Question 1184833
<br>
A standard formal algebraic setup for solving the problem....<br>
x kg at Php600 per kg plus (100-x) kg at Php900 per kg equals 100 kg at 720Php per kg:<br>
{{{600(x)+900(100-x)=720(100)}}}<br>
Large numbers to work with; but the algebra is basic....<br>
I leave it to you to finish the formal algebraic solution.<br>
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.<br>
(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<br>
ANSWER: 2/5 of 100kg, or 40kg, of the Php900 per kg candies; the other 60kg of the Php600 per kg candies<br>
CHECK:
40(900)+60(600)=36000+36000=72000
100(720)=72000<br>