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As part of an endorsement deal, a shoe company pays an athlete a fixed 05 Jan 2026, 22:00
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C
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85% (hard)

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35% (01:28) correct 65% (01:57) wrong based on 106 sessions

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As part of an endorsement deal, a shoe company pays an athlete a fixed yearly licensing fee and 1 percent of annual turnover above $10 million on the shoes he endorses. If the licensing fee was consistent across 2005 and 2006, how much higher was the revenue on the shoes in 2005 than in 2006?

(1) The total amount the company paid to the athlete was $460,000 in 2005 and $420,000 in 2006.
(2) The fixed licensing fee paid to the athlete each year was $400,000.


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C
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DaMa03
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As part of an endorsement deal, a shoe company pays an athlete a fixed Updated on: 07 Jan 2026, 08:01
Originally posted by DaMa03 on 06 Jan 2026, 06:34.
Last edited by DaMa03 on 07 Jan 2026, 08:01, edited 2 times in total.
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Quote:
As part of an endorsement deal, a shoe company pays an athlete a fixed yearly licensing fee and 1 percent of annual turnover above $10 million on the shoes he endorses. If the licensing fee was consistent across 2005 and 2006, how much higher was the revenue on the shoes in 2005 than in 2006?

(1) The total amount the company paid to the athlete was $460,000 in 2005 and $420,000 in 2006.
(2) The fixed licensing fee paid to the athlete each year was $400,000.
Show more


(1): We know that he got the same fixed fee, but not how much.
it could be $400.000, or $420.000 for example:
then he either earned $60.000 or $40.000 on his turnover in 2005 -> insufficient

or: let the fee be "f" and revenue "r1" and "r2" for years 2005 and 2006 respectively
in 2005 he earned: f + 0.01*r1 = $460.000
in 2006 he earned: f + 0.01*r2 = $400.000
-> 3 unknowns --> insufficient -> see edit

(2): let his total and his total salary be "s1" and "s2" for 2005 and 2006 respectively, and now f is fixed at $400.000,
in 2005 he earned: 400.000 + 0.01*r1 = s1
in 2005 he earned: 400.000 + 0.01*r2 = s2
-> now even 4 unknowns --> insufficient


(together):
f is $400.000, s1 is $460.000, s2 is $420.000
in 2005 he earned: 400.000 + 0.01*r1 = $460.000
in 2005 he earned: 400.000 + 0.01*r2 = $420.000
-> only 1 unknown per equation -> each can be solved to get r1 and r2
difference in revenue is r1-r2 (since we want 2005-2006) -> sufficient

Answer C



Edit: After reviewing some other answers, I looked at (1) again.
The full equations are:
in 2005 he earned: f + max(0 ; 0.01*(r1-10mill)) = $460.000
in 2006 he earned: f + max(0 ; 0.01*(r2-10mil)) = $400.000
subtracting both, we get:
max(0 ; 0.01*(r1-10mill)) - max(0 ; 0.01*(r2-10mill)) = $60.000
to solve this, we need to know whether r1 and r2 are > or < 10 million. Otherwise, it's insufficient
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As part of an endorsement deal, a shoe company pays an athlete a fixed 06 Jan 2026, 10:08
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total payment=licensing fee + 0.01(revenue in year - $10million)
given licensing fee was same for the two years
need to find (R2005-R2006) where these values denote the total revenues in the two years?
1. $420K=licensing fee + 0.01(R2005 - $10 million) and $400k=licensing fee + 0.01(R2006 - $10 million)
by subtracting these two equations value (R2005-R2006) can be found----sufficient
2. given the value of licensing fee only --non sufficient
so A
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As part of an endorsement deal, a shoe company pays an athlete a fixed 06 Jan 2026, 12:33
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As part of an endorsement deal, a shoe company pays an athlete a fixed yearly licensing fee and 1 percent of annual turnover above $10 million on the shoes he endorses. If the licensing fee was consistent across 2005 and 2006, how much higher was the revenue on the shoes in 2005 than in 2006?

(1) The total amount the company paid to the athlete was $460,000 in 2005 and $420,000 in 2006.
(2) The fixed licensing fee paid to the athlete each year was $400,000.


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Let the fixed license fee be denoted as F.

1% above annual turnover above $10 million is paid to the athlete apart from F.

So, total revenue = F + 1%*(amount exceeding 10 million).

Revenue of 2005 - Revenue of 2006 = ?

Statement 1:

(1) The total amount the company paid to the athlete was $460,000 in 2005 and $420,000 in 2006.

F + 1% * ( X 2005 - $10 million) = $460000.

F + 1%*( X2006 - $10 million) = $ 420000.

subtracting both, we get

X 2005 - $10 million - (X2006 - $10 million) = $460000 - $420000.

X2005 - X2006 = $40,00,000

So, the difference in revenue = $40,00,000.

Hence, Sufficient.

Statement 2:

(2) The fixed licensing fee paid to the athlete each year was $400,000.

As ,we don’t know the amount paid in excess of $10 million.

This statement is Insufficient

Option A
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As part of an endorsement deal, a shoe company pays an athlete a fixed 11 Jan 2026, 04:45
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ExpertsGlobal5
As part of an endorsement deal, a shoe company pays an athlete a fixed yearly licensing fee and 1 percent of annual turnover above $10 million on the shoes he endorses. If the licensing fee was consistent across 2005 and 2006, how much higher was the revenue on the shoes in 2005 than in 2006?

(1) The total amount the company paid to the athlete was $460,000 in 2005 and $420,000 in 2006.
(2) The fixed licensing fee paid to the athlete each year was $400,000.
Show more
Explanation:

Let the fixed yearly licensing fee be L.
Let the annual turnover in the year 2005 and year 2006 be A1 and A2, respectively.
If A1 ≤ $10 million, then the total amount paid to the athlete in the year 2005 = L
If A1 > $10 million, then the total amount paid to the athlete in the year 2005 = L + 0.01(A1 – 10,000,000) (Equation I)
If A2 ≤ $10 million, then the total amount paid to the athlete in the year 2006 = L
If A2 > $10 million, then the total amount paid to the athlete in the year 2006 = L + 0.01(A2 – 10,000,000) (Equation II)
Difference between the revenue on the shoes in 2005 and in 2006 = A1 – A2
We need to find whether the value of (A1 – A2) can be determined.

Statement (1)

The total amount paid to the athlete in 2005 = $460,000 (Equation III)
The total amount paid to the athlete in 2006 = $420,000 (Equation IV)

If A2 ≤ $10 million, then L = 420,000, and then since the total amount paid to the athlete in 2005 was greater than L, it follows that A1 ≥ $10 million.
After that, using Equations I and III we can determine the value of A1, however, it is not possible to calculate the value of A2.

Since no information is provided that suggests that A2 > $10 million, it is NOT possible to determine the value of (A1 – A2). Hence, Statement (1) is insufficient.

Statement (2)

L = 400,000 (Equation V)

Since no information is provided regarding the total amount paid to the athlete, it is NOT possible to determine the value of (A1 – A2). Hence, Statement (2) is insufficient.

As Statement (1) alone as well as Statement (2) alone is insufficient to answer the question, we need to now combine the two statements.

Statement (1) and Statement (2) combined

From Equation V, we know that L = 400,000, and since Equation III and Equation IV show that the total amount paid to the athlete is greater than L for both years, it follows that A1 > $10 million and A2 > $10 million.

From Equations I and III: L + 0.01(A1 – 10,000,000) = $460,000 (Equation VI)
From Equations II and IV: L + 0.01(A2 – 10,000,000) = $420,000 (Equation VII)

From Equations V, VI, and VII we have 3 equations with 3 unknown variables that can be solved to determine the value of A1 and A2, which in turn can be used to determine the value of A1 – A2. Hence, Statement (1) and Statement (2) combined are sufficient.

C is the correct answer choice.
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