The Mathematics 11
Competency Test
Percent Composition
When two or more types of things (ingredients) are mixed together, it is common to speak of the
individual ingredients as making up a certain percentage of the whole. The formal relationship
can be written
percent concentration
=
of ingredient A
Amount of A
× 100
Total Quantity of Mixture
We have shown the multiplication by 100 necessary to get the actual percent value. When you
do calculations involving this percent, you ignore the 100 and work directly with the decimal or
fractional equivalent of the percent.
Note particularly that the quantity forming the denominator on the right-hand side is the entire
quantity of material present – everything including the amount of ingredient A.
Problems involving percent composition can be quite complicated, and a few of the more involved
types of percent problems are illustrated in the section on mixtures word problems later in these
notes. In this section we’ll focus more on directly applying the basic formula given above.
Example 1: A storage tank contains 750 kg of a solution which is 8.5% salt by weight. If 380 kg
of pure water is added to the tank, what will be the percent salt of the resulting solution?
solution:
We are asked to calculate the percent salt in the final solution. Applying the property shown
above, we can write
percent concentration of
salt in the final solution
=
amount of salt in
the final solution
× 100
total quantity of
final solution
The amounts and quantities here will have units of kilograms.
Now, the total number of kilograms of final solution will be just
750 kg + 380 kg = 1130 kg
since 380 kg of water was added to the existing 750 kg of salt solution to form the final solution.
The kilograms of salt in the final solution is just the kilograms of salt in the initial solution:
kg salt = kg solution x percent salt in solution
= (750 kg) (0.085)
= 63.75 kg.
The 380 kilograms of pure water added to the original solution contains no salt, and so the only
salt in the final solution must come from the initial solution. Thus
David W. Sabo (2003)
Percent Composition
Page 1 of 5
percent concentration of
salt in the final solution
=
63.75 kg salt
× 100
1130 kg solution
= 5.6416%
rounded to four decimal places. (This is a plausible answer, since we expect the percent salt to
decrease when water is added to dilute the original solution, so we are expecting an answer
somewhat smaller than 8.5%.)
Example 2: Suppose a tank contains 1200 kg of a solution which is 6.39% salt by weight. If
another 50 kg of salt are dissolved in this tank, what will be the percent salt of the resulting
solution?
solution:
This problem is very similar to the previous one. Again
percent concentration of
salt in the final solution
=
amount of salt in
the final solution
× 100
total quantity of
final solution
In this situation, 50 kg of salt are added to 1200 kg of an initial salt solution, so that the total
quantity of final solution is
1200 kg + 50 kg = 1250 kg
Now, the salt in the final solution is the 50 kg of salt which was added, plus the number of
kilograms of salt already in the original solution. Since the original solution has a mass of 1200
kg and is 6.3% salt, then
kg of salt in the
percent salt in the
= kg solution ×
original solution
original soltuion
= (1200 kg )( 0.063 ) = 75.60 kg
So, the total kilograms of salt in the final solution is
75.60 kg + 50 kg = 125.60 kg
So, finally,
percent concentration of
salt in the final solution
=
125.60 kg
× 100
1250 kg
= 10.048%
David W. Sabo (2003)
Percent Composition
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(unrounded). After addition of the 50 kg of salt to the initial solution, the concentration will have
increased to 10.048% by weight.
Example 3: A tank contains 2500 kg of a solution which is 7.6% salt. Over a period of time, 268
kilograms of water evaporate from the tank. What is the percent salt of the solution in the tank
now?
solution:
We begin with the same general formula as in the previous two examples:
percent concentration of
salt in the final solution
=
amount of salt in
the final solution
× 100
total quantity of
final solution
In this case, the final quantity of solution is
2500 kg – 268 kg = 2232 kg
because the evaporation of the 268 kg of water actually reduces the mass of the remaining
solution by 268 kg.
However, only water evaporates, so the kilograms of salt remaining in the 2232 kg of solution will
be the same as the kilograms of salt in the original solution:
kilograms of
kilograms of salt
=
salt remaining
originally present
=
(kilograms of solution )
 percent salt in 
× 

 original solution 
= ( 2500 kg ) × ( 0.076 ) = 190 kg
So
percent concentration of
salt in the final solution
=
190 kg
× 100
2232 kg
= 8.5125%
rounded to four decimal places.
Thus, due to the evaporation of the 268 kg of water, the concentration of salt in this solution
increases to 8.5125%.
David W. Sabo (2003)
Percent Composition
Page 3 of 5
Example 4: A tank contains 4300 kg of a solution which is 5.8% salt. To this tank is added 2200
kg of a solution which is 9.3% salt. What is the percent salt in the resulting solution?
solution:
Once again, we use the general principle:
percent concentration of
salt in the final solution
=
amount of salt in
the final solution
× 100
total quantity of
final solution
In this case, the kilograms of solution of the final mixture is just the sum of the number of
kilograms of each of the two solutions mixed:
4300 kg + 2200 kg = 6500 kg
However, in this example, salt is contributed to the final solution from both of the solutions which
are mixed together. The first component is 4300 kg of a 5.8% salt solution, so it contributes
(4300 kg) x (0.058) = 249.4 kg
of salt. The second component is 2200 kg of a 9.3% salt solution, so it contributes
(2200 kg) x (0.093) = 204.6 kg
of salt. Thus, the total number of kilograms of salt in the final solution is
249.4 kg + 204.6 kg = 454 kg
So
percent concentration of
salt in the final solution
=
454 kg
× 100
6500 kg
= 6.9846%
rounded to four decimal places. Combining these two solutions in the amounts stated results in a
final solution which is 6.9846% salt.
Note that in Example 4, we could not get the percent salt for a combination of two solutions by
adding the original percents (5.8% + 9.3% = 15.1% ≠ 6.9846%) or even by averaging
(
5.8% + 9.3%
= 7.55% ≠ 6.9846% ) the percent concentrations of the original two solutions.
2
This is because it is quantities of material (the kilograms of salt, and the kilograms of solution)
which can be added. You can’t simply combine percent concentrations by themselves, since
percent numbers have no real meaning until they are related to actual amounts of mixture or
solutions.
David W. Sabo (2003)
Percent Composition
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Example 5: In a survey, it was found that 512 of the people contacted wore eyeglasses, and the
remaining 428 of the people questioned did not wear eyeglasses. What percentage of these
people questioned wore eyeglasses?
solution:
This is a very simple problem. The general principle here will be
number of people who
wear eyeglasses
× 100
total number of people
percent of people who
=
wear eyeglasses
=
512
× 100
512 + 428
=
512
× 100
940
= 54.47%
rounded to two decimal places.
David W. Sabo (2003)
Percent Composition
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