A framed picture is shown above. The frame, shown shaded, is 6 inches

hi,

let the length be x, so dimensions of bigger rectangle = 6 and x...
Also let the width around be y, so dimensions are 6-2y and x-2y....

For the Q asked, we require value of x and y....

lets see the statements-

(1) The area of the shaded region is 24 square inches.
Area of shaded region = \(6*x - (6-2y)*(x-2y) = 6x - 6x+12y+2xy-4y^2 = 12y+2xy-4y^2 = 24\)
We can't solve for two variables through one equation...
Insuff

(2) The frame is 8 inches tall.
we know the dimensions of bigger frame...
nothing about the inner picture
Insuff

Combined-
12y+2xy-4y^2 = 24.......... x=8
so \(12y+16y-4y^2 = 24......................4y^2-28y+24 = 0...................y^2-7y+6 = 0.........(y-1)(y-6) = 0..\)
either y=1 or y=6.......
But if y=6, sides will become -ive, 6-2y = 6-2*6 = -6, so y=1.....
sides are 6-1*2 = 4 and 8-2*1=6.......
Suff

C
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For x = height of the shaded region, the area of the shaded region can be expressed as:


24 = 6x - (Area of viewable portion V) => 24 = 6x - V

we have two unknowns so (1) and (2) they give one of the two variables x, V

Combining both we get the right answer hence C

Any chance to resolve it without going into equations? If the smaller quadrilateral with the area of 24 is inside of the larger quadrilateral with the area of 48 and the border is uniform that these two quadrilaterals should have the same proportion? or something like that? the pair of 6 and 4 immediately comes to mind... Is it a good tactic here just to realize that using facts A and B must give us a solution and move to the next problem?

Hi Erjan,

I found this explanation helpful- https://www.gmatquantum.com/gmat-quantit ... view-2017/

Check it out as it might help you too.

Aiena.

Hi chetan2u niks18 amanvermagmat gmatbusters






For the Q asked, we require value of x and y....




Do I need to perform the said step by opening the brackets? Can I halt by seeing: A quadratic equation with two unknowns , clearly insufficient .



We know x, but have no info about another variable. Insuff

Combining both statements, I have a quadratic equation (which opens the possibility of even a negative
value as a root of equation, but a side can not be negative) and I do have an unique value of x from St 2.
Clearly suff, Let me know if my approach is valid.
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Reply: Sometimes, the terms cancel out and equation can be solved. But yes careful examination can show that it is unsolvable. But under exam pressure sometimes, one tends to miss small aspects which can lead to mistake, So it is better to open the brackets. Believe me , it will not take more than 5 secs. One need to improve the calculation speed.



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Let the dimensions of the viewable portion of the picture be L and B. Let x be the uniform width of the border.
So dimensions of the frame would be L+2x and B+2x.
We need to find L and B.
Also we know that L+2x= 6

1) Area of the shaded region= Area of the frame- Area of the viewable portion
(L+2x)* (B+2x) - L*B= 24
6* (B+2x) - L*B= 24
Not sufficient to find L and B.

2) Height of the frame is 8 inches.
So B+2x= 8
Area of frame is 6*8= 48
Not enough information to find x.

1&2) Substituting information from statement 2 in 1, we get-
6*8- L*B= 24
L*B= 24

Also L+2x=6
2x= 6-L
Similarly, 2x= 8-B
So 6-L= 8-B
B-L=2
and L*B= 24
We have two unique equations for two different variables.
This is enough to find L and B.
Sufficient.

Answer: C

You get a quadratic equation at the end which gives 2 unique values. You need to test for both. It's just luck that in this case one value gives garbage result and the other doesn't. It was either C or E.

Agreed. There could have been a chance where both values for the quadratic would have been positive.
But since this is a rectangle with fixed dimensions, we cannot have multiple possible valid values, correct? A rectangle has to have a set of unique values. I reasoned that even if we have such a scenario, we would have to reject one of the values because it would violate some condition. GMATBusters Could you please suggest if such an approach where we don't solve for the quadratic particularly for such questions is correct?

The Answer of this DS is a value that has to come from solving the equation ( Many a time Answer for DS question is in Yes or No ).
To know the dimension We need.
1. Length of Frame
2. dimension of border

Now lets check the options:

A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture?

(1) The area of the shaded region is 24 square inches.
(2) The frame is 8 inches tall.

1. The are of the shaded reason does not give me unique value of Length or dimension of border

2. Now I have the length of frame and width. i.e. 8 and 6 , but still i am missing dimension of border

1+2 , Now I can subtract total area from 2 to The area of the shaded region from 1.
Now the dimension will be 6 x 4 , As the only possibility to get 24 with condition length is less than 8 and width is less then 6 and the reduction have to be same.

Solution:

We need to determine the dimensions of the viewable portion of the picture. If we let y be the length of the frame and x be the width of the uniform border of the frame and A be the area of the viewable portion of the picture, we can create the equation:

A = (6 - 2x)(y - 2x)

Therefore, to determine the dimensions of the viewable portion of the picture, we either need to know the values of x and y or the value of A and y (so that we can determine the value of x).

Statement One Alone:

This does not allow us to determine the value of A, x, or y. Statement one alone is not sufficient.

Statement Two Alone:

This means y = 8. However, this does not allow us to determine the value of A or x. Statement two alone is not sufficient.

Statements One and Two Together:

Notice that A, the area of the viewable portion of the picture, is the difference between the overall area and the area of the frame (i.e., the shaded region). The overall area is 6 x 8 = 48 and since the area of the frame is 24, A = 48 - 24 = 24. Therefore, we rewrite the equation from our stem analysis:

24 = (6 - 2x)(8 - 2x)

We see that we can determine the value of x. Therefore, both statements together are sufficient.

(Note: since the equation is quadratic, there are two solutions to the equation, i.e., two values of x. However, only one value of x makes sense since it’s really a geometry problem, i.e., only that value of x will make the dimensions positive (the other value of x will make the dimensions negative, which is not possible).)

Answer: C
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Why you have assumed that it's a rectangle . It is not given in the question , that it's a rectangle