Simplification questions for class 5

Class 5 students frequently struggle with questions that require them to simplify complex mathematical expressions to their simplest form. Many students often get confused by seeing complex mathematical expressions. Simplification questions for class 5 are designed to help students to solve complex mathematical expressions by simplifying them into simpler forms. These questions are essential in building a strong foundation in mathematics and are often included in primary school level examinations.

In this blog, we have given different types of simplification questions for class 5, that includes basic addition and subtraction, multiplication and division. By solving this question , you can improve you simplification skill and you can solve difficult simplification problem. We will provide step-by-step solutions and explanations to help students understand the concepts involved and build their concept in solving mathematical problems. In addition to providing helpful tips and examples, this blog will also include practice questions with answers , a quiz based on simplification questions for class 5 and a PDF in which some simplification questions are given.

To simplify a numerical expression having two or more operations, we use ‘BODMAS’ , where the letters B, O, D, M, A, S stand for Brackets , orders(powers/indices or roots) , Division, Multiplication, Addition and Subtraction respectively. The following questions on simplifications has different types of questions that can be practiced by the students to get more ideas to simplify the expressions.

Contents

Simplification questions for class 5 (solved examples)

Simplify :

(I) 14 x 7+36÷4+14
sol. 14 x 7+36÷4+14= 14 x 7+9 +14 (performing division )
= 98+9+14 (performing multiplication)
=121

(II) 100-56÷7+15 x 2
Sol. 100-56÷7+15 x 2= 100-8+15 x 2 (simplifying division)
= 100-8+30 ( simplifying multiplication)
= 100+30-8
= 130-8 = 122

$19 - [4 + {16 - (12 - 2)}]$ Sol. 19 – [4 + {16 – (12 – 2)}] = 19 – [4 + {16 – 10}] (Removing small brackets)
= 19 – [4 + 6] (Removing curly brackets)
= 19 – 10 (Removing square brackets)
= 9

(IV ) 12-[20 ÷{8-2(9-5-2)}]
Sol. 12-[20 ÷{8-2(9-5-2)}]= 12-[20 ÷{8-2 x 2}] (removing small brackets)
=12-[20 ÷{8-4}]
=12-[20 ÷4] (removing braces)
= 12-5 (removing big brackets)
=7

(V) [24 ÷ {10- (8- $\overline{6-2}$ )}]
Sol. [24 ÷ {10- (8- $\overline{6-2}$

)}] = [24 ÷ {10- (8- 4)}] (removing bar)
=[24 ÷ {10- 4}] (removing small brackets)
=[24 ÷ 6] (removing braces)
=4 (removing big brackets)

(VI) $\mathbf{60 -48 \div 6 \times 4 +8}$
Sol. $60 -48 \div 6 \times 4 +8$ = $60 -8 \times 4 +8$ (solving division)
= $60- 32 +8$ (solving multiplication)
= 68 -32 ( solving addition)
= 36

(VII) $Simplification questions for class 5$
Sol. $56\div 14 \times 3-10\div 5+1 = 4 \times 3-2+1$ (solving division)
= 12-2+1 (solving multiplication)
=13 -2 (solving addition)
=11

(VIII) $\mathbf{\left [ \left \{ \left ( 30- \overline{9-6} \right )\div 3 \right \}\times 6+6 \right ]}$
Sol. $\left [ \left \{ \left ( 30- \overline{9-6} \right )\div 3 \right \}\times 6+6 \right ]=\left [ \left \{ \left ( 30- 3 \right )\div 3 \right \}\times 6+6 \right ]$ (removing bar)
= $\left [ \left \{27 \div 3 \right \}\times 6+6 \right ]$ (removing small brackets)
= $\left [9\times 6+6 \right ]$ (removing curly braces)
= 54 +6 (solving multiplication)
= 60

(IX) $\mathbf{\left [ 40\div \left \{ 19-3\left ( 6-\overline{4-1} \right ) \right \} \right ]}$
Sol. $\left [ 40\div \left \{ 19-3\left ( 6-\overline{4-1} \right ) \right \} \right ]= \left [ 40\div \left \{ 19-3\left ( 6-3 \right ) \right \} \right ]$ (removing bar)
= $\left [ 40\div \left \{ 19-3\times 3 \right \} \right ]$ (removing small brackets)
= $\left [ 40\div \left \{ 19-9 \right \} \right ]$ (solving multiplication under curly braces}
= $\left [ 40\div 10 \right ]$ (removing curly braces)
= 4

(X) $\mathbf{100 \times 10 -100 +2000\div 100 }$
Sol. $100 \times 10 -100 +2000\div 100 =100 \times 10 -100 +20$ (solving division)
= 1000 -100 +20 (solving multiplication)
= 1020 -100 (solving addition)
= 920

(XI) 6+6-6 ÷ 6 x 6
Sol. 6+6-6 ÷ 6 x 6 = 6+6-1 x 6
= 6+6-1 x 6
=6+6-6
=6

(XII) 21÷ 7+16-5 x 3
Sol. 21÷ 7+16-5 x 3=3+16-5 x 3
=3+16-15
=19-15
=4

(XIII) 5 x 50 +57-57 ÷ 57
Sol. 5 x 50 +57-57 ÷ 57 =5 x 50 +57-1
= 250+57-1
=307-1
=306

(XIV)5751 x {45-(90 ÷ 2)}
Sol. 5751 x {45-(90 ÷ 2)} =5751 x {45-45}
=5751 x 0
=0

(XV) $[105\div\left \{ 23+2(9-\overline{5-2}) \right \} ]$
Sol. $105\div\left \{ 23+2(9-\overline{5-2}) \right \} ]=105\div\left \{ 23+2(9-3) \right \} ]$
$=[105\div \left \{ 23+2 \times 6\right \} ]$
$=[105\div \left \{ 23+12\right \} ]$
$=[105\div 35 ]$
= 3

(xvi)17 + [24 – {30 ÷ 2 – (4 – 8 ÷ 2) ÷ 2}]
Solution: 17 + [24 – {30 ÷ 2 – (4 – 8 ÷ 2) ÷ 2}]= 17 + [24 – {30 ÷ 2 – (4 – 4) ÷ 2}]
= 17 + [24 – {30 ÷ 2 – 0 ÷ 2}]
= 17 + [24 – {30 ÷ 2 – 0}]
= 17 + [24 – {15 – 0}]
= 17 + [24 – 15]
= 17 + [24 – 15]
=17+9
=26

(xvii) 18 – [18 – {42 ÷ 7 – (20 – 10 ÷ 2) ÷ 5}]
Solution: 18 – [18 – {42 ÷ 7 – (20 – 10 ÷ 2) ÷ 5}] =18 – [18 – {42 ÷ 7 – (20 – 5) ÷ 5}]
=18 – [18 – {42 ÷ 7 – 15 ÷ 5}]
=18 – [18 – {6 – 3}]
= 18-[18-3]
=18- 15
=3

(xviii) 55 + [18 – {72 ÷ 4 – (5 – 15 ÷ 5) ÷ 2}]
Solution: 55 + [18 – {72 ÷ 4 – (5 – 15 ÷ 5) ÷ 2}]=55 + [18 – {72 ÷ 4 – (5 – 3) ÷ 2}]
=55 + [18 – {72 ÷ 4 – 2 ÷ 2}]
=55 + [18 – {72 ÷ 4 – 1}]
=55 + [18 – {18 – 1}]
=55 + [18 – 17]
=55+1
=56

(xix) 40 – [30 – {54 ÷ 9 – (8 – 12 ÷ 4) ÷ 5}]
Solution: 40 – [30 – {54 ÷ 9 – (8 – 12 ÷ 4) ÷ 5}] =40 – [30 – {54 ÷ 9 – (8 – 3) ÷ 5}]
=40 – [30 – {54 ÷ 9 – 5 ÷ 5}]
=40 – [30 – {6 – 1}]
=40 – [30 – 5]
=40-25
=15

(xx) 63 + [28 – {90 ÷ 6 – (7 – 14 ÷ 2) ÷ 3}]
Solution: 63 + [28 – {90 ÷ 6 – (7 – 14 ÷ 2) ÷ 3}] = 63 + [28 – {90 ÷ 6 – (7 – 7) ÷ 3}]
= 63 + [28 – {90 ÷ 6 – 0 ÷ 3}]
= 63 + [28 – {90 ÷ 6 – 0 }]
= 63 + [28 – {15-0 }]
= 63 + [28 – 15]
=63+13
=76

(xxi ) $\frac{4}{9}\times \frac{18}{5}\div \frac{24}{5}$

Solution: $\frac{4}{9}\times \frac{18}{5}\div \frac{24}{5}$ = $\frac{4}{9}\times \frac{18}{5}\times \frac{5}{24}$

= $\frac{4}{9}\times \frac{3}{4}$

= $\frac{1}{3}$

(xxii) $\left [ \frac{2}{5}+\frac{1}{7} \right ]\div \left [ \frac{1}{5}-\frac{1}{8} \right ]-\frac{5}{21}$

Solution: $\left [ \frac{2}{5}+\frac{1}{7} \right ]\div \left [ \frac{1}{5}-\frac{1}{8} \right ]-\frac{5}{21}= \left [ \frac{14+5}{35} \right ]\div \left [ \frac{8-5}{40} \right ]-\frac{5}{21}$

= $\frac{19}{35}\div \frac{3}{40}-\frac{5}{21}$

= $\frac{19}{35}\times \frac{40}{3}-\frac{5}{21}$

= $\frac{152}{21}-\frac{5}{21}$

= $\frac{147}{21}$

xxiii) $1\frac{3}{5}-\frac{2}{3}\div \frac{12}{13}+\frac{7}{5}\times \frac{1}{3}$

Solution: $1\frac{3}{5}-\frac{2}{3}\div \frac{12}{13}+\frac{7}{5}\times \frac{1}{3}$ = $\frac{8}{5}-\frac{2}{3}\div \frac{12}{13}+\frac{7}{5}\times \frac{1}{3}$

= $\frac{8}{5}-\frac{2}{3}\times \frac{13}{12}+\frac{7}{5}\times \frac{1}{3}$

= $\frac{8}{5}- \frac{13}{18}+ \frac{7}{15}$

= $\frac{(144-65+42)}{90}$

= $\frac{186-65}{90}$

= $\frac{121}{90}=1\frac{31}{90}$

xxiv) $\mathbf{\frac{7+7+\frac{7}{7}}{\frac{(7+7+7)}{7}}}$

Solution : $\mathbf{\frac{7+7+\frac{7}{7}}{\frac{(7+7+7)}{7}}}$ = $\mathbf{\frac{7+7+1}{\frac{21}{7}}}$

= $\frac{15}{3}$
=5

xxv) $2\frac{3}{4}\div 2\frac{2}{3}\div 1\frac{1}{12}$

Solution: $2\frac{3}{4}\div 2\frac{2}{3}\div 1\frac{1}{12}$ = $\frac{11}{4}\div \frac{8}{3}\div \frac{13}{12}$

= $\frac{11}{4}\times \frac{3}{8}\times \frac{12}{13}$

= $\frac{99}{104}$

xxvi) $2-[3-\left \{ 6-(5-\overline{4-3}) \right \}]$
Solution : $2-[3-\left \{ 6-(5-\overline{4-3}) \right \}]$ = $2-[3-\left \{ 6-(5-1 \right )\}]$
= $2-[3-\left \{ 6-4 \right\}]$
= 2-[3-2]
= 2-1
=1

(xxvii) [29 – (-2) {6 – (7 – 3)}] ÷ [3 x {5 + (-3) x (-2)}]
Solution : [29 – (-2) {6 – (7 – 3)}] ÷ [3 x {5 + (-3) x (-2)}] = $[29 -(-2) \{6 - 4\}] \div [3 \times \{5 + 6\}]$
= $[29 -(-2) \times 2] \div [3 \times 11]$
= $[29 +4] \div 33$
= $33 \div 33$
= 1

(xviii) 0.8{0.75 ÷ (1.35 – 0.6)} – 0.8
Solution: 0.8{0.75 ÷ (1.35 – 0.6)} – 0.8= 0.8{0.75 ÷ 0.75} – 0.8
= 0.8 x 1 – 0.8
= 0.8 – 0.8
=0

(xix)8{1 – {3 + (5 – 6 + 7)}}
Solution: 8{1 – {3 + (5 – 6 + 7)}} =8{1 – {3 + (12-6)}}
=8{1 – {3 + 6}}
=8{1 – 9}
= 8 x (-8)
=-64

(xxx) $3\frac{1}{2}+\frac{5}{8}\div \frac{3}{4}-\frac{1}{2}\times 2\frac{1}{2}$
Solution : $3\frac{1}{2}+\frac{5}{8}\div \frac{3}{4}-\frac{1}{2}\times 2\frac{1}{2}$ = $\frac{7}{2}+\frac{5}{8}\div \frac{3}{4}-\frac{1}{2}\times \frac{5}{2}$

= $\frac{7}{2}+\frac{5}{8}\times \frac{4}{3}-\frac{1}{2}\times \frac{5}{2}$

= $\frac{7}{2}+\frac{5}{6}- \frac{5}{4}$

= $\frac{42+10-15}{12}$

= $\frac{37}{12}$

(xxxi) $\frac{26\times 10-4\times 17 }{14\times 14+20}$

Solution: $\frac{26\times 10-4\times 17 }{14\times 14+20}=\frac{260-68 }{196+20}$

= $\frac{192}{216}$

= $\frac{8}{9}$

(xxxii) $2\times\left[\{(64+16)\div 5+26\}+3\right]$
Solution: $2\times\left[\{(64+16)\div 5+26\}+3\right]$ = $2\times\left[\{80\div 5+26\}+3\right]$
= $2\times\left[\{16+26\}+3\right]$
= $2\times\left[42+3\right]$
= 2 x 45
=90 $38 - 46 - (15 - 13 - 2)$

Simplify questions for class 5 (Unsolved problems for practice)

(i) 30÷6+10-2 x 5

(ii) 48÷16 x 2+17 – 9

(iii) (18+10)-(3 x 6)

(iv) (10 x 8) ÷ (20÷5)

(v) {20-(24- 10)}+7

(vi) {20+(15+5)}-5

(vii) 12 x 3 ÷3 x 4-2+6

(viii) 121 ÷ 11+ 29 – 2 x 10

(ix) 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]

(x) 37 – 6 × 4 + 32 ÷ 8

(xi) 27 − [18 − {16 − (5 − $\overline{4-1}$ )}]

(xii) 100 x 10 -100+2000÷100

(xiii) 5751 x {45-(90÷2)}

(xiv) 289+153÷17-8 x 19

(xv) 7823-128÷16 of 4 -3973

(xvi) [{64-(12+13)}÷3]+15

(xvii) 4+4+4+4 ÷ 4

(xviii)20- {18÷(7-2+1)}

(xix) 2[19-{7+(12÷4)}]

(xx) [{(30- $\overline{9-6}$ )÷3} x 6+6]

Ans.(i) 5 (ii) 14 (iii) 10 (iv) 20 (v) 13 (vi) 35 (vii) 52 (viii) 20 (ix) 26 (x) 17 (xi) 23 (xii) 920

(xiii) 0 (xiv)146 (xv) 3848 (xvi) 28 (xvii) 13 (xviii) 17 (xix) 18 (xx) 60

इस पोस्ट को हिंदी में पढ़ने क लिए यहाँ क्लिक करें :- सरलीकरण के सवाल class 5

Simplification questions for class 5 PDF

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Simplification questions for class 5 (Quiz)

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Created on October 20, 2021

Simplification Grade 5

1 / 10

Which of the following operation is performed first in simplifying a numerical expression?

+

-

x

÷

2 / 10

2. 100÷10+10 x 10

1

50

110

200

3 / 10

3. 49 ÷ 7 x 7+5 x 3 -2

6

14

54

62

4 / 10

4 . 3 x 8 ÷ 4

6

45

12

14

5 / 10

4+4+4+4÷ 4

11

12

16

13

6 / 10

6. 80+800÷ 8

180

70

90

100

7 / 10

7. 16+8÷ 4-2 x 3

14

12

13

19

8 / 10

5 x 50+57-57÷ 57

306

0

150

21

9 / 10

80+[20 x {20-(10÷5)}]

440

430

400

80

10 / 10

10. 10 x 10 +[400÷ {100-(50- $\overline{3 \times 10}$ )}]

100

105

107

50

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