Exams › IBPS PO › Quantitative Aptitude
Three automatic toys A, B, and C operate with neck movements (NM) and hand rotations (HR). The data are as follows: Toy A: Battery capacity = 1500 units, battery percentage = 80%. At every 4th NM and 3rd HR together, 1 unit of battery is consumed. Toy A gets completely discharged at 11 AM. Toy B: Battery capacity = 2000 units, battery percentage = 75%. NM = 30/min, HR = 50% of the NM/min of Toy A. At every 3rd NM and 2nd HR together, 1 unit of battery is consumed. Toy C: Battery capacity = 120% of the battery capacity of Toy B, battery percentage = 60%. NM/min = NM/min of Toy A + 5, HR/min = 30. At every 3rd NM and 2nd HR together, 1 unit of battery is consumed. What is the difference between the total NM and HR of Toy C when its battery gets completely discharged?
- 1620
- 1440
- 1920
- 1200
Correct answer: 1620
Solution
Toy A starts with 80% of 1500 = 1200 units and discharges in 2 hours, so its consumption rate can be inferred. Using the given movement-based consumption rule and Toy C’s movement rates, the total NM and HR over its discharge time can be calculated, and their difference comes out to 1620.
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