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Difference between revisions of "1988 AHSME Problems/Problem 11"

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==Solution==
 
==Solution==
 
We simply have to divide the present population by the original, and whichever one is the largest have the greatest percent change. City <math>\text{A}</math>  is  
 
We simply have to divide the present population by the original, and whichever one is the largest have the greatest percent change. City <math>\text{A}</math>  is  
−
<math>\frac{50}{40}=1.25</math>. City <math>\text{B}</math> is <math>\frac{70}{50}=1.4</math>. City <math>\text{C}</math> is <math>\frac{100}{70} \approx 1.42</math>. City <math>\text{D}</math> is <math>\frac{130}{100}=1.3</math>. City <math>\text{E}</math> is <math>\frac{160}{120} \approx 1.33</math>. City <math>\text{D}</math> has the greatest value, so the answer is <math>\boxed{\text{D}}</math>.
+
<math>\frac{50}{40}=1.25</math>. City <math>\text{B}</math> is <math>\frac{70}{50}=1.4</math>. City <math>\text{C}</math> is <math>\frac{100}{70} \approx 1.42</math>. City <math>\text{D}</math> is <math>\frac{130}{100}=1.3</math>. City <math>\text{E}</math> is <math>\frac{160}{120} \approx 1.33</math>. City <math>\text{C}</math> has the greatest value, so the answer is <math>\boxed{\text{C}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:25, 21 November 2017

Problem

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(5,0), B=(7,0), C=(10,0), D=(13,0), E=(16,0); pair F=(4,3), G=(5,3), H=(7,3), I=(10,3), J=(12,3); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(I); dot(J); draw((0,0)--(18,0)^^(0,3)--(18,3)); draw((0,0)--(0,.5)^^(5,0)--(5,.5)^^(10,0)--(10,.5)^^(15,0)--(15,.5)); draw((0,3)--(0,2.5)^^(5,3)--(5,2.5)^^(10,3)--(10,2.5)^^(15,3)--(15,2.5)); draw((1,0)--(1,.2)^^(2,0)--(2,.2)^^(3,0)--(3,.2)^^(4,0)--(4,.2)^^(6,0)--(6,.2)^^(7,0)--(7,.2)^^(8,0)--(8,.2)^^(9,0)--(9,.2)^^(10,0)--(10,.2)^^(11,0)--(11,.2)^^(12,0)--(12,.2)^^(13,0)--(13,.2)^^(14,0)--(14,.2)^^(16,0)--(16,.2)^^(17,0)--(17,.2)^^(18,0)--(18,.2)); draw((1,3)--(1,2.8)^^(2,3)--(2,2.8)^^(3,3)--(3,2.8)^^(4,3)--(4,2.8)^^(6,3)--(6,2.8)^^(7,3)--(7,2.8)^^(8,3)--(8,2.8)^^(9,3)--(9,2.8)^^(10,3)--(10,2.8)^^(11,3)--(11,2.8)^^(12,3)--(12,2.8)^^(13,3)--(13,2.8)^^(14,3)--(14,2.8)^^(16,3)--(16,2.8)^^(17,3)--(17,2.8)^^(18,3)--(18,2.8)); label("A", A, S); label("B", B, S); label("C", C, S); label("D", D, S); label("E", E, S); label("A", F, N); label("B", G, N); label("C", H, N); label("D", I, N); label("E", J, N); label("1970", (0,3), W); label("1980", (0,0), W); label("0", (0,1.5)); label("50", (5,1.5)); label("100", (10,1.5)); label("150", (15,1.5)); label("Population in thousands", (9,-3)); [/asy]

On each horizontal line in the figure below, the five large dots indicate the populations of cities $A, B, C, D$ and $E$ in the year indicated. Which city had the greatest percentage increase in population from $1970$ to $1980$?

$\textbf{(A)}\ A\qquad \textbf{(B)}\ B\qquad \textbf{(C)}\ C\qquad \textbf{(D)}\ D\qquad \textbf{(E)}\ E$


Solution

We simply have to divide the present population by the original, and whichever one is the largest have the greatest percent change. City $\text{A}$ is $\frac{50}{40}=1.25$. City $\text{B}$ is $\frac{70}{50}=1.4$. City $\text{C}$ is $\frac{100}{70} \approx 1.42$. City $\text{D}$ is $\frac{130}{100}=1.3$. City $\text{E}$ is $\frac{160}{120} \approx 1.33$. City $\text{C}$ has the greatest value, so the answer is $\boxed{\text{C}}$.

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
All AHSME Problems and Solutions

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