Category: System of linear equations

At Nuts supermarket, a discount is offered on Almonds, Cashew and Walnuts  if more than a certain weight of it is bought. On a particular grocery shopping trip, King Kong bought a total of 12 kg of nuts as follows:

Types of Nut	Price per kg ($)	Discount offered
Almond	14	20% for more than 6 kg bought
Cashew	9	15% for more than 3 kg bought
Walnuts	11	10% for more than 2 kg bought

King Kong bought at least 3.5 kg of each type of nut and spent $124.03 after an overall discount of $10.17. Find the weight of each type of nut King Kong bought from the supermarket.

Solution

We know that discount is obtained from the purchase of cashew and walnuts, since the amount bought is at least 3.5 kg of each type. However, we do not know whether any discount is obtained for the purchase of almonds.

Therefore, we make the assumption that the amount of almonds bought is less than or equal to 6kg, and hence no discount. If the solution turns out to be between 3.5 to 6kg inclusive, we accept the solution. Else we reject and change the assumption to almonds bought is greater than 6kg, reformulate and solve the equations again.

Let a, b and c be the number of kg of almonds, cashew and walnuts bought respectively.

Therefore a+b+c= 12 (Eqn 1)

Assume 

14a+b(0.85×9)+c(0.9×11)=124.03

14a+7.65b+9.9c=124.03 (Eqn 2)

0.15x9b+0.1x11c=10.17

1.35b+1.1c= 10.17 (Eqn 3)

Solving the 3 equations with GC,

a= 3.8 , b=4.6, c=3.6

Since a is between 3.5 to 6, our assumption and hence solution is valid.

King Kong bought 3.8 kg of almonds, 4.6 kg of cashew and 3.6 kg of walnuts.