Question 1184557
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First a traditional algebraic setup for solving the problem....<br>
x = hours at $105 per hour
20-x = hours at $50 per hour<br>
Total charge for 20 hours was $1825:<br>
{{{105(x)+50(20-x)=1825}}}<br>
Solve using basic algebra.<br>
And a solution method without algebra, using logical reasoning and some mental arithmetic....<br>
All 20 hours at $105 per hour would cost $2100; all 20 hours at $50 per hour would cost $1000; the actual cost was $1825.<br>
Look at the three costs on a number line -- 1000, 1825, and 2100 -- and observe/calculate that 1825 is 825/1100 = 3/4 of the way from 1000 to 2100.<br>
That means 3/4 of the hours were at the higher rate.<br>
ANSWER: 3/4 of 20 hours, or 15 hours, at $105 per hour; the other 5 hours at $50 per hour<br>
CHECK: 15(105)+5(50)=1575+250=1825<br>