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# In the figure, the area of rectangle EFGH is 3 units greater than the

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In the figure, the area of rectangle EFGH is 3 units greater than the  [#permalink]

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Updated on: 15 Jul 2019, 04:24
00:00

Difficulty:

35% (medium)

Question Stats:

78% (02:16) correct 22% (02:38) wrong based on 37 sessions

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In the figure, the area of rectangle EFGH is 3 units greater than the area of rectangle ABCD. What is the value of ab if a + b = 8?

(A) 9
(B) 12
(C) 15
(D) 18
(E) 21

Source: Nova GMAT
Difficulty Level: 600

Attachment:

recs.jpg [ 15.7 KiB | Viewed 750 times ]

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Originally posted by Bunuel on 09 Oct 2018, 23:41.
Last edited by SajjadAhmad on 15 Jul 2019, 04:24, edited 1 time in total.
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Re: In the figure, the area of rectangle EFGH is 3 units greater than the  [#permalink]

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10 Oct 2018, 00:02
area of rectangle EFGH = 3 + area of rectangle ABCD

(x+a)(x+b) = 3+(x+2)(x+6)

$$x^2$$+x(a+b)+ab = $$x^2$$+8x+15

Given a+b = 8.

Solving gives ab = 15.

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Re: In the figure, the area of rectangle EFGH is 3 units greater than the  [#permalink]

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10 Oct 2018, 04:57
1
Bunuel wrote:

In the figure, the area of rectangle EFGH is 3 units greater than the area of rectangle ABCD. What is the value of ab if a + b = 8?

(A) 9
(B) 12
(C) 15
(D) 18
(E) 21

Attachment:
recs.jpg

$$(x+a) * (x+b) = 3 + (x+6)*(x+2)$$

$$x^2 + xb + xa + ab = 3 + x^2 + 2x + 6x + 12$$

$$x^2$$ cancel from both side.

take x as a common factor

$$x (a+b) + ab = 3 + x (2+6) + 12$$. we are given a+b = 8
$$8x + ab = 15 +8x$$

$$ab = 15$$

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Re: In the figure, the area of rectangle EFGH is 3 units greater than the  [#permalink]

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10 Oct 2018, 07:47
Bunuel wrote:

In the figure, the area of rectangle EFGH is 3 units greater than the area of rectangle ABCD. What is the value of ab if a + b = 8?

(A) 9
(B) 12
(C) 15
(D) 18
(E) 21

Attachment:
recs.jpg

$$(x+a)(x+b) - (x+2)(x+6) = 3$$

Or, $$(ab +ax + bx +x^2) - (x^2 +8x + 12) = 3$$

Or, $$ab + x (a + b) - 8x - 12 = 3$$

Or, $$ab + 8x - 8x - 12 = 3$$

Or, $$ab = 15$$, Answer must be (C)
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Re: In the figure, the area of rectangle EFGH is 3 units greater than the  [#permalink]

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11 Oct 2018, 01:00

Formula used: Area(Rectangle) = Length * Breadth.
Given: Area(Rectangle EFGH) is 3 units greater than the Area(Rectangle ABCD)

Area(Rectangle ABCD) = $$(x+2)(x+6) = x^2 + 8x + 12$$ units.
Area(Rectangle EFGH) = $$(x+a)(x+b) = x^2 + (a+b)x + ab$$ units.

Therefore, the area of the rectangle ABCD is $$(x^2 + 8x + 12) + 3 = x^2 + 8x + 15$$ and value of ab is 15(Option C)
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Re: In the figure, the area of rectangle EFGH is 3 units greater than the  [#permalink]

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13 Oct 2018, 18:13
Bunuel wrote:

In the figure, the area of rectangle EFGH is 3 units greater than the area of rectangle ABCD. What is the value of ab if a + b = 8?

(A) 9
(B) 12
(C) 15
(D) 18
(E) 21

Attachment:
recs.jpg

We can create the equation:

3 + (x + 2)(x + 6) = (x + a)(x + b)

3 + x^2 + 8x + 12 = x^2 + ax + bx + ab

15 + 8x = ax + bx + ab

15 + 8x = x(a + b) + ab

Since a + b = 8, we have:

15 + 8x = 8x + ab

15 = ab

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Re: In the figure, the area of rectangle EFGH is 3 units greater than the   [#permalink] 13 Oct 2018, 18:13
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