Bunuel wrote:
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.
Given \(9n+3p\leq20\), question is \(12n+12p\leq40\) true? Or is \(6n+6p\leq20\) true? So basically we are asked whether we can substitute 3 notebooks with 3 pencils. Now if \(p<n\) we can easily substitute notebooks with pencils (equal number of notebooks with pencils ) and the sum will be lees than 20. But if \(p>n\) we won't know this for sure.
But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both cases (\(p<n\) or \(p>n\)) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil, but we won't be sure for 3 (more than 2).
(1) \(7n+5p\leq20\). We can substitute 2 notebooks with 2 pencils, but this not enough. Not sufficient.
(2) \(4n+8p\leq20\). We can substitute 5 notebooks with 5 pencils, so in any case (\(p<n\) or \(p>n\)) we can substitute 3 notebooks with 3 pencils. Sufficient.
Answer: B.
Responding to a pm:
First of all, it is a tricky question. What makes it different from the run of the mill similar questions is the use of "is enough". Had the question said 9N and 3P cost 20 Swiss francs, life would have been much easier. Then your complain "I just don't understand that if we can replace 2 notebooks with 2 pencils, then why can't be substitute 3 notebooks with 3 pencils" would be totally justified.
The problem here is that we know that 9N and 3P cost
less than or equal to 20 Swiss Francs.
So why does Bunuel say "We can substitute 2 notebooks with 2 pencils, but this not enough."? I can try to answer this question of yours using an example.
Say 1 N costs 1.5 SF and 1 P costs 1.85 SF
Then 9N and 3P cost 19.05 SF (which is less than 20 SF)
7N and 5P cost 19.75 SF. (So even though 1P costs more than 1N, 2P can substitute 2N because total cost is less than 20 SF. Obviously 1P can substitute 1N since there was enough leeway for even 2P in place of 2N)
But 6N and 6P cost 20.1 SF. (Now we see that 3P cannot substitute 3N. This time it crossed 20 SF)
I hope this answers why we can substitute 2P in place of 2N but not 3P in place of 3N.
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