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11 Jul 2024, 06:06
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x = (1/6)^p
y = (1/6)^q
1/x + 1/y = 7992
we have
6^p + 6^q = 7992 = 6^3 * 37 = 6^3 + 6^5
Because x < y, then P > Q
Answer: P = 5, Q = 3
y = (1/6)^q
1/x + 1/y = 7992
we have
6^p + 6^q = 7992 = 6^3 * 37 = 6^3 + 6^5
Because x < y, then P > Q
Answer: P = 5, Q = 3
RenB
Joined: 13 Jul 2022
Last visit: 06 Jul 2026
Posts: 388
Own Kudos:
Given Kudos: 302
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '28 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '28 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 388
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1,633
11 Jul 2024, 06:25
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Bunuel
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11 Jul 2024, 06:33
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X = (1/6)^P
Y = (1/6)^Q
(1/X) + (1/Y) = 7992 As given
From this, we can calculate 6^P + 6^Q = 7992
As X<Y, so P>Q
Testing for values of P & Q, 7776+216 = 7992
Therefore, P=5, Q=3
Y = (1/6)^Q
(1/X) + (1/Y) = 7992 As given
From this, we can calculate 6^P + 6^Q = 7992
As X<Y, so P>Q
Testing for values of P & Q, 7776+216 = 7992
Therefore, P=5, Q=3
Rahul_Sharma23
GMAT Focus 1: 695 Q87 V83 DI83 
Posts: 145
11 Jul 2024, 06:38
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x = (1/6)^p
y = (1/6)^q
y>x , which means that p > q
1/x + 1/y = 7992
substituting x and y
6^p + 6^q = 7992
as we know that p > q , and sum is greater than 7000 by hit and trial we get p = 5 and q= 3
y = (1/6)^q
y>x , which means that p > q
1/x + 1/y = 7992
substituting x and y
6^p + 6^q = 7992
as we know that p > q , and sum is greater than 7000 by hit and trial we get p = 5 and q= 3
