Question 1185996
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The responses from the other two tutors  both use formal algebraic methods to solve the problem, which is probably what the student was supposed to do.<br>
However, solving the problem with formal algebra requires solving a quadratic equation using either difficult factoring or the quadratic formula.<br>
So, if formal algebra is not required, the problem is solved much more quickly with a bit of estimation and mental arithmetic.<br>
Note also that, even if a formal algebraic solution is required for the assignment, solving the problem with logical reasoning and mental arithmetic gives you valuable brain exercise.<br>
It might go something like this.<br>
For a very rough estimate, since the length is "a bit more than 3 times the width", let the width be x and the length be "a bit more than 3x", so that the area (length times width) is 1482.<br>
1482 is a bit more than 3x^2
1482/3 = 496 is a bit more than x^2<br>
The square root of 496 is a bit more than 22; so the width should be either 22 or perhaps 21.<br>
Then mental arithmetic determines that 21 is a factor of 1428 while 22 is not, so the width is probably 21.<br>
Dividing 1428 by 21 gives 68; and 68 is 5 more than 3 times 21.  So we have the answer.<br>
ANSWERS: width 21, length 68<br>