CAT 2017 Question Paper | Quants Slot 1

This is a question from Algebra. A student can expect 4-5 questions from this topic in CAT exam. There are a few questions where the intuitive approach works like wonders than the robust algebraic approach. Make sure you try this question before watching the video solution where Rajesh solves this question elegantly.

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]

Video Explanation

Explanatory Answer

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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