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12 Jun 2024, 01:31
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B
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Difficulty:
45%
(medium)
Question Stats:
64% (01:25) correct
36%
(01:28)
wrong
based on 261
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As y increases from y = 247 to y = 248, which of the following decreases?
I. \(2y - 100\)
II. \(\frac{50}{y} + 80\)
III. \(100 - \frac{10}{y^2 - 3y}\)
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
I. \(2y - 100\)
II. \(\frac{50}{y} + 80\)
III. \(100 - \frac{10}{y^2 - 3y}\)
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
12 Jun 2024, 02:51
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I think its B, Hope i did it correctly if anything is wrong please correct.
Substitue values of y (247,248) and check for given statements. you will notice only in statement 2 value decreases.
Statement 1 - 2y-100
y=247, 2y-100 = (2*247) - 100 = 494-100 = 394 / y=248, 2y-100 = (2*248) - 100 = 496 - 100 = 396
statement 2 - (50/y) + 80
y=247, (50/y) + 80 = 50/247 + 80 / y=248, (50/y) + 80 = 50/248 + 80
here if we compare 50 / 247 with 50 / 248 we will see (50/247) > (50/248) Hence value decreases.
statement 3 - 100−(10/(y^2−3y))
similarly here as well putting values and check for it.
10 / (y (y-3)) = 10 / (247 * 244)
10 / (y (y-3)) = 10 / (248 * 245)
if we compare 10 / (247 * 244) > 10 / (248 * 245) now if we subtract these values from 100 the value will increase from y=247 to 248.
Substitue values of y (247,248) and check for given statements. you will notice only in statement 2 value decreases.
Statement 1 - 2y-100
y=247, 2y-100 = (2*247) - 100 = 494-100 = 394 / y=248, 2y-100 = (2*248) - 100 = 496 - 100 = 396
statement 2 - (50/y) + 80
y=247, (50/y) + 80 = 50/247 + 80 / y=248, (50/y) + 80 = 50/248 + 80
here if we compare 50 / 247 with 50 / 248 we will see (50/247) > (50/248) Hence value decreases.
statement 3 - 100−(10/(y^2−3y))
similarly here as well putting values and check for it.
10 / (y (y-3)) = 10 / (247 * 244)
10 / (y (y-3)) = 10 / (248 * 245)
if we compare 10 / (247 * 244) > 10 / (248 * 245) now if we subtract these values from 100 the value will increase from y=247 to 248.
12 Jun 2024, 04:12
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I. \(2y−100\)
As y increases from y = 247 to y = 248, \(2y\) becomes bigger while the amount being subtract stays the same, therefore the value of this equation will increase. X
II. \(\frac{50}{y}+80\)
As we are dealing with fractions where the numerator is remains constant in both instances, while the denominator in both exceeds the value of the numerator, then the larger value of the two will be the one with the smaller numerator. Thus, as y increases from y = 247 to y = 248, \(\frac{50}{y}\) becomes smaller.✔
III. \(100−\frac{10}{y^2−3y}\)
Taking out y as a common factor in the bottom: \(100−\frac{10}{y(y−3)}\).
As y increases from y = 247 to y = 248, \(\frac{10}{y(y−3)}\) becomes a smaller value as the denominator value exceeds the value of the numerator in both instances, in which case the value of the fraction with the larger denominator becomes the smaller value. As in both instances the respective fractions are being subtracted from 100, then as the value of y increases the smaller \(\frac{10}{y(y−3)}\) becomes which increases the value of \(100−\frac{10}{y(y−3)}\). X
Only II. increases, therefore
ANSWER B
As y increases from y = 247 to y = 248, \(2y\) becomes bigger while the amount being subtract stays the same, therefore the value of this equation will increase. X
II. \(\frac{50}{y}+80\)
As we are dealing with fractions where the numerator is remains constant in both instances, while the denominator in both exceeds the value of the numerator, then the larger value of the two will be the one with the smaller numerator. Thus, as y increases from y = 247 to y = 248, \(\frac{50}{y}\) becomes smaller.✔
III. \(100−\frac{10}{y^2−3y}\)
Taking out y as a common factor in the bottom: \(100−\frac{10}{y(y−3)}\).
As y increases from y = 247 to y = 248, \(\frac{10}{y(y−3)}\) becomes a smaller value as the denominator value exceeds the value of the numerator in both instances, in which case the value of the fraction with the larger denominator becomes the smaller value. As in both instances the respective fractions are being subtracted from 100, then as the value of y increases the smaller \(\frac{10}{y(y−3)}\) becomes which increases the value of \(100−\frac{10}{y(y−3)}\). X
Only II. increases, therefore
ANSWER B
