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Byju's Answer
Standard XII
Mathematics
Intersection
Larger of 9...
Question
Larger of
99
50
+
100
50
and
101
50
A
99
50
+
100
50
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B
101
50
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C
Cannot be determined
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D
None
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Solution
The correct option is
B
101
50
Binomial theorem is given by,
(
x
+
a
)
n
=
n
C
0
x
n
a
0
+
n
C
1
x
n
−
1
a
1
+
n
C
2
x
n
−
2
a
2
+
.
.
.
+
n
C
r
x
n
−
r
a
r
+
.
.
.
.
.
.
.
+
a
n
Therefore,
101
50
=
(
100
+
1
)
50
=
100
50
+
50.100
49
+
25.49.100
48
+
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
99
50
=
(
100
−
1
)
50
=
100
50
−
50.100
49
+
25.49.100
48
+
.
.
.
.
.
.
.
.
.
.
.
.
101
50
−
99
50
=
2
[
50.100
49
+
25.49.16.100
47
+
.
.
.
.
.
.
.
.
.
.
.
.
.
]
101
50
−
99
50
=
100
50
+
50.
49.
16.100
47
+
.
.
.
.
.
.
.
.
.
.
.
.
>
100
50
⟹
101
50
−
99
50
>
100
50
⟹
∴
101
50
>
99
50
+
100
50
Hence larger number
=
101
50
Suggest Corrections
0
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Q.
Larger of
99
50
+
100
50
and
101
50
is
Q.
The larger of
99
50
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100
50
a
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is . . .
Q.
The larger of
99
50
+
100
50
and
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50
is:
Q.
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+
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