If $y = \frac{|4n - 3|}{2n^2 + 5}$, for what value... | Free GMAT Practice Question | GMAT Panda

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If y = \frac{∣4 n - 3∣}{2 n ^{2} + 5}, for what value of n will the value of y be greatest?

Explanation	Calculations	Help
$We first evaluate y at n = - \frac{5}{2} .$	$n = - \frac{5}{2} ∣4 n - 3∣ = ∣4 (- \frac{5}{2}) - 3∣ = ∣ - 10 - 3∣ = 13 2 n^{2} + 5 = 2 (\frac{25}{4}) + 5 = \frac{25}{2} + \frac{10}{2} = \frac{35}{2} y = \frac{13}{\frac{35}{2}} = 13 \times \frac{2}{35} = \frac{26}{35} \approx 0.74$	Theory & Tactics Method Card FDPR0-D Click to view full details
$Next, we evaluate y at n = - 1 .$	$n = - 1 ∣4 n - 3∣ = ∣4 (- 1) - 3∣ = ∣ - 4 - 3∣ = 7 2 n^{2} + 5 = 2 (1) + 5 = 7 y = \frac{7}{7} = 1$
$Then, we evaluate y at n = 0 .$	$n = 0 ∣4 n - 3∣ = ∣0 - 3∣ = 3 2 n^{2} + 5 = 2 (0) + 5 = 5 y = \frac{3}{5} = 0.6$
$We also check y at n = \frac{3}{4} .$	$n = \frac{3}{4} ∣4 n - 3∣ = ∣3 - 3∣ = 0 y = \frac{0}{(positive)} = 0$
$Finally, we evaluate y at n = \frac{5}{2} .$	$n = \frac{5}{2} ∣4 n - 3∣ = ∣10 - 3∣ = 7 2 n^{2} + 5 = 2 (\frac{25}{4}) + 5 = \frac{35}{2} y = \frac{7}{\frac{35}{2}} = 7 \times \frac{2}{35} = \frac{14}{35} = \frac{2}{5} = 0.4$	Theory & Tactics Method Card FDPR0-D Click to view full details
$We compare the values of y and see that 1 is the largest among 0.74, 1, 0.6, 0, and 0.4, so y is greatest at n = - 1 .$	$1 > \frac{26}{35}, 1 > \frac{3}{5}, 1 > 0, 1 > \frac{2}{5} n = - 1$

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B

If y = \frac{∣4 n - 3∣}{2 n ^{2} + 5}, for what value of n will the value of y be greatest?

Explanation	Calculations	Help
$We first evaluate y at n = - \frac{5}{2} .$	$n = - \frac{5}{2} ∣4 n - 3∣ = ∣4 (- \frac{5}{2}) - 3∣ = ∣ - 10 - 3∣ = 13 2 n^{2} + 5 = 2 (\frac{25}{4}) + 5 = \frac{25}{2} + \frac{10}{2} = \frac{35}{2} y = \frac{13}{\frac{35}{2}} = 13 \times \frac{2}{35} = \frac{26}{35} \approx 0.74$	Theory & Tactics Method Card FDPR0-D Click to view full details
$Next, we evaluate y at n = - 1 .$	$n = - 1 ∣4 n - 3∣ = ∣4 (- 1) - 3∣ = ∣ - 4 - 3∣ = 7 2 n^{2} + 5 = 2 (1) + 5 = 7 y = \frac{7}{7} = 1$
$Then, we evaluate y at n = 0 .$	$n = 0 ∣4 n - 3∣ = ∣0 - 3∣ = 3 2 n^{2} + 5 = 2 (0) + 5 = 5 y = \frac{3}{5} = 0.6$
$We also check y at n = \frac{3}{4} .$	$n = \frac{3}{4} ∣4 n - 3∣ = ∣3 - 3∣ = 0 y = \frac{0}{(positive)} = 0$
$Finally, we evaluate y at n = \frac{5}{2} .$	$n = \frac{5}{2} ∣4 n - 3∣ = ∣10 - 3∣ = 7 2 n^{2} + 5 = 2 (\frac{25}{4}) + 5 = \frac{35}{2} y = \frac{7}{\frac{35}{2}} = 7 \times \frac{2}{35} = \frac{14}{35} = \frac{2}{5} = 0.4$	Theory & Tactics Method Card FDPR0-D Click to view full details
$We compare the values of y and see that 1 is the largest among 0.74, 1, 0.6, 0, and 0.4, so y is greatest at n = - 1 .$	$1 > \frac{26}{35}, 1 > \frac{3}{5}, 1 > 0, 1 > \frac{2}{5} n = - 1$

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B