hD13 wrote:
We can solve the equation and find Head count - Time and time again B is correct answer because it follows " Only integer Solution".
If you were to try finding next integer solutions, You'll have to go all the way over 100 .
Hence , How can we chose ( c ) as an answer in such Questions else where.
Or how do we know that the Soln we have got here is not UNIQUE
I recall one of my math teachers saying, after the class gave the wrong answer to a tricky question, that "math is not a democracy". And sometimes that's an issue with gmatclub's system, where the forum elevates posts with the most 'votes' (or kudos). Atop this thread, none of the posts is correct - one just ignores the issue of unique integer solutions altogether, and that's the only interesting feature of this question. Another says you need number theory "way beyond the scope of the GMAT" to see if you have unique solutions, and you never need math beyond the scope of the test to definitively prove an answer is correct on the GMAT. And another post claims we need to go past adults = 100 to find the second smallest solution here, which also is not the case.
Statement 1 is clearly insufficient alone. For Statement 2, we can write an equation -- notice that the numbers in the question are nice to work with, but if we reduce them by 10%, they become annoying. So we should leave them as is. If we have c children and a adults, then, multiplying by 0.9 to apply the 10% discount, Statement 2 tells us:
(0.9)(4800)*c - (0.9)(5500)*a = 4860
4800c - 5500a = 4860*10/9
and since 4500 + 360 = 4860, we can see that 4860/9 = 500 + 40 = 540, so multiplying that by the extra 10, we get
4800c - 5500a = 5400
48c - 55a = 54
48c = 55a + 54
Notice what this equation says, if a and c are integers: it just says "55a + 54 is a multiple of 48". Since this is DS, we absolutely know there is at least one solution to this equation (it is never true in DS that there are zero solutions). So we know there is at least one integer a for which 55a + 54 is a multiple of 48. But if we take that integer a, and add 48 to it, this is what we get:
55(a + 48) + 54 = 55a + 54 + 55*48
and since 55a + 54 is a multiple of 48, and 55*48 is also a multiple of 48, this sum must be a multiple of 48 too. So if a gives one integer solution to the equation, so will a + 48, and a + 96, and so on. So we'll have an infinite number of solutions. There's no reason to actually find any of those solutions, since it's a DS problem (as soon as we use Statement 1+2 together, we'll clearly get one solution and we don't care what it is), but if you did solve here, you'd find that the smallest solution is a = 6 and c = 8, but we'll also have a solution, adding 48, when a = 54 and c = 63, and another solution when a = 102, and so on.
In general, if you want to know when these types of equations have unique integer solutions, there are a few different techniques that you might use, depending on the numbers involved. So it's too big of a topic to get into in one post here, but my Word Problems book discusses issues like this in detail.
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