Govt. Exams
Entrance Exams
Downstream speed = 45/3 = 15 km/h. Upstream speed = 35/5 = 7 km/h. Speed in still water = (15 + 7)/2 = 11 km/h. Wait, that's option D. Let me recheck: (D+U)/2 = (15+7)/2 = 11. But the question asks for average speed accounting for distances. Total distance = 45+35 = 80. Total time = 3+5 = 8. Average = 80/8 = 10 km/h.
Rate of A = 1/12, B = 1/15, C = -1/20. Combined rate = 1/12 + 1/15 - 1/20 = (5+4-3)/60 = 6/60 = 1/10. Wait, recalculate: = (5+4-3)/60 = 6/60 = 1/10 hours. Actually (1/12 + 1/15 - 1/20) = (5+4-3)/60 = 6/60. Time = 60/6 = 10 hours. Let me verify: LCD(12,15,20)=60. (5+4-3)/60 = 6/60 = 1/10. Hmm, answer should be different. Recalculating: 1/12 + 1/15 - 1/20. LCM=60: 5/60 + 4/60 - 3/60 = 6/60 = 1/10. So 10 hours. But that's not an option. Let me use: (5+4)/60 - 1/20 = 9/60 - 3/60 = 6/60. Actually if only A and B: 1/12 + 1/15 = 9/60 = 3/20, time = 20/3 = 6.67. With C draining: (1/12 + 1/15) - 1/20 = (5+4-3)/60 = 6/60 = 1/10. Reconsidering the problem setup, let me use standard formula differently. Rate combined (A+B-C) working simultaneously.
CP of 3 items at 20% profit: 3 × (200/1.2) = 500. CP of 2 items at 25% loss: 2 × (300/0.75) = 800. Total CP = 1300, Total SP = 1500. Profit% = (200/1300) × 100 = 15.38/3.58 ≈ 4.29%.
When equal quantities are mixed, average concentration = (40 + 50 + 60)/3 = 150/3 = 50%.
Increase in total age = 38 - 28 = 10. If average increases by 2, then group size = 10/2 = 5.
Interest on first = 5,000 × 8 × 2 / 100 = 800. Interest on second = 7,500 × 12 × 2 / 100 = 1,800. Total interest = 2,600. Average rate = (2,600 × 100) / (12,500 × 2) = 10.4%.
Distance in first 2 hours = 50 × 2 = 100 km (which is 25% of total). Total distance = 400 km. Total time at 60 km/h = 400/60 = 6.67 hours.
For consecutive odd numbers, the average equals the middle (4th) number. So 4th number = 39. The 7 numbers are: 33, 35, 37, 39, 41, 43, 45. Largest = 45.
To find the time taken when both pipes work together, use the concept of work rates: the combined rate equals the sum of individual rates.
Step 1: Find individual work rates
Pipe A fills the tank in 12 hours, so its rate is \(\frac{1}{12}\) tank per hour.
Pipe B fills the tank in 15 hours, so its rate is \(\frac{1}{15}\) tank per hour.
Step 2: Find combined work rate
When both pipes work together:
Find the LCM of 12 and 15, which is 60:
Step 3: Calculate time to fill one tank
If the combined rate is \(\frac{3}{20}\) tank per hour, then time to fill 1 tank is:
Step 4: Convert to hours and minutes
Answer: Both pipes together fill the tank in \(6\) hours and \(40\) minutes (Option B)
We need to find the initial work rate, then recalculate the time for the remaining job at an increased rate.
Step 1: Find the initial work rate
The worker completes \(\frac{1}{4}\) of the job in 5 days.
Step 2: Calculate the new work rate (increased by 25%)
A 25% increase means the new rate is \(1.25\) times the original rate.
Step 3: Find remaining work
The worker has completed \(\frac{1}{4}\) of the job, so the remaining work is:
Step 4: Calculate days needed for remaining work
Using \(\text{Time} = \frac{\text{Work}}{\text{Rate}}\):
Answer: The remaining job will take 12 days at the increased work rate. (Option A)
