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

Hi Bunuel,

Can we do it in this way ? (Only to check that B alone is sufficient)
9n+3p ≤ 20 (given in question) -------(1)
4n + 8p ≤ 20 (from statement (b)) --------(2)

Add (1) and (2), we get
13n + 11p ≤ 40 -------(3)

Subtract (2) from (1), we get (is this step correct ?)
5n - 5p ≤0 which means
n-p ≤ 0 --------(4)

Now subtract (4) from (3), we get
12n + 12p ≤ 40

Therefore, B alone is sufficient.

We cannot subtract the inequalities the way you did.

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

For more check Manipulating Inequalities.

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 ≤ 20

Can 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 ≤ 10/3
Case b: N = 0 and P = 4 In this case, N + P = 4. So, the answer to the REPHRASED target question is NO, N + P > 10/3
Since 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 pencils
We can write: 4N + 8P ≤ 20
We also know that 9N + 3P ≤ 20
Let'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 ≤ 20

Now 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 ≤ 40

ADD 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 ≤ 10/3
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

I am not able to understand the meaning behind the logic of substituting number of notebooks with pencils. Can you explain in more simple language?

The solution is elaborated in several posts on the previous two pages. Alternatively, there are other solutions that might be of help. Please consider reviewing those as well. Hope this helps.