WRT Practice PS Set – 4
Work & Rate: Level 4
1.
A can do a job in 20 days. B can do it in 30 days. They work together for 6 days, then A leaves. How many days will B take to finish the remaining work?
Solution (C):
Combined Rate = \( \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12} \).
In 6 days: \( 6 \times \frac{1}{12} = \frac{1}{2} \) job done. Remaining = 1/2.
Time for B = \( \frac{1/2}{1/30} = \frac{30}{2} = 15 \) days.
2.
P and Q together can do a job in 6 days. P alone takes 10 days. They work together for 2 days, then P leaves. How long does Q take to finish?
Solution (A):
Find Rate Q: \( \frac{1}{Q} = \frac{1}{6} – \frac{1}{10} = \frac{5-3}{30} = \frac{2}{30} = \frac{1}{15} \).
Together in 2 days: \( 2/6 = 1/3 \) job done. Remaining = 2/3.
Time for Q = \( \frac{2/3}{1/15} = \frac{2 \times 15}{3} = 10 \) days.
3.
Tom starts painting a wall at 8 AM and can finish in 6 hours. Jerry starts at 9 AM and can finish in 4 hours. At what time will the wall be finished?
Solution (B):
By 9 AM, Tom worked 1 hr. \(1/6\) done. Remaining = \(5/6\).
Combined Rate = \(1/6 + 1/4 = 5/12\).
Time for remaining = \( \frac{5/6}{5/12} = \frac{5}{6} \times \frac{12}{5} = 2 \) hours.
9 AM + 2 hours = 11:00 AM.
4.
A pool is filled by pipe A in 10 hours and pipe B in 15 hours. Pipe C empties it in 20 hours. If all three are open, how long to fill?
Solution (D):
Rate = \( \frac{1}{10} + \frac{1}{15} – \frac{1}{20} \).
LCD = 60. \( \frac{6}{60} + \frac{4}{60} – \frac{3}{60} = \frac{7}{60} \).
Time = \( 60/7 \approx 8.57 \) hours.
5.
Machine A completes a job in \(x\) hours. Machine B takes \(y\) hours. If they work together, the time \(T\) is:
Solution (E):
\(\frac{1}{T} = \frac{1}{x} + \frac{1}{y} = \frac{y+x}{xy}\).
\(T = \frac{xy}{x+y}\).
6.
A can do a piece of work in 20 days. He starts alone and works for 5 days, then B joins him. Together they finish the remaining work in 5 days. How long would B alone take?
Solution (C):
A works 5 days: \(5/20 = 1/4\) done. Rem = \(3/4\).
A+B do \(3/4\) in 5 days. Rate(A+B) = \(\frac{3/4}{5} = \frac{3}{20}\).
Rate B = Rate(A+B) – Rate A = \(\frac{3}{20} – \frac{1}{20} = \frac{2}{20} = \frac{1}{10}\).
B takes 10 days.
7.
Two taps can fill a tank in 12 and 15 minutes respectively. If both are opened together and the first tap is turned off after 3 minutes, how much longer will the second tap take to fill the tank?
Solution (B):
Combined rate = \(\frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20}\).
In 3 mins: \(3 \times \frac{3}{20} = \frac{9}{20}\) filled. Rem = \(\frac{11}{20}\).
Tap 2 rate = \(1/15\).
Time = \(\frac{11/20}{1/15} = \frac{11}{20} \times 15 = \frac{33}{4} = 8.25\) min.
8.25 min = 8 min 15 sec.
8.
Painter A is 50% faster than Painter B. If B takes 12 hours to paint a room, how long does A take?
Solution (A):
Rate B = \(1/12\).
Rate A = \(1.5 \times (1/12) = 3/24 = 1/8\).
Time A = 8 hours.
9.
A task takes 24 days for 15 men working 8 hours a day. How many days for 12 men working 10 hours a day?
Solution (C):
Total work = \( 15 \text{ men} \times 24 \text{ days} \times 8 \text{ hours} = 2880 \) man-hours.
New capacity = \( 12 \text{ men} \times 10 \text{ hours} = 120 \) man-hours/day.
Days = \( 2880 / 120 = 24 \) days.
10.
A paint crew paints 80 houses. First \(y\) houses at rate \(x\). The rest at rate \(1.25x\). If time taken is same as if they worked at original rate \(x\) for all houses, what is equation for \(y\)? (This is a conceptual trick).
Solution (D):
If they speed up for part of the job, the total time MUST be less than if they worked at the slower original rate for the whole job.
Time Actual = \( \frac{y}{x} + \frac{80-y}{1.25x} \).
Time Hypothetical = \( \frac{80}{x} \).
Since \( \frac{1}{1.25x} < \frac{1}{x} \), the actual time is strictly less than hypothetical (unless \(y=80\), meaning they never sped up). If they sped up, time reduces. The prompt implies they sped up for the rest, so \(y < 80\). Thus, time cannot be the same. Impossible.
Score: 0 / 0