MathDB

1971 AMC 12/AHSME

Part of AMC 12/AHSME

Subcontests

(35)
35
1

Infinite sequence of circles

Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. The ratio of the area of the first circle to the sum of areas of all other circles in the sequence, is
<spanclass=′latex−bold′>(A)</span>(4+32):4<spanclass=′latex−bold′>(B)</span>92:2<spanclass=′latex−bold′>(C)</span>(16+122):1<span class='latex-bold'>(A) </span>(4+3\sqrt{2}):4\qquad<span class='latex-bold'>(B) </span>9\sqrt{2}:2\qquad<span class='latex-bold'>(C) </span>(16+12\sqrt{2}):1\qquad
<spanclass=′latex−bold′>(D)</span>(2+22):1<spanclass=′latex−bold′>(E)</span>3+22):1<span class='latex-bold'>(D) </span>(2+2\sqrt{2}):1\qquad <span class='latex-bold'>(E) </span>3+2\sqrt{2}):1
34
1
33
1

Geometric Progression

If PP is the product of nn quantities in Geometric Progression, SS their sum, and S′S' the sum of their reciprocals, then PP in terms of SS, S′S', and nn is
<spanclass=′latex−bold′>(A)</span>(SS′)12n<spanclass=′latex−bold′>(B)</span>(S/S′)12n<spanclass=′latex−bold′>(C)</span>(SS′)n−2<spanclass=′latex−bold′>(D)</span>(S/S′)n<spanclass=′latex−bold′>(E)</span>(S/S′)12(n−1)<span class='latex-bold'>(A) </span>(SS')^{\frac{1}{2}n}\qquad<span class='latex-bold'>(B) </span>(S/S')^{\frac{1}{2}n}\qquad<span class='latex-bold'>(C) </span>(SS')^{n-2}\qquad<span class='latex-bold'>(D) </span>(S/S')^n\qquad <span class='latex-bold'>(E) </span>(S/S')^{\frac{1}{2}(n-1)}
32
1

Simplify s

If s=(1+2−132)(1+2−116)(1+2−18)(1+2−14)(1+2−12)s=(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})(1+2^{-\frac{1}{2}}), then ss is equal to
<spanclass=′latex−bold′>(A)</span>12(1−2−132)−1<spanclass=′latex−bold′>(B)</span>(1−2−132)−1<spanclass=′latex−bold′>(C)</span>1−2−132<span class='latex-bold'>(A) </span>\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})^{-1}\qquad<span class='latex-bold'>(B) </span>(1-2^{-\frac{1}{32}})^{-1}\qquad<span class='latex-bold'>(C) </span>1-2^{-\frac{1}{32}}\qquad
<spanclass=′latex−bold′>(D)</span>12(1−2−132)<spanclass=′latex−bold′>(E)</span>12<span class='latex-bold'>(D) </span>\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})\qquad <span class='latex-bold'>(E) </span>\frac{1}{2}
31
1

Find CD

[asy] size(2.5inch); pair A = (-2,0), B = 2dir(150), D = (2,0), C; draw(A..(0,2)..D--cycle); C = intersectionpoint(A..(0,2)..D,Circle(B,arclength(A--B))); draw(A--B--C--D--cycle); label("AA",A,W); label("BB",B,NW); label("CC",C,N); label("DD",D,E); label("44",A--D,S); label("11",A--B,E); label("11",B--C,SE); //Credit to chezbgone2 for the diagram[/asy]
Quadrilateral ABCDABCD is inscribed in a circle with side ADAD, a diameter of length 44. If sides ABAB and BCBC each have length 11, then side CDCD has length
<spanclass=′latex−bold′>(A)</span>72<spanclass=′latex−bold′>(B)</span>522<spanclass=′latex−bold′>(C)</span>11<spanclass=′latex−bold′>(D)</span>13<spanclass=′latex−bold′>(E)</span>23<span class='latex-bold'>(A) </span>\frac{7}{2}\qquad<span class='latex-bold'>(B) </span>\frac{5\sqrt{2}}{2}\qquad<span class='latex-bold'>(C) </span>\sqrt{11}\qquad<span class='latex-bold'>(D) </span>\sqrt{13}\qquad <span class='latex-bold'>(E) </span>2\sqrt{3}
30
1

Fractional Transformation of x

Given the linear fractional transformation of xx into f1(x)=2x−1x+1f_1(x)=\dfrac{2x-1}{x+1}. Define fn+1(x)=f1(fn(x))f_{n+1}(x)=f_1(f_n(x)) for n=1,2,3,⋯n=1,2,3,\cdots. Assuming that f35(x)=f5(x)f_{35}(x)=f_5(x), it follows that f28(x)f_{28}(x) is equal to
<spanclass=′latex−bold′>(A)</span>x<spanclass=′latex−bold′>(B)</span>1x<spanclass=′latex−bold′>(C)</span>x−1x<spanclass=′latex−bold′>(D)</span>11−x<spanclass=′latex−bold′>(E)</span>None of these<span class='latex-bold'>(A) </span>x\qquad<span class='latex-bold'>(B) </span>\frac{1}{x}\qquad<span class='latex-bold'>(C) </span>\frac{x-1}{x}\qquad<span class='latex-bold'>(D) </span>\frac{1}{1-x}\qquad <span class='latex-bold'>(E) </span>\text{None of these}
29
1

Progression 10^(n/11)

Given the progression 10111,10211,10311,10411,…,10n1110^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}. The least positive integer nn such that the product of the first nn terms of the progression exceeds 100,000100,000 is
<spanclass=′latex−bold′>(A)</span>7<spanclass=′latex−bold′>(B)</span>8<spanclass=′latex−bold′>(C)</span>9<spanclass=′latex−bold′>(D)</span>10<spanclass=′latex−bold′>(E)</span>11<span class='latex-bold'>(A) </span>7\qquad<span class='latex-bold'>(B) </span>8\qquad<span class='latex-bold'>(C) </span>9\qquad<span class='latex-bold'>(D) </span>10\qquad <span class='latex-bold'>(E) </span>11
28
1

Multiple lines in a triangle

Nine lines parallel to the base of a triangle divide the other sides each into 1010 equal segments and the area into 1010 distinct parts. If the area of the largest of these parts is 3838, then the area of the original triangle is
<spanclass=′latex−bold′>(A)</span>180<spanclass=′latex−bold′>(B)</span>190<spanclass=′latex−bold′>(C)</span>200<spanclass=′latex−bold′>(D)</span>210<spanclass=′latex−bold′>(E)</span>240<span class='latex-bold'>(A) </span>180\qquad<span class='latex-bold'>(B) </span>190\qquad<span class='latex-bold'>(C) </span>200\qquad<span class='latex-bold'>(D) </span>210\qquad <span class='latex-bold'>(E) </span>240
27
1

Find the minimum number of red chips

A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least 5555. The minimum number of red chips is
<spanclass=′latex−bold′>(A)</span>24<spanclass=′latex−bold′>(B)</span>33<spanclass=′latex−bold′>(C)</span>45<spanclass=′latex−bold′>(D)</span>54<spanclass=′latex−bold′>(E)</span>57<span class='latex-bold'>(A) </span>24\qquad<span class='latex-bold'>(B) </span>33\qquad<span class='latex-bold'>(C) </span>45\qquad<span class='latex-bold'>(D) </span>54\qquad <span class='latex-bold'>(E) </span>57
26
1

Find BE:EC

[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("AA",A,N); label("BB",B,W); label("CC",C,dir(0)); label("EE",E,S); label("FF",F,NE); label("GG",G,SE); //Credit to chezbgone2 for the diagram[/asy]
In triangle ABCABC, point FF divides side ACAC in the ratio 1:21:2. Let EE be the point of intersection of side BCBC and AGAG where GG is the midpoints of BFBF. The point EE divides side BCBC in the ratio
<spanclass=′latex−bold′>(A)</span>1:4<spanclass=′latex−bold′>(B)</span>1:3<spanclass=′latex−bold′>(C)</span>2:5<spanclass=′latex−bold′>(D)</span>4:11<spanclass=′latex−bold′>(E)</span>3:8<span class='latex-bold'>(A) </span>1:4\qquad<span class='latex-bold'>(B) </span>1:3\qquad<span class='latex-bold'>(C) </span>2:5\qquad<span class='latex-bold'>(D) </span>4:11\qquad <span class='latex-bold'>(E) </span>3:8
25
1

Father and son's ages

A teen age boy wrote his own age after his father's. From this new four place number, he subtracted the absolute value of the difference of their ages to get 4,2894,289. The sum of their ages was
<spanclass=′latex−bold′>(A)</span>48<spanclass=′latex−bold′>(B)</span>52<spanclass=′latex−bold′>(C)</span>56<spanclass=′latex−bold′>(D)</span>59<spanclass=′latex−bold′>(E)</span>64<span class='latex-bold'>(A) </span>48\qquad<span class='latex-bold'>(B) </span>52\qquad<span class='latex-bold'>(C) </span>56\qquad<span class='latex-bold'>(D) </span>59\qquad <span class='latex-bold'>(E) </span>64
24
1

Elements in Pascal's Triangle

[asy] label("11",(0,0),S); label("11",(-1,-1),S); label("11",(-2,-2),S); label("11",(-3,-3),S); label("11",(-4,-4),S); label("11",(1,-1),S); label("11",(2,-2),S); label("11",(3,-3),S); label("11",(4,-4),S); label("22",(0,-2),S); label("33",(-1,-3),S); label("33",(1,-3),S); label("44",(-2,-4),S); label("44",(2,-4),S); label("66",(0,-4),S); label("etc.",(0,-5),S); //Credit to chezbgone2 for the diagram[/asy]
Pascal's triangle is an array of positive integers(See figure), in which the first row is 11, the second row is two 11's, each row begins and ends with 11, and the kthk^\text{th} number in any row when it is not 11, is the sum of the kthk^\text{th} and (k−1)th(k-1)^\text{th} numbers in the immediately preceding row. The quotient of the number of numbers in the first nn rows which are not 11's and the number of 11's is
<spanclass=′latex−bold′>(A)</span>n2−n2n−1<spanclass=′latex−bold′>(B)</span>n2−n4n−2<spanclass=′latex−bold′>(C)</span>n2−2n2n−1<spanclass=′latex−bold′>(D)</span>n2−3n+24n−2<spanclass=′latex−bold′>(E)</span>None of these<span class='latex-bold'>(A) </span>\dfrac{n^2-n}{2n-1}\qquad<span class='latex-bold'>(B) </span>\dfrac{n^2-n}{4n-2}\qquad<span class='latex-bold'>(C) </span>\dfrac{n^2-2n}{2n-1}\qquad<span class='latex-bold'>(D) </span>\dfrac{n^2-3n+2}{4n-2}\qquad <span class='latex-bold'>(E) </span>\text{None of these}
23
1

Odds favoring Team A

Teams A\text{A} and B\text{B} are playing a series of games. If the odds for either to win any game are even and Team A\text{A} must win two or Team B\text{B} three games to win the series, then the odds favoring Team A\text{A} to win the series are
<spanclass=′latex−bold′>(A)</span>11 to 5<spanclass=′latex−bold′>(B)</span>5 to 2<spanclass=′latex−bold′>(C)</span>8 to 3<spanclass=′latex−bold′>(D)</span>3 to 2<spanclass=′latex−bold′>(E)</span>13 to 6<span class='latex-bold'>(A) </span>11\text{ to }5\qquad<span class='latex-bold'>(B) </span>5\text{ to }2\qquad<span class='latex-bold'>(C) </span>8\text{ to }3\qquad<span class='latex-bold'>(D) </span>3\text{ to }2\qquad <span class='latex-bold'>(E) </span>13\text{ to }6
22
1

Imaginary root w

If ww is one of the imaginary roots of the equation x3=1x^3=1, then the product (1−w+w2)(1+w−w2)(1-w+w^2)(1+w-w^2) is equal to
<spanclass=′latex−bold′>(A)</span>4<spanclass=′latex−bold′>(B)</span>w<spanclass=′latex−bold′>(C)</span>2<spanclass=′latex−bold′>(D)</span>w2<spanclass=′latex−bold′>(E)</span>1<span class='latex-bold'>(A) </span>4\qquad<span class='latex-bold'>(B) </span>w\qquad<span class='latex-bold'>(C) </span>2\qquad<span class='latex-bold'>(D) </span>w^2\qquad <span class='latex-bold'>(E) </span>1
21
1

Nested Logarithms

If log⁡2(log⁡3(log⁡4x))=log⁡3(log⁡4(log⁡2y))=log⁡4(log⁡2(log⁡3z))=0\log_2(\log_3(\log_4 x))=\log_3(\log_4(\log_2 y))=\log_4(\log_2(\log_3 z))=0, then the sum x+y+zx+y+z is equal to
<spanclass=′latex−bold′>(A)</span>50<spanclass=′latex−bold′>(B)</span>58<spanclass=′latex−bold′>(C)</span>89<spanclass=′latex−bold′>(D)</span>111<spanclass=′latex−bold′>(E)</span>1296<span class='latex-bold'>(A) </span>50\qquad<span class='latex-bold'>(B) </span>58\qquad<span class='latex-bold'>(C) </span>89\qquad<span class='latex-bold'>(D) </span>111\qquad <span class='latex-bold'>(E) </span>1296
20
1

Clever Vieta Manipulation

The sum of the squares of the roots of the equation x2+2hx=3x^2+2hx=3 is 1010. The absolute value of hh is equal to
<spanclass=′latex−bold′>(A)</span>−1<spanclass=′latex−bold′>(B)</span>12<spanclass=′latex−bold′>(C)</span>32<spanclass=′latex−bold′>(D)</span>2<spanclass=′latex−bold′>(E)</span>None of these<span class='latex-bold'>(A) </span>-1\qquad<span class='latex-bold'>(B) </span>\textstyle\frac{1}{2}\qquad<span class='latex-bold'>(C) </span>\textstyle\frac{3}{2}\qquad<span class='latex-bold'>(D) </span>2\qquad <span class='latex-bold'>(E) </span>\text{None of these}
19
1

Find m^2

If the line y=mx+1y=mx+1 intersects the ellipse x2+4y2=1x^2+4y^2=1 exactly once, then the value of m2m^2 is
<spanclass=′latex−bold′>(A)</span>12<spanclass=′latex−bold′>(B)</span>23<spanclass=′latex−bold′>(C)</span>34<spanclass=′latex−bold′>(D)</span>45<spanclass=′latex−bold′>(E)</span>56<span class='latex-bold'>(A) </span>\textstyle\frac{1}{2}\qquad<span class='latex-bold'>(B) </span>\frac{2}{3}\qquad<span class='latex-bold'>(C) </span>\frac{3}{4}\qquad<span class='latex-bold'>(D) </span>\frac{4}{5}\qquad <span class='latex-bold'>(E) </span>\frac{5}{6}
18
1

River current

The current in a river is flowing steadily at 33 miles per hour. A motor boat which travels at a constant rate in still water goes downstream 44 miles and then returns to its starting point. The trip takes one hour, excluding the time spent in turning the boat around. The ratio of the downstream to the upstream rate is
<spanclass=′latex−bold′>(A)</span>4:3<spanclass=′latex−bold′>(B)</span>3:2<spanclass=′latex−bold′>(C)</span>5:3<spanclass=′latex−bold′>(D)</span>2:1<spanclass=′latex−bold′>(E)</span>5:2<span class='latex-bold'>(A) </span>4:3\qquad<span class='latex-bold'>(B) </span>3:2\qquad<span class='latex-bold'>(C) </span>5:3\qquad<span class='latex-bold'>(D) </span>2:1\qquad <span class='latex-bold'>(E) </span>5:2
17
1

Maximum number of regions

A circular disk is divided by 2n2n equally spaced radii(n>0n>0) and one secant line. The maximum number of non-overlapping areas into which the disk can be divided is
<spanclass=′latex−bold′>(A)</span>2n+1<spanclass=′latex−bold′>(B)</span>2n+2<spanclass=′latex−bold′>(C)</span>3n−1<spanclass=′latex−bold′>(D)</span>3n<spanclass=′latex−bold′>(E)</span>3n+1<span class='latex-bold'>(A) </span>2n+1\qquad<span class='latex-bold'>(B) </span>2n+2\qquad<span class='latex-bold'>(C) </span>3n-1\qquad<span class='latex-bold'>(D) </span>3n\qquad <span class='latex-bold'>(E) </span>3n+1
16
1

Ratio of second average to first

After finding the average of 3535 scores, a student carelessly included the average with the 3535 scores and found the average of these 3636 numbers. The ratio of the second average to the true average was
<spanclass=′latex−bold′>(A)</span>1:1<spanclass=′latex−bold′>(B)</span>35:36<spanclass=′latex−bold′>(C)</span>36:35<spanclass=′latex−bold′>(D)</span>2:1<spanclass=′latex−bold′>(E)</span>None of these<span class='latex-bold'>(A) </span>1:1\qquad<span class='latex-bold'>(B) </span>35:36\qquad<span class='latex-bold'>(C) </span>36:35\qquad<span class='latex-bold'>(D) </span>2:1\qquad <span class='latex-bold'>(E) </span>\text{None of these}
15
1

Depth of an Aquarium

An aquarium on a level table has rectangular faces and is 1010 inches wide and 88 inches high. When it was tilted, the water in it covered an 8"×10"8"\times 10" end but only three-fourths of the rectangular room. The depth of the water when the bottom was again made level, was
<spanclass=′latex−bold′>(A)</span>212"<spanclass=′latex−bold′>(B)</span>3"<spanclass=′latex−bold′>(C)</span>314"<spanclass=′latex−bold′>(D)</span>312"<spanclass=′latex−bold′>(E)</span>4"<span class='latex-bold'>(A) </span>2\textstyle{\frac{1}{2}}"\qquad<span class='latex-bold'>(B) </span>3"\qquad<span class='latex-bold'>(C) </span>3\textstyle{\frac{1}{4}}"\qquad<span class='latex-bold'>(D) </span>3\textstyle{\frac{1}{2}}"\qquad <span class='latex-bold'>(E) </span>4"
14
1

Find the two numbers

The number (248−1)(2^{48}-1) is exactly divisible by two numbers between 6060 and 7070. These numbers are
<spanclass=′latex−bold′>(A)</span>61,63<spanclass=′latex−bold′>(B)</span>61,65<spanclass=′latex−bold′>(C)</span>63,65<spanclass=′latex−bold′>(D)</span>63,67<spanclass=′latex−bold′>(E)</span>67,69<span class='latex-bold'>(A) </span>61,63\qquad<span class='latex-bold'>(B) </span>61,65\qquad<span class='latex-bold'>(C) </span>63,65\qquad<span class='latex-bold'>(D) </span>63,67\qquad <span class='latex-bold'>(E) </span>67,69
13
1
12
1

New Mathematical System

For each integer N>1N>1, there is a mathematical system in which two or more positive integers are defined to be congruent if they leave the same non-negative remainder when divided by NN. If 69,90,69,90, and 125125 are congruent in one such system, then in that same system, 8181 is congruent to
<spanclass=′latex−bold′>(A)</span>3<spanclass=′latex−bold′>(B)</span>4<spanclass=′latex−bold′>(C)</span>5<spanclass=′latex−bold′>(D)</span>7<spanclass=′latex−bold′>(E)</span>8<span class='latex-bold'>(A) </span>3\qquad<span class='latex-bold'>(B) </span>4\qquad<span class='latex-bold'>(C) </span>5\qquad<span class='latex-bold'>(D) </span>7\qquad <span class='latex-bold'>(E) </span>8
11
1

Reverse numbers in different bases

The numeral 4747 in base aa represents the same number as 7474 in base bb. Assuming that both bases are positive integers, the least possible value of a+ba+b written as a Roman numeral, is
<spanclass=′latex−bold′>(A)</span>XIII<spanclass=′latex−bold′>(B)</span>XV<spanclass=′latex−bold′>(C)</span>XXI<spanclass=′latex−bold′>(D)</span>XXIV<spanclass=′latex−bold′>(E)</span>XVI<span class='latex-bold'>(A) </span>\mathrm{XIII}\qquad<span class='latex-bold'>(B) </span>\mathrm{XV}\qquad<span class='latex-bold'>(C) </span>\mathrm{XXI}\qquad<span class='latex-bold'>(D) </span>\mathrm{XXIV}\qquad <span class='latex-bold'>(E) </span>\mathrm{XVI}
10
1

Brown-Eyed Brunettes

Each of a group of 5050 girls is blonde or brunette and is blue eyed of brown eyed. If 1414 are blue-eyed blondes, 3131 are brunettes, and 1818 are brown-eyed, then the number of brown-eyed brunettes is
<spanclass=′latex−bold′>(A)</span>5<spanclass=′latex−bold′>(B)</span>7<spanclass=′latex−bold′>(C)</span>9<spanclass=′latex−bold′>(D)</span>11<spanclass=′latex−bold′>(E)</span>13<span class='latex-bold'>(A) </span>5\qquad<span class='latex-bold'>(B) </span>7\qquad<span class='latex-bold'>(C) </span>9\qquad<span class='latex-bold'>(D) </span>11\qquad <span class='latex-bold'>(E) </span>13
9
1

Distance between two pulleys

An uncrossed belt is fitted without slack around two circular pulleys with radii of 1414 inches and 44 inches. If the distance between the points of contact of the belt with the pulleys is 2424 inches, then the distance between the centers of the pulleys in inches is
<spanclass=′latex−bold′>(A)</span>24<spanclass=′latex−bold′>(B)</span>2119<spanclass=′latex−bold′>(C)</span>25<spanclass=′latex−bold′>(D)</span>26<spanclass=′latex−bold′>(E)</span>435<span class='latex-bold'>(A) </span>24\qquad<span class='latex-bold'>(B) </span>2\sqrt{119}\qquad<span class='latex-bold'>(C) </span>25\qquad<span class='latex-bold'>(D) </span>26\qquad <span class='latex-bold'>(E) </span>4\sqrt{35}
8
1

Find the solution set

The solution set of 6x2+5x<46x^2+5x<4 is the set of all values of xx such that
<spanclass=′latex−bold′>(A)</span>−2<x<1<spanclass=′latex−bold′>(B)</span>−43<x<12<spanclass=′latex−bold′>(C)</span>−12<x<43<span class='latex-bold'>(A) </span>\textstyle -2<x<1\qquad<span class='latex-bold'>(B) </span>-\frac{4}{3}<x<\frac{1}{2}\qquad<span class='latex-bold'>(C) </span>-\frac{1}{2}<x<\frac{4}{3}\qquad
<spanclass=′latex−bold′>(D)</span>x<12 or x>−43<spanclass=′latex−bold′>(E)</span>x<−43 or x>12<span class='latex-bold'>(D) </span>x<\textstyle\frac{1}{2}\text{ or }x>-\frac{4}{3}\qquad<span class='latex-bold'>(E) </span>x<-\frac{4}{3}\text{ or }x>\frac{1}{2}
7
1

Simplify

2−(2k+1)−2−(2k−1)+2−2k2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k} is equal to
<spanclass=′latex−bold′>(A)</span>2−2k<spanclass=′latex−bold′>(B)</span>2−(2k−1)<spanclass=′latex−bold′>(C)</span>−2−(2k+1)<spanclass=′latex−bold′>(D)</span>0<spanclass=′latex−bold′>(E)</span>2<span class='latex-bold'>(A) </span>2^{-2k}\qquad<span class='latex-bold'>(B) </span>2^{-(2k-1)}\qquad<span class='latex-bold'>(C) </span>-2^{-(2k+1)}\qquad<span class='latex-bold'>(D) </span>0\qquad <span class='latex-bold'>(E) </span>2
6
1

* the binary operation

Let ∗\ast be the symbol denoting the binary operation on the set SS of all non-zero real numbers as follows: For any two numbers aa and bb of SS, a∗b=2aba\ast b=2ab. Then the one of the following statements which is not true, is
<spanclass=′latex−bold′>(A)</span>∗ is commutative over S<spanclass=′latex−bold′>(B)</span>∗ is associative over S<span class='latex-bold'>(A) </span>\ast\text{ is commutative over }S \qquad<span class='latex-bold'>(B) </span>\ast\text{ is associative over }S\qquad
<spanclass=′latex−bold′>(C)</span>12 is an identity element for ∗ in S<spanclass=′latex−bold′>(D)</span>Every element of S has an inverse for ∗<span class='latex-bold'>(C) </span>\frac{1}{2}\text{ is an identity element for }\ast\text{ in }S\qquad<span class='latex-bold'>(D) </span>\text{Every element of }S\text{ has an inverse for }\ast\qquad
<spanclass=′latex−bold′>(E)</span>12a is an inverse for ∗ of the element a of S<span class='latex-bold'>(E) </span>\dfrac{1}{2a}\text{ is an inverse for }\ast\text{ of the element }a\text{ of }S
5
1

Find the sum of the angles

Points A,B,Q,D,A,B,Q,D, and CC lie on the circle shown and the measures of arcs BQ^\widehat{BQ} and QD^\widehat{QD} are 42∘42^\circ and 38∘38^\circ respectively. The sum of the measures of angles PP and QQ is
<spanclass=′latex−bold′>(A)</span>80∘<spanclass=′latex−bold′>(B)</span>62∘<spanclass=′latex−bold′>(C)</span>40∘<spanclass=′latex−bold′>(D)</span>46∘<spanclass=′latex−bold′>(E)</span>None of these<span class='latex-bold'>(A) </span>80^\circ\qquad<span class='latex-bold'>(B) </span>62^\circ\qquad<span class='latex-bold'>(C) </span>40^\circ\qquad<span class='latex-bold'>(D) </span>46^\circ\qquad <span class='latex-bold'>(E) </span>\text{None of these}
[asy] size(3inch); draw(Circle((1,0),1)); pair A, B, C, D, P, Q; P = (-2,0); B=(sqrt(2)/2+1,sqrt(2)/2); D=(sqrt(2)/2+1,-sqrt(2)/2); Q = (2,0); A = intersectionpoints(Circle((1,0),1),B--P)[1]; C = intersectionpoints(Circle((1,0),1),D--P)[0]; draw(B--P--D); draw(A--Q--C); label("AA",A,NW); label("BB",B,NE); label("CC",C,SW); label("DD",D,SE); label("PP",P,W); label("QQ",Q,E); //Credit to chezbgone2 for the diagram[/asy]
3
1

Find the x-coordinate

If the point (x,−4)(x,-4) lies on the straight line joining the points (0,8)(0,8) and (−4,0)(-4,0) in the xy-plane, then xx is equal to
<spanclass=′latex−bold′>(A)</span>−2<spanclass=′latex−bold′>(B)</span>2<spanclass=′latex−bold′>(C)</span>−8<spanclass=′latex−bold′>(D)</span>6<spanclass=′latex−bold′>(E)</span>−6<span class='latex-bold'>(A) </span>-2\qquad<span class='latex-bold'>(B) </span>2\qquad<span class='latex-bold'>(C) </span>-8\qquad<span class='latex-bold'>(D) </span>6\qquad <span class='latex-bold'>(E) </span>-6
4
1

Amount of Interest

After simple interest for two months at 5%5\% per annum was credited, a Boy Scout Troop had a total of \textdollar 255.31 in the Council Treasury. The interest credited was a number of dollars plus the following number of cents
<spanclass=′latex−bold′>(A)</span>11<spanclass=′latex−bold′>(B)</span>12<spanclass=′latex−bold′>(C)</span>13<spanclass=′latex−bold′>(D)</span>21<spanclass=′latex−bold′>(E)</span>31<span class='latex-bold'>(A) </span>11\qquad<span class='latex-bold'>(B) </span>12\qquad<span class='latex-bold'>(C) </span>13\qquad<span class='latex-bold'>(D) </span>21\qquad <span class='latex-bold'>(E) </span>31
2
1

Men laying bricks

If bb men take cc days to lay ff bricks, then the number of days it will take cc men working at the same rate to lay bb bricks, is
<spanclass=′latex−bold′>(A)</span>fb2<spanclass=′latex−bold′>(B)</span>b/f2<spanclass=′latex−bold′>(C)</span>f2/b<spanclass=′latex−bold′>(D)</span>b2/f<spanclass=′latex−bold′>(E)</span>f/b2<span class='latex-bold'>(A) </span>fb^2\qquad<span class='latex-bold'>(B) </span>b/f^2\qquad<span class='latex-bold'>(C) </span>f^2/b\qquad<span class='latex-bold'>(D) </span>b^2/f\qquad <span class='latex-bold'>(E) </span>f/b^2
1
1