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[GMAT math practice question]
(Algebra) Thomas went to a grocery store and bought some fruit with \($25\). The prices of the fruit are \($5, $1\), and \($0.50\) for each watermelon, pear, and apple, respectively. How many apples did he buy?
1) He bought \(10\) pieces of fruit, and he didn’t get any change back.
2) He had at least one of each fruit.
(Algebra) Thomas went to a grocery store and bought some fruit with \($25\). The prices of the fruit are \($5, $1\), and \($0.50\) for each watermelon, pear, and apple, respectively. How many apples did he buy?
1) He bought \(10\) pieces of fruit, and he didn’t get any change back.
2) He had at least one of each fruit.
Updated on: 04 Jul 2021, 07:27
Originally posted by MathRevolution on 19 Jan 2020, 07:22.
Last edited by MathRevolution on 04 Jul 2021, 07:27, edited 1 time in total.
Last edited by MathRevolution on 04 Jul 2021, 07:27, edited 1 time in total.
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MathRevolution
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since we have \(3\) variables (\(x, y,\) and \(k\)) and \(0\) equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since \(x + y = 2\), we have \(y = 2 – x.\)
Substituting \(y = 2 - x\) into \(3x + 5y = k + 1\) gives us \(3x + 5(2 - x) = k + 1, 3x + 10 - 5x = k + 1, -2x + 10 = k + 1\) or \(2x + k = 9.\)
Substituting \(y = 2 - x\) into \(2x + 3y = k\) gives us \(2x + 3(2 - x) = k, 2x + 6 = 3x = k, -x + 6 = k\) or \(x + k = 6.\)
We now have \(2\) equations: \(2x + k = 9\) and \(x + k = 6\). Rewriting the first equation gives us \(k = 9 - 2x\). Substituting this into the second equation gives us \(x + 9 - 2x = 6, -x = -3\), and \(x = 3.\) Then \(x + k = 6\) becomes \(3 + k = 6,\) and \(k = 3.\)
Then we have \(x = 3\) and \(k = 3.\)
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
Updated on: 04 Jul 2021, 07:33
Originally posted by MathRevolution on 19 Jan 2020, 07:25.
Last edited by MathRevolution on 04 Jul 2021, 07:33, edited 1 time in total.
Last edited by MathRevolution on 04 Jul 2021, 07:33, edited 1 time in total.
Kudos
Bookmarks
MathRevolution
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Assume \(x, y\) and \(z\) are the number of watermelons, pears, and apples, respectively.
Then we have \(5x + y + 0.5z = 25\) or \(10x + 2y + z = 50.\)
Since we have \(3\) variables (\(x, y,\) and \(z\)) and \(1\) equation, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
We have \(x + y + z = 10\) and \(5x + y + 0.5z = 25\) or \(10x + 2y + z = 50\). Subtracting the first equation from the second equation gives us \((10x + 2y + z) – (x + y + z) = 50 - 10\), or \(9x + y = 40.\)
Then \(y = 40 – 9x.\)
Then the possible values of \((x, y)\) are \((1, 31), (2, 22), (3, 13)\), and \((4, 4).\)
If \(x = 1,\) and \(y = 31\), then \(z = 10 – x – y = -22\) doesn’t make sense, since \(z\) must be positive.
If \(x = 2\), and \(y = 22\), then \(z = 10 – x – y = -14\) doesn’t make sense, since \(z\) must be positive.
If \(x = 3,\) and \(y = 13\), then \(z = 10 – x – y = -6\) doesn’t make sense, since \(z\) must be positive.
If \(x = 4,\) and \(y = 4\), then we have \(z = 10 – x – y = 2.\)
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
