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CAT 2020 Slot 2 — QA Question 14

Number TheoryEasy

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1+1x)(1+1y) is

Answer & solution

Answer: 2704

Solution

Easy

Expand the product and write it in terms of the fixed sum x+y=102x+y=102 and the product xyxy. The expression decreases as xyxy grows, and for a fixed sum the product is largest when x=yx=y. Plug that in.

1

Simplify the bracket product. Combine the two factors over a common denominator xyxy, using x+y=102x+y=102.

(1+1x)(1+1y)=(x+1)(y+1)xy=1+x+y+xyxy⇒ =103+xyxy=1+103xy(x+y=102)\begin{aligned} &\left(1+\tfrac{1}{x}\right)\left(1+\tfrac{1}{y}\right) = \frac{(x+1)(y+1)}{xy} = \frac{1+x+y+xy}{xy}\\ &\Rightarrow\ = \frac{103 + xy}{xy} = 1 + \frac{103}{xy} \quad\text{(}x+y=102\text{)} \end{aligned}
2

Minimize. The value falls as xyxy rises. For a fixed sum, xyxy is maximal when x=yx=y.

x=y=1022=51⇒ xy=51⋅51=2601\begin{aligned} &x = y = \tfrac{102}{2} = 51\\ &\Rightarrow\ xy = 51\cdot 51 = 2601 \end{aligned}
3

Evaluate the full expression. Multiply by the leading 26012601. Notice 2601=5122601=51^2, which cancels neatly.

2601(1+151)(1+151)=2601⋅5251⋅5251⇒ =512⋅522512=522=2704\begin{aligned} &2601\left(1+\tfrac{1}{51}\right)\left(1+\tfrac{1}{51}\right) = 2601\cdot\frac{52}{51}\cdot\frac{52}{51}\\ &\Rightarrow\ = 51^2\cdot\frac{52^2}{51^2} = 52^2 = 2704 \end{aligned}
Minimum value=2704\text{Minimum value} = 2704
CAT 2020 Slot 2 QA Q14: If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601 1 + 1 x 1 — Solution | TheCATExam