RRB JE Stage I 30 May 2019 Shift 1 (HM)

© examsiri.com

Question : 36 of 100

Marks: +1, -0

सरल कीजिए:

\√{{(2.3)^{2}+(23.3)^{2}+(233.3)^{2}}/{(0.23)^{2}+(2.33)^{2}+(23.33)^{2}}}

${1}/{100}$
$0.1$
100
10

Solution:

माना,

2.3=a ; 23.3=b

एवं

233.3=c

\∴0.23={a}/{10} ; 2.33={b}/{10}

एवं

23.3={c}/{10}

\∴

व्यंजक

=\√{{a^{2}+b^{2}+c^{2}}/{({a}/{10})^{2}+({b}/{10})^{2}+({c}/{10})^{2}}}

=\√{{a^{2}+b^{2}+c^{2}}/{{1}/{100}(a^{2}+b^{2}+c^{2})}}

=\√{100}=10

© examsiri.com

Go to Question:

Prev Question

Next Question