IPMAT Indore 2023 · Quantitative Ability · Question 4

Step 1 — exploit the symmetry. Both curves are even in x (unchanged when x\to -x), so their intersection points come in a symmetric pair \left( \pm{x}_{0},{y}_{0}\right). The segment joining them is horizontal with length 2{x}_{0}.

Step 2 — solve for the positive intersection. For x\ge 0, \left| x \right|=x:

{x}^{2}=4970-x \Rightarrow{x}^{2}+x-4970=0

Step 3 — factor or use the formula.

x=\frac{-1+\sqrt{1+4\left( 4970 \right)}}{2}=\frac{-1+\sqrt{19881}}{2}

Now 19881={141}^{2} (since {140}^{2}=19600 and {141}^{2}=19881), so

x=\frac{-1+141}{2}=70

Step 4 — the length. The points are \left( 70,4900 \right) and \left( -70,4900 \right):

\text{length}=70-\left( -70 \right)=140

Answer: 140

Shortcut. Never compute y. The two curves are both even, so the intersections are mirror images across the y-axis and the segment is horizontal of length 2{x}_{0}. You only need {x}_{0}.

{x}^{2}+x-4970=0 \Rightarrow \left( x-70 \right)\left( x+71 \right)=0

Spotting the factorisation beats the quadratic formula: you need two numbers multiplying to -4970 and differing by 1, and 70\times 71=4970 is the giveaway. Answer 2\left( 70 \right)=140.

Recognising 4970=70\times 71 is the intended shortcut — the setter chose the number precisely so that consecutive integers appear.