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

Contents

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)

(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 $-$ )}] [Performing multiplication]
= 8 $-$ [28 ÷ {34 $-$ 20}] [Removing parentheses]
= 8 $-$ [28 ÷ 14] [Removing braces]
= 8 $-$ 2 [Removing square brackets]

(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) $5\frac{1}{7}-$ { $3\frac{3}{10}$ ÷ $simplification questions for class 7$ }

(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