Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RS Aggarwal Solutions: Simplification
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Page 1
Q. 4. 19 – [4 + {16 – (12 – 2)}]
Sol. 19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6] = 19 – 10 = 9. Ans.
Q. 5. 36 – [18 – {14 – (15 – 4 2 × 2)}]
Sol. 36 – [18 – {14 – (15 – 4 2 × 2)}]
= 36 – [18 – {14 – (15 – 2 × 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3] = 36 – 15 = 21. Ans.
Q. 6. 27 – [18 – {16 – (5 – 4 1)}]
Sol. 27 – [18 – {16 – (5 – 4 1)}]
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14] = 27 – 4 = 23. Ans.
Q. 7. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
Sol. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
24
5
3
5
4
5
3
10
1
5
of
5
1
24
5
3
4
5
3
10
1
5
24
5
1
3
4
5
3
10
1
5
8
5
12
50
1
5
80 12 10
50
92 10
50
82
50
82 2
50 2
41
25
1
16
25
. Ans.
Q. 8.
2
3
4
9
1
2
3
1
1
4
1
3
F
H
I
K
of
3
5
Points to Remember :
1. Simplifying the expression. For
simplifications the given expression,
involving brackets, of division,
multiplication, addition and subtraction,
we proceed according to the letter of
these words, i.e., VBODMAS, where V
stands for vinculum is bar B for brackets
O for of, D for division, M for
multiplication, A for addition and S for
subtraction.
2. Brackets. There three kinds of brackets:
(i) Round brackets or Small brackets ( )
(ii) Curly brackets or Braces { }
(iii) Big brackets or Square brackets [ ]
While simplifying there backets, we
simplify W remove the brackets in the
above given order, i.e., firstly ( ), then {}
and lastly [ ].
EXERCISE 6 A
Simplify :
Q. 1. 21 – 12 3 × 2
Sol. 21 – 12 3 × 2
= 21 – 4 × 2 = 21 – 8 = 13. Ans.
Q. 2. 16 + 8 4 – 2 × 3
Sol. 16 + 8 4 – 2 × 3
= 16 + 2 – 2 × 3
= 16 + 2 – 6 = 18 – 6 = 12. Ans.
Q. 3. 13 – (12 – 6 3)
Sol. 13 – (12 – 6 3)
= 13 – (12 – 2) = 13 – (10)
= 13 – 10 = 3 Ans.
Page 2
Q. 4. 19 – [4 + {16 – (12 – 2)}]
Sol. 19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6] = 19 – 10 = 9. Ans.
Q. 5. 36 – [18 – {14 – (15 – 4 2 × 2)}]
Sol. 36 – [18 – {14 – (15 – 4 2 × 2)}]
= 36 – [18 – {14 – (15 – 2 × 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3] = 36 – 15 = 21. Ans.
Q. 6. 27 – [18 – {16 – (5 – 4 1)}]
Sol. 27 – [18 – {16 – (5 – 4 1)}]
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14] = 27 – 4 = 23. Ans.
Q. 7. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
Sol. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
24
5
3
5
4
5
3
10
1
5
of
5
1
24
5
3
4
5
3
10
1
5
24
5
1
3
4
5
3
10
1
5
8
5
12
50
1
5
80 12 10
50
92 10
50
82
50
82 2
50 2
41
25
1
16
25
. Ans.
Q. 8.
2
3
4
9
1
2
3
1
1
4
1
3
F
H
I
K
of
3
5
Points to Remember :
1. Simplifying the expression. For
simplifications the given expression,
involving brackets, of division,
multiplication, addition and subtraction,
we proceed according to the letter of
these words, i.e., VBODMAS, where V
stands for vinculum is bar B for brackets
O for of, D for division, M for
multiplication, A for addition and S for
subtraction.
2. Brackets. There three kinds of brackets:
(i) Round brackets or Small brackets ( )
(ii) Curly brackets or Braces { }
(iii) Big brackets or Square brackets [ ]
While simplifying there backets, we
simplify W remove the brackets in the
above given order, i.e., firstly ( ), then {}
and lastly [ ].
EXERCISE 6 A
Simplify :
Q. 1. 21 – 12 3 × 2
Sol. 21 – 12 3 × 2
= 21 – 4 × 2 = 21 – 8 = 13. Ans.
Q. 2. 16 + 8 4 – 2 × 3
Sol. 16 + 8 4 – 2 × 3
= 16 + 2 – 2 × 3
= 16 + 2 – 6 = 18 – 6 = 12. Ans.
Q. 3. 13 – (12 – 6 3)
Sol. 13 – (12 – 6 3)
= 13 – (12 – 2) = 13 – (10)
= 13 – 10 = 3 Ans.
Sol.
2
3
4
9
1
2
3
1
1
4
1
3
F
H
I
K
of
3
5
F
H
I
K
2
3
4
9
5
3
5
4
1
3
of
3
5
6 4
9
5
3
5
4
1
3
of
3
5
10
9
5
3
5
4
1
3
of
3
5
2
3
5
3
5
4
1
3
2
3
3
5
5
4
1
3
1
2
1
3
3 2
6
1
6
Ans.
Q. 9.
7
1
3
2
3
1
3
8
2
3
4
1
1
2
of 2
1
5
Sol.
7
1
3
2
3
1
3
8
2
3
4
1
1
2
of 2
1
5
22
3
2
3
11
8
11
4
3
2
of
11
5
22
3
22
15
11
8
11
4
3
2
22
3
15
22
11
8
4
11
3
2
5
1
1
2
3
2
10 1 3
2
8
2
4
Ans.
Q. 10.
5
1
7
3
3
10
2
4
5
7
10
F
H
I
K
R
S
T
U
V
W
Sol.
5
1
7
3
3
10
2
4
5
7
10
F
H
I
K
R
S
T
U
V
W
F
H
I
K
R
S
T
U
V
W
36
7
33
10
14
5
7
10
R
S
T
U
V
W
36
7
33
10
21
10
R
S
T
U
V
W
36
7
33
10
10
21
36
7
11
7
36 11
7
25
7
3
4
7
Ans.
Working :
14
5
7
10
28 7
10
21
10
L
N
M
M
M
O
Q
P
P
P
?
Q. 11.
9
3
4
2
1
6
4
1
3
1
1
2
1
3
4
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
9
3
4
2
1
6
4
1
3
1
1
2
1
3
4
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
39
4
13
6
13
3
3
2
7
4
R
S
T
U
V
W
L
N
M
O
Q
P
39
4
13
6
13
3
13
4
L
N
M
O
Q
P
39
4
13
6
13
12
L
N
M
O
Q
P
39
4
39
12
39
4
12
39
3 Ans.
Working :
3
2
7
4
6 7
4
13
4
13
3
13
4
52 39
12
13
12
13
6
13
12
26 13
12
39
12
Q. 12.
4
1
10
2
1
2
5
6
2
5
3
10
4
15
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
4
1
10
2
1
2
5
6
2
5
3
10
4
15
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
41
10
5
2
5
6
2
5
3
10
4
15
Page 3
Q. 4. 19 – [4 + {16 – (12 – 2)}]
Sol. 19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6] = 19 – 10 = 9. Ans.
Q. 5. 36 – [18 – {14 – (15 – 4 2 × 2)}]
Sol. 36 – [18 – {14 – (15 – 4 2 × 2)}]
= 36 – [18 – {14 – (15 – 2 × 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3] = 36 – 15 = 21. Ans.
Q. 6. 27 – [18 – {16 – (5 – 4 1)}]
Sol. 27 – [18 – {16 – (5 – 4 1)}]
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14] = 27 – 4 = 23. Ans.
Q. 7. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
Sol. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
24
5
3
5
4
5
3
10
1
5
of
5
1
24
5
3
4
5
3
10
1
5
24
5
1
3
4
5
3
10
1
5
8
5
12
50
1
5
80 12 10
50
92 10
50
82
50
82 2
50 2
41
25
1
16
25
. Ans.
Q. 8.
2
3
4
9
1
2
3
1
1
4
1
3
F
H
I
K
of
3
5
Points to Remember :
1. Simplifying the expression. For
simplifications the given expression,
involving brackets, of division,
multiplication, addition and subtraction,
we proceed according to the letter of
these words, i.e., VBODMAS, where V
stands for vinculum is bar B for brackets
O for of, D for division, M for
multiplication, A for addition and S for
subtraction.
2. Brackets. There three kinds of brackets:
(i) Round brackets or Small brackets ( )
(ii) Curly brackets or Braces { }
(iii) Big brackets or Square brackets [ ]
While simplifying there backets, we
simplify W remove the brackets in the
above given order, i.e., firstly ( ), then {}
and lastly [ ].
EXERCISE 6 A
Simplify :
Q. 1. 21 – 12 3 × 2
Sol. 21 – 12 3 × 2
= 21 – 4 × 2 = 21 – 8 = 13. Ans.
Q. 2. 16 + 8 4 – 2 × 3
Sol. 16 + 8 4 – 2 × 3
= 16 + 2 – 2 × 3
= 16 + 2 – 6 = 18 – 6 = 12. Ans.
Q. 3. 13 – (12 – 6 3)
Sol. 13 – (12 – 6 3)
= 13 – (12 – 2) = 13 – (10)
= 13 – 10 = 3 Ans.
Sol.
2
3
4
9
1
2
3
1
1
4
1
3
F
H
I
K
of
3
5
F
H
I
K
2
3
4
9
5
3
5
4
1
3
of
3
5
6 4
9
5
3
5
4
1
3
of
3
5
10
9
5
3
5
4
1
3
of
3
5
2
3
5
3
5
4
1
3
2
3
3
5
5
4
1
3
1
2
1
3
3 2
6
1
6
Ans.
Q. 9.
7
1
3
2
3
1
3
8
2
3
4
1
1
2
of 2
1
5
Sol.
7
1
3
2
3
1
3
8
2
3
4
1
1
2
of 2
1
5
22
3
2
3
11
8
11
4
3
2
of
11
5
22
3
22
15
11
8
11
4
3
2
22
3
15
22
11
8
4
11
3
2
5
1
1
2
3
2
10 1 3
2
8
2
4
Ans.
Q. 10.
5
1
7
3
3
10
2
4
5
7
10
F
H
I
K
R
S
T
U
V
W
Sol.
5
1
7
3
3
10
2
4
5
7
10
F
H
I
K
R
S
T
U
V
W
F
H
I
K
R
S
T
U
V
W
36
7
33
10
14
5
7
10
R
S
T
U
V
W
36
7
33
10
21
10
R
S
T
U
V
W
36
7
33
10
10
21
36
7
11
7
36 11
7
25
7
3
4
7
Ans.
Working :
14
5
7
10
28 7
10
21
10
L
N
M
M
M
O
Q
P
P
P
?
Q. 11.
9
3
4
2
1
6
4
1
3
1
1
2
1
3
4
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
9
3
4
2
1
6
4
1
3
1
1
2
1
3
4
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
39
4
13
6
13
3
3
2
7
4
R
S
T
U
V
W
L
N
M
O
Q
P
39
4
13
6
13
3
13
4
L
N
M
O
Q
P
39
4
13
6
13
12
L
N
M
O
Q
P
39
4
39
12
39
4
12
39
3 Ans.
Working :
3
2
7
4
6 7
4
13
4
13
3
13
4
52 39
12
13
12
13
6
13
12
26 13
12
39
12
Q. 12.
4
1
10
2
1
2
5
6
2
5
3
10
4
15
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
4
1
10
2
1
2
5
6
2
5
3
10
4
15
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
41
10
5
2
5
6
2
5
3
10
4
15
R
S
T
U
V
W
L
N
M
O
Q
P
41
10
5
2
5
6
2
5
3
10
4
15
L
N
M
O
Q
P
41
10
5
2
5
6
2
5
3
10
4
15
41
10
5
2
5
6
2
5
3
10
4
15
246 150 50 24 18 16
60
[LCM of 4, 2, 6, 5, 10, 15 = 60]
246 50 16 150 24 18
60
312 192
60
120
60
= 2 Ans.
Q. 13.
1
5
6
2
2
3
3
3
4
3
4
5
9
1
2
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
1
5
6
2
2
3
3
3
4
3
4
5
9
1
2
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
11
6
8
3
15
4
19
5
19
2
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
11
6
8
3
15
4
19
5
2
19
R
S
T
U
V
W
L
N
M
O
Q
P
11
6
8
3
15
4
2
5
L
N
M
O
Q
P
11
6
8
3
3
2
11
6
8
3
3
2
11 16 9
6
27 9
6
18
6
= 3 Ans.
Q. 14.
4
4
5
2
1
5
1
2
1
1
4
1
4
1
5
F
H
G
I
K
J
R
S
T
U
V
W
Sol.
4
4
5
2
1
5
1
2
1
1
4
1
4
1
5
F
H
G
I
K
J
R
S
T
U
V
W
F
H
G
I
K
J
R
S
T
U
V
W
24
5
11
5
1
2
5
4
1
4
1
5
F
H
I
K
R
S
T
U
V
W
24
5
11
5
1
2
5
4
1
4
1
5
R
S
T
U
V
W
24
5
11
5
1
2
24
20
R
S
T
U
V
W
24
5
11
5
3
5
24
5
8
5
24
5
5
8
= 3 Ans.
Q. 15.
7
1
2
2
1
4
1
1
4
1
2
3
2
1
3
1
6
F
H
G
I
K
J
R
S
T
U
V
W
L
N
M
M
O
Q
P
P
Sol.
7
1
2
2
1
4
1
1
4
1
2
3
2
1
3
1
6
F
H
G
I
K
J
R
S
T
U
V
W
L
N
M
M
O
Q
P
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
15
2
9
4
5
4
1
2
3
2
1
3
1
6
R
S
T
U
V
W
L
N
M
O
Q
P
15
2
9
4
5
4
1
2
4
3
R
S
T
U
V
W
L
N
M
O
Q
P
15
2
9
4
5
4
2
3
L
N
M
O
Q
P
15
2
9
4
7
12
L
N
M
O
Q
P
15
2
9
4
12
7
15
2
27
7
105 54
14
51
14
3
9
14
Ans.
EXERCISE 6 B
Mark ( ) against the correct answer in each
of the following :
1. 8 + 4 2 × 5 = ?
(a) 30 (b) 50
(c) 18 (d) none of these
Sol. (c)
.
.
.
8 + 4 2 × 5 8 4
1
2
5
= 8 + 10 = 18.
Page 4
Q. 4. 19 – [4 + {16 – (12 – 2)}]
Sol. 19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6] = 19 – 10 = 9. Ans.
Q. 5. 36 – [18 – {14 – (15 – 4 2 × 2)}]
Sol. 36 – [18 – {14 – (15 – 4 2 × 2)}]
= 36 – [18 – {14 – (15 – 2 × 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3] = 36 – 15 = 21. Ans.
Q. 6. 27 – [18 – {16 – (5 – 4 1)}]
Sol. 27 – [18 – {16 – (5 – 4 1)}]
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14] = 27 – 4 = 23. Ans.
Q. 7. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
Sol. 4
4
5
3
5
3
10
1
5
of 5 +
4
5
24
5
3
5
4
5
3
10
1
5
of
5
1
24
5
3
4
5
3
10
1
5
24
5
1
3
4
5
3
10
1
5
8
5
12
50
1
5
80 12 10
50
92 10
50
82
50
82 2
50 2
41
25
1
16
25
. Ans.
Q. 8.
2
3
4
9
1
2
3
1
1
4
1
3
F
H
I
K
of
3
5
Points to Remember :
1. Simplifying the expression. For
simplifications the given expression,
involving brackets, of division,
multiplication, addition and subtraction,
we proceed according to the letter of
these words, i.e., VBODMAS, where V
stands for vinculum is bar B for brackets
O for of, D for division, M for
multiplication, A for addition and S for
subtraction.
2. Brackets. There three kinds of brackets:
(i) Round brackets or Small brackets ( )
(ii) Curly brackets or Braces { }
(iii) Big brackets or Square brackets [ ]
While simplifying there backets, we
simplify W remove the brackets in the
above given order, i.e., firstly ( ), then {}
and lastly [ ].
EXERCISE 6 A
Simplify :
Q. 1. 21 – 12 3 × 2
Sol. 21 – 12 3 × 2
= 21 – 4 × 2 = 21 – 8 = 13. Ans.
Q. 2. 16 + 8 4 – 2 × 3
Sol. 16 + 8 4 – 2 × 3
= 16 + 2 – 2 × 3
= 16 + 2 – 6 = 18 – 6 = 12. Ans.
Q. 3. 13 – (12 – 6 3)
Sol. 13 – (12 – 6 3)
= 13 – (12 – 2) = 13 – (10)
= 13 – 10 = 3 Ans.
Sol.
2
3
4
9
1
2
3
1
1
4
1
3
F
H
I
K
of
3
5
F
H
I
K
2
3
4
9
5
3
5
4
1
3
of
3
5
6 4
9
5
3
5
4
1
3
of
3
5
10
9
5
3
5
4
1
3
of
3
5
2
3
5
3
5
4
1
3
2
3
3
5
5
4
1
3
1
2
1
3
3 2
6
1
6
Ans.
Q. 9.
7
1
3
2
3
1
3
8
2
3
4
1
1
2
of 2
1
5
Sol.
7
1
3
2
3
1
3
8
2
3
4
1
1
2
of 2
1
5
22
3
2
3
11
8
11
4
3
2
of
11
5
22
3
22
15
11
8
11
4
3
2
22
3
15
22
11
8
4
11
3
2
5
1
1
2
3
2
10 1 3
2
8
2
4
Ans.
Q. 10.
5
1
7
3
3
10
2
4
5
7
10
F
H
I
K
R
S
T
U
V
W
Sol.
5
1
7
3
3
10
2
4
5
7
10
F
H
I
K
R
S
T
U
V
W
F
H
I
K
R
S
T
U
V
W
36
7
33
10
14
5
7
10
R
S
T
U
V
W
36
7
33
10
21
10
R
S
T
U
V
W
36
7
33
10
10
21
36
7
11
7
36 11
7
25
7
3
4
7
Ans.
Working :
14
5
7
10
28 7
10
21
10
L
N
M
M
M
O
Q
P
P
P
?
Q. 11.
9
3
4
2
1
6
4
1
3
1
1
2
1
3
4
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
9
3
4
2
1
6
4
1
3
1
1
2
1
3
4
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
39
4
13
6
13
3
3
2
7
4
R
S
T
U
V
W
L
N
M
O
Q
P
39
4
13
6
13
3
13
4
L
N
M
O
Q
P
39
4
13
6
13
12
L
N
M
O
Q
P
39
4
39
12
39
4
12
39
3 Ans.
Working :
3
2
7
4
6 7
4
13
4
13
3
13
4
52 39
12
13
12
13
6
13
12
26 13
12
39
12
Q. 12.
4
1
10
2
1
2
5
6
2
5
3
10
4
15
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
4
1
10
2
1
2
5
6
2
5
3
10
4
15
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
41
10
5
2
5
6
2
5
3
10
4
15
R
S
T
U
V
W
L
N
M
O
Q
P
41
10
5
2
5
6
2
5
3
10
4
15
L
N
M
O
Q
P
41
10
5
2
5
6
2
5
3
10
4
15
41
10
5
2
5
6
2
5
3
10
4
15
246 150 50 24 18 16
60
[LCM of 4, 2, 6, 5, 10, 15 = 60]
246 50 16 150 24 18
60
312 192
60
120
60
= 2 Ans.
Q. 13.
1
5
6
2
2
3
3
3
4
3
4
5
9
1
2
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
Sol.
1
5
6
2
2
3
3
3
4
3
4
5
9
1
2
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
11
6
8
3
15
4
19
5
19
2
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
11
6
8
3
15
4
19
5
2
19
R
S
T
U
V
W
L
N
M
O
Q
P
11
6
8
3
15
4
2
5
L
N
M
O
Q
P
11
6
8
3
3
2
11
6
8
3
3
2
11 16 9
6
27 9
6
18
6
= 3 Ans.
Q. 14.
4
4
5
2
1
5
1
2
1
1
4
1
4
1
5
F
H
G
I
K
J
R
S
T
U
V
W
Sol.
4
4
5
2
1
5
1
2
1
1
4
1
4
1
5
F
H
G
I
K
J
R
S
T
U
V
W
F
H
G
I
K
J
R
S
T
U
V
W
24
5
11
5
1
2
5
4
1
4
1
5
F
H
I
K
R
S
T
U
V
W
24
5
11
5
1
2
5
4
1
4
1
5
R
S
T
U
V
W
24
5
11
5
1
2
24
20
R
S
T
U
V
W
24
5
11
5
3
5
24
5
8
5
24
5
5
8
= 3 Ans.
Q. 15.
7
1
2
2
1
4
1
1
4
1
2
3
2
1
3
1
6
F
H
G
I
K
J
R
S
T
U
V
W
L
N
M
M
O
Q
P
P
Sol.
7
1
2
2
1
4
1
1
4
1
2
3
2
1
3
1
6
F
H
G
I
K
J
R
S
T
U
V
W
L
N
M
M
O
Q
P
P
F
H
I
K
R
S
T
U
V
W
L
N
M
O
Q
P
15
2
9
4
5
4
1
2
3
2
1
3
1
6
R
S
T
U
V
W
L
N
M
O
Q
P
15
2
9
4
5
4
1
2
4
3
R
S
T
U
V
W
L
N
M
O
Q
P
15
2
9
4
5
4
2
3
L
N
M
O
Q
P
15
2
9
4
7
12
L
N
M
O
Q
P
15
2
9
4
12
7
15
2
27
7
105 54
14
51
14
3
9
14
Ans.
EXERCISE 6 B
Mark ( ) against the correct answer in each
of the following :
1. 8 + 4 2 × 5 = ?
(a) 30 (b) 50
(c) 18 (d) none of these
Sol. (c)
.
.
.
8 + 4 2 × 5 8 4
1
2
5
= 8 + 10 = 18.
2. 54 3 of 6 + 9 = ?
(a) 117 (b) 12
(c)
6
5
(d) none of these
Sol. (b)
.
.
.
54 3 of 6 + 9 = 54 18 + 9
54
1
18
9 = 3 + 9 = 12.
3. 13 – (12 – 6 3) = ?
(a) 11 (b) 3
(c)
7
3
(d) none of these
(b)
.
.
.
13 – (12 – 6 3) = 13 – (12 – 2)
= 13 – 10 = 3.
4. 1001 11 of 13 = ?
(a) 7 (b) 1183
(c) 847 (d) none of these
Sol. (a)
.
.
.
1001 11 of 13 = 1001 143
1001
1
143
7 .
5. 133 + 28 7 – 8 × 2 = ?
(a) 7 (b) 121
(c) 30 (d) none of these
Sol. (b)
.
.
.
133 + 28 7 – 8 × 2 = 133 + 4 –
16
= 137 – 16 = 121.
6. 3640 – 14 7 × 2 = ?
(a) 3636 (b) 1036
(c) 1819 (d) none of these
Sol. (a)
.
.
.
3640 – 14 7 × 2 = 3640 – 2 × 2
= 3640 – 4 = 3636.
7. 100 × 10 – 100 + 2000 + 100 = ?
(a) 29 (b) 920
(c) none of these
Sol. (b)
.
.
.
100 × 10 – 100 + 2000 100
= 1000 – 100 + 20
= 920.
8. 27 – [18 – {16 – (5 – 4 1)] = ?
(a) 25 (b) 23
(c) none of these
Sol. (b)
.
.
.
27 – [18 – {16 – (5 – 4 1)}]
= 27 – [18 – {16 – (5 – 4 + 1)}]
= 27 – [18 – {16 – 5 + 4 – 1}]
= 27 – [18 – 16 + 5 – 4 + 1]
= 27 – 18 + 16 – 5 + 4 – 1 = 23
9. 32 – [48 {36 – (27 – 16 9 )}] = ?
(a) 29 (b)
520
17
(c) none of these
Sol. (a)
.
.
.
32 – [48 {36 – (27 – 16 9 )}]
= 32 – [48 {36 – (27 – 16 + 9)}]
= 32 – [48 {36 – 27 + 16 – 9}]
= 32 – [48 16] = 32 – 3 = 29
10. 8 – [28 {34 – (36 – 18 9 × 8)}] = ?
(a) 6 (b) 6
4
9
(c) none of these
Sol. (a)
.
.
.
8 – [28 {34 – (36 – 18 9 × 8)}]
F
H
G
I
K
J
R
S
T
U
V
W
L
N
M
M
O
Q
P
P
8 28 34 36 18
1
9
8
= 8 – [28 {34 – (36 – 16)}]
= 8 – [28 {34 – 20}]
= 8 – {28 14} = 8 – 2 = 6.
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FAQs on RS Aggarwal Solutions: Simplification - Mathematics (Maths) Class 6
| 1. What is the importance of simplification in Class 6 mathematics? |  |
Ans. Simplification is important in Class 6 mathematics as it helps students understand complex mathematical expressions or equations in a simpler form. It allows students to solve problems more efficiently and accurately.
| 2. How can I improve my simplification skills in Class 6 mathematics? | |
Ans. To improve simplification skills in Class 6 mathematics, practice is key. Regularly solving simplification problems, using shortcuts or tricks, and understanding the basic concepts of arithmetic operations will help improve simplification skills.
| 3. Are there any specific rules or techniques for simplification in Class 6 mathematics? | |
Ans. Yes, there are specific rules and techniques for simplification in Class 6 mathematics. Some of these include the order of operations (BODMAS), simplifying brackets, combining like terms, and canceling out common factors.
| 4. What are the common mistakes to avoid in simplification problems in Class 6 mathematics? | |
Ans. Common mistakes to avoid in simplification problems in Class 6 mathematics include forgetting to apply the order of operations, not simplifying brackets correctly, making calculation errors, and not checking the final answer for accuracy.
| 5. Can simplification be used in real-life situations? | |
Ans. Yes, simplification can be used in various real-life situations. For example, in financial calculations, budgeting, cooking recipes, and problem-solving situations, simplification helps in making complex calculations or tasks easier to understand and execute.
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