ALGEBRA - solved problems
  Linear equations in one variable
Time and travel problems - Distance, rate (or speed) and time relations
137.    A rider has to catch a pedestrian up who is already 7 hours on his route. How long it will take if
the rider travels at the rate of 12 kilometers per hour and the pedestrian travels at 5 kilometers per hour?
In time and travel problems we use the following distance, rate and time relations:
  distance = rate ´ time,  and  .  

Solution:   The same distance rider travels x hours, the pedestrian travels (x + 7) hours,

since,  distance = rate ´ time,  then     12 · x = 5 · (x + 7)

                                                             12x = 5x + 35
                                                               7x = 35
                                                                 x = 5 hours.
138    To travel the distance between two stations a passenger train, traveling at rate of 12 meters per
second, takes 16 minutes and 40 seconds less than a freight train moving at speed of 8 meters per second.
What is the distance?

Solution:    We equate the times denoting the distance by x,

  where, 16 minutes and 40 seconds = 1000 seconds
                                  2x + 2400 = 3x
                                                x = 24000 meters = 24 kilometers
139.    A walker walking at the rate of 1 kilometer in 12 minutes travels a distance from A to B and spend
the same time as a cyclist who travels 10 kilometers longer distance riding at speed of 1 km in four and the half minutes. What is the distance from A to B?

Solution:    Walker took a way of x kilometers at the rate of 1/12 kilometers per minute through the time of  x/ (1/12) minutes.

The cyclist travels (x + 10) kilometers at the rate of 1/(4 and 1/2) kilometers per minute through the time
   of    (x + 10) / [1/(4 and 1/2)] minutes.  
They traveled the same period of time, thus
Geometry word problems
The general procedure to solve a geometry word problem is:
1. Draw a sketch of the geometric figure.
2. Label its elements using information given in the problem.
3. Choose the correct formula for the given problem (ex: perimeter, area, volume etc.).
4. Substitute the data from the problem into the formula.
5. Solve it for the unknown.
140.    The angle g of a triangle is twice as large as the angle a, and the angle b is three-fourth of the
angle g. What are the angles measures?.
Solution: 
given,   g = 2a  and  b = 3/4g = 3/4 · 2a = 3/2a
          since,     a + b + g = 180°
          then       a + 3/2a + 2a = 180°
                       9/2a = 180°    =>    a = 40°
                       b = 3/2a = 3/2 · 40 = 60°
                       g = 2a = 2 · 40° = 80°
 
141.    The exterior angles that lie on the hypotenuse of a right triangle are in the ratio of  11 : 16, find the 
angles.
Solution:   
                      r2 = (r - 8)2 + (c/2)2
                      r2 = (r - 8)2 + 122
                      r2 = r2 - 16r + 64 + 144
                   16r = 208
                      r = 13 cm
 
142.     The height of a right triangle is 3 units longer than the orthogonal projection of the shorter leg on the
hypotenuse and 4 units smaller than the projection of the longer leg. Find lengths of the projections
(segments).
Solution:   Using similarity of the two right triangles,
                      (h + 4) : h = h : (h - 3)
                      h2 = (h + 4) · (h - 3) 
                      h2 = h2 + h - 12  =>  h = 12
   p = h - 3 = 12 - 3 = 9,
   q = h + 4 = 12 + 4 = 16,     c = p + q = 25
 
143.     Sides of a rectangle are in the ratio of 3 : 5. If the length is decreased by 4 cm and the width is
increased by 1 cm, its area will be decreased by 39 cm2. Find dimensions of the rectangle.
Solution:       a : b = 3 : 5   =>    a = (3/5) · b
                       A = a · b = (3/5)b · b
      [(3/5)b + 1] · (b - 4) = (3/5)b · b - 39
         (3/5)b2 - 7/5b - 4 = (3/5)b2 - 39
         7/5b = 35            a = (3/5) · b = (3/5) · 35
             b = 25 cm,       a = 15 cm,
144.     If number of sides of a polygon increases by 3 then the number of its diagonals increases by 18.
Which polygon has this properties?
Solution:  Since an n-sided polygon has n vertices, and from each vertex can be drawn (n - 3) diagonals, then the total number of diagonals that can be drawn is n · (n - 3). However, as each diagonal joins two vertices, this means that each diagonal will be drawn twice, so the expression must be divided by 2.
Thus, number of diagonals of an n-sided polygon,   dn = n · (n - 3) / 2.
By substituting the given condition
        
      (n + 3) · n =  n · (n - 3) + 36
         n2 + 3n = n2 - 3n + 36
                6n = 36      =>     n = 6 
145.     Of which regular polygon, the difference between the interior and exterior angle, is 132°.
Solution:   In a regular polygon a = 360°/n  and a + b = 180°,  therefore a = b', as shows the picture.
                                          180n - 360 - 360 = 132n
                                          48n = 720    =>     n = 15
 
146.    The ratio of the lateral area of a cylinder to its surface area is 5 : 8. Find the radius of the base if it
is 8 cm shorter than the height of the cylinder.
Solution:  The lateral area of the cylinder    Slat = 2rp · h = 2rp(r + 8),
                              and the surface area     S = 2B + Slat = 2r2p + 2rp(r + 8).
    given   Slat : S = 5 : 8
   2rp(r + 8) : [2r2p + 2rp(r + 8)] = 5 : 8
                                          
                                         8r + 64 = 10r + 40
                                         2r = 24   =>     r = 12
  
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