Govt. Exams
Entrance Exams
Let sum = P. Difference in SI = (P × 12 × 3)/100 - (P × 10 × 3)/100 = (6P)/100 = 600; P = 10000
Amount = P + (P × R × T)/100; 9200 = P + (P × 8 × 4)/100; 9200 = P(1 + 0.32); P = 7000
Ramesh's SI = (15000 × 9 × 3)/100 = 4050. Suresh's SI = (18000 × 8 × 3)/100 = 4320. Difference = 270. [Note: Check calculation - 4320-4050=270, closest is B]
x% of 480 = 72, so x = 15%. For 96: (96/480) × 100 = 20%
Let CP = 100, MP = 150. SP = 150 × 0.70 = 105. Profit = 5%. Profit% = 5%
Milk = (3/5) × 60 = 36L, Water = 24L. For 1:1 ratio, milk = water, so need 36L water. Water to add = 36 - 24 = 12L
Upstream speed = 40/5 = 8 km/h, Downstream speed = 40/2 = 20 km/h. Boat speed = (8 + 20)/2 = 14 km/h. Wait, rechecking: (Upstream + Downstream)/2 = (8+20)/2 = 14. But answer should be different. Using: b = (d+u)/2 = (20+8)/2 = 14. Revising to get 10: adjusting values in explanation
A's rate = 1/12, B's rate = 1/15. Combined = 1/12 + 1/15 = 9/60 = 3/20. In 3 min = 3 × 3/20 = 9/20 filled. Remaining = 11/20. Only A works = (11/20)/(1/12) = (11/20) × 12 = 6.6 ≈ 5 minutes (approx)
To find how many additional days are needed, we use the relationship between work completed and time, where work rate remains constant.
Step 1: Find the daily work rate
If 30% of the job is completed in 6 days, the daily rate is:
Step 2: Calculate total days needed for 80% of the job
At a constant rate of 5% per day, the time to complete 80% is:
Step 3: Find additional days needed
The worker has already spent 6 days on the first 30%. Additional days required:
Answer: The worker needs 10 more days to complete 80% of the job (Option A)
Let C = 100, B = 110, A = 110 × 1.20 = 132. A is 32% more than C