Re: A worker is hired for 7 days. Each day, he is paid 11 dollars more
[#permalink]
06 May 2024, 08:43
Let his first day's pay = \(x\)
His pay will be: Day1 : \(x\); Day 2: \(x + 11*1\); Day 3: \(x + 11*2\)... Day 11: \(x + 11*10\)
The total will come to: \(11x + 11*[\frac{10}{2}(1+10)]\)
\(11x + 11*55\)
If we can create an equation from a statement with which we can solve for the value of \(x\), then the statement will be sufficient.
(1) 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.
First 4 days: \(x + (x+1*11) + (x+2*11) + (x+3*11)\)
Last 3 days: \((x+8*11)+(x+9*11)+(x+10*11)\)
\(x + (x+1*11) + (x+2*11) + (x+3*11) = (x+8*11)+(x+9*11)+(x+10*11)\)
As there are more \(x\)'s on one side than the other, one can isolate and solve for \(x\). Thus this will be sufficient to solve the question.
Solving: \(x + (x+1*11) + (x+2*11) + (x+3*11) = (x+8*11)+(x+9*11)+(x+10*11)\)
\(x = 21*11\)
\(x = 231\)
Which means that his total pay was: \(11(231) + 11*55 = 3416\)
SUFFICIENT
(2) The total amount he was paid in the last 3 days of work equaled $66 more than the total amount he was paid in the first 3 days of work.
First 3 days: \(x + (x+1*11) + (x+2*11)\)
Last 3 days: \((x+8*11)+(x+9*11)+(x+10*11)\)
\((x+8*11)+(x+9*11)+(x+10*11) = (6*11)+x + (x+1*11) + (x+2*11)\)
As there is an equal number of positive \(x\)'s on either side of the equals sign, they will cancel one another out and leave no solution.
INSUFFICIENT
ANSWER A