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emmak
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Marty's Pizza Shop guarantees that their pizzas all have a Updated on: 27 Feb 2013, 05:19
Originally posted by emmak on 27 Feb 2013, 04:14.
Last edited by Bunuel on 27 Feb 2013, 05:19, edited 1 time in total.
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51% (02:43) correct 49% (02:32) wrong based on 674 sessions

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Marty's Pizza Shop guarantees that their pizzas all have at least 75% of the surface area covered with toppings, with a crust of uniform width surrounding them. If you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust?

A. 0.8 inches
B. 1.1 inches
C. 1.6 inches
D. 2.0 inches
E. 2.5 inches
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B
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Marty's Pizza Shop guarantees that their pizzas all have a 27 Feb 2013, 06:17
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emmak
Marty's Pizza Shop guarantees that their pizzas all have at least 75% of the surface area covered with toppings, with a crust of uniform width surrounding them. If you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust?

A. 0.8 inches
B. 1.1 inches
C. 1.6 inches
D. 2.0 inches
E. 2.5 inches
Show more

The area of 16 inches pizza is \(\pi{R^2}=8^2\pi=64\pi\).

The minimum area covered with toppings is \(\frac{3}{4}*64\pi=48\pi\) --> the radius of the toppings is \(\pi{r^2}=48\pi\) --> \(r=4\sqrt{3}\approx{6.9}\).

The maximum width for the crust possible = R - r = 8 - 6.9 = 1.1 inches.

Answer: B.
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Marty's Pizza Shop guarantees that their pizzas all have a 03 Mar 2013, 06:50
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Hmmm not an official question and the wording is tough on this
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Marty's Pizza Shop guarantees that their pizzas all have a 03 Mar 2013, 21:07
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emmak
Marty's Pizza Shop guarantees that their pizzas all have at least 75% of the surface area covered with toppings, with a crust of uniform width surrounding them. If you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust?

A. 0.8 inches
B. 1.1 inches
C. 1.6 inches
D. 2.0 inches
E. 2.5 inches
Show more

Total Area = 8 * 8 * pi
Radius = 64 pi

Surface = .75 * 64 * pi = 48 pi
Radius of surface = 4 sqrt (3) ~ 6.8

Radius width = 8 - 6.8 = 1.2

Answer: B
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Marty's Pizza Shop guarantees that their pizzas all have a Updated on: 23 Sep 2014, 06:26
Originally posted by TehMoUsE on 23 Sep 2014, 06:20.
Last edited by TehMoUsE on 23 Sep 2014, 06:26, edited 1 time in total.
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emmak
Marty's Pizza Shop guarantees that their pizzas all have at least 75% of the surface area covered with toppings, with a crust of uniform width surrounding them. If you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust?

A. 0.8 inches
B. 1.1 inches
C. 1.6 inches
D. 2.0 inches
E. 2.5 inches

The area of 16 inches pizza is \(\pi{R^2}=8^2\pi=64\pi\).

The minimum area covered with toppings is \(\frac{3}{4}*64\pi=48\pi\) --> the radius of the toppings is \(\pi{r^2}=48\pi\) --> \(r=4\sqrt{3}\approx\) 6,9

The maximum width for the crust possible = R - r = 8 - 6.9 = 1.1 inches.

Answer: B.
Show more

\(r=4\sqrt{3}=6.9282....\).

The maximum width for the crust possible = R - r = 8 - 6.9282... = 1.0717.... inches. -----> 1,1 width would make the surface area covered with topping less than 75%

Answer: A

I know this might be picky, but shouldn't OA be A ?
Does GMAT give these possible answers where rounding errors become very important
Please fill me in case Im missing something (which I probably do)
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Marty's Pizza Shop guarantees that their pizzas all have a 23 Sep 2014, 06:25
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TehMoUsE
Bunuel
emmak
Marty's Pizza Shop guarantees that their pizzas all have at least 75% of the surface area covered with toppings, with a crust of uniform width surrounding them. If you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust?

A. 0.8 inches
B. 1.1 inches
C. 1.6 inches
D. 2.0 inches
E. 2.5 inches

The area of 16 inches pizza is \(\pi{R^2}=8^2\pi=64\pi\).

The minimum area covered with toppings is \(\frac{3}{4}*64\pi=48\pi\) --> the radius of the toppings is \(\pi{r^2}=48\pi\) --> \(r=4\sqrt{3}\approx{6.9}\).

The maximum width for the crust possible = R - r = 8 - 6.9 = 1.1 inches.

Answer: B.

\(r=4\sqrt{3}\={6.9282....}\).

The maximum width for the crust possible = R - r = 8 - 6.9282... = 1.0717.... inches. -----> 1,1 width would make the surface area covered with topping less than 75%

Answer: A

I know this might be picky, but shouldn't OA be A ?
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The approximate maximum is 1.1 inches.
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Marty's Pizza Shop guarantees that their pizzas all have a 23 Sep 2014, 06:32
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Bunuel

The approximate maximum is 1.1 inches.
Show more

I understand that, but the question doesn't say anything about approximations.

Is it silently implied in GMAT that you round up to the closest number, even tho it would give you the "wrong" answer?
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Marty's Pizza Shop guarantees that their pizzas all have a 23 Sep 2014, 06:37
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TehMoUsE
Bunuel

The approximate maximum is 1.1 inches.

I understand that, but the question doesn't say anything about approximations.

Is it silently implied in GMAT that you round up to the closest number, even tho it would give you the "wrong" answer?
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Word "approximate" is missing there. It should be "what is the approximate maximum width you can expect to see for the crust?"
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Marty's Pizza Shop guarantees that their pizzas all have a 23 Sep 2014, 07:31
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Hi Bunuel,

I am not able to find the solution with this method.

x be the width of the surrounding

pi (8-x)^2/pi(8)^2 = 3/4

does not lead to the answer..I am not able to find whats wrong in this approach.

Please help.

Regards,
Ravi
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Marty's Pizza Shop guarantees that their pizzas all have a 24 Sep 2014, 05:29
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Hi Bunuel,

I am not able to find the solution with this method.

x be the width of the surrounding

pi (8-x)^2/pi(8)^2 = 3/4

does not lead to the answer..I am not able to find whats wrong in this approach.

Please help.

Regards,
Ravi
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There is nothing wrong with this approach.

\(\frac{\pi (8-x)^2}{\pi(8)^2} = \frac{3}{4}\) --> \((8-x)^2=48\) --> \(8-x=4\sqrt{3}=6.9\) --> \(x=8-6.9=1.1\).
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Marty's Pizza Shop guarantees that their pizzas all have a 31 May 2015, 11:01
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Bunuel

The approximate maximum is 1.1 inches.

I understand that, but the question doesn't say anything about approximations.

Is it silently implied in GMAT that you round up to the closest number, even tho it would give you the "wrong" answer?

Word "approximate" is missing there. It should be "what is the approximate maximum width you can expect to see for the crust?"
Show more

I think the keyword here is "at least" so, if 75% calculation comes to 1.08 we have no choice but to round up and select the closest, which in this case is 1.10 and not 1.06, which will breach the requirement of "at least 75%".. Lemme know your views. Thanks.
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Marty's Pizza Shop guarantees that their pizzas all have a 07 Mar 2019, 07:54
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emmak
Marty's Pizza Shop guarantees that their pizzas all have at least 75% of the surface area covered with toppings, with a crust of uniform width surrounding them. If you order their best seller – a circular pizza with a diameter of 16 inches – what is the maximum width you can expect to see for the crust?

A. 0.8 inches
B. 1.1 inches
C. 1.6 inches
D. 2.0 inches
E. 2.5 inches
Show more

Since the diameter is 16 inches, the radius is 8 inches, and the surface area is 8^2(π) = 64π. Since the toppings covers at least 75% of the surface area, the toppings covers at least 0.75 x 64π = 48π. Therefore, the radius of the toppings’ surface area is √48 ≈ 6.9 inches. Therefore, the maximum width of the crust is approximately 8 - 6.9 = 1.1 inches.

Answer: B
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Marty's Pizza Shop guarantees that their pizzas all have a 09 Jan 2021, 19:48
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Bunuel
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Bunuel

The area of 16 inches pizza is \(\pi{R^2}=8^2\pi=64\pi\).

The minimum area covered with toppings is \(\frac{3}{4}*64\pi=48\pi\) --> the radius of the toppings is \(\pi{r^2}=48\pi\) --> \(r=4\sqrt{3}\approx{6.9}\).

The maximum width for the crust possible = R - r = 8 - 6.9 = 1.1 inches.

Answer: B.

\(r=4\sqrt{3}\={6.9282....}\).

The maximum width for the crust possible = R - r = 8 - 6.9282... = 1.0717.... inches. -----> 1,1 width would make the surface area covered with topping less than 75%

Answer: A

I know this might be picky, but shouldn't OA be A ?

The approximate maximum is 1.1 inches.
Show more


\(\pi{R^2}=8^2\pi=64\pi\)

* 3/4 = 48pi

16 - 2.2 = 13.8 / 2 = 6.9^2 = 47.61pi

making 1.1 too big.

I don't see the word "approximate" in the question. It's "missing" but this math is easily done on scratch paper.
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