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# How many squares with sides 1/2 inch long are needed to cover a rectan

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Math Expert
Joined: 02 Sep 2009
Posts: 56303
How many squares with sides 1/2 inch long are needed to cover a rectan  [#permalink]

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02 Nov 2017, 23:08
1
4
00:00

Difficulty:

45% (medium)

Question Stats:

69% (01:56) correct 31% (01:52) wrong based on 159 sessions

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How many squares with sides 1/2 inch long are needed to cover a rectangle that is 6 feet long and 4 feet wide? (1 foot = 12 inches)

(A) 24
(B) 96
(C) 3,456
(D) 13,824
(E) 14,266

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Re: How many squares with sides 1/2 inch long are needed to cover a rectan  [#permalink]

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03 Nov 2017, 00:15
Bunuel wrote:
How many squares with sides 1/2 inch long are needed to cover a rectangle that is 4 feet long and 6 feet wide? (1 foot = 12 inches)

(A) 24
(B) 96
(C) 3,456
(D) 13,824
(E) 14,266

The first step is to convert the rectangle to inches.
Since the rectangle is 4 ft by 6 ft, the dimensions of the rectangle will be 48 inch by 72 inch.
We have been asked to find out how many square of side 0.5 inch are needed to cover this rectangle.

The area covered by the rectangle is 48*72 inch and we will be needing x squares to cover this rectangle.

$$0.5*0.5*x = 48*72$$

$$x = \frac{48}{0.5} * \frac{72}{0.5} = 96 * 144 = 13824$$(Option D)
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Re: How many squares with sides 1/2 inch long are needed to cover a rectan  [#permalink]

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04 Nov 2017, 01:36
1
Bunuel wrote:
How many squares with sides 1/2 inch long are needed to cover a rectangle that is 4 feet long and 6 feet wide? (1 foot = 12 inches)

(A) 24
(B) 96
(C) 3,456
(D) 13,824
(E) 14,266

Hi Bunuel,

As per my understanding, length is always greater than width.

Is it OK to say length is 4ft and width is 6ft? This implies that width is greater than length.
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Joined: 02 Sep 2009
Posts: 56303
Re: How many squares with sides 1/2 inch long are needed to cover a rectan  [#permalink]

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04 Nov 2017, 01:38
Jabjagotabhisavera wrote:
Bunuel wrote:
How many squares with sides 1/2 inch long are needed to cover a rectangle that is 4 feet long and 6 feet wide? (1 foot = 12 inches)

(A) 24
(B) 96
(C) 3,456
(D) 13,824
(E) 14,266

Hi Bunuel,

As per my understanding, length is always greater than width.

Is it OK to say length is 4ft and width is 6ft? This implies that width is greater than length.

You are right length > width. Edited.
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Re: How many squares with sides 1/2 inch long are needed to cover a rectan  [#permalink]

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24 Jan 2019, 14:04
Bunuel wrote:
How many squares with sides 1/2 inch long are needed to cover a rectangle that is 6 feet long and 4 feet wide? (1 foot = 12 inches)

(A) 24
(B) 96
(C) 3,456
(D) 13,824
(E) 14,266

Total area = 24 ft^2 converted to inches = 3456 inch^2

Area of square is 1/4 inches

Thus you can hold 3456*4 squares or 13,824 squares
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Joined: 12 Sep 2017
Posts: 297
How many squares with sides 1/2 inch long are needed to cover a rectan  [#permalink]

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20 Feb 2019, 19:44
Step one

Convert the units

1 foot = 12 inches
6 feet = 6 * 12 = 132 inches
4feet = 4 * 12 = 48 inches

Step two

Create the rectangle

W = $$48/(1/2)= 96$$
L = $$132/(1/2)= 264$$

Third step

Just multiply the units digits

6*4 = 24 ... 13,824

(A) 24 ... discard ---> too small
(B) 96
(C) 3,456
(D) 13,824
(E) 14,266

D
How many squares with sides 1/2 inch long are needed to cover a rectan   [#permalink] 20 Feb 2019, 19:44
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