For any three digit number, let
h = the hundreds digit
t = the tens digit and
u = the units digit
The number = 100h + 10t + u
The number with digits reversed = 100u + 10t + h
The sum of digits = h + t + u
The product of digits = htu
There are four common types of mixture in verbal problems of Algebra.
Solutions
Solution is a homogeneous mixture formed by dissolving a substance (solute) in another substance (solvent). A common example is the salt as solute and water as solvent forming into one phase called brine or saline water.
Alloys
An alloy is a solid solution formed by fusing two or more metallic elements. A common alloy is bronze which is the product of fusing iron and copper.
Addition
Expressions that can be translated to addition, ( + ): sum, plus, added to, in addition, increased by, and more than.
| Verbal expression | Algebraic equivalent |
|---|---|
| the sum of x and y | x + y or y + x |
| x plus y | x + y |
| x increased by y | x + y |
| x added to y | y + x |
| x in addition to y | y + x |
| x more than y | y + x |
The following is an attempt to classify the verbal problems.
Number-related problems
Number-related problems are considered as the most basic type of verbal problems. It is taken as the base point of analysis for more complex type of problems.
Problem 05
Find the inverse transform of $\dfrac{2s^2 + 5s - 6}{s^3 - 3s^2 - 13s + 15}$
Problem 04
Perform the indicated operation: $\mathcal{L}^{-1} \left[ \dfrac{s - 5}{s^2 + s - 6} \right]$
Problem 03
Find the inverse transform of $\dfrac{7}{s^2 + 6}$.
Problem
Solve for the spherical triangle whose parts are a = 73°, b = 62°, and C = 90°.
Definition of oblique spherical triangle
Spherical triangles are said to be oblique if none of its included angle is 90° or two or three of its included angles are 90°. Spherical triangle with only one included angle equal to 90° is a right triangle.
Sine law
$\dfrac{\sin a}{\sin A} = \dfrac{\sin b}{\sin B} = \dfrac{\sin c}{\sin C}$
Solution of right spherical triangle
With any two quantities given (three quantities if the right angle is counted), any right spherical triangle can be solved by following the Napier’s rules. The rules are aided with the Napier’s circle. In Napier’s circle, the sides and angle of the triangle are written in consecutive order (not including the right angle), and complimentary angles are taken for quantities opposite the right angle.
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