SOLUTION: Problem 1 Separate 102 into two parts such that the larger part divided by the smaller part will give a quotient of 6 and a remainder of 11. Problem 2 A man invested part of P

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Question 803066: Problem 1
Separate 102 into two parts such that the larger part divided by the smaller part will give a quotient of 6 and a remainder of 11.
Problem 2
A man invested part of P10, 000 at 6% and the rest at 7%. The annual income from the 7% investment was P35 less than seven times the annual income from the 6 % investment. How much did he invest at each rate?
Problem 3
A tank can fill by one pipe stream in 5 hours and by another pipe stream in 4 hours. It can be emptied ba third pipe stream in 7 hours. How long will it take to fill the tank if all pipes are open?
Problem 4
A retailer wishes to mix rice selling at P12 per kilogram with rice selling at P 15 per kilogram so as to make a mixture of 48 kilograms worth P13 per kilogram. How many kilograms of each kind must be used?

Answer by josgarithmetic(39615) About Me  (Show Source):
You can put this solution on YOUR website!
Most of those are uniform rates problems, but I will help with your "102 two parts" problem.


x and y two numbers. Assuming x%2By=102, and assuming y%3Ex, y%2Fx=6%2B11%2Fx

The system to solve is x%2By=102 and y%2Fx=6%2B11%2Fx. In second equation with the quotient & remainder, multiply both sides by x:
y=6x%2B11
and then try substituting this for y in the 102 sum equation:
x%2B6x%2B11=102
and you can continue from there? Solve for x first?