This is the solution to the algebra finance word problem that An anonymous poster recently asked which read:
"An employee's new salary is $19,110 after getting a 5% raise. What was the salary before the increase in pay?"
Ok easiest way to think about this is try something you do know for sure. Whats 5% of 100? 5 right--- Ok, so what if the problem said an employee's new salary was $105 after a 5% increase in pay what was the salary before... Ok you already know the answer 100.... So how would u come up with the new total? Well 5% of 100 is 5 and then you add it to the original amount of 100 which makes 105. So, you set up your problem like that:
5% can be written 5/100
so,
5/100 * x + x = 105 multiply by 100
5 * x + 100x = 10500
5x + 100x = 10500
105x = 10500
x = 10500/105 = 100 so 100 is the answer you were looking for and it checks out..
You solve this problem exactly the same way.
5/100 * x + x = 19110 multiply by 100
5x + 100x = 1911000
105x = 1911000
1911000/105 = x
x = 18200 So check whats 5% of 18200 ? its 910 Now add 910 to 18200 and you get 19110. So it checks out.
Saturday, January 23, 2010
Finance Problem 1 Solution
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thats wrong
ReplyDeleteYou take 19110 x .05 which gives you 5% of 19110, which is 955.5 so he received an increase of 955.5. Soooo in order to figure out what he made before you subtract 19110 - 955.5 and you get the salary made before the 5% increase!!!! 18145.5
ReplyDeleteNO THAT'S INCORRECT... That would be like having a problems saying the salary is now $105 after a 5% increase what was it before and you take 5% of 105 and get 5.25 and subtract that from 105 (105 - 5.25) and get 99.75 --- which is WRONG!!!! IT WOULD BE 100.
ReplyDeletethere is a formula to get the cost price/real price, its number / 1.(%).
ReplyDeleteso in this problem the price is 19110 WHICH is already included the 5% raise so to get the cost price/real price it will be 19110 / 1.05 = 18200.
i find this helpful because the author made in algebraic terms which proves that algebra can solve any question regarding unknown.