1. ## Word Problem

These make my brain hurt. I heard there are two kinds of people who get math. The ones that understand it and also are capable of 'original math thinking' and the ones that can understand after it has been explained over and over again. The ones that need the explaining get by through the memorization of equations that fit a certain type of problem. I'm in the latter group and the following question deviates from the original example so much that I could literally spend hours on it and not get the answer.

Here's two questions for any original thinker who is up for the challenge. If you can please break down your answer step by step.

1.) A grocer mixes two kinds of nuts that cost $2.49 and$3.89 per pound to make a 100 pounds of a mixture that costs $3.47 per pound. How many pounds of each kind of nut are put into the mixture? 2.) The cooling system in a truck contains 4 gallons of coolant that is 30% antifreeze. How much must be withdrawn and replaced with 100% antifreeze to bring the coolant in the system to 50% antifreeze? These may seem like no brainers to some people. They give me a real headache and I cringe every time I read a problem like this. Thanks in advance for your help. 2. Good day mate. You seem a little bit under and a bit hard on yourself. Everybody is capable of original math thinking, you just need hard work, practice, patience and a willingness to make and learn from mistakes. Inspirational message delivered, onto the question 1) We call the nuts for $2.49$ per pound be $X$ and the $3.89$ per pound $Y$. The grocer mixes a certain amount of each nut to make $100$ pounds of nut mixture. Let the number of type $X$ nuts be $A$ Let the number of type $Y$ nuts be $B$ To make $100$ pounds of nuts, $A+B=100$ If this mixture costs $3.47$ per pound, then $100$ pounds costs $347$. That means the total cost of A pounds of $X$ and $B$ pounds of $Y$ is $347$. $AX+BY=100\rightarrow2.49A+3.89B=347 $ So we have two simultaneous equations $A+B=100...(1)$ $2.49A+3.89B=347...(2)$ Solve to obtain your answer. 2)If the truck contains $4$ gallons of coolant of which $30$% is antifreeze, then the truck has $\frac{30}{100}*4=1.2$ $gallons$ of antifreeze. Since we are withdrawing coolant with pure antifreeze, the truck will still contain $4$ gallons of the coolant mixture. For $50$% of the coolant to be antifreeze, it must contain $\frac{50}{100}*4=2$ $gallons$ of antifreeze. Hence, we must place $2-1.2=0.8$ $gallons$ of antifreeze into the truck and take out $0.8$ gallons of coolant. 3. Originally Posted by suddenlysmarter These make my brain hurt. I heard there are two kinds of people who get math. The ones that understand it and also are capable of 'original math thinking' and the ones that can understand after it has been explained over and over again. The ones that need the explaining get by through the memorization of equations that fit a certain type of problem. I'm in the latter group and the following question deviates from the original example so much that I could literally spend hours on it and not get the answer. Here's two questions for any original thinker who is up for the challenge. If you can please break down your answer step by step. 1.) A grocer mixes two kinds of nuts that cost$2.49 and $3.89 per pound to make a 100 pounds of a mixture that costs$3.47 per pound. How many pounds of each kind of nut are put into the mixture?

2.) The cooling system in a truck contains 4 gallons of coolant that is 30% antifreeze. How much must be withdrawn and replaced with 100% antifreeze to bring the coolant in the system to 50% antifreeze?

These may seem like no brainers to some people. They give me a real headache and I cringe every time I read a problem like this.

Any problem needs to be separated into what you know, and what you are seeking.

1.)The costs for two kinds of nuts
$2.49 / pound$3.89 / pound

You need 100 pounds that will have have cost
\$3.47 / pound.

How many pounds x and how many pounds of y are needed to make 100 pounds?

2.49x + 3.89y = 3.47*100
x, y, & 100 are pounds.
so
x+y = 100
or
y = 100-x

replacing y in the above equation

2.49x + 3.89*(100-x) = 3.47*100

expanding

2.49x + 3.89*100 - 3.89x = 3.47*100

subtract 3.89*100 from both sides

2.49x - 3.89x = 3.47*100 - 3.89*100

(2.49 - 3.89)x = 3.47*100 - 3.89*100

dividing

$x = \frac{(3.47-3.89)100}{(2.49-3.89)}$

x being the pounds of the cheaper grade and that subtracted from 100 is the number of pounds of the more expensive grade of nut.

Itemize your data: What you know & what you seek to know. Practice is the key. Practice gives the experience required to bridge unfamiliar problems.

4. Hello, suddenlysmarter!

2) The cooling system in a truck contains 4 gallons of coolant that is 30% antifreeze.
How much must be withdrawn and replaced with 100% antifreeze to bring the coolant
in the system to 50% antifreeze?

It contains: . $30\% \times 4 \:=\:(0.30)(4) \:=\:1.2$ gallons of AF.

We remove $x$ gallons of the mixture.
. . It contains: . $30\% \times x \:=\:0.3x$ gallons of AF (removed).

We add $x$ gallons of 100% AF.
. . It contains: . $100\% \times x \:=\:x$ gallons of AF (added).

Hence, the final mixture contains: . $1.2 - 0.3x + x \:=\:1.2 + 0.7x$ gallons of AF. .[1]

But we know that the final mixture will be 4 gallons which is 50% AF.
. . It will contain: . $50\% \times 4 \:=\:2$ gallons of AF. .[2]

We just described the final amount of AF in two ways.

There is our equation: . $1.2 + 0.7x \:=\:2 \quad\Rightarrow\quad 0.7x \:=\:0.8$

Therefore: . $x \:=\:\frac{0.8}{0.7} \quad\Rightarrow\quad \boxed{x \:=\:\frac{8}{7}}$ gallons.