June 20, 2024

# Easy simplification questions for class 5 with answers and practice questions

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

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

(VII)  $\mathbf{56\div&space;14&space;\times&space;3-10\div&space;5+1}$
Sol. $56\div&space;14&space;\times&space;3-10\div&space;5+1&space;=&space;4&space;\times&space;3-2+1$                 (solving division)
= 12-2+1                                                                                           (solving multiplication)
=11

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

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

(X)  $\mathbf{100&space;\times&space;10&space;-100&space;+2000\div&space;100&space;}$
Sol. $100&space;\times&space;10&space;-100&space;+2000\div&space;100&space;=100&space;\times&space;10&space;-100&space;+20$                                   (solving division)
=  1000 -100 +20                                                   (solving multiplication)
= 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) $[105\div\left&space;\{&space;23+2(9-\overline{5-2})&space;\right&space;\}&space;]$
Sol. $105\div\left&space;\{&space;23+2(9-\overline{5-2})&space;\right&space;\}&space;]=105\div\left&space;\{&space;23+2(9-3)&space;\right&space;\}&space;]$
$=[105\div&space;\left&space;\{&space;23+2&space;\times&space;6\right&space;\}&space;]$
$=[105\div&space;\left&space;\{&space;23+12\right&space;\}&space;]$
$=[105\div&space;35&space;]$
= 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 )  $\frac{4}{9}\times&space;\frac{18}{5}\div&space;\frac{24}{5}$

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

=$\frac{4}{9}\times&space;\frac{3}{4}$

=$\frac{1}{3}$

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

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

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

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

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

=$\frac{147}{21}$

=7

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

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

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

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

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

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

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

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) $2\frac{3}{4}\div&space;2\frac{2}{3}\div&space;1\frac{1}{12}$

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

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

= $\frac{99}{104}$

xxvi)  $2-[3-\left&space;\{&space;6-(5-\overline{4-3})&space;\right&space;\}]$
Solution :$2-[3-\left&space;\{&space;6-(5-\overline{4-3})&space;\right&space;\}]$
= $2-[3-\left&space;\{&space;6-(5-1&space;\right&space;)\}]$
= $2-[3-\left&space;\{&space;6-4&space;\right\}]$
= 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)}] =$[29&space;-(-2)&space;\{6&space;-&space;4\}]&space;\div&space;[3&space;\times&space;\{5&space;+&space;6\}]$
=$[29&space;-(-2)&space;\times&space;2]&space;\div&space;[3&space;\times&space;11]$
= $[29&space;+4]&space;\div&space;33$
= $33&space;\div&space;33$
= 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) $3\frac{1}{2}+\frac{5}{8}\div&space;\frac{3}{4}-\frac{1}{2}\times&space;2\frac{1}{2}$
Solution : $3\frac{1}{2}+\frac{5}{8}\div&space;\frac{3}{4}-\frac{1}{2}\times&space;2\frac{1}{2}$ = $\frac{7}{2}+\frac{5}{8}\div&space;\frac{3}{4}-\frac{1}{2}\times&space;\frac{5}{2}$

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

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

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

=$\frac{37}{12}$

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

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

=$\frac{192}{216}$

=$\frac{8}{9}$

## 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

File Name: simplification-question-for-class-5.pdf

#### Simplification questions for class 5  (Quiz)

341
Created on

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

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2.  100÷10+10 x 10

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3.   49 ÷ 7 x 7+5 x 3 -2

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4 .    3 x 8 ÷ 4

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4+4+4+4÷ 4

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6.   80+800÷ 8

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7.  16+8÷ 4-2 x 3

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5 x 50+57-57÷ 57

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80+[20 x {20-(10÷5)}]

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