Category: Algebra
"Published in Newark, California, USA"
Noel invested ₱ 25,000 in 2 business ventures one earning 6% and the other 8% at the end of the year. If his proceeds amounted to ₱ 1,900.00, how much did he invest in each?
Solution:
The given word problem is about money and investment problem. Let's analyze the given word problem as follows
Let x = amount of money invested in business A
y = amount of money invested in business B
₱ 25,000.00 = total amount of money invested
6% = annual interest in business A
8% = annual interest in business B
0.06x = annual earning in business A
0.08y = annual earning in business B
₱ 1,900.00 = annual total earning
From the word statement, "Noel invested ₱ 25,000 in 2 business ventures....", then the working equation will be
x + y = 25,000
From the word statement, "If his proceeds amounted to ₱ 1,900.00.....", then the working equation will be
0.06x + 0.08y = 1,900
We can solve for the value of x and y using the two equations, two unknowns. Consider the first equation
x + y = 25,000
or y = 25,000 - x
Substitute the value of y to the second equation, we have
0.06x + 0.08y = 1,900
0.06x + 0.08(25,000 - x) = 1,900
0.06x + 2,000 - 0.08x = 1,900
- 0.02x = 1,900 - 2,000
- 0.02x = - 100
x = 5,000
Substitute the value of x to the first equation, we have
y = 25,000 - x
y = 25,000 - 5,000
y = 20,000
Therefore, Noel invested ₱ 5,000.00 in business A and ₱ 20,000.00 in business B.
Note: The monetary sign, ₱ means Philippine Pesos and the word problem was in 1964.
This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)
Saturday, March 9, 2013
Friday, March 8, 2013
Sphere - Circular Section Problem
Category: Solid Geometry
"Published in Newark, California, USA"
Find the area of a section cut from a sphere of radius R by a plane distant R/2 from the center of the sphere.
Solution:
From the given figure above, we need to get the radius of the circular section in order to get its area. Let's consider the following procedures in order to get the radius of a circular section.
From point O which is the center of a sphere, draw a vertical line and label its end as point A.
At the end of a circular section, label as point B and then connect to the center of a sphere at point O. OB is also the radius of a sphere. In this case, OB = R.
From point B, draw a horizontal line towards to line OA and label their intersection as point C. CB is the radius of a circular section. In this case, CB = r.
Finally, label further the above figure as follows
To solve for r which is the radius of the circular section, use Pythagorean Theorem as follows
Therefore, the area of a circular section is
"Published in Newark, California, USA"
Find the area of a section cut from a sphere of radius R by a plane distant R/2 from the center of the sphere.
 |
| Photo by Math Principles in Everyday Life |
Solution:
From the given figure above, we need to get the radius of the circular section in order to get its area. Let's consider the following procedures in order to get the radius of a circular section.
From point O which is the center of a sphere, draw a vertical line and label its end as point A.
At the end of a circular section, label as point B and then connect to the center of a sphere at point O. OB is also the radius of a sphere. In this case, OB = R.
From point B, draw a horizontal line towards to line OA and label their intersection as point C. CB is the radius of a circular section. In this case, CB = r.
Finally, label further the above figure as follows
 |
| Photo by Math Principles in Everyday Life |
To solve for r which is the radius of the circular section, use Pythagorean Theorem as follows
Therefore, the area of a circular section is
Thursday, March 7, 2013
Mixing - Non Reacting Fluids
Category: Chemical Engineering Math, Differential Equations
"Published in Newark, California, USA"
Consider a tank that initially contains 100 gallons of a solution in which 50 pounds of salt are dissolved. Suppose that 3 gallons of brine, each gallon containing 2 pounds of salt, run into the tank each minute, and that the mixture, kept uniform by stirring, runs out at the rate of 2 gallons per minute. Find the amount of salt in the tank at time t.
Solution:
The first thing that we have to do is to analyze and illustrate the given word problem as follows
Since the given word problem involves the mixture of non-reactive fluids, the working equation will be as follows
where
r1 = volumetric flow rate at the entrance
c1 = concentration of substance at the entrance
r2 = volumetric flow rate at the exit
c2 = concentration of substance at the exit
Since c2 is usually not given in the problem, we can rewrite the above equation as follows
where
x = the amount of salt at time t
V = final volume of a solution at time t
but
where
V0 is the initial volume of solution at t = 0
Therefore, the final working equation will be
In the given word problem, we know that
r1 = 3 gals/min
c1 = 2 lbs/gal
r2 = 2 gals/min
V0 = 100 gals
x0 = 50 lbs
then the above equation becomes
Since the above equation is a first order, first degree linear equation, the integrating factor will be equal to
The general solution of the above equation is
If x = 50 lbs of salt at t = 0, then the value of C is
Therefore, the particular solution of the above equation or the amount of salt in the tank at time t is
"Published in Newark, California, USA"
Consider a tank that initially contains 100 gallons of a solution in which 50 pounds of salt are dissolved. Suppose that 3 gallons of brine, each gallon containing 2 pounds of salt, run into the tank each minute, and that the mixture, kept uniform by stirring, runs out at the rate of 2 gallons per minute. Find the amount of salt in the tank at time t.
Solution:
The first thing that we have to do is to analyze and illustrate the given word problem as follows
 |
| Photo by Math Principles in Everyday Life |
Since the given word problem involves the mixture of non-reactive fluids, the working equation will be as follows
where
r1 = volumetric flow rate at the entrance
c1 = concentration of substance at the entrance
r2 = volumetric flow rate at the exit
c2 = concentration of substance at the exit
Since c2 is usually not given in the problem, we can rewrite the above equation as follows
where
x = the amount of salt at time t
V = final volume of a solution at time t
but
V0 is the initial volume of solution at t = 0
Therefore, the final working equation will be
In the given word problem, we know that
r1 = 3 gals/min
c1 = 2 lbs/gal
r2 = 2 gals/min
V0 = 100 gals
x0 = 50 lbs
then the above equation becomes
Since the above equation is a first order, first degree linear equation, the integrating factor will be equal to
The general solution of the above equation is
If x = 50 lbs of salt at t = 0, then the value of C is
Therefore, the particular solution of the above equation or the amount of salt in the tank at time t is
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