Simplification questions for class 7 PDF

Simplification questions for class 7

Different types of questions on simplifications are explained here step by step.

Simplification questions for class 7(Solved examples) 

(i)16+8 ÷ 4-2 x  3

Sol.  According to the  BODMAS’    rule  we have to perform division first followed by  multiplication , addition and subtraction.

16+8 ÷ 4-2 x  3 =16+2-2 x  3                    ( performing division)

=16+2-6                              (performing multiplication)

=18-6                                 (performing addition)

= 12

(ii) 23-[23-\left \{ 23-(23-\overline{23-23})\right \}]

Sol. First we have to remove vinculum (bar) , this implies

23-[23-\left \{ 23-(23-\overline{23-23})\right \}]=23-[23-\left \{ 23-(23-0)\right \}]

=23-[23-\left \{ 23-23\right \}]                       (removing small brackets)

=23-[23-0]                                            (removing curly braces)

= 23-23                                                            (removing big brackets)

=0

(iii)  25-\frac{1}{2}\left \{ 5+4-(3+2-\overline{1+3}) \right \}

Sol.   25-\frac{1}{2}\left \{ 5+4-(3+2-\overline{1+3}) \right \}= 25-\frac{1}{2}\left \{ 5+4-(3+2-4) \right \}             (removing bar)

=25-\frac{1}{2}\left \{ 5+4-1 \right \}                                (removing small brackets)

=25-\frac{1}{2}\times 8                                            (removing curly braces)

=25-4                                                            (performing multiplication)

=21

(iv) \mathbf{4+\frac{1}{5}}[(-10 x (25-\overline{13-3})) ÷ (-5)]

Sol. \mathbf{4+\frac{1}{5}}[(-10 x (25-\overline{13-3})) ÷ (-5)]= \mathbf{4+\frac{1}{5}}[(-10 x (25-10)) ÷ (-5)]                         (removing vinculum)

= \mathbf{4+\frac{1}{5}}[(-10 x (15)) ÷ (-5)]             

                                                                         =\mathbf{4+\frac{1}{5}}[(-150) ÷ (-5)]                                       (removing small brackets)

                                                                       =      \mathbf{4+\frac{1}{5}} x 30                                                      (removing big brackets)

                                                                       =  4+6                                                                       (performing  multiplication)

                                                                       =10          

(v) 8 − [28 ÷ {34 − (36 − 18 ÷ 9 × 8)}] 

Sol. 8 − [28 ÷ {34 − (36 − 18 ÷ 9 × 8)}] =  8  [28 ÷ {34  (36 2 ×  8)}]              [solving division]
=     8  [28 ÷ {34  (36 −16)}]                    [Performing multiplication]
= 8  [28 ÷ {34 20}]                                    [Removing parentheses]
= 8 [28 ÷ 14]                                                [Removing braces]
= 8 2                                                              [Removing square brackets]

=6

(vi) 16  2 ÷ 7 + 6 × 2

Sol. 16  2 ÷ 7 + 6 × 2= 16-\frac{2}{7}+6 \times 2

= 16-\frac{2}{7}+12

= 28-\frac{2}{7}

= \frac{196-2}{7}=\frac{194}{7}

=27\frac{5}{7}

(vii) 197 – [\mathbf{\frac{1}{9}}{42 + (56 – \mathbf{\overline{8+9}})} +108]

Sol.   197 – [\frac{1}{9} {42 + (56 – \overline{8+9})} +108] = 197 – [\frac{1}{9} {42 + (56 – 17)} + 108]       [Removing vinculum]

= 197 – [\frac{1}{9} {42 + 39} + 108]                [removing innermost brackets]

= 197 – [(\frac{1}{9} \times 81)+ 108]                          [Removing braces]

= 197 – [9 + 108]                                    [removing big brackets]

= 197 – 117

= 80

Students can practice these examples on simplification which are discussed here step by step and then proceed to the worksheet for more practice.

  • Simplification questions for class 7  ( Unsolved questions with answers)

(i) 36-2(20+12÷ 4 x 3 -2 x 2)+10

(ii) 18+(28-7) ÷ 3 -11

(iii) 5\frac{1}{7}-{3\frac{3}{10} ÷\left ( 2\frac{4}{5} -\frac{7}{10}\right )}

(iv) 22-\frac{1}{4} {-5-(-48) ÷ (-16)}

(v) 7\frac{1}{3}÷\frac{2}{3}\, of\, 2\frac{1}{5}+1\frac{3}{8} ÷2\frac{3}{4}-1\frac{1}{2}

(vi) \left ( \frac{2}{3}+\frac{4}{9} \right ) \, \, of \, \, \frac{3}{5} ÷1\frac{2}{3}\times 1\frac{1}{4}-\frac{1}{3}

(vii)  8-[28÷{34-(36-18÷9 x 8)}]

(viii)6-[\frac{5}{6}+\{3\frac{7}{8}-2\frac{1}{3}+1\frac{7}{9}\}]

(ix) 9\frac{3}{4} ÷\left [ 2\frac{1}{6}+\left \{ 4\frac{1}{3}-(1\frac{1}{2}+1\frac{3}{4}) \right \} \right ]

(x)  54 ÷3 of 6+9

(xi)19  [4 + {16  (12  2)}]

(xii) 36  [18  {14  (15  4 ÷2 ×2)}]

(xiii) 39 – [23 – {29 – (17 – \overline{9-3})}]

(xiv) 15 – (-5) {4 – \overline{7-3}} ÷ [3{5 + (-3) x (-6)}]

Ans. (i)  -4       (ii) 14    (iii) 3\frac{4}{7}     (iv) 24    (v) 4     (vi) \frac{1}{6}      (vii)  6    (viii) 1\frac{61}{72}           (ix) 3      (x) 12    (xi) 9     (xii) 21      (xiii)34     (xiv) 15

Bodmas questions for class 7 worksheet

 To download  PDF      bodmas questions for class 7 worksheet

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