
ALGEBRA
 solved problems 





Linear
equations
in one variable


Time and travel problems  Distance, rate (or speed) and time relations


60. 
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.


61 
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


62. 
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. 

63. 
The
angle g
of a triangle is twice as large as the
angle a,
and the angle b is
threefourth 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° 




64. 
The
exterior angles that lie on the hypotenuse of a right triangle are in the
ratio of 11 : 16, find the


angles.

Solution: 
r^{2}
= (r
 8)^{2}
+ (c/2)^{2}

r^{2} = (r  8)^{2}
+ 12^{2}

r^{2}
= r^{2}
 16r
+ 64
+ 144

16r
= 208 
r = 13 cm 




65. 
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)

h^{2} =
(h
+ 4) · (h  3)

h^{2} =
h^{2}
+ h
 12
=> h
= 12

p
= h  3
= 12  3
= 9, 
q
= h
+ 4
= 12
+ 4
= 16,
c
= p +
q
= 25 




66. 
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 cm^{2}. 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)b^{2
}  7/5b
 4
= (3/5)b^{2}
 39 
7/5b =
35
a = (3/5) ·
b = (3/5) · 35 
b = 25 cm,
a = 15 cm, 




67. 
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
nsided 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 nsided
polygon, d_{n}
= n
· (n  3)
/ 2.


By substituting the given condition


(n
+ 3) · n = n
· (n  3)
+ 36

n^{2
}
+ 3n = n^{2
}  3n
+ 36 
6n = 36 =>
n =
6 




68. 
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





69. 
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 S_{lat} =
2rp
· h = 2rp(r
+ 8),

and the surface area S =
2B + S_{lat} =
2r^{2}p
+ 2rp(r
+ 8).


given
S_{lat}
: S = 5 :
8

2rp(r
+ 8)
:
[2r^{2}p
+ 2rp(r
+ 8)] =
5 :
8


8r
+ 64 = 10r
+ 40

2r
= 24 => r
= 12














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