X is 45.6% larger than Y. What is the value of Y?
(1) X is 30% larger than Z. Z is 12% larger than Y.
(2) Z is 15 more than Y.
The OA is C.
How can I use both statements to get an answer? Experts, may you give me some help? Thanks. :roll:
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X is 45.6% larger than Y. What is the value of Y?
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Hi Vincen,
We're told that X is 45.6% larger than Y. We're asked for the value of Y.
1) X is 30% larger than Z. Z is 12% larger than Y.
Fact 1 gives us percentages that relate the three variables (X, Y and Z) to one another, but we do not know the value of any of the variables, so there's no way to determine the exact value of Y.
Fact 1 is INSUFFICIENT
(2) Z is 15 more than Y.
Fact 2 provides a distinct 'value relationship' between Z and Y, but since Z can be any value at this point, Y can also be any value.
Fact 2 is INSUFFICIENT
Combined, we have the information to create the following equations:
X = 1.456(Y)
X = 1.3(Z)
Z = 1.12(Y)
Z = Y+15
We can set the last two equations equal to one another...
1.12(Y) = Y+15
.12(Y) = 15
Y = 15/.12
At this point, we can stop working - there is clearly a distinct value for Y, so we have the answer to the given question.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that X is 45.6% larger than Y. We're asked for the value of Y.
1) X is 30% larger than Z. Z is 12% larger than Y.
Fact 1 gives us percentages that relate the three variables (X, Y and Z) to one another, but we do not know the value of any of the variables, so there's no way to determine the exact value of Y.
Fact 1 is INSUFFICIENT
(2) Z is 15 more than Y.
Fact 2 provides a distinct 'value relationship' between Z and Y, but since Z can be any value at this point, Y can also be any value.
Fact 2 is INSUFFICIENT
Combined, we have the information to create the following equations:
X = 1.456(Y)
X = 1.3(Z)
Z = 1.12(Y)
Z = Y+15
We can set the last two equations equal to one another...
1.12(Y) = Y+15
.12(Y) = 15
Y = 15/.12
At this point, we can stop working - there is clearly a distinct value for Y, so we have the answer to the given question.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Matt@VeritasPrep
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From the stem we've got x = 1.46y. We need one other equation (technically one other independent linear equation) in x and y to solve.
S1::
x = 1.3z and z = 1.12y
This takes us in the wrong direction! We wanted another equation in x and y, but we got a new variable, z. That said, we now have two equations in three variables, so if we could get one more equation we'd have three equations in three variables and be good to go. Not sufficient, but possibly helpful.
S2::
z = y + 15
By itself this doesn't help - what's z? We don't know anything about its relationship to x and y.
S1 + S2::
Now we've got three equations in three variables, so we can solve if we'd like (but it's DS, so we shouldn't bother).
S1::
x = 1.3z and z = 1.12y
This takes us in the wrong direction! We wanted another equation in x and y, but we got a new variable, z. That said, we now have two equations in three variables, so if we could get one more equation we'd have three equations in three variables and be good to go. Not sufficient, but possibly helpful.
S2::
z = y + 15
By itself this doesn't help - what's z? We don't know anything about its relationship to x and y.
S1 + S2::
Now we've got three equations in three variables, so we can solve if we'd like (but it's DS, so we shouldn't bother).
