Simplification questions for class 7 are mathematical problems that involve simplifying mathematical expressions or equations. To solve this type of questions the students need to know to apply basic arithmetic operations such as addition, subtraction, multiplication, and division, as well as knowledge of mathematical rules such as the order of operations, BODMAS and PEMDAS Rule. They will also need to carefully read and interpret the problem to identify the correct mathematical operations to apply. It is important for students to double-check their answers to ensure accuracy.

These questions may also require students to use concepts such as brackets, fractions, decimals, and percentages to simplify the given expression. Simplification questions may be presented in various formats, such as multiple-choice questions, fill-in-the-blank questions, or open-ended questions.

In general, simplification questions for class 7 pdf are made to test a student’s understanding of fundamental mathematical ideas and operations as well as their ability to use them to solve problems. These questions are important for building a strong foundation in mathematics, which will be useful in higher grades and in real-life situations.

In this article, different types of questions on simplifications are explained here step by step. To help students gain confidence and improve their skills, the blog post will include plenty of examples and practice questions. We will provide step-by-step solutions and explain the reasoning behind each step.

Students will have a solid understanding of simplification and the method that are used to solve difficult simplification questions by the end of this blog post. This blog post will be a valuable resource for class 7 students, whether they are preparing for an exam or simply looking to improve their maths skills.

Contents

Simplification Questions for Class 7 with Answers (Solved examples)

(1)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

(2) $23-[23-\left \{ 23-(23-\overline{23-23})\right \}]$

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

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

(4) $\mathbf{4+\frac{1}{5}}$ [(-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

(5) 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]
=6

(6) 16 − 2 ÷ 7 + 6 × 2
Sol. 16 − 2 ÷ 7 + 6 × 2= $16-\frac{2}{7}+6 \times 2$
= $16-\frac{2}{7}+12$
= $28-\frac{2}{7}$
= $\frac{196-2}{7}=\frac{194}{7}$
= $27\frac{5}{7}$

(7) 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

(8) $25-\frac{1}{2}\left \{ 5+4-\left ( 3+2-1+3 \right ) \right \}$
Sol. $25-\frac{1}{2}\left \{ 5+4-\left ( 3+2-1+3 \right ) \right \} =25-\frac{1}{2}\left \{ 5+4-\left ( 8-1 \right ) \right \}$
= $25-\frac{1}{2}\left \{ 5+4-7 \right \}$
= $25-\frac{1}{2}\left \{ 9-7 \right \}$
= $25-\frac{1}{2}\times 2$
= 25 -1
= 24

9) 27 – [38 – {46 – (15 – 13 – 2)}]
Sol. 27 – [38 – {46 – (15 – 13 – 2)}] =27 – [38 – {46 – (15 – 15)}]
= 27 – [38 – {46 – 0}]
= 27 – [38 – 46 ]
= 27 – [38 – 46 ]
=27 -(-8)
= 27+8
= 35

10) 36 – [18 – {14 – (15 – 4 ÷ 2 x 2)}]
Sol. 36 – [18 – {14 – (15 – 4 ÷ 2 x 2)}] = 36 – [18 – {14 – (15 – 2 x 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3]
= 36 – 15
=21

Fraction simplification questions for class 7

(1) $\left ( \frac{3}{4}+\frac{1}{2} \right )\div 2$
Solution : $\left ( \frac{3}{4}+\frac{1}{2} \right )\div 2$ = $\left ( \frac{3+2}{4} \right )\div 2$

= $\frac{5}{4} \times \frac{1}{2}$

= $\frac{5}{8}$

(2) $6+\left \{ \frac{4}{3} +(\frac{3}{4}-\frac{1}{3})\right \}$

Solution : $6+\left \{ \frac{4}{3} +(\frac{3}{4}-\frac{1}{3})\right \}$ = $6+\left \{ \frac{4}{3} +(\frac{9-4}{12})\right \}$

= $6+\left \{ \frac{4}{3} +\frac{5}{12}\right \}$

= $6+\left \{\frac{16+5}{12}\right \}$

= $6+\frac{21}{12}$

= $\frac{72+21}{12}$

= $\frac{93}{12}$

= $7\frac{3}{4}$

(3) $\left ( \frac{3}{4}+\frac{1}{2} \right )\div 2$

Solution: $\left ( \frac{3}{4}+\frac{1}{2} \right )\div 2= \left ( \frac{6+4}{8} \right )\div 2$

= $\frac{10}{8} \times \frac{1}{2}$

= $\frac{5}{8}$

(4) $\frac{2}{3}\times \left ( \frac{4}{5} -\frac{1}{5}\right )$

Solution: $\frac{2}{3}\times \left ( \frac{4}{5} -\frac{1}{5}\right )$ = $\frac{2}{3}\times \left ( \frac{4-1}{5}\right )$

= $\frac{2}{3}\times \frac{3}{5}$

= $\frac{2}{5}$

(5) $\frac{1}{2}+\frac{3}{4}\div \left ( \frac{2}{3}-\frac{1}{6} \right )$

Solution : $\frac{1}{2}+\frac{3}{4}\div \left ( \frac{2}{3}-\frac{1}{6} \right )=\frac{1}{2}+\frac{3}{4}\div \left (\frac{4-1}{6} \right )$

= $\frac{1}{2}+\frac{3}{4}\div \left (\frac{3}{6} \right )$

= $\frac{1}{2}+\frac{3}{4}\times \left (\frac{6}{3} \right )$

= $\frac{1}{2}+\frac{3}{2}$

$=\frac{4}{2}$

= 2

(6) $8-\left \{ \frac{3}{2}+\left ( \frac{3}{5}-\frac{1}{2} \right ) \right \}$

Solution : $8-\left \{ \frac{3}{2}+\left ( \frac{3}{5}-\frac{1}{2} \right ) \right \} = 8-\left \{ \frac{3}{2}+\left ( \frac{6-5}{10} \right ) \right \}$

= $8-\left \{ \frac{3}{2}+ \frac{1}{10} \right \}$

= $8-\left \{ \frac{15+1}{10} \right \}$

= $8-\frac{16}{10}$

= $8-\frac{8}{5}$

= $\frac{40-8}{5}$

= $\frac{32}{5}$

= $6\frac{2}{5}$

(7) $2\frac{3}{4}-\left [ 3\frac{1}{8}\div \left \{ 5-\left ( 4\frac{2}{3} -\frac{11}{12}\right ) \right \} \right ]$

Solution : $2\frac{3}{4}-\left [ 3\frac{1}{8}\div \left \{ 5-\left ( 4\frac{2}{3} -\frac{11}{12}\right ) \right \} \right ] =\frac{11}{4}-\left [ \frac{25}{8}\div \left \{ 5-\left ( \frac{14}{3}-\frac{11}{12} \right ) \right \} \right ]$

$=\frac{11}{4}-\left [ \frac{25}{8}\div \left \{ 5-\left ( \frac{56-11}{12} \right ) \right \} \right ]$

$=\frac{11}{4}-\left [ \frac{25}{8}\div \left \{ 5-\left ( \frac{45}{12} \right ) \right \} \right ]$

$=\frac{11}{4}-\left [ \frac{25}{8}\div \left \{ \frac{60-45}{12} \right \} \right ]$

$=\frac{11}{4}-\left [ \frac{25}{8}\div \frac{5}{4} \right ]$

$=\frac{11}{4}-\left [ \frac{25}{8}\times \frac{4}{5} \right ]$

= $\frac{11}{4}-\frac{5}{2}$

= $\frac{1}{4}$

(8) $1\frac{1}{5}\div \left \{ 2\frac{1}{3}-\left ( 5+\overline{2-3} \right ) \right \}-3\frac{1}{2}$

Solution : $1\frac{1}{5}\div \left \{ 2\frac{1}{3}-\left ( 5+\overline{2-3} \right ) \right \}-3\frac{1}{2} = \frac{6}{5}\div \left \{ \frac{7}{3}-\left ( 5+\overline{2-3} \right ) \right \}-\frac{7}{2}$

= $\frac{6}{5}\div \left \{ \frac{7}{3}-\left ( 5+(-1) \right ) \right \}-\frac{7}{2}$

= $\frac{6}{5}\div \left \{ \frac{7}{3}-4 \right \}-\frac{7}{2}$

= $\frac{6}{5}\div \left \{ \frac{7-12}{3} \right \}-\frac{7}{2}$

= $\frac{6}{5}\div \left \{ \frac{-5}{3} \right \}-\frac{7}{2}$

= $\frac{6}{5}\times \left \{ -\frac{3}{5} \right \}-\frac{7}{2}$

= $-\frac{18}{25}-\frac{7}{2}$

= $-\frac{211}{ 50}$

= $-4\frac{11}{50}$

(9) $\frac{6}{5} of \left ( 3\frac{1}{3}-2\frac{1}{2} \right )\div \left ( 2\frac{5}{21}-2 \right )$

Solution : $\frac{6}{5} of \left ( 3\frac{1}{3}-2\frac{1}{2} \right )\div \left ( 2\frac{5}{21}-2 \right )=\frac{6}{5} of \left ( \frac{10}{3}-\frac{5}{2} \right )\div \left ( \frac{47}{21}-2 \right )$

= $\frac{6}{5} of \left ( \frac{20-15}{6} \right )\div \left ( \frac{47 -42}{21} \right )$

= $\frac{6}{5} of \left ( \frac{5}{6} \right )\div \left ( \frac{5}{21} \right )$

= $\frac{6}{5} \times \left ( \frac{5}{6} \right )\times \left ( \frac{21}{5} \right )$

= $\frac{21}{5}$

= $4\frac{1}{5}$

(10 ) $\frac{2}{3}+\frac{1}{6}\times \frac{5}{4}$

Solution : $\frac{2}{3}+\frac{1}{6}\times \frac{5}{4} = \frac{2}{3}+\frac{5}{24}$

= $\frac{16+5}{ 24}$

= $\frac{21}{ 24}$

= $\frac{7}{8}$

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

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נערות ליווי באשקלון says:

April 16, 2023 at 4:49 am

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Simplification Questions for Class 7 PDF Free Download

Simplification Questions for Class 7 with Answers (Solved examples)

Simplification questions for class 7 ( Unsolved questions with answers)

Simplification Questions for Class 7 PDF Download

Bina singh

One thought on “Simplification Questions for Class 7 PDF Free Download”

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Simplification Questions for Class 7 with Answers (Solved examples)

Simplification questions for class 7 ( Unsolved questions with answers)

Simplification Questions for Class 7 PDF Download

Bina singh

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