John invest $2,000 at a constant interest rate of annually compound 5%. If the amount from the investment after n years is p, p=(2,000)1.05^n, is n>4?
1) p>2,100
2) p<3,000
* A solution will be posted in two days.
John invest $2,000 at a constant interest rate of annually c
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- Max@Math Revolution
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- Max@Math Revolution
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
John invest $2,000 at a constant interest rate of annually compound 5%. If the amount from the investment after n years is p, p=(2,000)1.05^n, is n>4?
1) p>2,100
2) p<3,000
In the original condition, there is 1 variable, which should match with the number of equations. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
Howver, 1) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
2) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
1) & 2) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
The final answer is E
John invest $2,000 at a constant interest rate of annually compound 5%. If the amount from the investment after n years is p, p=(2,000)1.05^n, is n>4?
1) p>2,100
2) p<3,000
In the original condition, there is 1 variable, which should match with the number of equations. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
Howver, 1) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
2) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
1) & 2) 2,000(1.05^4)=2,430 -> no, 2,000(1.05^5)=2,551 -> yes
The final answer is E
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- John fran kennedi
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Hi Max
I do not understand the way you combined the statement 1 and 2. Is there another way to determine that the answer choice C is insufficient?
I do not understand the way you combined the statement 1 and 2. Is there another way to determine that the answer choice C is insufficient?
- Max@Math Revolution
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In the original condition, there is 1 variable, which should match with the number of equations. So you need 1 equation, which is likely to make D the answer. Thus, !) and 2) are separately calculated and each of them is not sufficient. Therefore, according to definition of DS, 1) & 2) are combined, which is also yes and no. Thus, it is not sufficient and the answer is E.
-> For cases where we need 1 more equation, such as original conditions with "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
-> For cases where we need 1 more equation, such as original conditions with "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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