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Math Expert V
Joined: 02 Sep 2009
Posts: 58434
In the figure above, all angles are right angles. If the lengths a, b  [#permalink]

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16 00:00

Difficulty:   75% (hard)

Question Stats: 51% (01:43) correct 49% (02:07) wrong based on 307 sessions

HideShow timer Statistics In the figure above, all angles are right angles. If the lengths a, b, u, and v are integers, what is the perimeter of the six-sided figure?

(1) ab = 77

(2) u/v = 3/5

Attachment: 2957-1.gif [ 843 Bytes | Viewed 7787 times ]

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Marshall & McDonough Moderator D
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Re: In the figure above, all angles are right angles. If the lengths a, b  [#permalink]

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1
1
Line segment u and v can be moved to form a rectangle.
Perimeter of a rectangle = 2(l + w)

St1: ab = 77
11 * 7 = 77 --> Perimeter = 2(11 + 7) = 36
77 * 1 = 77 --> Perimeter = 2(77 + 1) = 156 --> This case can be ruled out because u and v are integers.
Sufficient.

St2: u/v = 3/5 --> 0.6v. Since u and v are integers, u can assume a minimum value of 3 and v can assume a minimum value of 5.
a = u + some value = 3 + some value
b = v + some value = 5 + some value
Perimeter cannot be calculated.
Not Sufficient.

Intern  Joined: 26 Nov 2015
Posts: 28
Re: In the figure above, all angles are right angles. If the lengths a, b  [#permalink]

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Statement 1

Firstly until and unless all angles are right angles, it doesn't matter the value of u & v, By the same amount of perimeter deduced from the piece due to u-v occupation - it adds that pieces by the same

So essentially the perimeter of the area will remain = 2 (a+b)
as area is 77 = a * b (as given a and b can be integer)
so two case a= 11,b=7 or a =7, b=11

Therefore the perimeter in both case 2(11+7) & 2(7+11) is the same, i.e. 36

Statement 2

Here you know Ration of u/v, it is not sufficient as we don't know the factor in comma, it's the ration of primes with no factor involved

Plus knowing u & v can't tell you about the perimeter 2 (a+b)

Hence Option A
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Re: In the figure above, all angles are right angles. If the lengths a, b  [#permalink]

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

In the figure above, all angles are right angles. If the lengths a, b, u, and v are integers, what is the perimeter of the six-sided figure?

(1) ab = 77

(2) u/v = 3/5

When you modify the original condition and the question, the length of the perimeter is 2(a+b). That is, you only need to figure out a+b. a and b are integers and (a,b)=(7,11),(11,7) from 1) ab=77. Also, a+b=18 is always derived and sufficient, which makes A the answer.
The reason (a,b)=(1,77) is not valid is if a=1, you cannot subtract u, which makes 1 impossible.
Thus, A is the answer.

 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: In the figure above, all angles are right angles. If the lengths a, b  [#permalink]

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@bunnel can you explain this one ??
I understood the concept of a and b . But when we calculate the perimeter i am getting some variables either u or v
I could only make out the OA if a-u=b
can you please explain ?
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GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 In the figure above, all angles are right angles. If the lengths a, b  [#permalink]

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Bunuel wrote: In the figure above, all angles are right angles. If the lengths a, b, u, and v are integers, what is the perimeter of the six-sided figure?

(1) ab = 77

(2) u/v = 3/5

Attachment:
The attachment 2957-1.gif is no longer available

Official Solution (Credit: Manhattan Prep)

Before tackling the statements, note that the desired perimeter is the same as that of a rectangle measuring a by b units. This fact can be seen perhaps most easily by completing the rectangle:

See attachment!

The large rectangle can be formed by relocating the segments with lengths u and v to the locations of the dotted lines, without changing anything else in the figure, so the two perimeters are the same.

The desired perimeter is thus the same as the perimeter of the large rectangle, i.e., 2a + 2b.

This expression can also be found by filling in the lengths of the two missing sides. By subtraction, the unlabeled horizontal side has length b – v, and the unlabeled vertical side has length a – u. Therefore, the perimeter of the figure is a + b + u + v + (b – v) + (a – u) = 2a + 2b.

(1) SUFFICIENT: The lengths a and b are positive integers whose product is 77, so a = 11 and b = 7, or else a = 7 and b = 11. (Note that a and b cannot be 1 and 77. If a = 1 then u cannot be an integer; likewise, if b = 1 then v cannot be an integer.)

In either case, the desired perimeter is 2a + 2b = 2(11) + 2(7) = 36.

(2) INSUFFICIENT: This statement gives no information about the values of a or b, so the desired perimeter cannot be found.

The correct answer is A.
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If you found my post useful, KUDOS are much appreciated. Giving Kudos is a great way to thank and motivate contributors, without costing you anything. In the figure above, all angles are right angles. If the lengths a, b   [#permalink] 17 Aug 2018, 18:47
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