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Byju's Answer
Standard VII
Mathematics
Types of Fractions
Solve: i 1 ...
Question
Solve:
(i)
1
2
3
−
3
5
6
(ii)
4
1
2
−
3
1
3
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Solution
(i)
1
2
3
−
3
5
6
=
(
3
×
1
)
+
2
3
−
(
6
×
3
)
+
5
6
=
3
+
2
3
−
18
+
5
6
=
5
3
−
23
6
=
10
−
23
6
=
−
13
6
=
−
2
1
6
∴
1
2
3
−
3
5
6
=
−
2
1
6
.
(ii)
4
1
2
−
3
1
3
=
(
2
×
4
)
+
1
2
−
(
3
×
3
)
+
1
3
=
8
+
1
2
−
9
+
1
3
=
9
2
−
10
3
=
27
−
20
6
=
7
6
=
1
1
6
∴
4
1
2
−
3
1
3
=
1
1
6
.
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Q.
Solve the following -:
√
4
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×
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=
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(a) 25/3
(b) 26/2
(c) 24/3
(d) 25/2
Q.
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Q.
Simplify:
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(ii)
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(iii)
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