Difference between revisions of "2000 AMC 8 Problems/Problem 4"

(Solution)
(Solution)
 
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==Solution==
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==Solution 1==
  
 
The data are <math>1960 (5\%)</math>, <math>1970 (8\%)</math>, <math>1980 (15\%)</math>, and <math>1990 (30\%)</math>. Only one of these graphs has the answer and that is choice <math>\boxed{E}</math>.
 
The data are <math>1960 (5\%)</math>, <math>1970 (8\%)</math>, <math>1980 (15\%)</math>, and <math>1990 (30\%)</math>. Only one of these graphs has the answer and that is choice <math>\boxed{E}</math>.
 +
 
==Solution 2==
 
==Solution 2==
 
We can look at the graphs and note that the only one that has all the correct points is <math>\boxed{E}</math>
 
We can look at the graphs and note that the only one that has all the correct points is <math>\boxed{E}</math>

Latest revision as of 16:29, 22 October 2020

Problem

In 1960 only 5% of the working adults in Carlin City worked at home. By 1970 the "at-home" work force had increased to 8%. In 1980 there were approximately 15% working at home, and in 1990 there were 30%. The graph that best illustrates this is:

[asy] unitsize(18);  draw((0,4)--(0,0)--(7,0)); draw((0,1)--(.2,1)); draw((0,2)--(.2,2)); draw((0,3)--(.2,3)); draw((2,0)--(2,.2)); draw((4,0)--(4,.2)); draw((6,0)--(6,.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a,b-.1)--(2*a,b+.1)); draw((2*a-.1,b)--(2*a+.1,b)); } } label("1960",(0,0),S); label("1970",(2,0),S); label("1980",(4,0),S); label("1990",(6,0),S); label("10",(0,1),W); label("20",(0,2),W); label("30",(0,3),W); label("$\%$",(0,4),N);  draw((12,4)--(12,0)--(19,0)); draw((12,1)--(12.2,1)); draw((12,2)--(12.2,2)); draw((12,3)--(12.2,3)); draw((14,0)--(14,.2)); draw((16,0)--(16,.2)); draw((18,0)--(18,.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a+12,b-.1)--(2*a+12,b+.1)); draw((2*a+11.9,b)--(2*a+12.1,b)); } } label("1960",(12,0),S); label("1970",(14,0),S); label("1980",(16,0),S); label("1990",(18,0),S); label("10",(12,1),W); label("20",(12,2),W); label("30",(12,3),W); label("$\%$",(12,4),N);  draw((0,12)--(0,8)--(7,8)); draw((0,9)--(.2,9)); draw((0,10)--(.2,10)); draw((0,11)--(.2,11)); draw((2,8)--(2,8.2)); draw((4,8)--(4,8.2)); draw((6,8)--(6,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a,b+7.9)--(2*a,b+8.1)); draw((2*a-.1,b+8)--(2*a+.1,b+8)); } } label("1960",(0,8),S); label("1970",(2,8),S); label("1980",(4,8),S); label("1990",(6,8),S); label("10",(0,9),W); label("20",(0,10),W); label("30",(0,11),W); label("$\%$",(0,12),N);  draw((12,12)--(12,8)--(19,8)); draw((12,9)--(12.2,9)); draw((12,10)--(12.2,10)); draw((12,11)--(12.2,11)); draw((14,8)--(14,8.2)); draw((16,8)--(16,8.2)); draw((18,8)--(18,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a+12,b+7.9)--(2*a+12,b+8.1)); draw((2*a+11.9,b+8)--(2*a+12.1,b+8)); } } label("1960",(12,8),S); label("1970",(14,8),S); label("1980",(16,8),S); label("1990",(18,8),S); label("10",(12,9),W); label("20",(12,10),W); label("30",(12,11),W); label("$\%$",(12,12),N);  draw((24,12)--(24,8)--(31,8)); draw((24,9)--(24.2,9)); draw((24,10)--(24.2,10)); draw((24,11)--(24.2,11)); draw((26,8)--(26,8.2)); draw((28,8)--(28,8.2)); draw((30,8)--(30,8.2)); for (int a = 1; a < 4; ++a) { for (int b = 1; b < 4; ++b) { draw((2*a+24,b+7.9)--(2*a+24,b+8.1)); draw((2*a+23.9,b+8)--(2*a+24.1,b+8)); } } label("1960",(24,8),S); label("1970",(26,8),S); label("1980",(28,8),S); label("1990",(30,8),S); label("10",(24,9),W); label("20",(24,10),W); label("30",(24,11),W); label("$\%$",(24,12),N);  draw((0,9)--(2,9.25)--(4,10)--(6,11)); draw((12,8.5)--(14,9)--(16,10)--(18,10.5)); draw((24,8.5)--(26,8.8)--(28,10.5)--(30,11)); draw((0,0.5)--(2,1)--(4,2.8)--(6,3)); draw((12,0.5)--(14,.8)--(16,1.5)--(18,3));  label("(A)",(-1,12),W); label("(B)",(11,12),W); label("(C)",(23,12),W); label("(D)",(-1,4),W); label("(E)",(11,4),W); [/asy]


Solution 1

The data are $1960 (5\%)$, $1970 (8\%)$, $1980 (15\%)$, and $1990 (30\%)$. Only one of these graphs has the answer and that is choice $\boxed{E}$.

Solution 2

We can look at the graphs and note that the only one that has all the correct points is $\boxed{E}$

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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
All AJHSME/AMC 8 Problems and Solutions

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