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In Episode 4 of our GMAT Ninja CR series, we tackle the most intimidating CR question type: Boldface & "Legalese" questions. If you've ever stared at an answer choice that reads, "The first is a consideration introduced to counter a position that...- May 14
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18 Apr 2018, 18:18
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[GMAT math practice question]
Which of the following is equal to x!/(x+1)! + (x+1)!/(x+2)!?
A. 1+1/x
B. 1+1/(x+1)
C. (2x+3)/(x+1)(x+2)
D. 1/{ x(x+1) }
E. (2x-1)/{(x+1)(x+2)}
=>
x! / ( x + 1 )! + ( x + 1 )! / ( x + 2 )!
= x! / {x! * (x+1)} + (x+1)! / { (x+1)! * (x+2) }
= 1/(x+1) + 1/(x+2) = (2x+3)/(x+1)(x+2)
Therefore, C is the answer.
Answer: C
Which of the following is equal to x!/(x+1)! + (x+1)!/(x+2)!?
A. 1+1/x
B. 1+1/(x+1)
C. (2x+3)/(x+1)(x+2)
D. 1/{ x(x+1) }
E. (2x-1)/{(x+1)(x+2)}
=>
x! / ( x + 1 )! + ( x + 1 )! / ( x + 2 )!
= x! / {x! * (x+1)} + (x+1)! / { (x+1)! * (x+2) }
= 1/(x+1) + 1/(x+2) = (2x+3)/(x+1)(x+2)
Therefore, C is the answer.
Answer: C
19 Apr 2018, 18:21
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[GMAT math practice question]
Is the total of the sales prices of 3 products greater than $23?
1) The price of the cheapest of the 3 products is at least $8
2) The price of the second cheapest of the 3 products is at least $12.
=>
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.
If a question includes the expression “greater than”, we should determine the smallest possible value, since all other possibilities are greater than the minimum. Each of the conditions includes the expression “at least”, so this gives us some minimum values.
Condition 1)
Since the cheapest of the three products costs at least $8, the minimum total cost of the three products is
$8 + $8 + $8 = $24 > $23.
Thus, condition 1) is sufficient.
Condition 2)
Since the second cheapest of the three products costs at least $12, the minimum total cost of the three products is
$0 + $12 + $12 = $24 > $23.
Thus, condition 2) is sufficient.
Answer: D
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.
Is the total of the sales prices of 3 products greater than $23?
1) The price of the cheapest of the 3 products is at least $8
2) The price of the second cheapest of the 3 products is at least $12.
=>
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.
If a question includes the expression “greater than”, we should determine the smallest possible value, since all other possibilities are greater than the minimum. Each of the conditions includes the expression “at least”, so this gives us some minimum values.
Condition 1)
Since the cheapest of the three products costs at least $8, the minimum total cost of the three products is
$8 + $8 + $8 = $24 > $23.
Thus, condition 1) is sufficient.
Condition 2)
Since the second cheapest of the three products costs at least $12, the minimum total cost of the three products is
$0 + $12 + $12 = $24 > $23.
Thus, condition 2) is sufficient.
Answer: D
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.
22 Apr 2018, 18:31
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[GMAT math practice question]
Each of three consecutive positive integers is less than 100. Their sum of is a multiple of 10. What is the smallest of the three integers?
1) Their median is a multiple of 9
2) Two of the integers are prime numbers.
=>
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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Let the integers be x – 1, x and x + 1 where x is a positive integer. Since their sum, x – 1 + x + x + 1 = 3x is a multiple of 10, x must be a multiple of 10.
Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.
Condition 1)
The median, x, of the three integers is a multiple of 9 and a multiple of 10.
Since x < 100, we must have x = 90.
Therefore, the smallest of the integers is x – 1 = 89.
Thus, condition 1) is sufficient.
Condition 2)
If x = 30, the integers are 29, 30, 31 and the smallest integer is 29.
If x = 90, the integers are 89, 90, 91 and the smallest integer is 89.
Since we do not obtain a unique answer, condition 2) is not sufficient.
Therefore, the answer is A.
Answer: A
Each of three consecutive positive integers is less than 100. Their sum of is a multiple of 10. What is the smallest of the three integers?
1) Their median is a multiple of 9
2) Two of the integers are prime numbers.
=>
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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Let the integers be x – 1, x and x + 1 where x is a positive integer. Since their sum, x – 1 + x + x + 1 = 3x is a multiple of 10, x must be a multiple of 10.
Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.
Condition 1)
The median, x, of the three integers is a multiple of 9 and a multiple of 10.
Since x < 100, we must have x = 90.
Therefore, the smallest of the integers is x – 1 = 89.
Thus, condition 1) is sufficient.
Condition 2)
If x = 30, the integers are 29, 30, 31 and the smallest integer is 29.
If x = 90, the integers are 89, 90, 91 and the smallest integer is 89.
Since we do not obtain a unique answer, condition 2) is not sufficient.
Therefore, the answer is A.
Answer: A
