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?

  1. A.

    50 min

  2. B.

    90 min

  3. C.

    0 min

  4. D.

    15 min

  5. 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.

  1. Available units at 80% = 0.80 x 1500 = 1200 units.

  2. Time to discharge = 11 AM - 9 AM = 120 minutes.

  3. 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.

  1. Via NM: drain = NMB / 3 = 30 / 3 = 10 units per minute.

  2. Via HR: drain = HRB / 2 = 20 / 2 = 10 units per minute.

  3. 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.

  1. Toy A at 100% = 1500 units; time = 1500 / 10 = 150 minutes.

  2. Toy B at 100% = 2000 units; time = 2000 / 10 = 200 minutes.

Step 4 - Required difference

  1. 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.

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