This question was previously asked in

DSSSB PRT General Section Official Paper - 14 Nov 2019 Shift 3

Option 4 : 24/17

**Given:**

\(A=\frac{31}{10}\times\frac{3}{10}+\frac{7}{5}\div20\)

\(B=\frac{16-6\times2+3}{23-3\times2}\)

**Concept used:**

**Calculation:**

Calculating values

\(A=\frac{31}{10}\times\frac{3}{10}+\frac{7}{5}\div20\)

\(A=\frac{31}{10}\times\frac{3}{10}+\frac{7}{5}\times\frac{1}{20}\)

\(A=\frac{93}{100}+\frac{7}{100} = \frac{100}{100} \)

A = 1

\(B=\frac{16-6\times2+3}{23-3\times2} \)

\(B=\frac{19-12}{23-6}\)

\(B=\frac{7}{17}\)

So now,

A + B = 1 + \(\frac{7}{17}\)

**A + B = 24/17**