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Bunuel
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Let first day work pay be x

So, he was paid x, x+2, x+4,...x+10

x=?

S1
His total wages for the 6 days were $150.
150 = 6*(x+x+10)/2
x=20
Sufficient

S2
He was paid 150 percent of his first day’s pay for the sixth day
1.5x = x+10
x=20
Sufficient

Bunuel
A worker is hired for 6 days. He is paid $2 more for each day of work than he was paid for the preceding day of work. How much was he paid for the first day of work?

(1) His total wages for the 6 days were $150.
(2) He was paid 150 percent of his first day’s pay for the sixth day


­
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
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if day 1 he is paid X , then day :2 x+2 . day 3 x+4 and so on ....

now he was hired for 6 days .

Option 1 at the end of 6 days he was paid 150.

So we can say X+(X+2)+(X+4)+(x+6)+(x+8)+(x+10)=150
6x+30=150 . We can solve for X (hence satisfied)

Option 2
Day 6 salary =x+10
Day 1 salary X
150% of X=x+10
3/2(x)=x+10

Here also we can get the value of X

Hence Option D. Each option alone is sufficient
Bunuel
A worker is hired for 6 days. He is paid $2 more for each day of work than he was paid for the preceding day of work. How much was he paid for the first day of work?

(1) His total wages for the 6 days were $150.
(2) He was paid 150 percent of his first day’s pay for the sixth day


­
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
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