Patricia purchased x meters of fencing. She originally intended to use

Question

Patricia purchased x meters of fencing. She originally intended to use all of the fencing to enclose a square region, but later decided to use all of the fencing to enclose a rectangular region with length y meters greater than its width. In square meters, what is the positive difference between the area of the square region and the area of the rectangular region?

(1) xy = 256
(2) y = 4

DS37571.01

(1) xy = 256
(2) y = 4

DS37571.01

shridhar786 · Accepted Answer

length of the fencing = x
originally intended to use all of the fencings to enclose a square region
let the side of square = a
perimeter of square = 4a
4a = x
a =  \frac{x}{4} --(side of square)

but later decided to use all of the fencings to enclose a rectangular region with length y meters greater than its width
let the width of rectangle be = t
length of rectangle = t+y
perimeter of rectangle = 2(t+t+y)
2(t+t+y) = x
t =  \frac{x-2y}{4} --width of rectangle
lenght of rectangle =  \frac{x+2y}{4}

positive difference between the area of the square region and the area of the rectangular region =  \frac{x^2}{16} - \frac{x-2y}{4} * \frac{x+2y}{4} 
=  \frac{x^2}{16} - \frac{(x^2-4y^2)}{16} 
= \frac{4y^2}{16}

--basically we need to know the value of y

STATEMENT (1)-xy = 256
If x = 32 then y = 8 {x must be greater than 2y since x = 2(t+t+y)}
if x = 64 then y = 4

we cant find the definite value of y 
so, we can't find the definite positive difference of area
INSUFFICIENT

STATEMENT (2)-y = 4
The positive difference of area =  \frac{4y^2}{16}  = 4
SUFFICIENT

B is the correct answer

BelalHossain046 · Answer

Decoding the prompt

Rectangle:   
Width=a & Length=a+y, 
 So, Area= a(a+y)=a^2+ay. 
Perimeter= 2(a+a+y)=2(2a+y)=4a+2y

Square  
Perimeter= 4a+2y
Each side= (4a+2y)/4=a+y/2.
 Area = (a+y/2)^2=a^2+2*a*y/2+(y/2)^2= a^2+a*y+(y/2)^2

Positive Difference in Areas between Square and Rectangle=  a^2+a*y+(y/2)^2  -  (a^2+ay) =(y/2)^2

So, if we know the value of y, We can easily find out the vale of Positive Difference in Areas between Square and Rectangle i.e. (y/2)^2.

The question ask: y=?

Only statement 2 is sufficient.

BelalHossain046 · Answer

One important concept: Since we are talking about the ENCLOSING, we will bother about the PERIMETER. 
So, the same length of rope/fencing will be used as perimeters of SQUARE and the rectangle.

BelalHossain046 · Answer

St-1: 256 can be factorized many ways. So, Not sufficient

St-2:
In case of Square, Area = S*S
In case of Rectangle, Area = (S-2)*(S+2)
Now, difference = S*S - (S-2)*(S+2) = 2^2 = 4
So, sufficient.
Then ans. is B

zmtalha · Answer

Fence is  x . So, for the square, each side is  \frac{x}{4} . So area of the square is  \frac{x^2}{16} 
(we can cut  \frac{y}{2}  meter from any two sides of the  \frac{x}{4}  square and add  \frac{y}{2}  meter to the other sides to form rectangle.)
For the rectangle,
width =  \frac{x}{4 } - \frac{y}{2} .
Then, length =  \frac{x}{4 }+\frac{y}{2} 
 area= length*width=   (\frac{x}{4}+\frac{y}{2})(\frac{x}{4}-\frac{y}{2}) = \frac{x^2}{16} - \frac{y^2}{4}; Since, {a^2-b^2 =(a+b)(a-b)} 
Then the difference is  |\frac{x^2}{16} - \frac{x^2}{16} + \frac{y^2}{4} | 
which is  \frac{y^2}{4} 
so, for any  x  and  y  value, the difference of the area will always be  \frac{y^2}{4} .
we need only  y  value to know the difference.
From statement 1,  xy=256  : we need to know by heart that this statement will never give any unique value of  y  and  x .
Statement 2 is giving exactly what we need.  y=4 .
so, the difference in area is 4.
Answer: B

CrackverbalGMAT · Answer

The x metres of fencing that Patricia purchased represents the perimeter of the geometrical objects mentioned. From this information, we need to find the areas and hence the difference between the areas.

If we assume the side of the square as ‘a’, the perimeter would be 4a and we can say x = 4a.

If we assume the width of the rectangle as b, its length will be b+y. Therefore, the perimeter would be 4b + 2y and so we can say x = 4b + 2y.

Equating the two perimeters (since both represent x) and simplifying, we get,
a = b +  \frac{y}{2} . 
Therefore, area of the square =  a^2  =  (b + \frac{y}{2})^2  =  b^2  +  y^2  + by.

Area of the rectangle = b ( b+y) =  b^2  + by.

Clearly,  the area of the square is bigger  and  hence the positive difference in the areas  = Area of the square – Area of the rectangle =  y^2 . This way, we now know that we need the value of y to answer the question.
 
 The value of y is given in statement II. Therefore, statement II alone is sufficient .

Statement I alone is insufficient since it does not give us a unique value for y and hence doesn’t give us a unique answer to the question.
 The correct answer option is B .

At the start, I focussed on breaking down the question statement and used the question data to construct an equation which told me what to look for in the statements. You need to do this on all word problems where a lot of data is given as part of the question. Without doing this, if you start analyzing the statements directly, you’d be missing out on the most important data in a DS question.

Hope that helps!

suminha · Answer

Information given is; 4a = 2w+2w(w+y)

What do we want to know; a^2 — w(w+y), which is the difference between the area of the square region and the area of the rectangular region.

Posted from my mobile device

Pritishd · Answer

1. Let the side of the square be  b . Now, as the fencing was bought to enclose the square hence we can safely conclude that  4b = x  i.e. perimeter of the square is equal to  x

2. Let the width of the rectangle be  a  and length be  y + a . Now, as the fencing was bought to enclose the rectangle hence we can safely conclude that  4a + 2y = x  i.e. perimeter of the rectangle is equal to  x

3. The difference between the areas:
 b^2 - a(y+a) = b^2 - a^2 - ay

4. Replacing  b  with  \frac{x}{4}  and  a  with  \frac{x-2y}{4}

(\frac{x}{4})^2 - (\frac{x-2y}{4})^2 - (\frac{x-2y}{4})*y

\frac{x^2}{16} - (\frac{x^2 - 4xy + 4y^2}{16}) - (\frac{xy - 2y^2}{4})

Equalizing the denominators to work out the fractions:- 
 \frac{x^2}{16} - (\frac{x^2 - 4xy + 4y^2}{16}) - (\frac{4xy - 8y^2}{16})

\frac{x^2 - x^2 - 4y^2 + 4xy - 4xy + 8y^2}{16}

Cancelling out terms:- 
 \frac{x^2 - x^2 + 4xy - 4xy + 8y^2 - 4y^2}{16}   =   \frac{4y^2}{16}

Multiplying the numerator and denominator with 4:- 
 \frac{y^2}{4}

We need the value of  y  and option B provides it. Ans.  B

shridhar786 · Answer

hi guys 
I edited the solution to the question as the cases I took to prove statement 1 was wrong

thanks,  ZoltanBP  for pointing out the mistake

Bambi2021 · Answer

I get the same result as you, y^2, but I see that many in this thread get y^2/4.

Here is my reasoning:

Side of the square = x/4

Width of rectangle = (x/4)-y

Length of rectangle = (x/4)+y

What we add from a square side to the length of the rectangle will be equal to what we subtract from a square side to get the width of the rectangle.

If x = 100 m, then sides of square = 25 m.
We still have 100 m of fencing for the rectangle. If the length is 25+15 m, then the width will be 25-15 m.

Area of rectangle = ((x/4)+y) * ((x/4)-y) = (a+b)(a-b) = x^2/16 - y^2

The positive difference will be y^2 = 16, not y^2/4.

GmatKnightTutor · Answer

GMAT Data Sufficiency - Squares and Rectangles - How are they connected sometimes?

https://www.youtube.com/watch?v=UuQXfmvOlxM

bumpbot · Answer

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Patricia purchased x meters of fencing. She originally intended to use

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GMAT Data Sufficiency (DS) Questions

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