IPMAT Indore 2026 · Quantitative Ability · Question 1

Step 1 — factor the first equation.

{x}^{2}-12x+35=\left( x-5 \right)\left( x-7 \right)=0 \Rightarrow x=5\text{or}7

Step 2 — the common root must be one of these. Substitute each into the second equation.

If x=5: 25+5a+105=0\Rightarrow 5a=-130\Rightarrow a=-26

If x=7: 49+7a+105=0\Rightarrow 7a=-154\Rightarrow a=-22

Step 3 — minimise the y part. Complete the square:

{y}^{2}+4y={\left( y+2 \right)}^{2}-4 \ge -4

so the smallest value of {y}^{2}+4y is -4, at y=-2.

Step 4 — combine. The expression is {y}^{2}+4y-5a, minimised as -4-5a:

a=-26: -4+130=126

a=-22: -4+110=106

The question asks for the minimum possible value over everything allowed, so take the smaller:

\boxed{106}

Answer: 106

Shortcut. Two separate minimisations, done independently: the y part bottoms out at -4 regardless of a, and then you simply pick whichever a makes -5a smaller. Since a is negative, -5a is positive, so you want the a closer to zero — that is -22.