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Bunuel
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Bunuel
A worker is hired for 7 days. Each day, he is paid 10 dollars more than what he is paid for the preceding day of work. The total amount he was paid in the first 4 days of work equaled the total amount he was paid in the last 3 days. What was his starting pay?

(A) 90
(B) 138
(C) 153
(D) 160
(E) 163


Salary for the 1st 4 days:

x

x + 10

x + 20

x + 30

Salary for the last 3 days:

x + 40

x + 50

x + 60.

As per the instruction :

sum of the salary of the 1st 4 days = sum of the salary of the last 3 days

4x + 60 = 3x + 150

x = 90.

A is the correct answer.
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We can eliminate fastly C, B and E because the last digits have to be equal.

Ex.

B) 138, 148, 158, 168... 8*4 = 32 Is not equal to 178, 188, 198 = 8*3 = 24

From A and D we try first for A, 420 = 420.

90 + 100 + 110 + 120 = 130 + 140 +150

A
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