Bodhee Prep-CAT Online Preparation

CAT 2021 Quant Question [Slot 1] with Solution 16

Question
If \(5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}\frac{1}{{\sqrt {1 - {x^2}} }}\), then 100x equals
Option: 99
Solution:
>\(5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}\frac{1}{{\sqrt {1 - {x^2}} }}\)
We can re-write the equation as: \(5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = {\log _{10}}{\left( {\sqrt {1 + x} \times \;\sqrt {1 - x} } \right)^{ - 1}}\)
\(5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = \left( { - 1} \right){\log _{10}}\left( {\sqrt {1 + x} } \right) + \left( { - 1} \right){\log _{10}}\left( {\sqrt {1 - x} } \right)\)
\(5 = - {\log _{10}}\sqrt {1 + x} + {\log _{10}}\sqrt {1 + x} - {\log _{10}}\sqrt {1 - x} - 4{\log _{10}}\sqrt {1 - x} \)
\(5 = - 5{\log _{10}}\sqrt {1 - x} \)
\(\sqrt {1 - x} = \frac{1}{{10}}\)
Squaring both sides: \({\left( {\sqrt {1 - x} } \right)^2} = \frac{1}{{100}}\)
\(\therefore \;\) \(x = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}\)
Hence, \(100\;x\; = 100 \times \;\frac{{99}}{{100}} = 99\)
CAT Online Course @ INR 9999 only

CAT 2021 Quant questions with Solutions

CAT 2023
Classroom Course

We are starting classroom course for CAT 2023 in Gurugram from the month of December.
Please fill the form to book your seat for FREE Demo Classes

CAT 2023 Classroom Course starts in Gurgaon