Last year Publisher X published 1,100 books, consisting of first editi

Solution

Steps 1 & 2: Understand Question and Draw Inferences
We are given that:

• Last year publisher X published 1,100 books, consisting of first editions, revised editions, and reprints.
• Number of First edition + Revised Editions + Reprints published = 1100

We need to find:

• The number of first editions published by publisher X last year.

Since we do not have any relevant information to infer from, let us now analyse the individual statements.

Step 3: Analyse Statement 1
“The number of first editions published was 50 more than twice the number of reprints published.”

• Number of First Editions = 2 × Number of reprints published + 50

However, from this statement we cannot find the number of first editions as we do not know the number of reprints published.

Thus, statement 1 is not sufficient to answer the question.

Step 4: Analyse Statement 2
“The number of revised editions published was half the number of reprints published.”

• Number of revised editions = (Number of reprints published)/2
• Number of reprints published = 2 × Number of revised editions

However, from this statement we cannot find the number of first editions.

Thus, statement 2 is not sufficient to answer the question.

Step 5: Combine Both Statements Together (If Needed)
From Statement 1:

• Number of First Editions = 2 × Number of reprints published + 50

From Statement 2:

• Number of reprints published = 2 × Number of revised editions

So, we have the number of first editions published and number of revised editions published in terms of number of reprints published.

• Hence, we can substitute these in the equation: “Number of First edition + Revised Editions + Reprints published = 1100” and we can find the number of reprints published.
• And, then we can find the number of first editions published.

Hence, we can find the answer by combining both the statements together.

Thus, the correct answer choice is option C.

Straightforward...
No solving needed.
Each of 1 and 2 gives you one extra equation. So two equations with three unknowns doesn't lead to solution.
With BOTH equations and that in the stem we have three equations with three unknowns.
Both is sufficient.


C

Math is, unfortunately, not that simple. Sometimes with only two equations in three unknowns, you can solve for one of your unknowns. For example, from these equations:

a + b + c = 5
a + b = 3

c must be equal to 2. And sometimes with three equations in three unknowns, you can't solve at all. For example with these equations:

a + b = 3
b + c = 5
a + 2b + c = 8

c can be equal to literally anything.

The equations genuinely matter, the GMAT very often tests if you know why that's true, and anyone who is merely counting equations and counting unknowns without thinking about what kinds of equations they're seeing will get roughly 2/3 of higher-level official DS questions wrong (when that 'strategy' could potentially be used), as anyone can confirm just by trying to apply that 'strategy' to a batch of higher level official problems.

In this question, each Statement alone tells us almost nothing about one type of book - using Statement 1, for example, we only know we have a positive integer number of revised editions, and no more than 1047 of them (since we must have at least 1 reprint and at least 52 first editions). There are still going to be lots of possibilities for every number. Using both Statements, we do get 3 equations in 3 unknowns. Because the three relationships are fundamentally different (we could not deduce one of them from the other two), and because they are linear (we aren't multiplying unknowns together, and we have no exponents) then we can be certain we can solve for all three unknowns. So because of that, we can be sure the answer is C.

Let,
\(F=\) First Edition
\(R=\)Revised Edition
\(P=\)Reprintss

\(F+R+P=1100\)

(1) \(F=2P+50;\)

\(2P+50+R+P=1100\)

\(3P+R=1050\); Gives no information about the value R and P, Insufficient.

(2) \(R=\frac{1}{2}P; P=2R;\) Still R is unknown to find the value of F. Insufficient.

Considering Both:
\(F=2P+50\)

\(P=2R\)


\(So, 3*2R+R=1050\)

\(7R=1050; R=150 \)

\(P=150*2=300\)

\(F=2P+50=600+50=650 \ Sufficient. \)

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