Question

Three men, Alpha, Beta, and Gamma, working together, do a job in $6$ hours less time than Alpha alone, in $1$ hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let $h$ be the number of hours needed by Alpha and Beta, working together, to do the job. Then $h$ equals:

$\dfrac52$

$\dfrac32$

$\dfrac43$

$\dfrac54$

$\dfrac34$

Answer: C

Concepts:rate rational equation system of equations

Difficulty rating: 1850

Small Hint:

Let $t$ be the time all three take together

Big Hint:

Their individual times are $t+6,$ $t+1,$ and $2t$ ; add reciprocal rates

Solution:

If all three together take $t$ hours, then $\frac1{t+6}+\frac1{t+1}+\frac1{2t}=\frac1t.$ This simplifies to $3t^2+7t-6=0,$ so $t=\frac{2}{3}.$ Alpha and Beta alone take $\frac{20}{3}$ and $\frac{5}{3}$ hours, so their combined rate is $\frac{3}{20}+\frac{3}{5}=\frac{3}{4}.$ Hence $h=\frac{4}{3}.$

Therefore, the correct answer is C.

1966 AMC 12 Problem 37

Answer: C

Difficulty rating: 1850

Small Hint:

Big Hint:

Solution:

Problem 37 in Other Years