Math Is Fun Forum

paulb203

Member

Registered: 2023-02-24

Posts: 136

Ratio. Direct Proportion. Amounts change causing ratio to change, etc.

Jane and Kim own some marbles in the ratio 5:6
Jane gains 2 more marbles and the ratio is now 7:8
How many marbles do each of them own initially?

I can solve these problems using trial and error with the numbers but I don't know the simplest way of solving them using basic algebra.
Please don't tell me how to do that fully. Just a hint or two would be helpful initially.

Do please tell me the answer though (how many they each own initially), to check with my trial and error answer.

Last edited by paulb203 (2023-10-12 04:47:22)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Ratio. Direct Proportion. Amounts change causing ratio to change, etc.

hi

When I was at school the recommended method was to call one unknown x and attempt to build an equation for x.

So if Jane has x, then Kim has 6x/5

Jane gets 2 more so she has x+2 and 7:8 implies Kim has (x+2).8/7

But Kim hasn't changed her number so (x+2).8/7 = 6x/5 ...... solve for x

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Ratio. Direct Proportion. Amounts change causing ratio to change, etc.

Beautiful. Thanks, Bob.

So,

(x+2).8/7 = 6x/5

8/7(x+2) = 6x/5

(8x+16)/7 = 6x/5

Now what?

Times both sides by 7?

Times both sides by 5?

I’m not sure how to do either.

Try times both sides by 7:

(8x+16) = ?

How to do, 7(6x/5)?

Try: 7/1 x 6x/5 = 42x/5 ?

So, (8x+16)= 42x5 (I hope!)

Times boths sides by 5?

**90x+80 = 42x**

90x+80-42x=0

48x+80=0

48x=(-80)

x=(-80/48)

x= -5/3?

That can’t be right

Where have I gone wrong.....?

**EDIT**

Should've been 40x+80=42x
42x-40x = 80
2x=80
x=40

Which is what I got using trial and error; and then again when amnkb helped me do it his ingenious way.

Last edited by paulb203 (2023-10-12 10:56:01)

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Ratio. Direct Proportion. Amounts change causing ratio to change, etc.

Jane and Kim own some marbles in the ratio 5:6
Jane gains 2 more marbles and the ratio is now 7:8
How many marbles do each of them own initially?

the ratios tell you that there is some factor that divides out of jane/kim
the first ratio simplifies as jane/kim = 5/6
the second ratio simplifies as jane/kim = 7/8
try common denominators
5/6 = 20/24
7/8 = 21/24
not 2 apart
try other common denominators

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Ratio. Direct Proportion. Amounts change causing ratio to change, etc.

Thanks, amnkb. I think I've got it. 48 as a common denominator gives, 40/48, and, 42/48. Which is what I got when I did it another way, using trial and error.

Jane and Kim own some marbles in the ratio 5:6
Jane gains 2 more marbles and the ratio is now 7:8
How many marbles do each of them own initially?

the ratios tell you that there is some factor that divides out of jane/kim
the first ratio simplifies as jane/kim = 5/6
the second ratio simplifies as jane/kim = 7/8
try common denominators
5/6 = 20/24
7/8 = 21/24
not 2 apart
try other common denominators

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Ratio. Direct Proportion. Amounts change causing ratio to change, etc.

I learned a nice way to solve this kind of problem today:

Initial ratio 5:6
J = 5x
K = 6x
Then, after J gains 2:
Ratio becomes 7:8
J = 5x+2
K = 6x
Therefore:
J/K = 7/8
and:
J/K = 5x+2/6x
Therefore:
5x+2/6x = 7/8
Cross multiply to give
42x = 40x+16
2x=16
x=16/2
x=8
5x=40
6x=48
Jane owned 40 initially
Kim owned 48 initially

I know I've ended up with the correct answer, but is the working and the method all correct?