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CAT 2018 Slot 1 — QA Question 21

IndicesEasy

Given that x2018y2017 = 1/2 and x2016y2019 = 8, the value of x2 + y3 is

Answer & solution

Correct answer: 33/4

  • A

    37/4

  • 33/4

  • C

    35/4

  • D

    31/4

Solution

Easy

Dividing the two equations cancels most of the exponents and relates xx to yy. Substitute back to pin down yy, then xx, and finally evaluate x2+y3x^2+y^3.

1

Record the equations.

x2018y2017=12(1)x2016y2019=8(2)\begin{aligned} &x^{2018}y^{2017} = \tfrac12 \quad\text{(1)}\\ &x^{2016}y^{2019} = 8 \quad\text{(2)} \end{aligned}
2

Divide (1) by (2).

x2018y2017x2016y2019=1/28⇒ x2y2=116[(1)÷(2)]⇒ x2=y216\begin{aligned} &\frac{x^{2018}y^{2017}}{x^{2016}y^{2019}} = \frac{1/2}{8}\\ &\Rightarrow\ \frac{x^2}{y^2} = \frac{1}{16} \quad\text{[(1)} \div \text{(2)]}\\ &\Rightarrow\ x^2 = \frac{y^2}{16} \end{aligned}
3

Find yy. Write x2018=(x2)1009x^{2018}=(x^2)^{1009} and substitute step 2 into (1):

(y216)1009y2017=12⇒ y2018⋅y2017161009=12⇒ y4035=12⋅24×1009=2−1⋅24036=24035⇒ y=2\begin{aligned} &\left(\frac{y^2}{16}\right)^{1009} y^{2017} = \tfrac12\\ &\Rightarrow\ \frac{y^{2018}\cdot y^{2017}}{16^{1009}} = \tfrac12\\ &\Rightarrow\ y^{4035} = \tfrac12\cdot 2^{4\times 1009} = 2^{-1}\cdot 2^{4036} = 2^{4035}\\ &\Rightarrow\ y = 2 \end{aligned}
4

Find x2x^2. Put y=2y=2 into (1):

x2018⋅22017=12⇒ x2018=2−2018⇒ x=±12  ⇒  x2=14\begin{aligned} &x^{2018}\cdot 2^{2017} = \tfrac12\\ &\Rightarrow\ x^{2018} = 2^{-2018}\\ &\Rightarrow\ x = \pm\tfrac12 \;\Rightarrow\; x^2 = \tfrac14 \end{aligned}
5

Evaluate the target.

x2+y3=14+23=14+8=334\begin{aligned} &x^2 + y^3 = \tfrac14 + 2^3 = \tfrac14 + 8 = \tfrac{33}{4} \end{aligned}
x2+y3=334x^2 + y^3 = \mathbf{\dfrac{33}{4}}

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CAT 2018 Slot 1 QA Q21: Given that x 2018 y 2017 = 1/2 and x 2016 y 2019 = 8, the value of x 2 + y 3 is — Solution | TheCATExam