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nick13
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In the formula R= x/y^2, x and y are both positive. If x is decreased Updated on: 22 Oct 2023, 23:38
Originally posted by nick13 on 22 Oct 2023, 16:47.
Last edited by Bunuel on 22 Oct 2023, 23:38, edited 3 times in total.
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66% (02:12) correct 34% (02:07) wrong based on 1133 sessions

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In the formula \(R = \frac{x}{y^2}\), \(x\) and \(y\) are both positive. If \(x\) is decreased by 30 percent and \(y\) is increased by 10 percent by approximately what percent will \(R\) be decreased?

A. 36%

B. 42%

C. 48%

D. 58%

E. 64%

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In the formula R= x/y^2, x and y are both positive. If x is decreased 28 Mar 2024, 12:30
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ashdank94
MartyMurray is there an easier way to do this problem?
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In the formula \(R = \frac{x}{y^2}\), \(x\) and \(y\) are both positive. If \(x\) is decreased by 30 percent and \(y\) is increased by 10 percent by approximately what percent will \(R\) be decreased?

If \(x\) is decreased by 30 percent and \(y\) is increased by 10 percent, the new \(R\) will be \(\frac{0.7x}{(1.1y)^2}\).

So, the new \(R\) is \(R × \frac{0.7}{1.1^2}\).

\(11 × 11 = 121\)

So, \(1.1^2 = 1.21\).

Thus, the new R is \(R × \frac{0.7}{1.21}\), or approximately \(R × \frac{7}{12}\).

Thus, the percentage decrease is \(\frac{5}{12}\).

So, \(\frac{100}{12}\) is a little over 8, \(\frac{5}{12}\) is a little over 5 × 0.08, or a little over 40%.

A. 36%

B. 42%

C. 48%

D. 58%

E. 64%­

Correct answer: B­
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In the formula R= x/y^2, x and y are both positive. If x is decreased 22 Oct 2023, 18:25
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In the formula \(R =\frac{ x}{y^2}\), \(x\) and \(y\) are both positive. If \(x\) is decreased by 30 percent and \(y\) is increased by 10 percent by approximately what percent will \(R\) be decreased?

A. 36%

B. 42%

C. 48%

D. 58%

E. 64%

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If \(x\) is decreased by 30 percent

The new value of x = \(0.7x\)

\(y\) is increased by 10 percent

The new value of y = \(1.1y\)

\(R_1 = \frac{0.7x}{(1.1y)^2}\)

Assume \(R * f = R_1\) ⇒ \(f =\) change factor

\(\frac{ x}{y^2} * f= \frac{0.7x}{(1.1y)^2}\)

\(\frac{ x}{y^2} * f= \frac{0.7x}{1.21y^2}\)

Multiplying both sides by \(y^2\) and dividing both sides by \(x\), we get

\(f= \frac{7}{12.1}\approx = \frac{7}{12} \approx 0.583\)

We have to find the percentage change, i.e. the value of \(1 - f \approx 1 - 0.58 \approx 0.42\)

\(\approx 42\)%

Option B
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In the formula R= x/y^2, x and y are both positive. If x is decreased 03 Jan 2024, 05:02
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gmatophobia
nick13
In the formula \(R =\frac{ x}{y^2}\), \(x\) and \(y\) are both positive. If \(x\) is decreased by 30 percent and \(y\) is increased by 10 percent by approximately what percent will \(R\) be decreased?

A. 36%

B. 42%

C. 48%

D. 58%

E. 64%


If \(x\) is decreased by 30 percent

The new value of x = \(0.7x\)

\(y\) is increased by 10 percent

The new value of y = \(1.1y\)

\(R_1 = \frac{0.7x}{(1.1y)^2}\)

Assume \(R * f = R_1\) ⇒ \(f =\) change factor

\(\frac{ x}{y^2} * f= \frac{0.7x}{(1.1y)^2}\)

\(\frac{ x}{y^2} * f= \frac{0.7x}{1.21y^2}\)

Multiplying both sides by \(y^2\) and dividing both sides by \(x\), we get

\(f= \frac{7}{12.1}\approx = \frac{7}{12} \approx 0.583\)

We have to find the percentage change, i.e. the value of \(1 - f \approx 1 - 0.58 \approx 0.42\)

\(\approx 42\)%

Option B
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thanks!
how did you approximate this please - f=7 / 12.1≈=7/ 12≈0.583 ?
I would've thought its = ~1/2 therefore answer = C= 48%
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In the formula R= x/y^2, x and y are both positive. If x is decreased 03 Jan 2024, 05:47
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nick13
In the formula \(R =\frac{ x}{y^2}\), \(x\) and \(y\) are both positive. If \(x\) is decreased by 30 percent and \(y\) is increased by 10 percent by approximately what percent will \(R\) be decreased?

A. 36%

B. 42%

C. 48%

D. 58%

E. 64%


If \(x\) is decreased by 30 percent

The new value of x = \(0.7x\)

\(y\) is increased by 10 percent

The new value of y = \(1.1y\)

\(R_1 = \frac{0.7x}{(1.1y)^2}\)

Assume \(R * f = R_1\) ⇒ \(f =\) change factor

\(\frac{ x}{y^2} * f= \frac{0.7x}{(1.1y)^2}\)

\(\frac{ x}{y^2} * f= \frac{0.7x}{1.21y^2}\)

Multiplying both sides by \(y^2\) and dividing both sides by \(x\), we get

\(f= \frac{7}{12.1}\approx = \frac{7}{12} \approx 0.583\)

We have to find the percentage change, i.e. the value of \(1 - f \approx 1 - 0.58 \approx 0.42\)

\(\approx 42\)%

Option B

thanks!
how did you approximate this please - f=7 / 12.1≈=7/ 12≈0.583 ?
I would've thought its = ~1/2 therefore answer = C= 48%
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With reasonably close percentages given in the answer choices (42,48), it's risky to merely eyeball the choices and more work needs to be done.

Since you're trying to determine the equivalent divided by 100 instead of 12, multiply numerator and denominator by 8.

That yields 56/96.

Since 96 is a bit less than 100, we know that the actual numerator will be a bit more than 56 (about half as much, right ?).

Meaning that 1-required numerator will be a bit less than 44.

42 is a bit less.

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In the formula R= x/y^2, x and y are both positive. If x is decreased 23 Mar 2024, 15:23
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MartyMurray is there an easier way to do this problem?
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In the formula R= x/y^2, x and y are both positive. If x is decreased 24 Mar 2024, 01:12
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ashdank94
MartyMurray is there an easier way to do this problem?
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­1-(0.7/1.21)=1-7/12=5/12.

you know 12*4=48.. so ans should be slightly greater than 40%
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In the formula R= x/y^2, x and y are both positive. If x is decreased 07 Jul 2024, 06:30
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­We make this chart:

                   \(Assumed\) \(value\)              \(Final\) \(Value\)
\(x\)                           10                          7 (30% decrease)

\(y\)                           10                          11 (10% increase)


Thus, \(\frac{x}{y^2}\)          \(\frac{10}{100}\) = \(\frac{1}{10}\)                  \(\frac{7}{121}\)


Thus, percentage change = \((\frac{7}{121} - \frac{1}{10})/\frac{1}{10}\)­

                                     = \(\frac{-51}{121}\) = \(-0.42\) = reduction of \(42\) \(%\)

The answer must be \(B\).­
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In the formula R= x/y^2, x and y are both positive. If x is decreased 27 Jul 2025, 06:11
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Hi all,

Just a quick question. Could this problem be considered an EASY one?
I encountered it in one of my mocks as the first question, so technically an easy one, but here it classified as Medium, range 655-705.

I am honestly quite confused right now. Anyone could help me understand it please?

Thank you
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In the formula R= x/y^2, x and y are both positive. If x is decreased 27 Jul 2025, 09:21
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SBLight
Hi all,

Just a quick question. Could this problem be considered an EASY one?
I encountered it in one of my mocks as the first question, so technically an easy one, but here it classified as Medium, range 655-705.

I am honestly quite confused right now. Anyone could help me understand it please?

Thank you
Show more

The difficulty level of a question on the site, after sufficient attempts, is determined automatically based on various parameters collected from users' attempts via timer, such as the percentage of correct answers and the time taken to answer the question.



Also, just because it appeared as the first question in your mock doesn’t necessarily mean it’s easy.
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In the formula R= x/y^2, x and y are both positive. If x is decreased 27 Jul 2025, 10:11
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Thank you so much Bunuel, I really appreciate it.
So technically the mock could present an hardest first question (or considered hard by the test makers) while in the real exam the first questions should be esier.
May I assume this?

Thanks again,
SBLight
Bunuel
SBLight
Hi all,

Just a quick question. Could this problem be considered an EASY one?
I encountered it in one of my mocks as the first question, so technically an easy one, but here it classified as Medium, range 655-705.

I am honestly quite confused right now. Anyone could help me understand it please?

Thank you

The difficulty level of a question on the site, after sufficient attempts, is determined automatically based on various parameters collected from users' attempts via timer, such as the percentage of correct answers and the time taken to answer the question.



Also, just because it appeared as the first question in your mock doesn’t necessarily mean it’s easy.
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