Hussain15 wrote:

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.

(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Target question: Is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?This is a great candidate for

rephrasing the target question.

Let N = the cost of 1 notebook (in Swiss francs)

Let P = the cost of 1 pencil (in Swiss francs)

So, 12 notebooks cost 12N and 12 pencils cost 12P

So, we want to know whether 12N + 12P ≤ 40 (francs)

We can divide both sides by 12 to get: N + P ≤ 40/12

Simplify to get: N + P ≤ 10/3

We can now REPHRASE the target question....

REPHRASED target question: Is N + P ≤ 10/3? Given: 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils We can write:

9N + 3P ≤ 20 Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils We can write: 7N + 5P ≤ 20

We also know that

9N + 3P ≤ 20Can we use these two inequalities to answer the

REPHRASED target question?

It's hard to tell.

Let's add the inequalities to get: 16N + 8P ≤ 40

Divide both sides by 8 to get: 2N + P ≤ 5

In other words (N + P) + N ≤ 5

ASIDE: This inequality looks similar to our

REPHRASED target question.

If we subtract N from both sides we get: (N + P) ≤ 5 - N

The REPHRASED target question asks

Is N + P ≤ 10/3?The answer to that question depends on the value of N.

So, let's test some (extreme) values that satisfy the given information:

Case a: N = 2 and P = 0 In this case, N + P = 2. So, the answer to the REPHRASED target question is

YES, N + P ≤ 10/3Case b: N = 0 and P = 4 In this case, N + P = 4. So, the answer to the REPHRASED target question is

NO, N + P > 10/3Since we cannot answer the

REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencilsWe can write: 4N + 8P ≤ 20

We also know that

9N + 3P ≤ 20Let's add the inequalities to get: 13N + 11P ≤ 40

So close!!!

Too bad we don't have the same number of N's and P's!!

Wait. Perhaps we CAN have the same number of N's and P's if we create some EQUIVALENT inequalities.

First notice that 4N + 8P ≤ 20 is the same as 4(N + P) + 4P ≤ 20

And

9N + 3P ≤ 20 is the same as

3(N + P) + 6N ≤ 20Now take 4(N + P) + 4P ≤ 20 and multiply both sides by 3 to get: 12(N + P) + 12P ≤ 60

And take

3(N + P) + 6N ≤ 20 and multiply both sides by 2 to get:

6(N + P) + 12N ≤ 40ADD the two inequalities to get: 18(N + P) + 12P + 12N ≤ 100

[aha!! We now have the SAME number of P's and Q's]Rewrite as: 18(N + P) + 12(P + N) ≤ 100

Simplify: 30(P + N) ≤ 100

Divide both sides by 30 to get: (P + N) ≤ 100/30

Simplify: (P + N) ≤ 10/3

Perfect! The answer to the REPHRASED target question is

YES, N + P ≤ 10/3Since we can answer the

REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,

Brent

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