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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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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
[Reveal] Spoiler: OA

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

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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.

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

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04 Feb 2014, 10:17
Hi ,

I would request to you Bunuel, please do not post the answer on the same dat you post the questions.

Let others try to solve the questions otherwise it is a human tendency after first attempt a person look for solutions. If he finds it he never give the problem a thought.
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Re: A rectangular wall is covered entirely with two kinds of dec [#permalink]

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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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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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16 Apr 2014, 14:25
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.

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.

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

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16 Oct 2014, 11:31
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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30 Oct 2014, 09:40
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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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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11 Jul 2015, 03:32
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.

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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11 Jul 2015, 03:35
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.

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

Thanks a lot Bunuel..

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

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

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26 Jul 2016, 15:00
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

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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26 Jul 2016, 18:06
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

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

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Re: A rectangular wall is covered entirely with two kinds of dec   [#permalink] 10 Aug 2017, 01:25
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# A rectangular wall is covered entirely with two kinds of dec

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