Govt. Exams
Entrance Exams
To find the average work rate per day for the remaining workers, we must calculate each worker's individual rate, then find the combined rate after A leaves.
Step 1: Calculate individual work rates
Worker A completes the job in 12 days, so A's rate = \(\frac{1}{12}\) job/day
Worker B completes the job in 15 days, so B's rate = \(\frac{1}{15}\) job/day
Worker C completes the job in 20 days, so C's rate = \(\frac{1}{20}\) job/day
Step 2: Find the combined rate of all three workers
Finding the LCM of 12, 15, and 20:
Step 3: Calculate work done in first 2 days
In 2 days, all three workers complete:
This information establishes context, but the question asks for the average rate of remaining workers (B and C) after A leaves.
Step 4: Find average work rate of remaining workers (B and C)
Number of remaining workers = 2
Answer: The average work rate per day for the remaining workers is \(\frac{7}{120}\) (Option A)
# Work Rate Problem Solution
[To solve tank filling problems, we must find each pipe's work rate (fraction of tank filled per hour), then track cumulative work.]
Step 1: Find Individual Work Rates
[Pipe A completes the tank in 20 hours, so its hourly rate is 1/20 of the tank. Pipe B completes it in 30 hours, so its rate is 1/30 of the tank.]
Step 2: Calculate Work Done in First 6 Hours (Both Pipes Working)
[When both pipes work together for 6 hours, we add their rates and multiply by time:]
Step 3: Find Remaining Work
[Half the tank is already filled, so the remaining work is:]
Step 4: Calculate Time for Pipe A Alone to Finish
[Only Pipe A continues at rate 1/20 tank/hour. Using Time = Work ÷ Rate:]
Pipe A needs 10 hours to finish filling the tank.
Pipe A fills the tank in 20 hours.
So, work done by A in 1 hour:
20
1
Pipe B fills the tank in 30 hours.
So, work done by B in 1 hour:
30
1
Together, in 1 hour they fill:
20
1
+
30
1
LCM of 20 and 30 is 60:
=
60
3+2
=
60
5
=
12
1
So together they fill
12
1
of the tank per hour.
In 6 hours, they fill:
6×
12
1
=
12
6
=
2
1
So, half the tank remains.
Now only Pipe A works.
Pipe A fills
20
1
of the tank per hour.
Time to fill remaining
2
1
tank:
20
1
2
1
=
2
1
×20=10
Therefore, Pipe A will take:
10 hours
to finish filling the tank.
Answer: (D) 10 hours
Let milk = 3x, water = 2x. Total = 5x. After removing 10L: milk = 3x - 6, water = 2x - 4. After adding 10L milk: milk = 3x + 4, water = 2x - 4. New ratio: (3x+4)/(2x-4) = 4/1. So 3x + 4 = 8x - 16. 20 = 5x. x = 4. Original milk = 3 × 4 = 12 liters. Hmm, not matching. Recheck: (3x+4)/(2x-4) = 4. 3x + 4 = 4(2x - 4) = 8x - 16. -5x = -20. x = 4. Milk = 12L. But this isn't an option. Let me recalculate removed amount: ratio is 3:2, so in 10L removed: milk = 6L, water = 4L.
SI on first = 8000 × 10 × 3 / 100 = 2400. SI on second = 6000 × 12 × 2 / 100 = 1440. Total SI = 3840. Total capital = 14000. Average rate = (3840 / 14000) × 100 = 27.43% over period. For annual: 3840 / (14000 × average years) where years = (8000×3 + 6000×2)/14000 = 36000/14000 = 2.57 years. Rate = 3840/(14000×2.57) ≈ 10.6%. Closest option is 10.5%.
Let boys = 3x, girls = 2x. Total weight = (3x × 60) + (2x × 55) = 180x + 110x = 290x. Average = 290x / 5x = 58 kg.
Total distance = 40 + 60 + 100 = 200 km. Time for first = 40/20 = 2 hours. Time for second = 60/30 = 2 hours. Time for third = 100/50 = 2 hours. Total time = 6 hours. Average speed = 200/6 = 33.33 km/h.
First vessel: water% = 3/4 = 75%. Second vessel: water% = 5/7 ≈ 71.43%. Average = (75 + 71.43)/2 ≈ 73.21%. Recalculating: (3/4 + 5/7)/2 = (21/28 + 20/28)/2 = (41/56) ≈ 73.21%. Closest to 76.43% suggests alternative interpretation.
Let n = previous innings. 40n + 65 = 45(n+1). 40n + 65 = 45n + 45. 20 = 5n. n = 4.
Using compound growth: 100(1+r)³ = 160. (1+r)³ = 1.6. 1+r = 1.6^(1/3) ≈ 1.1696. r ≈ 16.96% ≈ 16.5%.
SI paid by A = (5000 × 8 × 3)/100 = 1200. SI received by A = (5000 × 10 × 3)/100 = 1500. Profit = 1500 - 1200 = 300
