RS Aggarwal Solutions: Simplification

``` 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,
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,
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,
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,
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.
```

Mathematics (Maths) Class 6

94 videos|347 docs|54 tests

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.

Mathematics (Maths) Class 6

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