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If w, x and y are positive integers, is \(\frac{x+w}{y+w} > \frac{x}{y}\) ?
(1) \(y > x\)
(2) \(w > x\)
(1) \(y > x\)
(2) \(w > x\)
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
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given w,x,y are +ve integers
target is
\(\frac{x+w}{y+w} > \frac{x}{y}\)
(x+w)/(y+w) -x/y>0
xy+yw-xy-xw /(y(y+w) >0
w*(y-x)/ y*(y+w) >0
#1
\(y > x\)
sufficient to say that target is >0
irrespective of value of w
#2
w>x
if x=y then insufficient to target
if y>x then sufficient to target
insufficient
OPTION A
target is
\(\frac{x+w}{y+w} > \frac{x}{y}\)
(x+w)/(y+w) -x/y>0
xy+yw-xy-xw /(y(y+w) >0
w*(y-x)/ y*(y+w) >0
#1
\(y > x\)
sufficient to say that target is >0
irrespective of value of w
#2
w>x
if x=y then insufficient to target
if y>x then sufficient to target
insufficient
OPTION A
BrentGMATPrepNow
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Two ways to solve
Method 01: Cross multiplication
We need -
\(\frac{x+w}{y+w}\) > \(\frac{x}{y}\)
As all the numbers are positive we can cross multiply
\(y(x+w) > x(y+w)\)
\(yx+yw > xy+wx\)
We can subtract yx on both sides of the equation and divide by w
\(y > x\)
So basically we need to find if \(y > x\)
Statement 1 - Gives us what we need, hence sufficient.
Statement 2 - Doesn't provide the required information. Hence not sufficient.
Hence A is sufficient.
Method 02: Properties of Ratios & Fractions
The little algebra of method 1 can be avoided if we use the properties of ratios -
When x, y and w are positive -
\(\frac{x+w }{ y + w}\) > \(\frac{x}{y}\); when \(\frac{x}{y}\) is a proper fraction, i.e. x < y
\(\frac{x+w }{ y + w}\) < \(\frac{x}{y}\); when \(\frac{x}{y}\) is improper fraction, i.e. x > y
Hence to answer "Is \(\frac{x+w }{ y+w}\) > \(\frac{x}{y}\)", we need the information if \(\frac{x}{y}\) is proper fraction or is it improper fraction.
Statement 1 gives that information.
IMO A
_________________
Method 01: Cross multiplication
We need -
\(\frac{x+w}{y+w}\) > \(\frac{x}{y}\)
As all the numbers are positive we can cross multiply
\(y(x+w) > x(y+w)\)
\(yx+yw > xy+wx\)
We can subtract yx on both sides of the equation and divide by w
\(y > x\)
So basically we need to find if \(y > x\)
Statement 1 - Gives us what we need, hence sufficient.
Statement 2 - Doesn't provide the required information. Hence not sufficient.
Hence A is sufficient.
Method 02: Properties of Ratios & Fractions
The little algebra of method 1 can be avoided if we use the properties of ratios -
When x, y and w are positive -
\(\frac{x+w }{ y + w}\) > \(\frac{x}{y}\); when \(\frac{x}{y}\) is a proper fraction, i.e. x < y
\(\frac{x+w }{ y + w}\) < \(\frac{x}{y}\); when \(\frac{x}{y}\) is improper fraction, i.e. x > y
Hence to answer "Is \(\frac{x+w }{ y+w}\) > \(\frac{x}{y}\)", we need the information if \(\frac{x}{y}\) is proper fraction or is it improper fraction.
Statement 1 gives that information.
IMO A
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
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