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UPSC CDS Math Law of Indices Questions
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© examsnet.com
Question : 11
Total: 13
If
9
x
3
y
=
2187
and
2
3
x
2
2
y
−
4
x
y
=
0
,
then what can be the value of (x + y) ?
[2017 CDS-I]
1
3
5
7
Validate
Solution:
9
x
3
y
=
2187
⇒
3
2
x
+
y
=
3
7
⇒
2
x
+
y
=
7
....(1)
2
3
x
2
2
y
−
4
x
y
=
0
⇒
2
3
x
+
2
y
=
2
2
x
y
⇒
3
x
+
2
y
=
2
x
y
....(2)
From (1), we have
4
x
+
2
y
=
14
....(3)
Subtracting (2) from (3), we have
x
=
14
−
2
x
y
⇒
x
(
2
y
+
1
)
=
2
×
7
⇒
x
(
2
y
+
1
)
=
2
×
(
2
×
3
+
1
)
∴
x
=
2
and
y
=
3
⇒
x
+
y
=
5
© examsnet.com
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