SOLUTION: A storekeeper has candies that set for Php600 and Php 900 per kilogram. How many kilograms of each must be mixed together to make 100 kilograms of candy that would be sold for Php

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A storekeeper has candies that set for Php600 and Php 900 per kilogram. How many kilograms of each must be mixed together to make 100 kilograms of candy that would be sold for Php       Log On


   



Question 1184833: A storekeeper has candies that set for Php600 and Php 900 per kilogram. How many kilograms of each must be mixed together to make 100 kilograms of candy that would be sold for Php 720 per kilogram?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


A standard formal algebraic setup for solving the problem....

x kg at Php600 per kg plus (100-x) kg at Php900 per kg equals 100 kg at 720Php per kg:

600%28x%29%2B900%28100-x%29=720%28100%29

Large numbers to work with; but the algebra is basic....

I leave it to you to finish the formal algebraic solution.

If a formal algebraic solution is not required, here is a fast and easy way to solve any two-part "mixture" problem like this informally.

(1) Look at the three costs per pound -- 600, 720, and 900 -- (perhaps on a number line) and observe/calculate that 720 is 120/300 = 2/5 of the way from 600 to 900
(2) That means 2/5 of the mixture is the more expensive ingredient

ANSWER: 2/5 of 100kg, or 40kg, of the Php900 per kg candies; the other 60kg of the Php600 per kg candies

CHECK:
40(900)+60(600)=36000+36000=72000
100(720)=72000