Amount of Money Withdrawn

[Guest]

Two men, A and B, have in possession a money box containing $210. Each man removes a certain sum daily. The sum is fixed for each, but different for each. The box was emptied after 6 weeks. Determine the sum each removed daily. A alone would have emptied the box 5 weeks earlier than B alone.

How do I solve ?

[Guest]

Sort out equations using the information given in the question......

Probably easier to Work in units of days, as the money is removed daily.
6 weeks = 42 days
5 weeks = 35 days

42 days times the total taken out by A and B per day and the box is empty.....$210 out

$210 divided by "quantity A only takes per day" equals the number of days the money lasts for A only.
$210 divided by "quantity B only takes per day" equals the number of days the money lasts for B only.
the money lasts 5 weeks or 35 days longer for B than for A

That should give you 2 simultaneous equations to solve for A and B.

[Guest]

I don't know how.

[Guest]

42(A + B) = 210

210/A = number of days money lasts for A

210/B = number of days money lasts for B and this is 35 days more than it was for A

210/B = 210/A + 35

[Guest]

I still can't solve it.

[Guest]

42(A + B) = 210 .......eqn1

210/B = 210/A + 35 .....eqn2

two simultaneous equations, two unknowns, so can be solved OK

Re-arrange one of them to find A in terms of B OR if you want B in terms of A

Then substitute into the other equation to get an equation with only 1 unknown, then solve this equation.

[Guest]

I can't solve.

[Guest]

42(A + B) = 210 .......eqn1

divide both sides by 42

gives A + B = 5

[Guest]

I still can't solve.

[Guest]

210/B = 210/A + 35 .....eqn2

Divide both sides by 35 gives....

6/B = 6/A + 1

[Guest]

B = 5 - A

6 / 5 - A = 6 /A + 1

I am stuck.

[Guest]

B = 5 - A

6 /(5 - A) = (6 /A) + 1

you have used substitution...thats OK.....
I put brackets in to show what you have better.....
There are various things you can do now.....
You have a fraction on LHS and both a fraction and a number on RHS......the idea is to get rid on the mixtures....
Multiply both sides by A that will get rid of fraction on RHS and the 1 will become A.....then put all of RHS over 1 and cross multiply

[Guest]

6/ (5 - A) = (6 / A ) + 1

6 / (5 - A) * A = (6 / A) + 1 * A

6/ 5 = 6 / A

6A = 30

6A / 6 = 30 / 6

A = 5


B = 5 - A = 5 - 5 = 0

I don't know.

[Guest]

Please reply, Thanks.

[Guest]

This will let you see it laid out as you would be doing it by hand....

[Guest]

I get A = 5

42(A + B) = 210

42(5 + B) = 210

210 + 42B = 210

42B = 0

I still don't know.

[Guest]

Look at the attachment on my previous post

I have multiplied by A both sides

All you have to do is continue with the last line and cross multiply as indicated on the picture attachment.

[Guest]

6A / (5 - A) = (6 + A) / A + 1

6A = 30 + A - A

6A = 30

6A / 5 = 30 / 6

A = 5

How do I continue ?

[Guest]

You need to follow the steps in the sum1.jpg
The last step at the bottom shows what you do to cross multiply

[Guest]

6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A ) * (6 + A)

6A = 30 - A + A

6A = 30

6A / 6 = 30 / 6

A = 5

Now what?