Algebra Pre-Quiz Notes (Problem Solving w/ Solutions) [3pts. each]
1. What is the greatest common factor of 108 and 60?
a. 12
b. 18
c. 6
d. 24
Solution:

Use continuous division until the numbers have no more common factors. Take
note of the divisors (green line).

Since 9 and 5 have no more common factor, multiply all the divisors (common
factors).
2. What is the remainder if the polynomial
a. 60
b. 54
c. 3
d. 16
is divided by x-2?
Solution:

Using remainder theorem – if a f(x) is divided by x-a, then the remainder is f(a).
3. The sum of Kim’s and Kevin ages is 18. In 3 years, Kim will be twice as old as Kevin.
How old is Kevin now?
a. 13
b. 5
c. 9
d. 10
Solution:

Create a present & future table. Use any symbol to represent Kim and Kevin.
Present
Future
Kim
y
y+3
Kevin
x
x+3

Rewrite the condition statement into mathematical expressions.
Condition 1 (present):
Condition 2 (future):

Solve for Kevin’s age using elimination (x).

You can also solve for Kim’s age by substituting x to any of the condition
equation.
4. Mary is 24 years old. Mary was twice as old as Ann was when Mary was as old as Ann is
now. How old is Ann?
a. 16
b. 18
c. 20
d. 12
Solution:

Create a table with the given statements.
NOTE: The statement “Mary was as old as Ann is now” says that the age of
Mary in the past is the same age of Ann in the present, which can be represented
as “x”.
Present
Past
Mary
24
x
Ann
x
?

Since no more information can be gained from the statements, we can use Mary’s
data to complete the past age of Ann. Determine how many years have passed
from Mary’s data and do the same to Ann’s past to complete the table.
NOTE: The statement “Mary was twice as old as Ann was…” is a condition
statement which is used to calculate the missing value.

To determine how many years have passed, just subtract the present and the past
values in Mary’s data, that is, (24-x) years. Now, we can use this to determine
Ann’s past data by subtracting her present age to the years passed (x-(24-x)).
Mary
Ann
Present
24
x

Rewrite the condition statement to identify Ann’s age.
Condition 1 (past):

Solve for Ann’s age (x).
Past
x
x-(24-x)
5. The age of Diophantus, a Greek mathematician, may be calculated from the epitaphs which
reads as follows: “Diophantus passed 1/6 of his life in childhood, 1/12 in youth, and 1/7 as
bachelor. Five years after his marriage, was born a son who died 4 years before his father
at half his father’s final age”. How old was Diophantus when he died?
a. 80
b. 84
c. 76
d. 90
Solution:
(next page)

Create a timeline.

Diophantus’ years can be solved through the summation of his life events.
NOTE: The statement “…was born a son who died 4 years before his father at half
1
his father’s final age.” can be translated as 2x since his son died half his father’s
age.

Solve for x.
6. Two thousand kg of steel containing 8% nickel is to be made by mixing steel containing
14% nickel with another 6% nickel. How much of the 14% nickel is needed?
a. 1500
b. 1305
c. 500
d. 1200
Solution:

Identify the components using algebraic expressions.

Rewrite the condition statements to solve how much kilograms of % nickel is need
for the 14% and 6%. Make sure that the statements have the same unit.
Condition 1 (in kilograms):
Condition 2 (in % kilograms):

Use systems of linear equation to solve for the amount of x kg needed.
NOTE: You can multiply (-0.06) on the first equation to eliminate the “y” and solve
the value of x.
7. Solve for x in
a. -1
b. 0.734
c. 1
d. 0.618
Solution:

Since it is an infinite expression, you must find a “stop” to the equation.
NOTE: If you analyze the equation, the value of x is inside the equation itself.

Substitute the value of x inside the equation.

Solve for x.
NOTE: You can use quadratic equation.
8. The orange bulbs of a light flash every 8 seconds, the red bulbs every 9 seconds, and the
green bulb flash every 10 seconds. If they all start flashing together, in every how many
seconds they can flash simultaneously?
a. 300
b. 360
c. 270
d. 540
Solution:
 Use continuous division to get the LCM. Divide until the numbers have no multiple
left.

Multiply the divisor by the last common multiples.
9. A book is bought by a company for 200 pesos per copy. What will be the selling price if a
discount of 20% and a profit of 30% is to be made?
a. 225
b. 300
c. 325
d. 200
Solution:

Use profit formula.

Solve for the selling price (sales).
10. A painting job can be done by 72 men in 100 days. There were 80 men at the start of the
project but after 40 days, 30 of them had to be transferred to another project. If the contract
time is only for 80 days, how many men should be hired 20 days after the 30 men had to
be transferred to complete the job on time?
a. 150
b. 100
c. 50
d. 120
Solution:

1
Let 𝑋 be the rate of work of each man. “x” is for the time to complete “1” job. Use
general equation for completing jobs.

Compute for the general rate of work of 72 men.

Compute for the value of number of men needed to complete the work (y).
NOTE: You can formulate an equation based on the problem.