X is 45.6% larger than Y. What is the value of Y?

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Fri Nov 10, 2017 6:03 am

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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Source: Beat The GMAT — Data Sufficiency | Fri Nov 10, 2017 6:03 am

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by [email protected] » Fri Nov 10, 2017 10:59 am

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

Contact Rich at [email protected]

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Fri Nov 10, 2017 1:04 pm

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).

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