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CAT 2023 Slot 3 — QA Question 1

Number TheoryEasy

For some real numbers a and b, the system of equations x + y = 4 and (a + 5)x + (b2 -15)y = 8b has infinitely many solutions for x and y. Then, the maximum possible value of ab is?

Answer & solution

  • A

    15

  • B

    25

  • 33

  • D

    55

Solution

Easy

Two linear equations have infinitely many solutions only when one is a scalar multiple of the other — i.e. the ratios of the xx-coefficients, yy-coefficients and constants are all equal. Set up that triple ratio, solve for bb, then for aa, and pick the pair giving the largest abab.

1

Write the proportionality condition for x+y=4x+y=4 and (a+5)x+(b2−15)y=8b(a+5)x+(b^2-15)y=8b:

1a+5=1b2−15=48b\frac{1}{a+5}=\frac{1}{b^2-15}=\frac{4}{8b}
2

Use the 2nd and 3rd ratios to find bb:

1b2−15=48b=12b⇒ b2−15=2b⇒ b2−2b−15=0⇒ (b−5)(b+3)=0 ⇒ b=5 or b=−3\begin{aligned} &\frac{1}{b^2-15}=\frac{4}{8b}=\frac{1}{2b}\\ &\Rightarrow\ b^2-15=2b\\ &\Rightarrow\ b^2-2b-15=0\\ &\Rightarrow\ (b-5)(b+3)=0\ \Rightarrow\ b=5 \text{ or } b=-3 \end{aligned}
3

Use the 1st and 2nd ratios to link aa and bb:

1a+5=1b2−15 ⇒ a+5=b2−15 ⇒ a=b2−20\frac{1}{a+5}=\frac{1}{b^2-15}\ \Rightarrow\ a+5=b^2-15\ \Rightarrow\ a=b^2-20
4

Evaluate abab for each bb:

b=5: a=25−20=5 ⇒ ab=25b=−3: a=9−20=−11 ⇒ ab=(−11)(−3)=33\begin{aligned} &b=5:\ a=25-20=5\ \Rightarrow\ ab=25\\ &b=-3:\ a=9-20=-11\ \Rightarrow\ ab=(-11)(-3)=33 \end{aligned}

The larger value is 3333.

Maximum possible value of ab=33ab=\mathbf{33}.

Don't stop at the "nice" root b=5b=5. The negative root b=−3b=-3 makes both aa and bb negative, so their product flips positive and beats 2525 — always test every root when a problem asks for a maximum.

CAT 2023 Slot 3 QA Q1: For some real numbers a and b, the system of equations x + y = 4 and (a + 5)x + (b 2 -15)y = 8b has infinitely — Solution | TheCATExam