CAT 2020 Slot 2QA Question 6

Number TheoryEasy

If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals

Answer & solution

Answer: 23

Solution

Easy

Express xx and yy in terms of zz using the two equalities, feed them into the inequality to bound zz, then write 2x+y2x+y as a linear function of zz and use the largest allowed integer zz.

1

Express in terms of zz. From the two equations:

x=z9,y=z1\begin{aligned} &x = z - 9,\qquad y = z - 1 \end{aligned}
2

Apply the inequality. Substitute into x+y<z+5x+y < z+5.

(z9)+(z1)<z+5(from step 1) 2z10<z+5 z<15\begin{aligned} &(z-9)+(z-1) < z+5 \quad\text{(from step 1)}\\ &\Rightarrow\ 2z - 10 < z + 5\\ &\Rightarrow\ z < 15 \end{aligned}

Also x=z90x=z-9\ge 0 requires z9z\ge 9, so zz is an integer with the largest value 1414.

3

Maximise 2x+y2x+y. Write it in zz and use z=14z=14.

2x+y=2(z9)+(z1)=3z19 at z=14: 3(14)19=23\begin{aligned} &2x + y = 2(z-9) + (z-1) = 3z - 19\\ &\Rightarrow\ \text{at } z=14:\ 3(14) - 19 = 23 \end{aligned}
2x+y=232x+y = 23
CAT 2020 Slot 2 QA Q6: If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possib — Solution | TheCATExam