Difference between revisions of "1977 AHSME Problems/Problem 1"
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| + | == Problem 1 == | ||
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| + | If <math>y = 2x</math> and <math>z = 2y</math>, then <math>x + y + z</math> equals | ||
| + | |||
| + | <math>\text{(A)}\ x \qquad | ||
| + | \text{(B)}\ 3x \qquad | ||
| + | \text{(C)}\ 5x \qquad | ||
| + | \text{(D)}\ 7x \qquad | ||
| + | \text{(E)}\ 9x</math> | ||
| + | |||
| + | |||
==Solution== | ==Solution== | ||
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
<math>x+y+z = x+(2x)+(4x) = \boxed{\text{(D)}\ 7x}</math> | <math>x+y+z = x+(2x)+(4x) = \boxed{\text{(D)}\ 7x}</math> | ||
Latest revision as of 12:24, 21 November 2016
Problem 1
If 
and 
, then 
equals
Solution
Solution by e_power_pi_times_i
