IPMAT Indore 2023 - QA (MCQ) Q27 Explanation

Explanation:

(d)

Let [x] = I, an Integer & {x} = f, a proper fraction

Given: {x} = x –[x]

Or, x = [x] + {x}

Or, x = I + f

Also given that: 2 [x] = x + n{x}, where n is a natural number.

We can write:

2I = I + f + n.f

= I = f(n+1)

= f = I/(n+1)

= We know that: 0< f < 1

= 0< I/(n+1) <1

= 0< I < (n+1) ‘I’ can have values 1, 2, 3,…n & corresponding values of ‘f’ can be 1 n+1 , 2 n+1 , 3 n+1 ,….. n n+1 .

Thus, there can be ‘n’ possible values of x which are:

x1= 1 + 1/n+1 = n+2/n+1

x2 = 2 + 2/n+1 = 2(n+2/n+1 )

x3 = 3 + 3/n+1 = 3(n+2/n+1 )

.

.

x𝑛 = 𝑛 + 𝑛/𝑛+1 = 𝑛(𝑛+2/𝑛+1)

Sum of all these values of x = 𝑛+2/𝑛+1 [1+2+3+⋯…+𝑛]

= 𝑛+2/𝑛+1 [𝑛(𝑛+1)/2]

= 𝑛(𝑛+2)/2 Ans.