The value of 27+[3(50-20)+168 \\div4+2-11 \times2] is

SSC CHSL 15 Oct 2020 Shift 1 Paper

Question : 72 of 100

Marks: +1, -0

The value of

27+[3(50-20)+168 \÷4+2-11 ×2]

Solution:

Given,

27+[3(50-20)+168 \÷4+2-11 ×2]

Solving by BODMAS, we get

27+[3 ×30+42+2-22]

=27+112=139

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