Sherman
Who remembers Algebra? Even more to the point, who wants to?

The GED contains a number of Algebra problems. Not enough to flunk the test if thatÕs all you missed, but enough to need to get at least some of them right. HereÕs a nifty way to solve multiple-choice Algebra problems without having to do the Algebra. ItÕs called backsolving. It works like this.

Basically backsolving involves plugging in the numbers until you find which one of the five choices is the correct one. The trick is to be methodically in your solution so that you only have to solve for two answers. LetÕs take an example and see how this works.

Sample Problem #1:

If there are four times as many men as women employed at Acme Woodworking, how many of the 75 employees are men?

(1) 65
(2) 40
(3) 45
(4) 15
(5) 60

The fastest way to solve this is plug in the most reasonable answer first. Looking at the possibilities we can take the middle number, 45, and see how that works. So if there are 45 men, and there are four times as many men as women, then we know there has to be 1/4 of 45 women, or 11.25 women. If we add 11.25 women and 45 men the answer has to equal 75, the total number of employees. So obviously, 45 is not right and is too small a number because the total, 11.25 + 45 = 56.25 is way less than 75. So our answer has to be larger than 45. That leaves answer (5) 60, and answer (1) 65. I would take 60 and plug in the numbers. LetÕs see, 60 divided by 4 is 15, and 15 plus 60 is 75. Right on! The answer is (5).

Did you follow this? LetÕs try another.


Sample Problem #2
Which of the values for x below will make the equation 3x < 12 true?

(1) 7
(2) 6
(3) 9
(4) 2
(5) 4

LetÕs start with the middle number in the list, 6 and plug it in the equation. 3 X 6 = 18. So this canÕt be the answer because x needs to be less than 12 when multiplied by 3. My next choice could be 4, but I know that 3 X 4 = 12, so I will rule that out and choose number (4) 2 as the answer.

HereÕs another. It doesnÕt need the backsolving method, just some common sense.

Sample Problem #3
A company charges $60 per day plus $0.50 per mile for car rentals. If Sam rents a car for three days and drives 150 miles, what will he pay in rental fees?

(1) $255
(2) $115
(3) $300
(4) $230
(5) $135

You could make an Algebraic formula to solve this, or you could use the backsolving method. Of the three numbers, which is in the middle? Number (4), $230. So letÕs start with that and see if we can plug in the numbers to see if this is the correct answer. 3 days at $60 a day equals $180. Then we need to add the mileage at $0.50 for 150 miles which is $75. $180 plus $75 is

Deleting Attachment...