CAT 2022 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like **Speed-time-distance**, **Mixture and Alligations**. This year, there was a surprise. The questions from **Geometry** were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 22 : Let \(0 \leq a \leq x \leq 100\) and \(f(x)=|x-a|+|x-100|+|x-a-50|\). Then the maximum value of \(f(x)\) becomes 100 when \(a\) is equal to

- 100
- 25
- 0
- 50

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f(x) = |x − a| + |x − 100| + |x − a − 50|

f(x) be re-written as |x − a| + |x − 100| + |x − (a+50)|

| x - a | signifies the distance between x and a.

| x - 100 | signifies the distance between x and 100.

| x - (a + 50) | signifies the distance between x and a + 50.

Since \( a \leq x \leq 100 \),

**|x − a| + |x − 100| is just 100 - a and is independent of x**

f(x) = |x − a| + |x − 100| + |x − (a+50)|

f(x) = (100 - a) + |x − (a+50)|

Now, let’s talk about |x − (a+50)|.

We know that \( x \geq a \), let us try to understand the maximum value that |x − (a+50)| can take.

x can be between a and (a + 50) or x can be to the right of (a + 50)

Case (i)

x is between a and (a + 50)

If this is the case, the maximum distance between a and (a+50) is just 50.

And this happens when x = a.

**For any given value of ‘a’, We can just put ‘x’ at ‘a’ and make |x − (a+50)| equal to 50.**

Case (ii)

x is to the right of (a + 50)

To find the maximum value that |x − (a+50)| can take, let us push x to the right most possible position and push (a + 50) to the left-most possible position.

The right-most value that x can take is 100.

It looks like there is no hold on the left-most value that (a + 50) can take, but there is!

The left-most value that (a + 50) can take is dependent on the left-most value that ‘a’ can take. And the left-most value that ‘a’ can take is 0.

So in this extreme case of maximizing the distance between x and (a + 50),

a is at 0, (a+50) will be at 50, x is at 100

And the distance between x and (a + 50) is 50.

But why is understanding the maximum value that |x − (a+50)| can take important?

Once we establish the following two facts,

**(i)** **|x − a| + |x − 100| is just 100 - a and is independent of x**

**(ii) For any given value of ‘a’, We can just put ‘x’ at ‘a’ and make |x − (a+50)| equal to 50.**

For any given ‘a’, we can think of **f(x) = |x − a| + |x − 100| + |x − (a+50)|** as having a fixed part and a variable part.

**Fixed part:** **|x − a| + |x − 100| = 100 - a**

**Whose value is fixed and does not change with the value of ‘x’**

**Variable part: |x − (a+50)|**

**Whose value can range from 0 to 50, based on the value of ‘x’**

f(x) = (100 - a) + (something that reaches from 0 to 50)

If we need to fix the maximum value of f(x) at 100, we need to tackle the “worst-case scenario” that the ‘variable part’ can cause…

That is, we need to see f(x) = (100 - a) + (something that reaches from 0 to 50) as f(x) = (100 - a) + (something that is 50)

Now to cap f(x) at 100, (and also allow it take the value 100) we just need to set (100 - a) to 50, which happens when a = 50.

Therefore, **a = 50**.

The question is **" Let \(0 \leq a \leq x \leq 100\) and \(f(x)=|x-a|+|x-100|+|x-a-50|\). Then the maximum value of \(f(x)\) becomes 100 when \(a\) is equal to " **

Choice D is the correct answer.

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