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A rectangular wall is covered entirely with two kinds of dec

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A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 04 Feb 2014, 06:30
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A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560

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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 04 Feb 2014, 06:46
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Bunuel wrote:
A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560


The number of jumbo tiles = x.
The number of regular tiles = 2x.

Assume the ratio of the dimensions of a regular tile is a:a --> area = a^2.
The dimensions of a jumbo tile is 3a:3a --> area = 9a^2.

The area of regular tiles = 2x*a^2 = 80.
The area of jumbo tiles = x*9a^2 = 4.5(2x*a^2 ) = 4.5*80 = 360.

Total area = 80 + 360 = 440.

Answer: D.
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 04 Mar 2014, 03:15
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Let small tile dimension = 1 x 1 = 1 Square feet.......... (1)

So big tile dimension would be = 3 x 3 = 9 Square feet

80 small tiles are required to cover 80sq feet area

2/3rd of the tiles are small, so 2/3 are 80, then remaining 1/3 would be 40

So 40 big tiles are used = 40 * 9 = 360 Sq feets

Total Area = 360+80 = 440 = Answer = D
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 11 Apr 2014, 01:00
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easiest way :

2/3 is equal to 80 , so 1/3 will be equal to 40 . As given lenght and width is 3 times so 3*3(area of rectangle l*h) ; are would be 9 times 40*9=360

total area would be =360+80=440
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 16 Apr 2014, 14:25
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Bunuel wrote:
Bunuel wrote:
A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560


The number of jumbo tiles = x.
The number of regular tiles = 2x.

Assume the ratio of the dimensions of a regular tile is a:a --> area = a^2.
The dimensions of a jumbo tile is 3a:3a --> area = 9a^2.

The area of regular tiles = 2x*a^2 = 80.
The area of jumbo tiles = x*9a^2 = 4.5(2x*a^2 ) = 4.5*80 = 360.

Total area = 80 + 360 = 440.

Answer: D.


Hi B,

Though it will not change the answer, should not we consider the tiles to be rectangular, instead of square.
as the ratio could have been different, and that would require us to consider Length and Width differently. Suggest if there is something I'm missing out on!

The number of jumbo tiles = x.
The number of regular tiles = 2x.

Assume the ratio of the dimensions of a regular tile is L:B --> area = L*B.
The dimensions of a jumbo tile is 3L:3B --> area = 9(L*B).

The area of regular tiles = 2x*L*B = 80.
The area of jumbo tiles = x*9(L*B) = 4.5(2x*L*B ) = 4.5*80 = 360.

Total area = 80 + 360 = 440.

Answer: D
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 16 Oct 2014, 11:31
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A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560

Let the area covered by regular tiles be x.
The ratio of number of jumbo tile to regular tiles = 1 / 2
The length and width of the jumbo tiles are 3 times that of the regular tiles.
Thus area of jumbo tiles = 9x/2

x = 80
9x/2 = 360

Thus total area = 360 + 80 = 440
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 30 Oct 2014, 09:40
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Bunuel wrote:
A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560



I am not sure how you guys are assuming it to be a square tile. Here is my sol

Regular 2x/3
length L
width w

lw*2x/3=80

lwx= 120--------<1>


Jumbo x/3
length : 3l
width :3w (since ratio of l:w of jumbo is same as regular)

area covered will be 3(lwx) = 3(120)= 360

total area =360+80= 440
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A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 10 Nov 2014, 09:03
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Here's a simple straight forward solution:

Let Number of tiles be \(T.\)

Jumbo tiles:-
no. of tiles: \(j\)
Length: \(Lj\)
Width: \(Wj\)

Regular tiles:-
no.: \(r\)
Length: \(Lr\)
Width: \(Wr\)

Given
a) \(j = T/3, r = 2T/3\)
b) \(Lj = 3Lr\) also \(Lj/Wj = Lr/Wr\) , from this follows -> \(Wj = 3 Wr\)


c) Area of Regular tiles = \(80\)
no. of regular tiles * Area of one Regular tile


\((r) * (Lr * Wr) = 80\) --------> \((Lr * Wr) = 80/r\)

Now to get Area of jumbo tiles, we will do little and only work:
no. of Jumbo tiles * Area of one Jumbo tile

\(Area Jumbo =\) \((j) * (Lj * Wj)\)

Using our deductions from " b) " and using values of 'j' and 'r' above, and placing it in the equation above -:
\(Area Jumbo =\) \(T/3* (3Lr * 3Wr)\)

->\(T/3* 9 (80/r)\)

place\(r = 2T/3\) above we get

360

Total area = 80 + 360 = 440

P.S. Tried to simplify it too much for everyone, otherwise its a 3 step process.
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 11 Jul 2015, 03:32
1
Bunuel wrote:
Bunuel wrote:
A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560


The number of jumbo tiles = x.
The number of regular tiles = 2x.

Assume the ratio of the dimensions of a regular tile is a:a --> area = a^2.
The dimensions of a jumbo tile is 3a:3a --> area = 9a^2.

The area of regular tiles = 2x*a^2 = 80.
The area of jumbo tiles = x*9a^2 = 4.5(2x*a^2 ) = 4.5*80 = 360.

Total area = 80 + 360 = 440.

Answer: D.


can anyone pls. tell me how the ratio of jumbo to regular tiles became 1:2?

Is it given? not able to figure out. :(
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 11 Jul 2015, 03:35
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robinpallickal wrote:
Bunuel wrote:
Bunuel wrote:
A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560


The number of jumbo tiles = x.
The number of regular tiles = 2x.

Assume the ratio of the dimensions of a regular tile is a:a --> area = a^2.
The dimensions of a jumbo tile is 3a:3a --> area = 9a^2.

The area of regular tiles = 2x*a^2 = 80.
The area of jumbo tiles = x*9a^2 = 4.5(2x*a^2 ) = 4.5*80 = 360.

Total area = 80 + 360 = 440.

Answer: D.


can anyone pls. tell me how the ratio of jumbo to regular tiles became 1:2?

Is it given? not able to figure out. :(


1/3 of the tiles are jumbo tiles and the remaining, so 2/3, are regular tile:

(jumbo)/(regular) = (1/3):(2/3) = 1:2.
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 11 Jul 2015, 03:51
Oh...missed that part....got it.

Thanks a lot Bunuel..
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A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post Updated on: 26 Jul 2016, 18:31
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ratio of r/j tile #=2:1
ratio of r/j tile size=(3*2):(9*6)=1:9
80+(r/2)(9*80/r)=440 square feet

Originally posted by gracie on 26 Jul 2016, 15:00.
Last edited by gracie on 26 Jul 2016, 18:31, edited 1 time in total.
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 26 Jul 2016, 18:06
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Bunuel wrote:
A rectangular wall is covered entirely with two kinds of decorative tiles: regular and jumbo. 1/3 of the tiles are jumbo tiles, which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles. If regular tiles cover 80 square feet of the wall, and no tiles overlap, what is the area of the entire wall?

A. 160
B. 240
C. 360
D. 440
E. 560


2/3rd of the tiles are regular tiles with an area of 80 sq ft

2/3*l*b= 80
l*b= 80*3/2= 120

Since the ratio of L/B = l/b, and L is 3l. Therefore, B=3b

1/3rd of the area covered by jumbo tiles:-
1/3 L*B= 1/3 *3l *3b= 3lb= 360

Total area= area covered by jumbo tiles+ area covered by regular tiles= 360+80= 440

D is the answer
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 03 Feb 2018, 16:31
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Hi All,

This question is loaded with little details. You'll need to take lots of notes, stay organized and do some basic math to get the correct answer.

We're told that there are two types of tiles: REGULAR and JUMBO.

Now come all of the little math facts you have to take note of (and deduce)…..
1) 1/3 of the tiles are JUMBO tiles.

Since 1/3 are JUMBO, the other 2/3 are REGULAR.
This means that the ratio of JUMBO:REGULAR = 1:2; in other words, there are twice as many REGULAR tiles as JUMBO tiles.

2) JUMBO tiles have a length that is 3 TIMES that of the REGULAR tiles AND has the same ratio of length:width as the REGULAR tiles.

We'll come back to this information in a moment.

3) REGULAR tiles cover 80 square-feet of the wall.

Since we weren't given any information about the dimensions of the REGULAR tiles, we can TEST VALUES. Let's say….

1 REGULAR tile = 1ft x 1ft.

Since REGULAR tiles cover 80 square feet, that means there are 80 REGULAR tiles.

Now, let's go back to the 2nd piece of information…..

JUMBO TILES have 3 times the length and the same ratio of length:width as REGULAR tiles.

1 REGULAR tile = 1ft x 1ft.
1 JUMBO tile = 3ft x 3ft.

NOW, let's go back to the 1st piece of information…..

The ratio of JUMBO:REGULAR = 1:2

80 REGULAR tiles (each 1ft x 1ft)
40 JUMBO tiles (each 3ft x 3ft).

Total area of the wall = 80(1)(1) + 40(3)(3) = 80 + 360 = 440 square feet

Final Answer:

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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 06 Feb 2018, 07:54
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Number of Tiles R:J = \(2/3\) : \(1/3\) ---- J=\(R/2\) ===> 1

Dimensions of the tiles R=LW & J=3L*3W=9LW ===> 2

Total area of the wall= (LW) * Total number of tiles
From 1&2: Total area of the wall= (LW)*[R+9\(R/2\)] = 80 + 9\((90/2)\)= 80+360= 440
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 22 Oct 2019, 22:55
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Please tell me where am I understanding the problem wrong...

Area covered by rectangle tile is 80, also that is 2/3 of the total area since 1/3 is covered by jumbo.
therefore 2/3 of total area is 80, which makes total area 120. But the total area OA is 440. Can anyone guide me where I am wrong.
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 23 Oct 2019, 10:16
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Hi mayur278,

I have provided a full explanation for this question (it's a few posts above your post). The issue with your approach is that you're confusing the number of TILES with the TOTAL AREA.

We know that the regular-sized tiles cover 80 square-feet; we do NOT know how much space the jumbo-sized tiles take up yet. We know that 1/3 of the tiles are jumbo-sized tiles (which means that 2/3 of the tiles are regular-sized tiles).

If you reread the prompt, you'll see a description of how much BIGGER the jumbo-sized tiles are compared to the regular-sized tiles. You need to consider that additional information to get the correct answer.

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A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 23 Oct 2019, 12:58
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EMPOWERgmatRichC wrote:
Hi All,

This question is loaded with little details. You'll need to take lots of notes, stay organized and do some basic math to get the correct answer.

We're told that there are two types of tiles: REGULAR and JUMBO.

Now come all of the little math facts you have to take note of (and deduce)…..
1) 1/3 of the tiles are JUMBO tiles.

Since 1/3 are JUMBO, the other 2/3 are REGULAR.
This means that the ratio of JUMBO:REGULAR = 1:2; in other words, there are twice as many REGULAR tiles as JUMBO tiles.

2) JUMBO tiles have a length that is 3 TIMES that of the REGULAR tiles AND has the same ratio of length:width as the REGULAR tiles.

We'll come back to this information in a moment.

3) REGULAR tiles cover 80 square-feet of the wall.

Since we weren't given any information about the dimensions of the REGULAR tiles, we can TEST VALUES. Let's say….

1 REGULAR tile = 1ft x 1ft.

Since REGULAR tiles cover 80 square feet, that means there are 80 REGULAR tiles.

Now, let's go back to the 2nd piece of information…..

JUMBO TILES have 3 times the length and the same ratio of length:width as REGULAR tiles.

1 REGULAR tile = 1ft x 1ft.
1 JUMBO tile = 3ft x 3ft.

NOW, let's go back to the 1st piece of information…..

The ratio of JUMBO:REGULAR = 1:2

80 REGULAR tiles (each 1ft x 1ft)
40 JUMBO tiles (each 3ft x 3ft).

Total area of the wall = 80(1)(1) + 40(3)(3) = 80 + 360 = 440 square feet

Final Answer:

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Rich



Hi EMPOWERgmatRichC

I'm confused on why we assume the tile is square not rectangular? How can I correctly interpret if to consider it rectangular :
2) JUMBO tiles have a length that is 3 TIMES that of the REGULAR tiles AND has the same ratio of length:width as the REGULAR tiles.


Thanks in advance for your help
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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 23 Oct 2019, 20:46
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Hi Mo2men,

To start, the prompt states that the WALL is a rectangle, but it does NOT state whether the tiles are squares or rectangles. Thus, we can choose either shape as long as the dimensions fit the rest of the information in the prompt.

For the sake of argument though, we can make the tiles rectangles...

IF.... a regular-sized tile has a length of 2 feet and a width of 1 foot, then its area is (2)(1) = 2 square-feet. Since these tiles cover 80 square-feet of wall, then we know that there are 80/2 = 40 regular-sized tiles. These 40 tiles represent 2/3 of the total TILES, so the remaining 1/3 of the tiles (re: 20 TILES) are jumbo-sized tiles.

A jumbo-sized tile has a length that is 3 times that of a regular-sized tile and a ratio of length-to-width that matches. Thus, each jumbo-sized tile has a length of 6 feet and a width of 3 feet and covers (6)(3) = 18 square-feet. The total area covered by the 20 jumbo-sized tiles is (20)(18) = 360 square-feet.

The 40 regular-tiles and 20 jumbo-sized tiles would cover... 80 + 360 = 440 square-feet... the same answer we get if we chose for the tiles to be squares.

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Re: A rectangular wall is covered entirely with two kinds of dec  [#permalink]

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New post 24 Oct 2019, 04:01
EMPOWERgmatRichC wrote:
Hi Mo2men,

To start, the prompt states that the WALL is a rectangle, but it does NOT state whether the tiles are squares or rectangles. Thus, we can choose either shape as long as the dimensions fit the rest of the information in the prompt.

For the sake of argument though, we can make the tiles rectangles...

IF.... a regular-sized tile has a length of 2 feet and a width of 1 foot, then its area is (2)(1) = 2 square-feet. Since these tiles cover 80 square-feet of wall, then we know that there are 80/2 = 40 regular-sized tiles. These 40 tiles represent 2/3 of the total TILES, so the remaining 1/3 of the tiles (re: 20 TILES) are jumbo-sized tiles.

A jumbo-sized tile has a length that is 3 times that of a regular-sized tile and a ratio of length-to-width that matches. Thus, each jumbo-sized tile has a length of 6 feet and a width of 3 feet and covers (6)(3) = 18 square-feet. The total area covered by the 20 jumbo-sized tiles is (20)(18) = 360 square-feet.

The 40 regular-tiles and 20 jumbo-sized tiles would cover... 80 + 360 = 440 square-feet... the same answer we get if we chose for the tiles to be squares.

GMAT assassins aren't born, they're made,
Rich


Thanks EMPOWERgmatRichC for taking the time for answering my question. Very informative.
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Re: A rectangular wall is covered entirely with two kinds of dec   [#permalink] 24 Oct 2019, 04:01
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