Help! Stuck! Math Problem!

[marko2434]

Hi everyone,

I feel really stupid about this, but I wonder if you could help with a Math word problem.

"John has an electric meter. The meter has 2 tariffs. He pays £0.35p more for his night units than he does for his day units. He used 3 times more day units than he does night units. He has just had his bill for 620 units. The bill was £234.65

How much are his night units?"

My thinking is to divide £234.65 by 3 to give me the night units used (£78.22) but got stuck from there.

Help!!

 

[wjm11]

Hi everyone,

I feel really stupid about this, but I wonder if you could help with a Math word problem.

"John has an electric meter. The meter has 2 tariffs. He pays £0.35p more for his night units than he does for his day units. He used 3 times more day units than he does night units. He has just had his bill for 620 units. The bill was £234.65

How much are his night units?"

My thinking is to divide £234.65 by 3 to give me the night units used (£78.22) but got stuck from there.

Help!!

Assign variables:
D = number of day units used
N = number of night units used
T = total units used

Write some simple equations that make sense. Total units used is the sum of the day and night units, right?

T = D + N

" He used 3 times more day units than he does night units."

D = 3N

Substituting, we get

T = 3N + N

Make sense? This is the process you want to use: 1) Assign variables; 2) Write simple equations; 3) Combine equations.
You will need to write equations regarding the different rates, however. The above equations are only good for figuring out units used -- not costs.
What cost equation can you come up with?

 

[Novice]

Let x be the amount he pays for his day units.
Then, for night , he pays 0.35+x.

Let John uses y night units.
Then he uses 3y day units.

Therefore,
y*(.35+x) + 3y*(x) = 234.65

Since this is the bill for 620 units,
3y+y= 620

Solve for x and y.

 

[marko2434]

Ref: Help! cont...

I'm gonna get this if it kiils me! (I assume that once I've cracked this it will apply on lots of word problems - as you can see, not my strong point!)

So,

n+d = 620
p=r+0.35

And I know that:

n+d = £234.65
dx3=n

I think these are my four equations? Is that right?

I know this will seem dumb to you guys, but I am starting over with my maths at the moment, so all your advice is gratefully received!



----------------------

In word problems, almost always the first thing to do is to WRITE DOWN a unique letter to stand for each unknown. Name things.

n = night units used.
d = day units used.
p = price of night units.
r = price of day units.

The second thing to do is to WRITE DOWN in mathematical form using your letters the facts specified in your problem plus any relevant facts of general knowledge.

I'll start this step for you.

n + d = 620.

p = r + 0.35.

What are the rest?

When you are done with step 2, you will have a purely mathematical problem to solve, in this case 4 equations with 4 unknowns. Step 3 is to solve the math problem. Step 4 is to check your answer. This four step process works every time.

 

[marko2434]

First forum post, so don't know equette, so I'll try and ask in each reply as you've been good enough to help.

I see your thinking, but what numbers do I substitute for X & Y?

Let x be the amount he pays for his day units.
Then, for night , he pays 0.35+x.

Let John uses y night units.
Then he uses 3y day units.

Therefore,
y*(.35+x) + 3y*(x) = 234.65

Since this is the bill for 620 units,
3y+y= 620

Solve for x and y.

 

[marko2434]

Help cont...

Following on from your equation, I worked out that:

Y = 155

and,

X = 0.291

Which I think led to £135.315 for the night units.

The only thing is I did trial and error to get X & Y. How would you do this mathematically using your equations?

Thanks again

First forum post, so don't know equette, so I'll try and ask in each reply as you've been good enough to help.

I see your thinking, but what numbers do I substitute for X & Y?