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K-Mart and Wal-Mart bought different size square buildings for their

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Bunuel
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K-Mart and Wal-Mart bought different size square buildings for their [#permalink] New post  10 Jul 2020, 04:18
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K-Mart and Wal-Mart bought different size square buildings for their new stores. The length of K-mart's store is four times as long as the length of Wal-Mart's store. If the sum of the perimeters of both storefronts is 6,000 feet, what is the area of the K-Mart store (in square feet)?

A. 60,000
B. 72,000
C. 144,000
D. 720,000
E. 1,440,000
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Re: K-Mart and Wal-Mart bought different size square buildings for their [#permalink] New post  10 Jul 2020, 04:36
(E). length of kmart = l, length of walmart = L. So, l = 4L.

6000 = 4*l + 4*L = 4*4*L + 4*L = 20L

L = 300, l = 4*L = 4*300 = 1200.

Area od Kmart = l*l = 1200 * 1200 = 1,440,000.
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Re: K-Mart and Wal-Mart bought different size square buildings for their [#permalink] New post  10 Jul 2020, 05:29
Let side of K mart = K
Side of M mart = M
Therefore
4k+4m = 6000
4(K+M) = 6000
K+M = 1500

The length of K-mart's store is four times as long as the length of Wal-Mart's store.
4M = K
M+4M= 1500
5M = 1500.
M = 300
K = 300*4 = 1200

Area of K mart = 1200*1200 = 1,440,000

Answer is E

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Re: K-Mart and Wal-Mart bought different size square buildings for their [#permalink] New post  10 Jul 2020, 06:26
Bunuel wrote:
K-Mart and Wal-Mart bought different size square buildings for their new stores. The length of K-mart's store is four times as long as the length of Wal-Mart's store. If the sum of the perimeters of both storefronts is 6,000 feet, what is the area of the K-Mart store (in square feet)?

A. 60,000
B. 72,000
C. 144,000
D. 720,000
E. 1,440,000


Given that length of K-mart's store is four times as long as the length of Wal-Mart's store.

Let the length of K-mart's store be K and length of Wal-Mart's store be W

Means, K = 4W

As per the question stem, 4K = 16W

And, 4K+4W = 5K = 6000
Means, K = 1200

So, Area of K-Mart store = K^2 = 1200^2 = 1440000

Option : E
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K-Mart and Wal-Mart bought different size square buildings for their [#permalink] New post  10 Jul 2020, 06:43
Bunuel wrote:
K-Mart and Wal-Mart bought different size square buildings for their new stores. The length of K-mart's store is four times as long as the length of Wal-Mart's store. If the sum of the perimeters of both storefronts is 6,000 feet, what is the area of the K-Mart store (in square feet)?

A. 60,000
B. 72,000
C. 144,000
D. 720,000
E. 1,440,000


K-Mart and Wal-Mart bought different size square buildings
length of k-mart=4X
length of wal-mart=X
the sum of the perimeters of both storefronts is 6,000 feet ,
means 4X+4X+4X+4X+X+X+X+X=6000
20X=6000
X=300 feet

Length of k-mart = 4X = 4*300 = 1200 feet
AREA of k-mart = 1200^2=1,440,000
ANS:E
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warrior1991
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Re: K-Mart and Wal-Mart bought different size square buildings for their [#permalink] New post  10 Jul 2020, 20:24
1) Both are square builidngs

Let each side of a M mart building be a

Kmart's length is 4 times-->4a

Given sum of parameter=6000

Kmart's sum of perimeter=4*4a(Since it is a square building so all sides are equal)

Wmart's sum of perimeter=4a(Since it is a square building so all sides are equal)

Total sum=6000
16a+4a=6000
a=300

We have been asked to find the area of Kmart store
Each side is 4a=1200

Area=Side^2
Area=1200*1200--->1440000

Therefore E
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Re: K-Mart and Wal-Mart bought different size square buildings for their [#permalink]
10 Jul 2020, 20:24
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