RenB wrote:
A boy working alone at 50 percent of his normal speed took 3 more days than a man working alone at 75 percent of his normal speed to produce n units of a product. The boy and the man worked together at their respective normal speeds for 9 days, after which the man left the job but the boy continued working for 6 more days. If a total of 3n units of the product were produced during this period, how many days would it take the boy working alone at 75 percent of his normal speed to produce 2n units?
A. 10
B. 15
C. 20
D. 30
E. 40
No, this is not representative of a GMAT question. Too many steps involved
SolutionAlgebraic method
Man at 3/4 his capability takes x days to produce n products, so at his full capability, he would take 3x/4 days.
The boy at 1/2 his capability takes x+3 days to produce n products, so at his full capability, he would take (x+3)/2 days.
This means man and boy take 3x/4 and (x+3)/2 days respectively to produce n products. 9 days work of man = \(9*\frac{4}{3x}=\frac{12}{x}\), while 9+6 or 15 days work of boy = \(15*\frac{2}{x+3}=\frac{30}{x+3}\)
If the outcome of above work is 3n, that is thrice, we get the equation as \(\frac{12}{x}+\frac{30}{x+3}=3\)
\(\frac{12x+36+30x}{x(x+3)}=3………42x+36=3(x^2+3x)………x^2+3x=14x+12……..x^2-11x+12=0\)
Solving, x=12 or -1. But x>0, so x=12
Days taken by boy for n products = \(\frac{12+3}{2}\)
Days taken by boy for 2n products = \(2*\frac{12+3}{2}=15\)
If the speed was 75%, time taken = 15/0.75 or 15*4/3 or 20 days.