# RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A

## RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A.

Other Exercises

Simplify

Question 1.
Solution:
21 – 12 ÷ 3 x 2
= 21 – 4 x 2
= 21 – 8
= 13. Ans

Question 2.
Solution:
16 + 8 ÷ 4 – 2 x 3
= 16 + 2 – 2 x 3
= 16 + 2 – 6
= 18 – 6
= 12. Ans.

Question 3.
Solution:
13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – (10)
= 13 – 10
= 3 Ans.

Question 4.
Solution:
19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6]
= 19 – 10
= 9. Ans

Question 5.
Solution:
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 – 3]
= 36 – 15
= 21. Ans.

Question 6.
Solution:
$$27-[18-\{ 16-(5-\overline { 4-1 } )\} ]$$
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14]
= 27 – 4
= 23. Ans.

Question 7.
Solution:
$$4\frac { 4 }{ 3 } \div \frac { 3 }{ 5 } of5+\frac { 4 }{ 5 } \times \frac { 3 }{ 10 } -\frac { 1 }{ 5 }$$

Question 8.
Solution:
$$\left( \frac { 2 }{ 3 } +\frac { 4 }{ 9 } \right) of\frac { 3 }{ 5 } \div 1\frac { 2 }{ 3 } \times 1\frac { 1 }{ 4 } -\frac { 1 }{ 3 }$$

Question 9.
Solution:
$$7\frac { 1 }{ 3 } \div \frac { 2 }{ 3 } of2\frac { 1 }{ 5 } +1\frac { 3 }{ 8 } \div 2\frac { 3 }{ 4 } -1\frac { 1 }{ 2 }$$

Question 10.
Solution:
$$5\frac { 1 }{ 7 } -\left\{ 3\frac { 3 }{ 10 } \div \left( 2\frac { 4 }{ 5 } -\frac { 7 }{ 10 } \right) \right\}$$

Question 11.
Solution:
$$9\frac { 3 }{ 4 } \div \left[ 2\frac { 1 }{ 6 } +\left\{ 4\frac { 1 }{ 3 } -\left( 1\frac { 1 }{ 2 } +1\frac { 3 }{ 4 } \right) \right\} \right]$$

Question 12.
Solution:
$$4\frac { 1 }{ 10 } -\left[ 2\frac { 1 }{ 2 } -\left\{ \frac { 5 }{ 6 } -\left( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } -\frac { 4 }{ 15 } \right) \right\} \right]$$

Question 13.
Solution:
$$1\frac { 5 }{ 6 } +\left[ 2\frac { 2 }{ 3 } -\left\{ 3\frac { 3 }{ 4 } \left( 3\frac { 4 }{ 5 } \div 9\frac { 1 }{ 2 } \right) \right\} \right]$$

Question 14.
Solution:
$$4\frac { 4 }{ 5 } \div \left\{ 2\frac { 1 }{ 5 } -\frac { 1 }{ 2 } \left( 1\frac { 1 }{ 4 } -\overline { \frac { 1 }{ 4 } -\frac { 1 }{ 5 } } \right) \right\}$$

Question 15.
Solution:
$$7\frac { 1 }{ 2 } -\left[ 2\frac { 1 }{ 4 } \div \left\{ 1\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \left( \frac { 3 }{ 2 } -\overline { \frac { 1 }{ 3 } -\frac { 1 }{ 6 } } \right) \right\} \right]$$

Hope given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A are helpful to complete your math homework.

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