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In the formula R= x/y^2, x and y are both positive. If x is decreased

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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 [#permalink] New post   Updated on: 22 Oct 2023, 23:38
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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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B

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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gmatophobia
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Re: In the formula R= x/y^2, x and y are both positive. If x is decreased [#permalink] New post   22 Oct 2023, 18:25
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nick13 wrote:
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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tickledpink001
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Re: In the formula R= x/y^2, x and y are both positive. If x is decreased [#permalink] New post   03 Jan 2024, 05:02
gmatophobia wrote:
nick13 wrote:
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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Re: In the formula R= x/y^2, x and y are both positive. If x is decreased [#permalink] New post   03 Jan 2024, 05:47
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tickledpink001 wrote:
gmatophobia wrote:
nick13 wrote:
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%


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.

Posted from my mobile device
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ashdank94
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Re: In the formula R= x/y^2, x and y are both positive. If x is decreased [#permalink] New post   23 Mar 2024, 15:23
MartyMurray is there an easier way to do this problem?
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Oppenheimer1945
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Re: In the formula R= x/y^2, x and y are both positive. If x is decreased [#permalink] New post   24 Mar 2024, 01:12
ashdank94 wrote:
MartyMurray is there an easier way to do this problem?

­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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Re: In the formula R= x/y^2, x and y are both positive. If x is decreased [#permalink] New post   28 Mar 2024, 12:30
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ashdank94 wrote:
MartyMurray is there an easier way to do this problem?

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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Re: In the formula R= x/y^2, x and y are both positive. If x is decreased [#permalink]
28 Mar 2024, 12:30
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