# RS Aggarwal Class 6 Solutions Chapter 6 - Simplification

## RS Aggarwal Class 6 Chapter 6 - Simplification Solutions Free PDF

Before solving the questions of simplification let us understand what is a simplification. Simplification is the process where a mathematical expression is replaced by a similar or equivalent one. Learn the concepts of simplification from RS Aggarwal for class 6. While solving, if you face any difficulties you can refer to RS Aggarwal class 6 solutions chapter 6 simplification. To help students we have provided the RS Aggarwal solutions here.

Exercise 6A

Simplify:

Q1. 21 â€“ 12 Ã· 3 Ã— 2

Given expression:

= 21 â€“ 12 Ã· 3 Ã— 2

= 21 â€“ 4 Ã— 2 [performing division]

= 21 â€“ 8 [performing multiplication]

= 13 [performing subtraction]

Q2. 16 + 8 Ã· 4 â€“ 2 Ã— 3

Given expression:

= 16 + 8 Ã· 4 â€“ 2 Ã— 3

= 16 + 2 â€“ 2 Ã— 3 [performing division]

= 16 + 2 â€“ 6 [performing multiplication]

= 18 â€“ 6 [performing addition]

= 12 [performing subtraction]

Q3. 13 â€“ (12 â€“ 6 Ã· 3)

Given expression:

= 13 â€“ (12 â€“ 6 Ã· 3)

= 13 â€“ (12 – 2) [performing division]

= 13 â€“ 10 = 3 [performing subtraction]

Q4. 19 â€“ [4 + {16 â€“ (12 â€“ 2)}]

Given expression:

= 19 â€“ [4 + {16 â€“ (12 â€“ 2)}]

= 19 â€“ [4 + {16 â€“ 10}] [removing parentheses]

= 19 â€“ [4 + 6] [removing braces]

= 19 â€“ 10 [removing square brackets]

= 9

Q5. 36 â€“ [18 â€“ {14 â€“ (15 â€“ 4 Ã· 2 Ã— 2)}]

Given expression:

= 36 â€“ [18 â€“ {14 â€“ (15 â€“ 4 Ã· 2 Ã— 2)}]

= 36 â€“ [18 â€“ {14 â€“ (15 â€“ 2 Ã— 2)}] [performing division]

= 36 â€“ [18 â€“ {14 â€“ (15 â€“ 4)}] [performing multiplication]

= 36 â€“ [18 â€“ {14- 11}] [removing parentheses]

= 36 â€“ [18 â€“ 3] [removing braces]

= 36 â€“ 15 [removing square brackets]

= 21

Q6. 27 â€“ [18 â€“ {16 â€“ (5 â€“ $\overline {4 – 1}$)}]

Given expression:

= 27 â€“ [18 â€“ {16 â€“ (5 â€“ $\overline {4 – 1}$)}]

= 27 â€“ [18 â€“ {16 â€“ (5 â€“ 3)}] [removing bar]

= 27 â€“ [18 â€“ {16 â€“ 2}] [removing parentheses]

= 27 â€“ [18 â€“ 14] [removing braces]

= 27 â€“ 4 [removing square brackets]

= 23

Q7. $4\frac {4} {5} \div \frac {3} {5} \; of \; 5 + \frac {4} {5} \times \frac {3} {10} – \frac {1} {5}$

Given expression:

= $4\frac {4} {5} \div \frac {3} {5} \; of \; 5 + \frac {4} {5} \times \frac {3} {10} – \frac {1} {5}$

= $4\frac {4} {5} \div \frac {3} {5} \times \frac {5} {1} + \frac {4} {5} \times \frac {3} {10} – \frac {1} {5}$

= $\frac {24} {5} \div \frac {3} {1} + \frac {4} {5} \times \frac {3} {10} – \frac {1} {5}$

= $\frac {24} {5} \times \frac {1} {3} + \frac {4} {5} \times \frac {3} {10} – \frac {1} {5}$

= $\frac {8} {5} + \frac {4} {5} \times \frac {3} {10} – \frac {1} {5}$

= $\frac {8} {5} + \frac {6} {25} – \frac {1} {5}$

= $\frac {40 + 6 – 5} {25} = \frac {41} {25} = 1 \frac {16}{25}$

Q8. $\left (\frac {2}{3} + \frac{4}{9} \right ) of \frac{3}{5} \div \frac{2}{3} \times \frac{1}{4} – \frac{1}{3}$

Given expression:

= $\left (\frac {2}{3} + \frac{4}{9} \right ) of \frac{3}{5} \div 1\frac{2}{3} \times 1\frac{1}{4} – \frac{1}{3}$

= $\left ( \frac {6 + 4} {9} \right ) of \frac {3} {5} \div 1 \frac {2} {3} \times 1 \frac {1} {4} – \frac {1} {3}$

= $\frac {10} {9} \times \frac {3} {5} \div 1 \frac {2} {3} \times 1 \frac {1} {4} – \frac {1} {3}$

= $\frac {2} {3} \div \frac {5} {3} \times 1 \frac {1} {4} – \frac {1} {3}$

= $\frac {2} {3} \times \frac {3} {5} \times \frac {5} {4} – \frac {1} {3}$

= $\frac {2} {5} \times \frac {5} {4} – \frac {1} {3}$

= $\frac {1} {2} – \frac {1} {3}$

= $\frac {3 – 2} {6} = \frac {1} {6}$

Q9. $7\frac {1} {3} \div \frac {2} {3} \; of \; 2 \frac {1} {5} + 1 \frac {3} {8} \div 2 \frac {3} {4} â€“ 1 \frac {1} {2}$

The given expression

= $7\frac {1} {3} \div \frac {2} {3} \; of \; 2 \frac {1} {5} + 1 \frac {3} {8} \div 2 \frac {3} {4} â€“ 1 \frac {1} {2}$

= $\frac {22} {3} \div \frac {2} {3} \; of \; \frac {11} {5} + \frac {11} {8} \div \frac {11} {4} – \frac {3} {2}$

= $\frac {22} {3} \div \frac {2} {3} \; \times \; \frac {11} {5} + \frac {11} {8} \div \frac {11} {4} – \frac {3} {2}$

= $\frac {22} {3} \div \frac {22} {15} \; + \frac {11} {8} \div \frac {11} {4} – \frac {3} {2}$

= $\frac {22} {3} \times \frac {15} {22} \; + \frac {11} {8} \div \frac {11} {4} – \frac {3} {2}$

= $5 + \frac {11}{8} \times \frac {4}{11} – \frac {3} {2}$

= $5 + \frac {1} {2} – \frac {3} {2}$

= $\frac {10 + 1 – 3} {2} = \frac {8} {2} = 4$

Q10. $5\frac {1} {7}- \left [ 3\frac{3}{10}\div \left ( 2\frac{4}{5}-\frac{7}{10} \right ) \right ]$

Given expression:

= $5\frac {1} {7}- \left [ 3\frac{3}{10}\div \left ( 2\frac{4}{5}-\frac{7}{10} \right ) \right ]$

= $\frac {36} {7}- \left [ \frac{33}{10}\div \left ( \frac{14}{5}-\frac{7}{10} \right ) \right ]$

= $\frac {36} {7} – \left [ \frac {33} {10} \div \left ( \frac {28 – 7} {10}\right ) \right ]$

= $\frac {36} {7} – \left [ \frac {33} {10} \div \frac {21} {10} \right ]$

= $\frac {36} {7} – \left [ \frac {33} {10} \times \frac {10} {21}\right ]$

= $\frac {36} {7} – \frac {11} {7}$

= $\frac {36 – 11} {7} = \frac {25} {7} = 3 \frac {4} {7}$

Q11. $9\frac {3} {4} \div \left [ 2 \frac {1} {6} + \left [ 4 \frac {1} {3} – \left ( 1 \frac {1} {2} + 1 \frac {3} {4} \right ) \right ] \right ]$

Given expression

= $9\frac {3} {4} \div \left [ 2 \frac {1} {6} + \left [ 4 \frac {1} {3} – \left ( 1 \frac {1} {2} + 1 \frac {3} {4} \right ) \right ] \right ]$

= $\frac {39} {4} \div \left [ \frac {13} {6} + \left [ \frac {13} {3} – \left ( \frac {3} {2} + \frac {7} {4} \right ) \right ] \right ]$

= $\frac {39} {4} \div \left [ \frac {13} {6} + \left [ \frac {13} {3} – \left ( \frac {6 + 7} {4} \right ) \right ] \right ]$

= $\frac {39} {4} \div \left [ \frac {13} {6} + \left [ \frac {13} {3} – \frac {13} {4} \right ] \right ]$

= $\frac {39} {4} \div \left [ \frac {13} {6} + \left [ \frac {59 – 39} {12} \right ] \right ]$

= $\frac {39} {4} \div \left [ \frac {13} {6} + \frac {13} {12} \right ]$

= $\frac {39} {4} \div \left [ \frac {26 + 13} {12} \right ]$

= $\frac {39} {4} \div \frac {39} {12}$

= $\frac {39} {4} \times \frac {12} {39} = 3$

Q12. $4\frac {1} {10} – \left [ 2 \frac {1} {2} – \left [ \frac {5} {6} – \left ( \frac {2} {5} + \frac {3} {10} – \frac {4} {15} \right ) \right ] \right ]$

Given expression:

= $4\frac {1} {10} – \left [ 2 \frac {1} {2} – \left [ \frac {5} {6} – \left ( \frac {2} {5} + \frac {3} {10} – \frac {4} {15} \right ) \right ] \right ]$

= $\frac {41} {10} – \left [ \frac {5} {2} – \left [ \frac {5} {6} – \left ( \frac {2} {5} + \frac {3} {10} – \frac {4} {15} \right ) \right ] \right ]$

= $\frac{41} {10} – \left [ \frac {5} {2} – \left [ \frac {5} {6} – \left ( \frac {12 + 9 – 8 } {30} \right ) \right ] \right ]$

= $\frac {41} {10} – \left [ \frac {5} {2} – \left [ \frac {5} {6} – \frac {13} {30} \right ] \right ]$

= $\frac {41} {10} – \left [ \frac {5} {2} – \left [ \frac {25 – 13} {30} \right ] \right ]$

= $\frac {41} {10} – \left [ \frac {5} {2} – \frac {12} {30} \right ]$

= $\frac {41} {10} – \left [ \frac {75 – 12} {30} \right ]$

= $\frac {41} {10} – \frac {63} {30}$

= $\frac{123 – 63} {30} = \frac {60} {30} = 2$

Q13. $1 \frac {5} {6} + \left [ 2 \frac {2} {3} – \left [ 3 \frac {3} {4} \left ( 3 \frac {4} {5} \div 9 \frac {1} {2} \right ) \right ]\right ]$

Given expression:

= $1 \frac {5} {6} + \left [ 2 \frac {2} {3} – \left [ 3 \frac {3} {4} \left ( 3 \frac {4} {5} \div 9 \frac {1} {2} \right ) \right ]\right ]$

= $\frac {11} {6} + \left [ \frac {8} {3} – \left [ \frac {15} {4} \left ( \frac {19} {5} \div \frac {19} {2} \right ) \right ]\right ]$

= $\frac {11} {6} + \left [ \frac {8} {3} – \left [ \frac {15} {4} \left ( \frac {19} {5} \times \frac {2} {19} \right ) \right ]\right ]$

= $\frac {11} {6} + \left [ \frac {8} {3} – \left [ \frac {15} {4} \times \frac {2} {5} \right ] \right ]$

= $\frac {11} {6} + \left [ \frac {8} {3} – \frac {3} {2} \right ]$

= $\frac {11} {6} + \left [ \frac {16 – 9} {6} \right ]$

= $\frac {11} {6} + \frac {7} {6}$

= $\frac {18} {6} = 3$

Q14. $4 \frac {4} {5} \div \left [ 2\frac{1}{5} -\frac{1}{2}\left ( 1\frac{1}{4}-\overline{\overline{\frac{1}{4}}-\frac{1}{5} }\right )\right ]$

Given expression:

= $4 \frac {4} {5} \div \left [ 2\frac{1}{5} -\frac{1}{2}\left ( 1\frac{1}{4}-\overline{\overline{\frac{1}{4}}-\frac{1}{5} }\right )\right ]$

= $\frac {24} {5} \div \left [\frac{11}{5} -\frac{1}{2}\left (\frac{5}{4}-\overline{\overline{\frac{1}{4}}-\frac{1}{5} }\right )\right ]$

= $\frac {24} {5} \div \left [ \frac {11} {5} – \frac {1} {2} \left ( \frac {5} {4} – \frac {1} {20} \right ) \right ]$

= $\frac {24} {5} \div \left [ \frac {11} {5} – \frac {1} {2} \left ( \frac {25 – 1} {20} \right ) \right ]$

= $\frac {24} {5} \div \left [ \frac {11} {5} – \frac {1} {2} \times \frac {24} {20} \right ]$

= $\frac {24} {5} \div \left [ \frac{11} {5} – \frac {12} {20} \right ]$

= $\frac {24} {5} \div \left [ \frac {44 – 12} {20} \right ]$

= $\frac {24} {5} \div \frac {32} {20}$

= $\frac {24} {5} \times \frac {20} {32}$

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

Q15. $7 \frac {1} {2} – \left [ 2\frac {1} {4} \div \left [ 1 \frac {1} {4} – \frac {1} {2} \left ( \frac {3} {2} – \overline {\overline {\frac {1} {3}} – \frac {1} {6}} \right ) \right ] \right ]$

Given expression:

= $7 \frac {1} {2} – \left [ 2\frac {1} {4} \div \left [ 1 \frac {1} {4} – \frac {1} {2} \left ( \frac {3} {2} – \overline {\overline {\frac {1} {3}} – \frac {1} {6}} \right ) \right ] \right ]$

= $\frac {15} {2} – \left [ \frac {9} {4} \div \left [ \frac {5} {4} – \frac {1} {2} \left ( \frac {3} {2} – \overline {\overline {\frac {1} {3}} – \frac {1} {6}} \right ) \right ] \right ]$

= $\frac {15} {2} – \left [ \frac {9} {4} \div \left [ \frac {5} {4} – \frac {1} {2} \left ( \frac {3} {2} – \frac {1} {6} \right ) \right ] \right ]$

= $\frac {15} {2} – \left [ \frac {9} {4} \div \left [ \frac {5} {4} – \frac {1} {2} \left ( \frac {9 – 1} {6} \right ) \right ] \right ]$

= $\frac {15} {2} – \left [ \frac {9} {4} \div \left [ \frac {5} {4} – \frac {1} {2} \times \frac {4} {3} \right ] \right ]$

= $\frac {15} {2} – \left [ \frac {9} {4} \div \left [ \frac {5} {4} – \frac {2} {3} \right ] \right ]$

= $\frac {15} {2} – \left [ \frac {9} {4} \div \left [ \frac {15 – 8} {12} \right ] \right ]$

= $\frac {15} {2} -\left [ \frac {9} {4} \div \frac {7} {12} \right ]$

= $\frac {15} {2} – \left [ \frac {9} {4} \times \frac {12} {7} \right ]$

= $\frac {15} {2} – \frac {27} {7}$

= $\frac {105 – 54} {14} = \frac {51} {14} = 3 \frac {9} {14}$

Exercise 6B

OBJECTIVE QUESTIONS

Mark correct against the correct answer in each of the following:

Q1. 8 + 4 Ã· 2 Ã— 5 =?

(a) 30 (b) 50 (c) 18 (d) none of these

(c) 18

Explanation:

= 8 + 4 Ã· 2 Ã— 5

= 8 + 2 Ã— 5

= 8 + 10 = 18

Q2. 54 Ã· 3 of 6 + 9 =?

(a) 117 (b) 12 (c) 6/5 (d) none of these

(b) 12

Explanation:

= 54 Ã· 3 of 6 + 9

= 54 Ã· (3 Ã· 6) + 9

= 54 Ã· 18 + 9

= 3 + 9 = 12

Q3. 13 â€“ (12 â€“ 6 Ã· 3) =?

(a) 11 (b) 3 (c) 7/3 (d) none of these

(b) 3

Explanation:

= 13 â€“ (12 â€“ 6 Ã· 3)

= 13 â€“ (12 – 2)

= 13 â€“ 10 = 3

Q4. 1001 Ã· 11 of 13 =?

(a) 7 (b) 1183 (c) 847 (d) none of these

(a) 7

Explanation:

= 1001 Ã· 11 of 13

= 1001 Ã· (11 Ã— 13)

= 1001 Ã· 143 = 7

Q5. 133 + 28 Ã· 7 â€“ 8 Ã— 2 =?

(a) 7 (b) 121 (c) 30 (d) none of these

(b) 121

Explanation:

Given expression:

= 133 + 28 Ã· 7 â€“ 8 Ã— 2

= 133 + 4 â€“ 8 Ã— 2 [performing division]

= 133 + 4 â€“ 16 [performing multiplication]

= 137 â€“ 16 [performing addition]

= 121 [performing subtraction]

Q6. 3640 â€“ 14 Ã· 7 Ã— 2 =?

(a) 3636 (b) 1036 (c) 1819 (d) none of these

(a) 3636

Explanation:

Given expression:

= 3640 â€“ 14 Ã· 7 Ã— 2

= 3640 â€“ 2 Ã— 2 [performing division]

= 3640 â€“ 4 [performing multiplication]

= 3636 [performing subtraction]

Q7. 100 Ã— 10 â€“ 100 + 2000 Ã· 100 =?

(a) 29 (b) 920 (c) none of these

(b) 920

Explanation:

Given expression:

= 100 Ã— 10 â€“ 100 + 2000 Ã· 100

= 100 Ã— 10 â€“ 100 + 20 [performing division]

= 1000 â€“ 100 + 20 [performing multiplication]

= 1020 â€“ 100 [performing addition]

= 920 [performing subtraction]

Q8. $27 – \left [ 18 – \left [ 16 – \left ( 5 – \overline{4 – 1} \right ) \right ] \right ]$ =?

(a) 25 (b) 23 (c) none of these

(b) 23

Explanation:

Given expression:

= $27 – \left [ 18 – \left [ 16 – \left ( 5 – \overline{4 – 1} \right ) \right ] \right ]$

= 27 â€“ [18 â€“ {16 â€“ (5 â€“ 3)}] [removing bar]

= 27 â€“ [18 â€“ {16 â€“ 2}] [removing parentheses]

= 27 â€“ [18 â€“ 14] [removing braces]

= 27 â€“ 4 [removing square brackets]

= 23

Q9. $32 – \left [ 48 \div \left [ 36 – \left ( 27 – \overline{16 – 9} \right ) \right ] \right ]$ =?

(a) 29 (b) 520/17 (c) none of these

(a) 29

Explanation:

Given expression:

= $32 – \left [ 48 \div \left [ 36 – \left ( 27 – \overline{16 – 9} \right ) \right ] \right ]$

= 32 â€“ [48 Ã· {36 â€“ (27 â€“ 7)}] [removing bar]

= 32 â€“ [48 Ã· {36 â€“ 20}] [removing parentheses]

= 32 â€“ [48 Ã· 16] [removing braces]

= 32 â€“ 3 [removing square bracket]

= 29

Q10. 8 â€“ [28 Ã· {34 â€“ (36 â€“ 18 Ã· 9 Ã— 8)}] =?

(a) 6 (b) $6 \frac {4} {9}$ (c) none of these

(a) 6

Explanation:

Given expression:

= 8 â€“ [28 Ã· {34 â€“ (36 â€“ 18 Ã· 9 Ã— 8)}] [performing division]

= 8 â€“ [28 Ã· {34 â€“ (36 â€“ 2 Ã— 8)}] [performing multiplication]

= 8 â€“ [28 Ã· {34 â€“ (36 – 16)}]

= 8 â€“ [28 Ã· {34 â€“ 20}] [removing parentheses]

= 8 â€“ [28 Ã· 14] [removing braces]

= 8 â€“ 2 = 6 [removing square brackets]

#### Practise This Question

Sushweta wants to present her friend a gift made of plant fibre. Which one of the following will she select?