Difference between revisions of "2000 AMC 8 Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | Shea has grown <math>20\%</math>, so she was originally <math>\frac{60}{1.2}=50</math> inches tall which is a <math>60 - 50 = 10</math> inch increase. Ara also started off at <math>50</math> inches. Since Ara grew half as much as Shea, Ara grew <math>10 | + | Shea has grown <math>20\%</math>, if x was her original height, then <math>1.2x = 60</math>, so she was originally <math>\frac{60}{1.2}=50</math> inches tall which is a <math>60 - 50 = 10</math> inch increase. Ara also started off at <math>50</math> inches. Since Ara grew half as much as Shea, Ara grew <math>\frac{10}{2} = 5</math> inches. Therefore, Ara is now <math>50+5=55</math> inches tall which is choice <math>\boxed{E}.</math> |
==See Also== | ==See Also== | ||
Latest revision as of 00:27, 8 January 2020
Problem
Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
Solution
Shea has grown 
, if x was her original height, then 
, so she was originally 
inches tall which is a 
inch increase. Ara also started off at 
inches. Since Ara grew half as much as Shea, Ara grew 
inches. Therefore, Ara is now 
inches tall which is choice 
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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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