7$ , then $m$ equals:

$\textbf{(A)}\ -\frac{1}{2}\qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ -\frac{3}{2} \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 13

A triangle $ABC$ is to be constructed given a side $a$ (oppisite angle $A$ ). angle $B$ , and $h_c$ , the altitude from $C$ . If $N$ is the number of noncongruent solutions, then $N$

$\textbf{(A)}\ \text{is} \; 1\qquad \textbf{(B)}\ \text{is} \; 2\qquad \textbf{(C)}\ \text{must be zero}\qquad \textbf{(D)}\ \text{must be infinite}\qquad \textbf{(E)}\ \text{must be zero or infinite}$

Solution

Problem 14

Let $f(t)=\frac{t}{1-t}$ , $t \not= 1$ . If $y=f(x)$ , then $x$ can be expressed as

$\textbf{(A)}\ f\left(\frac{1}{y}\right)\qquad \textbf{(B)}\ -f(y)\qquad \textbf{(C)}\ -f(-y)\qquad \textbf{(D)}\ f(-y)\qquad \textbf{(E)}\ f(y)$

Solution

Problem 15

The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is:

$\textbf{(A)}\ 12\quad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 6\sqrt{2}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 3\sqrt{2}$

Solution

Problem 16

Let the product $(12)(15)(16)$ , each factor written in base $b$ , equal $3146$ in base $b$ . Let $s=12+15+16$ , each term expressed in base $b$ . Then $s$ , in base $b$ , is

$\textbf{(A)}\ 43\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 47$

Solution

Problem 17

If $r_1$ and $r_2$ are the distinct real roots of $x^2+px+8=0$ , then it must follow that:

$\textbf{(A)}\ |r_1+r_2|>4\sqrt{2}\qquad \textbf{(B)}\ |r_1|>3 \; \text{or} \; |r_2| >3 \\ \textbf{(C)}\ |r_1|>2 \; \text{and} \; |r_2|>2\qquad \textbf{(D)}\ r_1<0 \; \text{and} \; r_2<0\qquad \textbf{(E)}\ |r_1+r_2|<4\sqrt{2}$

Solution

Problem 18

If $x^2-5x+6<0$ and $P=x^2+5x+6$ then

$\textbf{(A)}\ P \; \text{can take any real value} \qquad \textbf{(B)}\ 20<P<30\\ \textbf{(C)}\ 0<P<20 \qquad \textbf{(D)}\ P<0 \qquad \textbf{(E)}\ P>30$

Solution

Problem 19

The area of a rectangle remains unchanged when it is made $2 \frac{1}{2}$ inches longer and $\frac{2}{3}$ inch narrower, or when it is made $2 \frac{1}{2}$ inches shorter and $\frac{4}{3}$ inch wider. Its area, in square inches, is:

$\textbf{(A)}\ 30\qquad \textbf{(B)}\ \frac{80}{3}\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ \frac{45}{2}\qquad \textbf{(E)}\ 20$

Solution

Problem 20

A circle is inscribed in a square of side $m$ , then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If $S_n$ is the sum of the areas of the first $n$ circles so inscribed, then, as $n$ grows beyond all bounds, $S_n$ approaches:

$\textbf{(A)}\ \frac{\pi m^2}{2}\qquad \textbf{(B)}\ \frac{3\pi m^2}{8}\qquad \textbf{(C)}\ \frac{\pi m^2}{3}\qquad \textbf{(D)}\ \frac{\pi m^2}{4}\qquad \textbf{(E)}\ \frac{\pi m^2}{8}$

Solution

Problem 21

In right triangle $ABC$ the hypotenuse $\overline{AB}=5$ and leg $\overline{AC}=3$ . The bisector of angle $A$ meets the opposite side in $A_1$ . A second right triangle $PQR$ is then constructed with hypotenuse $\overline{PQ}=A_1B$ and leg $\overline{PR}=A_1C$ . If the bisector of angle $P$ meets the opposite side in $P_1$ , the length of $PP_1$ is:

$\textbf{(A)}\ \frac{3\sqrt{6}}{4}\qquad \textbf{(B)}\ \frac{3\sqrt{5}}{4}\qquad \textbf{(C)}\ \frac{3\sqrt{3}}{4}\qquad \textbf{(D)}\ \frac{3\sqrt{2}}{2}\qquad \textbf{(E)}\ \frac{15\sqrt{2}}{16}$

Solution

Problem 22

For natural numbers, when $P$ is divided by $D$ , the quotient is $Q$ and the remainder is $R$ . When $Q$ is divided by $D'$ , the quotient is $Q'$ and the remainder is $R'$ . Then, when $P$ is divided by $DD'$ , the remainder is:

$\textbf{(A)}\ R+R'D\qquad \textbf{(B)}\ R'+RD\qquad \textbf{(C)}\ RR'\qquad \textbf{(D)}\ R\qquad \textbf{(E)}\ R'$

Solution

Problem 23

If $x$ is real and positive and grows beyond all bounds, then $\log_3{(6x-5)}-\log_3{(2x+1)}$ approaches:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{no finite number}$

Solution

Problem 24

The number of solution-pairs in the positive integers of the equation $3x+5y=501$ is:

$\textbf{(A)}\ 33\qquad \textbf{(B)}\ 34\qquad \textbf{(C)}\ 35\qquad \textbf{(D)}\ 100\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 25

For every odd number $p>1$ we have:

$\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad \textbf{(B)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p\\ \textbf{(C)}\ (p-1)^{\frac{1}{2}(p-1)} \; \text{is divisible by} \; p\qquad \textbf{(D)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p+1\\ \textbf{(E)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-1$

Solution

Problem 26

If one uses only the tabular information $10^3=1000$ , $10^4=10,000$ , $2^{10}=1024$ , $2^{11}=2048$ , $2^{12}=4096$ , $2^{13}=8192$ , then the strongest statement one can make for $\log_{10}{2}$ is that it lies between:

$\textbf{(A)}\ \frac{3}{10} \; \text{and} \; \frac{4}{11}\qquad \textbf{(B)}\ \frac{3}{10} \; \text{and} \; \frac{4}{12}\qquad \textbf{(C)}\ \frac{3}{10} \; \text{and} \; \frac{4}{13}\qquad \textbf{(D)}\ \frac{3}{10} \; \text{and} \; \frac{40}{132}\qquad \textbf{(E)}\ \frac{3}{11} \; \text{and} \; \frac{40}{132}$

Solution

Problem 27

Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in $3$ hours and the other in $4$ hours. At what time P.M. should the candles be lighted so that, at 4 P.M., one stub is twice the length of the other?

$\textbf{(A)}\ 1:24\qquad \textbf{(B)}\ 1:28\qquad \textbf{(C)}\ 1:36\qquad \textbf{(D)}\ 1:40\qquad \textbf{(E)}\ 1:48$

Solution

Problem 28

Given the two hypotheses: $\text{I}$ Some Mems are not Ens and $\text{II}$ No Ens are Veens. If "some" means "at least one," we can conclude that:

$\textbf{(A)}\ \text{Some Mems are not Veens}\qquad \textbf{(B)}\ \text{Some Vees are not Mems}\\ \textbf{(C)}\ \text{No Mem is a Vee}\qquad \textbf{(D)}\ \text{Some Mems are Vees}\\ \textbf{(E)}\ \text{Neither} \; \textbf{(A)} \; \text{nor} \; \textbf{(B)} \; \text{nor} \; \textbf{(C)} \; \text{nor} \; \textbf{(D)} \; \text{is deducible from the given statements}$

Solution

Problem 29

$\overline{AB}$ is a diameter of a circle. Tangents $\overline{AD}$ and $\overline{BC}$ are drawn so that $\overline{AC}$ and $\overline{BD}$ intersect in a point on the circle. If $\overline{AD}=a$ and $\overline{BD}=b$ , $a \not= b$ , the diameter of the circle is:

$\textbf{(A)}\ |a-b|\qquad \textbf{(B)}\ \frac{1}{2}(a+b)\qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \frac{ab}{a+b}\qquad \textbf{(E)}\ \frac{1}{2}\frac{ab}{a+b}$

Solution

Problem 30

A dealer bought $n$ radios for $d$ dollars, $d$ a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of $8 on each radio sold. If the overall profit was $72, then the least possible value of $n$ for the given information is:

$\textbf{(A)}\ 18\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 15\qquad \textbf{(D)}\ 12\qquad \textbf{(E)}\ 11$

Solution

Problem 31

Let $D=a^2+b^2+c^2$ , where $a$ , $b$ , are consecutive integers and $c=ab$ . Then $\sqrt{D}$ is:

$\textbf{(A)}\ \text{always an even integer}\qquad \textbf{(B)}\ \text{sometimes an odd integer, sometimes not}\\ \textbf{(C)}\ \text{always an odd integer}\qquad \textbf{(D)}\ \text{sometimes rational, sometimes not}\\ \textbf{(E)}\ \text{always irrational}$

Solution

Problem 32

In quadrilateral $ABCD$ with diagonals $\overline{AC}$ and $\overline{BD}$ intersecting at $O$ , $\overline{BO}=4$ , $\overline{OD}=6$ , $\overline{AO}=8$ , $\overline{OC}=3$ , and $\overline{AB}=6$ . The length of $\overline{AD}$ is:

$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

Solution

Problem 33

$[asy] fill(circle((4,0),4),grey); fill((0,0)--(8,0)--(8,-4)--(0,-4)--cycle,white); fill(circle((7,0),1),white); fill(circle((3,0),3),white); draw((0,0)--(8,0),black+linewidth(1)); draw((6,0)--(6,sqrt(12)),black+linewidth(1)); MP("A", (0,0), W); MP("B", (8,0), E); MP("C", (6,0), S); MP("D",(6,sqrt(12)), N); [/asy]$

In this diagram semi-circles are constructed on diameters $\overline{AB}$ , $\overline{AC}$ , and $\overline{CB}$ , so that they are mutually tangent. If $\overline{CD} \bot \overline{AB}$ , then the ratio of the shaded area to the area of a circle with $\overline{CD}$ as radius is:

$\textbf{(A)}\ 1:2\qquad \textbf{(B)}\ 1:3\qquad \textbf{(C)}\ \sqrt{3}:7\qquad \textbf{(D)}\ 1:4\qquad \textbf{(E)}\ \sqrt{2}:6$

Solution

Problem 34

Points $D$ , $E$ , $F$ are taken respectively on sides $AB$ , $BC$ , and $CA$ of triangle $ABC$ so that $AD:DB=BE:CE=CF:FA=1:n$ . The ratio of the area of triangle $DEF$ to that of triangle $ABC$ is:

$\textbf{(A)}\ \frac{n^2-n+1}{(n+1)^2}\qquad \textbf{(B)}\ \frac{1}{(n+1)^2}\qquad \textbf{(C)}\ \frac{2n^2}{(n+1)^2}\qquad \textbf{(D)}\ \frac{n^2}{(n+1)^2}\qquad \textbf{(E)}\ \frac{n(n-1)}{n+1}$

Solution

Problem 35

The roots of $64x^3-144x^2+92x-15=0$ are in arithmetic progression. The difference between the largest and smallest roots is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \frac{1}{2}\qquad \textbf{(D)}\ \frac{3}{8}\qquad \textbf{(E)}\ \frac{1}{4}$

Solution

Problem 36

Given a geometric progression of five terms, each a positive integer less than $100$ . The sum of the five terms is $211$ . If $S$ is the sum of those terms in the progression which are squares of integers, then $S$ is:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 91\qquad \textbf{(C)}\ 133\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 211$

Solution

Problem 37

Segments $AD=10$ , $BE=6$ , $CF=24$ are drawn from the vertices of triangle $ABC$ , each perpendicular to a straight line $RS$ , not intersecting the triangle. Points $D$ , $E$ , $F$ are the intersection points of $RS$ with the perpendiculars. If $x$ is the length of the perpendicular segment $GH$ drawn to $RS$ from the intersection point $G$ of the medians of the triangle, then $x$ is:

$\textbf{(A)}\ \frac{40}{3}\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ \frac{56}{3}\qquad \textbf{(D)}\ \frac{80}{3}\qquad \textbf{(E)}\ \text{undetermined}$

Solution

Problem 38

Given a set $S$ consisting of two undefined elements "pib" and "maa", and the four postulates: $P_1$ : Every pib is a collection of maas, $P_2$ : Any two distinct pibs have one and only one maa in common, $P_3$ : Every maa belongs to two and only two pibs, $P_4$ : There are exactly four pibs. Consider the three theorems: $T_1$ : There are exactly six maas, $T_2$ : There are exactly three maas in each pib, $T_3$ : For each maa there is exactly one other maa not in the same pid with it. The theorems which are deducible from the postulates are:

$\textbf{(A)}\ T_3 \; \text{only}\qquad \textbf{(B)}\ T_2 \; \text{and} \; T_3 \; \text{only} \qquad \textbf{(C)}\ T_1 \; \text{and} \; T_2 \; \text{only}\\ \textbf{(D)}\ T_1 \; \text{and} \; T_3 \; \text{only}\qquad \textbf{(E)}\ \text{all}$

Solution

Problem 39

Given the sets of consecutive integers $\{1\}$ , $\{2, 3\}$ , $\{4,5,6\}$ , $\{7,8,9,10\}$ , $\; \cdots \;$ , where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let $S_n$ be the sum of the elements in the nth set. Then $S_{21}$ equals:

$\textbf{(A)}\ 1113\qquad \textbf{(B)}\ 4641 \qquad \textbf{(C)}\ 5082\qquad \textbf{(D)}\ 53361\qquad \textbf{(E)}\ \text{none of these}$

Solution

Problem 40

Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ , $PB=6$ , and $PC=10$ . To the nearest integer the area of triangle $ABC$ is:

$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$

Solution

@@ Line 1: / Line 1: @@
+{{AHSC 40 Problems
+|year = 1967
+}}
 ==Problem 1==
@@ Line 6: / Line 9: @@
 [[1967 AHSME Problems/Problem 1|Solution]]
 ==Problem 2==
@@ Line 330: / Line 333: @@
 ==Problem 30==
-A dealer bought <math>n</math> radios for <math>d</math> dollars, <math>d</math> a positive integer.  He contributed two radios to a community bazaar at half their cost.  The rest he sold at a profit of <math>&#36;8</math> on each radio sold.  If the overall profit was <math>&#36;72</math>, then the least possible value of <math>n</math> for the given information is:
+A dealer bought <math>n</math> radios for <math>d</math> dollars, <math>d</math> a positive integer.  He contributed two radios to a community bazaar at half their cost.  The rest he sold at a profit of \$8 on each radio sold.  If the overall profit was \$72, then the least possible value of <math>n</math> for the given information is:
 <math>\textbf{(A)}\ 18\qquad
@@ Line 354: / Line 357: @@
 ==Problem 32==
-In quadrilateral <math>ABCD</math> with diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersecting at <math>O</math>, <math>\overline{BO}=4</math>, <math>\overline{AO}=8</math>, <math>\overline{OC}=3</math>, and <math>\overline{AB}=6</math>.  The length of <math>\overline{AD}</math> is:
+In quadrilateral <math>ABCD</math> with diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersecting at <math>O</math>, <math>\overline{BO}=4</math>, <math>\overline{OD}=6</math>, <math>\overline{AO}=8</math>, <math>\overline{OC}=3</math>, and <math>\overline{AB}=6</math>.  The length of <math>\overline{AD}</math> is:
 <math>\textbf{(A)}\ 9\qquad
@@ Line 448: / Line 451: @@
 ==Problem 39==
-Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set.  Let <math>S_n</math> be the sum of the elements in the nth set.  Then <math>S_n</math> equals:
+Given the sets of consecutive integers <math>\{1\}</math>,<math> \{2, 3\}</math>,<math> \{4,5,6\}</math>,<math> \{7,8,9,10\}</math>,<math>\; \cdots \; </math>, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set.  Let <math>S_n</math> be the sum of the elements in the nth set.  Then <math>S_{21}</math> equals:
 <math>\textbf{(A)}\ 1113\qquad
@@ Line 469: / Line 472: @@
 [[1967 AHSME Problems/Problem 40|Solution]]
+== See also ==
+* [[AMC 12 Problems and Solutions]]
+* [[Mathematics competition resources]]
+{{AHSME 40p box|year=1967|before=[[1966 AHSME|1966 AHSC]]|after=[[1968 AHSME|1968 AHSC]]}}
+{{MAA Notice}}

1967 AHSC (Problems • Answer Key • Resources)
Preceded by 1966 AHSC	Followed by 1968 AHSC
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All AHSME Problems and Solutions

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Difference between revisions of "1967 AHSME Problems"

Latest revision as of 13:14, 20 February 2020

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

See also