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t>1
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23 May 2020, 17:15
23 May 2020, 17:15
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Question Stats:
81% (00:32) correct
18% (00:23) wrong based on 33 sessions
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\(t>1\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
I solved as picking up some values(decimal and whole numbers) for t which was higher than 1. And I got an answer as D. But solution shows something like this: First, multiply both sides by 3 to clear the fraction. Multipying by a positive number does not affect the comparison:
2t + 5 9t
Now, subtract 2t from both columns (again, not affecting the comparison):
5 7t
Because t > 1, 7t is greater than 5. Therefore, Quantity B is greater.
Why is this way is the right thing to do?
Quantity A |
Quantity B |
\(\frac{2t+5}{3}\) |
\(3t\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
I solved as picking up some values(decimal and whole numbers) for t which was higher than 1. And I got an answer as D. But solution shows something like this: First, multiply both sides by 3 to clear the fraction. Multipying by a positive number does not affect the comparison:
2t + 5 9t
Now, subtract 2t from both columns (again, not affecting the comparison):
5 7t
Because t > 1, 7t is greater than 5. Therefore, Quantity B is greater.
Why is this way is the right thing to do?
Re: Quant Comparsion
[#permalink]
23 May 2020, 17:50
23 May 2020, 17:50
t>1
Quantity A is [2t + 5]/[3]
Quantity B is 3t.
Set them equal to each other:
[2t + 5]/[3] =?= 3t
2t + 5 =?= 9t
5 =?= 7t
So when t = 5/7, they're equal, BUT t is > 1, therefore B is greater.
It's the right thing to do to set them "equal" to each other as a hypothetical equality because that's what you have to do to figure out what the solution is for all values of t. But the caveat that t >1 changes the answer and is an important detail to see.
Quantity A is [2t + 5]/[3]
Quantity B is 3t.
Set them equal to each other:
[2t + 5]/[3] =?= 3t
2t + 5 =?= 9t
5 =?= 7t
So when t = 5/7, they're equal, BUT t is > 1, therefore B is greater.
It's the right thing to do to set them "equal" to each other as a hypothetical equality because that's what you have to do to figure out what the solution is for all values of t. But the caveat that t >1 changes the answer and is an important detail to see.
Re: Quant Comparsion
[#permalink]
23 May 2020, 18:08
23 May 2020, 18:08
Mudslide53 wrote:
t>1
Quantity A is [2t + 5]/[3]
Quantity B is 3t.
Set them equal to each other:
[2t + 5]/[3] =?= 3t
2t + 5 =?= 9t
5 =?= 7t
So when t = 5/7, they're equal, BUT t is > 1, therefore B is greater.
It's the right thing to do to set them "equal" to each other as a hypothetical equality because that's what you have to do to figure out what the solution is for all values of t. But the caveat that t >1 changes the answer and is an important detail to see.
Quantity A is [2t + 5]/[3]
Quantity B is 3t.
Set them equal to each other:
[2t + 5]/[3] =?= 3t
2t + 5 =?= 9t
5 =?= 7t
So when t = 5/7, they're equal, BUT t is > 1, therefore B is greater.
It's the right thing to do to set them "equal" to each other as a hypothetical equality because that's what you have to do to figure out what the solution is for all values of t. But the caveat that t >1 changes the answer and is an important detail to see.
Is it safe bet to choose some values for t and see which one is coming higher for frequent?
