Question 1206174: HI
An egg tart costs $1.60. Each person is entitled to buy another 2 egg tarts at $1.30 each after buying 3 egg tarts. Find the maximum you can buy with $24.00 and how much money would be left over.
Thanks
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! An egg tart costs $1.60. Each person is entitled to buy another 2 egg tarts at $1.30 each after buying 3 egg tarts. Find the maximum you can buy with $24.00 and how much money would be left over.
Let m be the number of sets of 3 tarts
Let n be the number of sets of 2 tarts
one set -3tarts will cost 4.80
1one set -2 tarts will cost 2.60
4.80m +2.60 n <=24
You can maximise the number of tarts by buying maximum sets of three.
lets try 4 sets of m and 2 sets of n
4.80*4+2.60*2
19.20 +5.20 >24 so not possible
lets try 4 sets of m and 1 set of n
19.20 +2.60 = 21.80 possible
3*4+1*2= 14 tarts for 21.80 <24
24-21.80
balance $2.20
Answer by ikleyn(52780)  (Show Source):
You can put this solution on YOUR website! .
The solution in the post by @mananth is inconsistent with the problem
and therefore is incorrect and irrelevant.
I came to bring a correct solution.
In accordance with the problem's logic, you should combine 3 egg tarts at $1.60 with 2 egg tarts at $1.30
and do it as many times until the restriction of $24.00 allows you do it.
Each such elementary group costs 3*1.60 + 2*1.30 = 7.40 dollars.
Next you divide 24 by 7.40 and round it to have integer quotient.
Obviously, this integer quotient is 3; so, you may have 3 elementary groups
that cost 7.40*3 = 22.20 dollars,
leaving 24.00 - 22.20 = 1.80 dollars.
For this $1.80, you can buy one egg tart at $1.60.
So, at the end, you will have maximum (3*(3+2) + 1) = 3*5+1 = 16 egg tarts and $0.20 remaining in your pocket.
Solved.
Actually, it is quite primitive task/problem/assignment.
The solution is straightforward and does not require "mind game".
Any house-wife can solve it mentally, based on common sense,
even without having any mathematical education.
One similar problem was solved for you today by Edwin.
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