An Elementary School Maths Problem Sum

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July 13th 2012, 04:22 AM
liukawa

An Elementary School Maths Problem Sum

Hi Guys,

I am stuck with this elementary school maths problem sum which looks relatively easy but i just simply cannot solve it !

Algebra method cannot be used. The school only allows me to use other methods other than algebra =(

Here goes the question :

John , Peter and Alex bought a watch. Alex paid 1/4 of what Peter and John Paid for. John Paid 1/5 of what Peter and Alex paid. If Peter paid $56 more than John, how much is the watch?

Please help !

My Brain cells are dying !!
July 13th 2012, 12:52 PM
Soroban

Re: An Elementary School Maths Problem Sum

Hello, liukawa!

Quote:

Algebra method cannot be used. . What?
The school only allows me to use other methods other than algebra.
I'd like to see the school's solution!

John , Peter and Alex bought a watch.
Alex paid 1/4 of what Peter and John paid.
John Paid 1/5 of what Peter and Alex paid.
Peter paid $56 more than John.
How much is the watch?

Without algebra, trial-and-error is the only way.
Why would ANYONE want us to do that?

After several minutes (hours? days?) of scribbling,

. . we might find: . $\begin{Bmatrix} A&=&6 \\P&=&19 \\ J&=&5\end{Bmatrix}$

This satisfies the first two restrictions:

. . $A \:=\:\tfrac{1}{4}(P + J) \quad\Rightarrow\quad 6 \:=\:\tfrac{1}{4}(19+5)$

. . $J \:=\:\tfrac{1}{5}(A + P) \quad\Rightarrow\quad 5 \:=\:\tfrac{1}{5}(6+19)$

It does not satisfy the third restriction: $P$ is 56 more than $J.$

In our solution, $P$ is only 14 more than $J.$

Maybe if we multiplied everything by 4, we might get the answer.

So, we try: . $\begin{Bmatrix}A \:=\:24 \\ P \:=\:76 \\ J \:=\:20 \end{Bmatrix}\;\hdots\;\text{ and it works!}$

I can't imagine ANYONE assigning a problem this complicated
. . and expecting one to solve it by arithmetic? .guessing?
July 17th 2012, 05:21 AM
Wilmer

Re: An Elementary School Maths Problem Sum

Quote:

Originally Posted by liukawa

Alex paid 1/4 of what Peter and John Paid for. [1]
John Paid 1/5 of what Peter and Alex paid. [2]
Peter paid $56 more than John. [3]

Soroban seems to be in an unusually grouchy mood! Wife had a headache?

[1][3]: J + P + A = "60", 64, 68, 72 ........ 116, "120", 124 ....
[2][3]: A + P + J = "60", 65, 70, 75 ........ 115, "120", 125
60 too low (since P paid over 56), so watch = 120
July 18th 2012, 10:57 PM
zzephod

Re: An Elementary School Maths Problem Sum

Quote:

John , Peter and Alex bought a watch.
Alex paid 1/4 of what Peter and John paid.
John Paid 1/5 of what Peter and Alex paid.
Peter paid $56 more than John.
How much is the watch?

So A paid 1/5 of total, J paid 1/6 of total and P paid 19/30 of total.

Now set up a spreadsheet with total in the first column running upwards from $60 in steps of $1, and with columns for what A,J and P paid and a further column with P-J. This last column increases as you move down, look for where it comes to $56, and you have the answer. (Yes, it is $120)