Some days back I came across a question which was a slight twist on a regular question type. The usual active voice of the sentence had been changed to passive but in such a way that the meaning had been altered. It was a lesson in DS as well as SC – read every word carefully. One word could change a 600 level to a 750 level one, a mundane everyday question to a smart question. We often see this interesting transformation in P&C questions but for that to happen in algebra was quite a delight. Let’s discuss that particular question today.

First let’s look at the mundane version.

**Question**: If 9 notebooks and 3 pencils cost 20 Swiss Francs, do 12 notebooks and 12 pencils cost 40 Swiss Francs?

Statement 1: 7 notebooks and 5 pencils cost 20 Swiss Francs.

Statement 2: 4 notebooks and 8 pencils cost 20 Swiss Francs.

**Solution**: Both statements give very similar information. It looks like the answer will be (D). That is, if one statement is enough to answer the question alone, the other will probably be enough to answer alone too. Also it seems that we will have two simultaneous equations in two variables so we will be able to solve for the variables.

Let’s quickly review how we actually solve this

*Given: If 9 notebooks and 3 pencils cost 20 Swiss Francs* –> 9N + 3P = 20 (assuming N is the cost of each notebook and P is the cost of each pencil)……..(I)

*Question: do 12 notebooks and 12 pencils cost 40 Swiss Francs* – > Is 12N + 12P = 40? OR Is 6N + 6P = 20?

*Statement 1: 7 notebooks and 5 pencils cost 20 Swiss Francs.*

7N + 5P = 20 ……… (II)

Equating (I) and (II), we get N = P = 20/12. This is sufficient to answer whether 6N + 6P is equal to 20.

*Statement 2: 4 notebooks and 8 pencils cost 20 Swiss Francs.*

4N + 8P = 20 ……… (III)

Equating (I) and (III), we get N = P = 20/12. This is sufficient to answer whether 6N + 6P is equal to 20.

So as expected, **answer is (D)** in this case.

The problem arises when the question is changed a bit.

**Question 2**: 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?

Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.

Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

**Solution**: Note that the numbers are unchanged. Does the changed wording convey the same meaning? At first glance, you may think so but that is not true. Now, 9 notebooks and 3 pencils may cost less than 20 SF too. All that the statement tells us is that 20 SF is enough – whether it is just enough or comfortably enough, we don’t know. So we don’t have the actual cost of 9N and 3 pencils. We just know that 9N + 3P <= 20. So here we will have to solve inequalities.

But it still seems that both statements give very similar information and so if one alone is sufficient, the other alone should be sufficient too.

*Given: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils* –> 9N + 3P <= 20 ……. (I)

*Question: is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils* – > “Is 12N + 12P <= 40?” OR “Is 6N + 6P <= 20?” OR “Is N + P < 10/3?”

*Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.*

7N + 5P <= 20 ……..(II)

Adding (I) and (II), we get 2N + P <= 5. We want the coefficients of N and P to be the same in the resulting inequality. Since coefficient of N is greater in both inequalities, we cannot have the same coefficient of N and P in the resultant inequality. So we cannot say whether N + P < 10/3 so this statement alone is not sufficient.

*Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.*

4N + 8P <= 20 …………(III)

Again, we need the coefficient of N and P to be the same in the resultant inequality. You can get this as

2N + 4P <= 10 (from III)

3N + P <= 20/3 (from I)

Adding them, we get 5N + 5P <= 50/3 OR N + P <= 10/3

This statement alone is sufficient to answer the question.

**Answer (B)**

As opposed to our instinct, we find that the second statement alone is sufficient while the first is not.

Let’s try to understand why this is so using a logical solution.

Given: 9N + 3P <= 20

**Question**: Is 12N + 12P <= 40? OR Is 6N + 6P <= 20?

We know that 9 notebooks and 3 pencils cost 20 SF or less. We want to find whether 6 notebooks and 6 pencils will cost 20 SF or less i.e. if you drop 3 notebooks but take another 3 pencils, will your total cost still not exceed 20 SF? That depends on the relative cost of notebooks and pencils. If pencils are cheaper than (or have same cost as) notebooks, then obviously the total cost will stay less than or equal to 20. If pencils are more expensive than notebooks, we need to know how much more expensive they are to be able to judge whether the cost of 6 notebooks and 6 pencils will exceed 20 SF.

Statement 1: *20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.*

Now we know that we can substitute 2 notebooks with 2 pencils i.e. pencils may be cheaper than notebooks, may have the same cost or may be a little more expensive but there is enough leeway in our total cost for us to bear the extra cost of 2 pencils in place of 2 notebooks. But do we have enough leeway in our total cost to replace 3 notebooks with 3 pencils, we don’t know. Let me explain this using an example:

Say 1 notebook costs 1 SF and 1 pencil costs 1 SF too. So 9 notebooks and 3 pencils costs 12 SF. 7 notebooks and 5 pencils cost 12 SF. 6 notebooks and 6 pencils will cost 12 SF.

Take a different case – say 1 notebook costs 1.5 SF and 1 pencil costs 1.85 SF

Then 9 notebooks and 3 pencils cost 19.05 SF (which is less than 20 SF). 7 notebooks and 5 pencils cost 19.75 SF. (So even though 1 pencil costs more than 1 notebook, 2 pencils can substitute 2 notebooks because total cost is less than 20 SF. Obviously 1 pencil can substitute 1 notebook since there was enough leeway for even 2 pencils in place of 2 notebooks)

But 6 notebooks and 6 pencils cost 20.1 SF. (Now we see that 3 pencils cannot substitute 3 notebooks because there isn’t enough leeway. This time it crossed 20 SF)

Hence this statement alone is not sufficient to answer the question.

*Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.*

Now we know that we can drop 5 notebooks and buy 5 extra pencils in their place and the total cost will still stay below 20 SF. Hence there is enough leeway for 5 replacements. This obviously means that there is enough leeway for 3 replacements and hence the cost of 6 notebooks and 6 pencils will stay below 20 SF. This statement alone is sufficient to answer the question.

**Answer (B)**

We hope all this made sense. If you are reeling after reading all these numbers, give it another try. The question is a good 750 level question and certainly not easy.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

Simply too awesome ! Thank you for the effort and pain you have taken to make this clear.

I fell for D and couldn’t understand the explanation given in GC.

Now its clear

In my world you’d get the Nobel prize for this.

Thanks

Thanks! That’s flattering as well as amusing!

Hi Karishma,

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In Question #2, I understand your “logical solution”, but I have a couple questions about your “algebric solution.”

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Why do you need “the coefficients of N and P to be the same in the resulting inequality”? N and P can be different and still add up to be < 10/3.

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In Statement 2, how did you know that you need to convert "4N + 8P <= 20" to "2N + 4P <= 10" AND "9N + 3P <= 20" to "3N + P <= 20/3 "? I'm sure you have a reasoning for those steps but those steps look so random. My first instinct was to add 4N + 8P <= 20 to 9N + 3P <= 20 resulting : 13N + 11P <= 40. That inequality did not help. Thank you.

Mike