Directions : Given data is regarding three automatic toys on two types of…
2019
Directions : Given data is regarding three automatic toys on two types of movements: Neck movements (NM) and Hand rotation (HR). It starts recording from 9 AM onwards on 12 June. Each toy has different battery percentage and battery capacity.
Toy A: Battery Capacity = 1500 units, Battery Percent = 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 percent = 75%
NM = 30/min, HR/min = 50% of 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 battery capacity of toy B, Battery Percent = 60% NM/min = NM/min of toy A + 5, HR = 30/min. at every 3rd NM and 2nd HR together. 1 unit of battery is consumed.
If toy B & A had been charged completely (100%), then what would be the difference between time taken by both the toys to get discharged completely?
- A.
50 min
- B.
90 min
- C.
0 min
- D.
15 min
- E.
10 min
Show answer & explanation
Correct answer: A
Concept
Battery life of a toy = (total battery units available) divided by (units consumed per minute). A consumption rule of the form "at every k-th NM and m-th HR together, 1 unit is consumed" means the per-minute drain rate equals (NM per minute)/k, which equals (HR per minute)/m. So the whole problem reduces to finding each toy's units-per-minute drain, then dividing the FULL capacity by that drain to get the discharge time.
Step 1 - Find Toy A's drain rate
Toy A holds 1500 units. At 80% it carries 0.80 x 1500 = 1200 units, and these 1200 units last from 9 AM to 11 AM, i.e. 120 minutes.
Available units at 80% = 0.80 x 1500 = 1200 units.
Time to discharge = 11 AM - 9 AM = 120 minutes.
Drain rate of A = 1200 / 120 = 10 units per minute.
From A's rule (1 unit per 4th NM and 3rd HR together), drain = NMA/4, so NMA = 4 x 10 = 40 NM per minute. This NMA is needed for the other toys.
Step 2 - Find Toy B's drain rate
Toy B does 30 NM per minute, and its HR per minute = 50% of Toy A's NM per minute = 0.50 x 40 = 20 HR per minute. B's rule consumes 1 unit per 3rd NM and 2nd HR together.
Via NM: drain = NMB / 3 = 30 / 3 = 10 units per minute.
Via HR: drain = HRB / 2 = 20 / 2 = 10 units per minute.
Both agree, so Toy B's drain rate = 10 units per minute.
Step 3 - Discharge time at 100% for each toy
Charged to 100%, each toy starts from its FULL capacity, then drains at the rate found above.
Toy A at 100% = 1500 units; time = 1500 / 10 = 150 minutes.
Toy B at 100% = 2000 units; time = 2000 / 10 = 200 minutes.
Step 4 - Required difference
Difference = time for B - time for A = 200 - 150 = 50 minutes.
Hence the difference in full-charge discharge time between the two toys is 50 minutes.
Cross-check
The drain rates do not depend on the starting charge, so re-deriving them from full capacity changes nothing: A still drains 10/min and B still drains 10/min. Dividing the larger capacity (2000) and the smaller capacity (1500) by the same 10/min gives 200 and 150 minutes, a gap of exactly 50 minutes - confirming the result.