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x is a positive integer.
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19 Dec 2022, 14:54
19 Dec 2022, 14:54
Expert Reply
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Question Stats:
67% (01:54) correct
32% (02:02) wrong based on 28 sessions
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x is a positive integer.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\( \frac{\frac{5x}{7}}{\frac{8}{2x+3}}\) |
\(\frac{x^2}{4 }+ \frac{8x}{7}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: x is a positive integer.
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22 Jan 2023, 10:03
22 Jan 2023, 10:03
Expert Reply
A can be written as
\(\frac{10x^2+15x}{56}=\)
B is \(\frac{7x^2+32x}{28}\)
making a common denominator
B is \(\frac{14x^2+64x}{56}\)
Clearly B is greater
\(\frac{10x^2+15x}{56}=\)
B is \(\frac{7x^2+32x}{28}\)
making a common denominator
B is \(\frac{14x^2+64x}{56}\)
Clearly B is greater
x is a positive integer.
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19 Jan 2024, 19:40
19 Jan 2024, 19:40
1
Quantity A
\(\dfrac{\frac{5x}{7}}{\frac{8}{2x+3}}\)
\(\frac{5x}{7} \times \frac{2x+3}{8}\)
\(\frac{10x^2+15x}{56}\)
Quantity B
\(\frac{x^2}{4} + \frac{8x}{7}\)
\(\frac{7x^2 + 32x}{28}\)
dividing both sides by 28
Quantity A = \(\frac{10x^2+15x}{2} = 5x^2 + 7.5x\)
Quantity B = \(7x^2 + 32x\)
Since \(x\) is a positive integer
Clearly Quantity B is greater
\(\dfrac{\frac{5x}{7}}{\frac{8}{2x+3}}\)
\(\frac{5x}{7} \times \frac{2x+3}{8}\)
\(\frac{10x^2+15x}{56}\)
Quantity B
\(\frac{x^2}{4} + \frac{8x}{7}\)
\(\frac{7x^2 + 32x}{28}\)
dividing both sides by 28
Quantity A = \(\frac{10x^2+15x}{2} = 5x^2 + 7.5x\)
Quantity B = \(7x^2 + 32x\)
Since \(x\) is a positive integer
Clearly Quantity B is greater
