The length of rectangle B is x percent less than the length of rectang

Question

FRESH GMAT CLUB TESTS QUESTION

The length of rectangle B is x percent less than the length of rectangle A, and the width of rectangle B is y percent, greater than the width of rectangle A (where both x and y are greater than 0 and less than 100). Is the area of rectangle A greater than the area of rectangle B?

(1) x = y
(2) x + y = 20

M36-49

CareerGeek · Answer

Let the length of rectangle A = a
So, length of rectangle B = a(1-x/100)

Let the breadth of rectangle A = b
So, breadth of rectangle B = b(1+y/100)

Area of A = a*b
Area of B = a(1-x/100)*b(1+y/100)
 = ab(1 + (y-x)/100 - xy/10000)

(1) x = y

Area of B = ab(1 + (x-x)/100 - x^2/10000)
 = ab(1 - x^2/10000)

Since, 0<x<100. 0<x^2<10000
So Area of B is less than A

Sufficient

(2) x+y = 20. 
Nothing can be said.
Eg: If x=19,y=1
Area of B = ab(0.81)(1.01) < ab
If x=1,y=19
Area of B = ab(0.99)(1.19) > ab

Insufficient

So, option A

Posted from my mobile device

Archit3110 · Answer

#1
let l of A = 10 and w= 5
area = 50
x=y ; 2=2
so B ;l = 8 and w ; 6
area of b ; 48
sufficient
#2
x+y=20 now value of x & y would vary over ranges 1,19 , 2,18 , 3,17 so the relation of area of A wont be > than B always.. 
IMO A

exc4libur · Answer

Rectangle A: L=a, W=b; Rectangle B: L=c=a(1-x/100), W=d=b(1+y/100) Question:ab>a(1-x/100)b(1+y/100)?…ab>(ab/100^2)(100-x)(100+y)… Question:100^2>(100-x)(100+y)? (1) x = y sufic. Question: 100^2>(100-x)(100+x)…100^2>(100^2-x^2)…0>-x^2…00? 0<(x,y)<100…x>0?…Answer:yes (2) x + y = 20 insufic. if:x=y…Answer:yes if:x≠y…Answer:yes/no Answer (A)

Bunuel · Answer

Official Solution:

The length of rectangle B is  x  percent less than the length of rectangle A, and the width of rectangle B is  y  percent greater than the width of rectangle A (where both  x  and  y  are greater than 0 and less than 100). Is the area of rectangle A greater than the area of rectangle B? 
 
The length of rectangle B is  x  percent less than the length of rectangle A (say  m ):  length_B=m(1-\frac{x}{100}) 
 
The width of rectangle B is  y  percent greater than the width of rectangle A (say  n ):  width_B=n(1+\frac{y}{100}) 
 
The question: is  mn > m(1-\frac{x}{100})*n(1+\frac{y}{100}) ?
 
Reduce by  mn : is  1 > (1-\frac{x}{100})(1+\frac{y}{100}) ? 
 
Simplify: is  100^2 > (100-x)(100+y) ?
 
(1)  x = y 
 
The question becomes: is  100^2 > (100-x)(100+x) ?
 
Is  100^2 > (100-x)(100+x) ?
 
Is  100^2 > 100^2-x^2 ?
 
Is  x^2>0 ?
 
Since given that  x > 0 , then the answer to this question is YES. Sufficient.
 
(2)  x + y = 20 
 
Test extreme cases: 
 
If  x  very close to 0 and  y  is very close to 20, then  (100-x)(100+y) \approx 100*120 > 100^2 . So, in this case we'd have an NO answer to the question.
 
If  x  very close to 20 and  y  is very close to 0, then  (100-x)(100+y) \approx 80*100 < 100^2 . So, in this case we'd have an YES answer to the question.
 
Not sufficient.

Answer: A

bumpbot · Answer

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The length of rectangle B is x percent less than the length of rectang

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