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CAT 2023 Slot 2 Solved Paper
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© examsnet.com
Question : 65
Total: 66
If
p
2
+
q
2
−
29
=
2
p
q
−
20
=
52
−
2
p
q
, then the difference between the maximum and minimum possible value of
(
p
3
−
q
3
)
243
486
378
189
Validate
Solution:
Given that
2
p
q
−
20
=
52
−
2
p
q
⇒
4
p
q
=
72
⇒
p
q
=
18
→(1)
Now,
p
2
+
q
2
−
29
=
2
p
q
−
20
⇒
p
2
+
q
2
−
2
p
q
=
9
⇒
(
p
−
q
)
2
=
9
⇒
p
−
q
=
±
3
Also,
p
2
+
q
2
−
29
=
2
p
q
−
20
⇒
p
2
+
q
2
=
2
p
q
+
9
=
2
(
18
)
+
9
=
45
Now,
p
3
−
q
3
=
(
p
−
q
)
(
p
2
+
p
q
+
q
2
)
=
(
p
−
q
)
(
45
+
18
)
=
(
p
−
q
)
(
63
)
When
p
−
q
=
−
3
⇒
The value is
63
(
−
3
)
=
−
189
and when
p
−
q
=
3
⇒
The value is
63
(
3
)
=
189
The difference
⇒
=
189
−
(
−
189
)
=
378
© examsnet.com
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