# CAT 2017 Question Paper | Quants Slot 1

###### CAT Previous Year Paper | CAT Algebra Questions | Question 28

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Question 28 : If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a - b)2 + (a - c)2 + (a - d)2 is [TITA]

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##### Method of solving this CAT Question from Algebra

Given that a , b , c and d are integers such that a + b + c + d = 30
The minimum possible value of (a - b)2 + (a - c)2 + (a - d)2 has to be found.
For that a , b , c , d should be having equal or closer values such that (a – b) = (a - c) = (a – d) = 0 but this is not possible because 30 is not a multiple of 4.
We should find a as close to b , c , d as possible.
⟹ $$frac{30}{4}$ = 7.5 So a , b , c , d should take the values of 7 or 8. Hence taking two 7 and two 8 can work such that a , b , c , d is equal to 7 , 7 , 8 , 8 respectively. So$a - b)2 + (a - c)2 + (a - d)2 = (7 - 7)2 + (7 - 8)2 + (7 -8)2
(a - b)2 + (a - c)2 + (a - d)2 = 0 + 1 + 1 = 2
If we are having the difference to be 2, on squaring it becomes 4 so it is not possible.
So let us try for values less than 2 i.e 0 + 0 + 1 but it doesn’t work.
So if a , b , c and d are integers such that a + b + c + d = 30,
the minimum possible value of (a - b)2 + (a - c)2 + (a - d)2 is 2.

The question is "If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a - b)2 + (a - c)2 + (a - d)2 is [TITA]"

##### Hence, the answer is 2

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