word problem

[Adriane]

just need to know how to set this up...
Lilly and Sara each had an equal amount of money at first. After Lilly spent $18 and Sara spent $25, Lilly had twice as
much as Sara. How much money did each have at first?

 

[JeffM]

just need to know how to set this up...
Lilly and Sara each had an equal amount of money at first. After Lilly spent $18 and Sara spent $25, Lilly had twice as
much as Sara. How much money did each have at first?

This is not an easy problem to solve by arithmetic. If you are not studying pre-algebra or algebra, let us know and we can address the problem arithmetically.

If you are studying algebra or pre-algebra, here is how to do word problems.

[1] In writing, assign a letter to each unknown.

x = amount of money each young lady had at first.

y = amount Lily had left after shopping spree.

z = amount Sara had left after shopping spree.

[2] In writing, translate the relevant facts from the problem into mathematical form. (In some problems some of those facts may be implied.)

I'll do one.

y = 2z.

There are two more facts given. What are they in mathematical form?

Once this step is done, the word part of the problem is done. The rest is a pure math problem.

[3] Simplify and then solve the math problem.

[4] Check your answer.

 

[Deleted member 4993]

just need to know how to set this up...
Lilly and Sara each had an equal amount of money at first. After Lilly spent $18 and Sara spent $25, Lilly had twice as
much as Sara. How much money did each have at first?

I t can be done with arithmetic logic - but it will be somewhat convoluted.

Now Lilly has 7 dollars more than what Sara has(now).

Now Lilly has twice as much as Sara has(now).

So 1/2 of Lilly' s money now = 7 dollars

So Lilly has $14 now → Lilly started with (14 + 18 =) $ 32

So Sara has $7 now → Sara started with (7 + 25 =) $ 32

Now check back .....

 

[Adriane]

Thank you for the reply! I'm sorry I did not mention that we have not learned algebra yet. If this could be addressed using basic math only that would be great. Thank you!

 

[stapel]

Lilly and Sara each had an equal amount of money at first. After Lilly spent $18 and Sara spent $25, Lilly had twice as much as Sara. How much money did each have at first?

If Lilly ends up with twice as much as Sara, then their total can be split into three equal parts; Lilly gets two of those parts and Sara gets one. So draw three equal-size squares, with two labelled "Lilly" and one labelled "Sara". (Draw the two "Lilly" squares as sharing a side, so it looks like a rectangle split into two halves.)

To the "Lilly" squares, add another square-ish shape (sharing a side with the second original square, so you've got a "bar" with three square-ish parts) and fill this in with "18". This is what Lilly had started with.

To the "Sara" square, add another square-ish shape (sharing a side with the one original square) and fill this in with "25". This is what Sara had started with.

You know that, at the start, the two bars had equal lengths. So you've got 2 original squares plus 18 equalling 1 original square plus 25. If you cross off one original square from each bar, what do you have? What then must be the number that goes inside one original box?

Back-solve to get the numerical answer.

 

[JeffM]

Thank you for the reply! I'm sorry I did not mention that we have not learned algebra yet. If this could be addressed using basic math only that would be great. Thank you!

There is nothing wrong with guess-and-check except that it is slow. When mathematicians do it, they call it the method of successive approximations, which sounds more impressive.

You know that the young ladies had at least $25 to start with because that is how much Sara spent.

So let's guess each started $30.
30 - 25 = 5.
30 - 18 = 12.
But 2 * 5 = 10, not 12.

So 30 is too low, but not by much.

Let's guess 31 next.
31 - 25 = 6.
31 - 18 = 13.
But 2 * 6 = 12, not 13.

So 31 is still too low, but not by much at all.

Let's guess 32 next.
32 - 25 = 7,
32 - 18 = 14.
2 * 7 = 14.

Basically algebra is a way to solve problems like this much more quickly.