David M. answered • 02/20/23
Finding the patterns that work
Your units of currency is #s, not $s, so I am presuming this is in British currency and you are using # for "pound Sterling". I am also using decimal number format 000.00 where anything to the right of "." is less than 1. I am taking the presumption that pound sterling is divided into 1/100, or 0.01, increments.
First step in any word problem is to identify what is being asked. After you think you've finished the problem, go back to this first step to make sure you don't have any more work to do.
1.) What is the least value of y, which is the amount Dauda earns for charity, in #s.
Second step is to identify what is given, keeping in mind what is being asked so that you don't spend a whole lot of time on obviously irrelevant information. Here is a good spot to abbreviate and in an easily referred to format.
2.) Obi earns half as much as Dauda
Chidi earns three times as much as Obi,
wale earns #250 less than all the other three
Total earned is more than #4300
A third step is to assign variables for an algebraic solution. Use meaningful variables to help keep track easier. I would have assigned D for what Dauda earns, but the problem assigns "y", so let's stick with y in that case. I would avoid "O" for Obi, because too close to the number "0", so I'll use "b". C for what Chidi earns and w for what wale earns seems innocent enough.
3) y= what Dauda earns
C= what Chidi earns
b= what Obi earns.
w= what wale earns
The fourth step is to rewrite step 2 in terms of the variables
4.) b=(1/2)y
C=3b
w=y+b+C-250
y+b+C+w>4300
The 5th step is to try and simplify the information algebraically to get down to the inequality in one variable, y.
4.) y + b + C + w > 4300
y + (1/2)y + 3b + y+ b + C -250 >4300
y + (1/2)y + 3(1/2)y + y+(1/2)y+ 3b -250 >4300
y + (1/2)y + 3(1/2)y + y+(1/2)y+3(1/2)y - 250>4300
In the 6th step is to solve for y NOTING that if you have to multiply or divide both sides of the equation by a negative number, you have to "flip" the inequality, > becoming < and visa versa. HOWEVER, such is not the case here.
6.) 6y-250>4300
6y >4300+250
6y >4550
6y/6>4550/6
y> 758 and 1/3rd
In the 7th step, we compare our answer, thus far, to the question being asked, in context of the whole word problem and the question being asked. We recognize that the answer has to be in #s, which is currency and, therefore, likely needs to be in decimal format to 2 places past the decimal point and, therefore, increments of 1/100 or 0.01 are presumed. So we have to ask our selves "What is the least value that y can be in increments of 0.01.
7.) 758.33<(758 and 1/3)<758.34 where 758.34-758.33=0.01
So the smallest value for y that is greater than 758 and 1/3, in #s currency, is 756.34
8.) Make sure the question asked is being answered by step 7.
