Solve Diophantine Eq.: 14 Bananas & Pears, £1.52

  • Thread starter Mathematicsresear
  • Start date
In summary, John bought Bananas (m=14), Pears (n=14-m=7), and bought a total of £1.52 worth of fruit. If pears cost 5p more than bananas, John bought a total of £1.57 worth of fruit from the three fruits.
  • #1
Mathematicsresear
66
0

Homework Statement


14 bananas and pears were bought in total by John from the supermarket.
The total cost was £1.52
if Banannas that John purchased cost 5p more than pears, how much did John buy from each fruit?

Homework Equations

The Attempt at a Solution


number of bananas bought = m
cost of bananas bought = n
number of pears bought = 14-m
cost of pears bought = n -5

I am unsure as to what follows
 
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  • #2
Mathematicsresear said:
I am unsure as to what follows
Time to write the equation with those variables you've defined and relate that to the total cost...
 
  • #3
berkeman said:
Time to write the equation with those variables you've defined and relate that to the total cost...
mn+(14-m)(n-5)=14n+5m = 1.52, so does that mean there are no solutions? This does not make sense.
 
  • #4
Mathematicsresear said:

Homework Statement


14 bananas and pears were bought in total by John from the supermarket.
The total cost was £1.52
if Banannas that John purchased cost 5p more than pears, how much did John buy from each fruit?

Homework Equations

The Attempt at a Solution


number of bananas bought = m
cost of bananas bought = n
number of pears bought = 14-m
cost of pears bought = n -5

I am unsure as to what follows

Please clarify for those of us who are not very familiar with English money: is there nowadays 100 p per pound?
 
  • #5
Ray Vickson said:
Please clarify for those of us who are not very familiar with English money: is there nowadays 100 p per pound?
Yes
 
  • #6
Ray Vickson said:
100 p per pound
Mathematicsresear said:
mn+(14-m)(n-5)=14n+5m = 1.52
Then 1.52 is not quite correct, right? :smile:
 
  • #7
Mathematicsresear said:
mn+(14-m)(n-5)=14n+5m
And I'm not understanding this step...
 
  • #8
berkeman said:
Then 1.52 is not quite correct, right? :smile:
So, should 1.52 be 152? If so, does that mean I need to multiply all sides by 100?
 
  • #9
Mathematicsresear said:
So, should 1.52 be 152? If so, does that mean I need to multiply all sides by 100?
Just use consistent units on both sides of all equations. Either use pounds or pence on both sides.
 
  • #10
berkeman said:
Then 1.52 is not quite correct, right? :smile:

(1) ##m n + (14-m) (n-5) \neq 14 n + 5 m.##
(2) Since you want prices to be in integer numbers of pence, it is convenient to use pence as the monetary unit throughout; otherwise (with the pound as the monetary unit) you would need to find prices that are precise to exactly two decimal places.
(3) When I do it I get three possible solutions to the resulting equation, but two of those are not valid solutions to the original problem
 
Last edited:

1. How do you solve a Diophantine equation?

In order to solve a Diophantine equation, you need to find integer solutions for the variables in the equation. This can be done using various methods such as trial and error, substitution, and algebraic manipulation.

2. What is a Diophantine equation?

A Diophantine equation is a polynomial equation where the variables and coefficients are restricted to integers. These types of equations are named after the ancient Greek mathematician Diophantus.

3. What is the given Diophantine equation trying to solve?

The given Diophantine equation, 14x + y = 152, is trying to find the number of bananas (x) and pears (y) that can be bought for £1.52, where a banana costs 14p and a pear costs 1p.

4. Is there a general method for solving Diophantine equations?

Unfortunately, there is no universal method for solving all Diophantine equations. The approach for solving each equation will depend on its specific form and variables.

5. Why are Diophantine equations important in mathematics?

Diophantine equations have applications in various fields such as number theory, cryptography, and coding theory. They also provide interesting and challenging problems for mathematicians to solve.

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