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11. Mixture Problems
11.1. A jeweler had an alloy that was 24% silver by weight. After 200 grams of a metal containing no silver were mixed in, the percentage of silver in the alloy decreased by one-third. What was the weight of the original alloy?
A. 96 grams
B. 100 grams
C. 200 grams
D. 400 grams
E. 600 grams
Let the weight of the original alloy be x grams. Then the amount of silver in it is
0.24x
If the percentage decreased by one-third, then the new percentage is
24% * 2/3 = 16%
After 200 grams of a metal containing no silver are added, the total weight becomes
x + 200
Since no silver was added, the amount of silver stays the same. So we get
0.24x = 0.16(x + 200)
Solving:
0.24x = 0.16x + 32
0.08x = 32
x = 400
Therefore, the weight of the original alloy was 400 grams.
Answer: E
Takeaway
In mixture problems like this one, the key is to track the amount of the substance of interest. If something containing none of that substance is added, the amount of that substance stays the same, even though the total weight increases and the concentration falls.
What This Question Tests
This question tests mixture and concentration reasoning, together with algebraic translation. It also tests whether you can work with a fractional decrease in percentage and recognize that the amount of the key substance remains unchanged while the total weight changes.
11.1. A jeweler had an alloy that was 24% silver by weight. After 200 grams of a metal containing no silver were mixed in, the percentage of silver in the alloy decreased by one-third. What was the weight of the original alloy?
A. 96 grams
B. 100 grams
C. 200 grams
D. 400 grams
E. 600 grams
Let the weight of the original alloy be x grams. Then the amount of silver in it is
0.24x
If the percentage decreased by one-third, then the new percentage is
24% * 2/3 = 16%
After 200 grams of a metal containing no silver are added, the total weight becomes
x + 200
Since no silver was added, the amount of silver stays the same. So we get
0.24x = 0.16(x + 200)
Solving:
0.24x = 0.16x + 32
0.08x = 32
x = 400
Therefore, the weight of the original alloy was 400 grams.
Answer: E
Takeaway
In mixture problems like this one, the key is to track the amount of the substance of interest. If something containing none of that substance is added, the amount of that substance stays the same, even though the total weight increases and the concentration falls.
What This Question Tests
This question tests mixture and concentration reasoning, together with algebraic translation. It also tests whether you can work with a fractional decrease in percentage and recognize that the amount of the key substance remains unchanged while the total weight changes.
15 Apr 2026, 06:04
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11. Mixture Problems
11.2. A container is full of a solution that is 55% salt. Part of the solution is removed and replaced with the same amount of a 15% salt solution. If the resulting solution is 27% salt, what fraction of the original solution was replaced?
A. 15%
B. 28%
C. 30%
D. 40%
E. 70%
Let x be the fraction of the original solution that was replaced, and assume the container originally held 1 liter.
Then x liters of the 55% salt solution were removed and replaced with x liters of the 15% salt solution.
After the removal, the amount of salt left from the original solution is
0.55(1 - x)
The replacement solution adds
0.15x
Since the final solution is 27% salt, the final amount of salt is 0.27 liters. So we get
0.55(1 - x) + 0.15x = 0.27
Now solve:
0.55 - 0.55x + 0.15x = 0.27
0.55 - 0.40x = 0.27
0.40x = 0.28
x = 0.70
So 70% of the original solution was replaced.
Answer: E.
Takeaway
In problem-solving questions where no actual quantities are fixed and the information is given only in percentages and fractions, assuming a smart convenient total is often better than working with abstract variables alone. It makes the setup cleaner and the relationships easier to track. In mixture problems, the key is to focus on how much of the substance of interest is being removed, kept, or added, and then translate that change into a clean equation using the given concentrations.
What This Question Tests
This question tests a specific type of mixture problem, namely a mixture with replacement, together with algebraic translation and manipulation.
11.2. A container is full of a solution that is 55% salt. Part of the solution is removed and replaced with the same amount of a 15% salt solution. If the resulting solution is 27% salt, what fraction of the original solution was replaced?
A. 15%
B. 28%
C. 30%
D. 40%
E. 70%
Let x be the fraction of the original solution that was replaced, and assume the container originally held 1 liter.
Then x liters of the 55% salt solution were removed and replaced with x liters of the 15% salt solution.
After the removal, the amount of salt left from the original solution is
0.55(1 - x)
The replacement solution adds
0.15x
Since the final solution is 27% salt, the final amount of salt is 0.27 liters. So we get
0.55(1 - x) + 0.15x = 0.27
Now solve:
0.55 - 0.55x + 0.15x = 0.27
0.55 - 0.40x = 0.27
0.40x = 0.28
x = 0.70
So 70% of the original solution was replaced.
Answer: E.
Takeaway
In problem-solving questions where no actual quantities are fixed and the information is given only in percentages and fractions, assuming a smart convenient total is often better than working with abstract variables alone. It makes the setup cleaner and the relationships easier to track. In mixture problems, the key is to focus on how much of the substance of interest is being removed, kept, or added, and then translate that change into a clean equation using the given concentrations.
What This Question Tests
This question tests a specific type of mixture problem, namely a mixture with replacement, together with algebraic translation and manipulation.
15 Apr 2026, 06:04
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Bookmarks
12. Multiples and Factors
12.1. How many integers are there between -1000 and 1000, inclusive, that are not divisible by either 12 or 27?
A. 1777
B. 1778
C. 1779
D. 1780
E. 1781
The total number of integers from -1000 to 1000, inclusive, is
1000 - (-1000) + 1 = 2001 (There are 1000 negative integers, 1000 positive integers, and 0.)
Now let us determine how many integers in this range are divisible by 12 or 27, and subtract that number from 2001.
The number of multiples of an integer within a range can be calculated using the formula
(last multiple in the range - first multiple in the range) / multiple + 1
Thus:
• The number of multiples of 12 in the given range is
(996 - (-996))/12 + 1 = 1992/12 + 1 = 166 + 1 = 167
To find the greatest multiple of 12 in the range, note that 12 = 3 * 4, so the number must be divisible by both 3 and 4. Since 1000 is divisible by 4 but not by 3, we look at the next smaller multiple of 4, which is 996. Since 996 is also divisible by 3, 996 is the greatest multiple of 12 in the range.
• The number of multiples of 27 in the given range is
(999 - (-999))/27 + 1 = 1998/27 + 1 = 74 + 1 = 75
To find the greatest multiple of 27 in the range, note that 27 = 9 * 3. The largest multiple of 9 below 1000 is 999. Also, 999/9 = 111, and 111 is divisible by 3 since the sum of its digits is divisible by 3. Therefore, 999 is divisible by 27, so it is the greatest multiple of 27 in the range.
• The number of multiples of both 12 and 27 is the number of multiples of their least common multiple.
LCM(12, 27) = 108
To find the LCM, use prime factorization: 12 = 2^2 * 3 and 27 = 3^3. Then take the highest power of each prime that appears. So LCM(12, 27) = 2^2 * 3^3 = 108.
So the number of multiples of both 12 and 27, that is, the number of multiples of 108 in the range, is
(972 - (-972))/108 + 1 = 1944/108 + 1 = 18 + 1 = 19 (9 positive multiples of 108, 9 negative multiples of 108, and 0. Recall that 0 is a multiple of every integer.)
To find the greatest multiple of 108 in the range, note that 108 * 9 = 972, while 108 * 10 = 1080, which is greater than 1000. So 972 is the greatest positive multiple of 108 between -1000 and 1000.
Observe that these 19 numbers are included in both the count of multiples of 12 and the count of multiples of 27. To avoid double-counting, we subtract them once.
Therefore, the number of integers divisible by either 12 or 27 is
167 + 75 - 19 = 223
Consequently, the number of integers that are not divisible by either 12 or 27 is
2001 - 223 = 1778
Answer: B.
Takeaway
It is important to know how to find multiples of an integer in a given range. It is also important to know basic divisibility rules, such as divisibility by 2, 3, 5, and 9. You should also keep in mind basic properties of 0, such as the fact that 0 is a multiple of every integer and a divisor of none.
What This Question Tests
This question tests divisibility, counting multiples in a range, and the inclusion-exclusion principle. It also tests whether you can apply a systematic counting formula correctly when the interval includes negative integers, positive integers, and 0.
12.1. How many integers are there between -1000 and 1000, inclusive, that are not divisible by either 12 or 27?
A. 1777
B. 1778
C. 1779
D. 1780
E. 1781
The total number of integers from -1000 to 1000, inclusive, is
1000 - (-1000) + 1 = 2001 (There are 1000 negative integers, 1000 positive integers, and 0.)
Now let us determine how many integers in this range are divisible by 12 or 27, and subtract that number from 2001.
The number of multiples of an integer within a range can be calculated using the formula
(last multiple in the range - first multiple in the range) / multiple + 1
Thus:
• The number of multiples of 12 in the given range is
(996 - (-996))/12 + 1 = 1992/12 + 1 = 166 + 1 = 167
To find the greatest multiple of 12 in the range, note that 12 = 3 * 4, so the number must be divisible by both 3 and 4. Since 1000 is divisible by 4 but not by 3, we look at the next smaller multiple of 4, which is 996. Since 996 is also divisible by 3, 996 is the greatest multiple of 12 in the range.
• The number of multiples of 27 in the given range is
(999 - (-999))/27 + 1 = 1998/27 + 1 = 74 + 1 = 75
To find the greatest multiple of 27 in the range, note that 27 = 9 * 3. The largest multiple of 9 below 1000 is 999. Also, 999/9 = 111, and 111 is divisible by 3 since the sum of its digits is divisible by 3. Therefore, 999 is divisible by 27, so it is the greatest multiple of 27 in the range.
• The number of multiples of both 12 and 27 is the number of multiples of their least common multiple.
LCM(12, 27) = 108
To find the LCM, use prime factorization: 12 = 2^2 * 3 and 27 = 3^3. Then take the highest power of each prime that appears. So LCM(12, 27) = 2^2 * 3^3 = 108.
So the number of multiples of both 12 and 27, that is, the number of multiples of 108 in the range, is
(972 - (-972))/108 + 1 = 1944/108 + 1 = 18 + 1 = 19 (9 positive multiples of 108, 9 negative multiples of 108, and 0. Recall that 0 is a multiple of every integer.)
To find the greatest multiple of 108 in the range, note that 108 * 9 = 972, while 108 * 10 = 1080, which is greater than 1000. So 972 is the greatest positive multiple of 108 between -1000 and 1000.
Observe that these 19 numbers are included in both the count of multiples of 12 and the count of multiples of 27. To avoid double-counting, we subtract them once.
Therefore, the number of integers divisible by either 12 or 27 is
167 + 75 - 19 = 223
Consequently, the number of integers that are not divisible by either 12 or 27 is
2001 - 223 = 1778
Answer: B.
Takeaway
It is important to know how to find multiples of an integer in a given range. It is also important to know basic divisibility rules, such as divisibility by 2, 3, 5, and 9. You should also keep in mind basic properties of 0, such as the fact that 0 is a multiple of every integer and a divisor of none.
What This Question Tests
This question tests divisibility, counting multiples in a range, and the inclusion-exclusion principle. It also tests whether you can apply a systematic counting formula correctly when the interval includes negative integers, positive integers, and 0.
