Eve steps into a supermarket to buy a biro. A biro costs 29 Kobe. She has #2note, and the shopkeeper has 5-kobo and 6-kobo coins in his drawer. If eve is to get her change in these denomination, What is the largest product she can obtain after multiplying the 5-kobo coins by 6-kobo coin in Her change
I know the Kobo is a currency unit of Liberia, but what is a Kobe? What is a #2note ?
Are there 100 Kobo in 1 Naira?
yes
it was a typo I meant Kobo not kobe
100kobo=1naira
2# is 2naira
How do you multiply "5-kobo coins by 6-kobo coin"?
Ok then:
so I will skip writing the units ...
cost = 29
pay with 200
so the change will be 171
number of 5 kobo coins ---- x
number of 6 kobo coins ---- y
5x + 6y = 171
we need whole number solutions for this equation, (called a diophantine equation)
I actually have a method to solve this type using repeating fractions,
but that would be hard to explain here, so ...
by trial and error:
x ---- y
3 26
9 21
15 16
21 11
...
notice since the slope of my line is -6/5, increasing the x by 6 and
decreasing the y by 5 gives me new points.
So we want the largest product of x and y which would be when they
are closest to each other , or
15 and 16 ---> xy = 240
check
5(15) + 6(16) = 171, the amount needed for your change
To find the largest product Eve can obtain after multiplying the 5-kobo coins by 6-kobo coins, we need to determine the maximum number of 5-kobo coins and the maximum number of 6-kobo coins that Eve can receive as change.
To begin, let's analyze the given information:
- The price of a biro is 29 kobos.
- Eve has a #2 note, which is equivalent to 200 kobos (assuming # represents the Nigerian currency symbol).
- The shopkeeper has 5-kobo and 6-kobo coins.
First, we need to calculate Eve's change. Since she paid with a #2 note (200 kobos) for a biro that costs 29 kobos, her change would be 200 - 29 = 171 kobos.
Now, let's determine the maximum number of 5-kobo coins Eve can receive in her change:
- Since each 5-kobo coin has a value of 5 kobos, the maximum number of 5-kobo coins that can be used to make up 171 kobos is (171 / 5) = 34.2 coins.
- However, we can't have a fraction of a coin, so the maximum number of 5-kobo coins is 34.
Next, let's determine the maximum number of 6-kobo coins Eve can receive in her change:
- Since each 6-kobo coin has a value of 6 kobos, the maximum number of 6-kobo coins that can be used to make up 171 kobos is (171 / 6) = 28.5 coins.
- Similarly, we can't have a fraction of a coin, so the maximum number of 6-kobo coins is 28.
To find the largest product, we need to calculate 34 (maximum number of 5-kobo coins) multiplied by 28 (maximum number of 6-kobo coins):
34 x 28 = 952
Therefore, the largest product Eve can obtain by multiplying the 5-kobo coins by the 6-kobo coins in her change is 952.