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CAT 2020 Exam Shift 2 Solved Paper
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© examsnet.com
Question : 56
Total: 76
If
x
and
y
are positive real numbers satisfying
x
+
y
=
102
, then the minimum possible value of
2601
(
1
+
1
x
)
(
1
+
1
y
)
is
Your Answer:
Validate
Solution:
A
M
≥
G
M
≥
H
M
x
+
y
2
≥
√
x
y
≥
2
1
x
+
1
y
Given
x
+
y
=
102
⇒
x
y
≤
51
2
or
1
x
y
≥
1
2601
⇒
1
x
+
1
y
≥
2
51
The minimum value of
2601
(
1
+
1
x
)
(
1
+
1
y
)
=
(
2601
)
(
1
+
1
x
+
1
y
+
1
x
y
)
=
2601
(
1
+
2
51
+
1
2601
)
=
2704
© examsnet.com
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