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 C would be 100% charged then at what time battery of toy C will drain completely?

  1. A.

    11:30 AM

  2. B.

    11:45 AM

  3. C.

    11:35 AM

  4. D.

    11:40 AM

  5. E.

    11:50 AM

Show answer & explanation

Correct answer: D

Concept

For any device, the run-time equals its total available battery units divided by the units consumed per minute. Battery units = Battery Capacity x Battery Percent. The per-minute consumption is found by counting how many complete "trigger sets" occur each minute, where one set (a fixed number of neck movements NM together with a fixed number of hand rotations HR) drains exactly 1 unit. Because a set needs BOTH its NM quota and its HR quota, the number of sets per minute is limited by whichever quota is reached first: sets/min = min(NM per min / NM-per-set, HR per min / HR-per-set).

Step 1 - Fix Toy A's movement rates from its known life

Toy A is the anchor: its NM/min and HR/min are not given directly but are recoverable from the fact that it runs out exactly between 9 AM and 11 AM.

  1. Battery units of A = 1500 x 80% = 1200 units.

  2. It runs 9 AM to 11 AM = 120 minutes, so consumption = 1200 / 120 = 10 units per minute.

  3. Each unit needs a set of 4 NM and 3 HR, so 10 sets per minute means NM = 4 x 10 = 40 per minute and HR = 3 x 10 = 30 per minute for Toy A.

Step 2 - Derive Toy C's parameters

  1. Capacity of C = 120% of capacity of B = 1.2 x 2000 = 2400 units. At 100% charge the usable units are the full 2400.

  2. NM of C = NM of A + 5 = 40 + 5 = 45 per minute. HR of C = 30 per minute (given).

Step 3 - Consumption rate and drain time for Toy C

For Toy C one unit is drained per set of 3 NM and 2 HR.

  1. Sets allowed by NM = 45 / 3 = 15 per minute; sets allowed by HR = 30 / 2 = 15 per minute. Both quotas finish together, so consumption = 15 units per minute.

  2. Full-charge run-time = 2400 / 15 = 160 minutes.

  3. Recording starts at 9 AM, so 9 AM + 160 minutes = 9 AM + 2 hours 40 minutes = 11:40 AM.

Cross-check

160 minutes x 15 units/min = 2400 units, exactly the full capacity of Toy C - so the battery empties precisely at 11:40 AM. Note the NM and HR quotas both yield 15 sets/min, so neither movement is the sole bottleneck; this consistency confirms the rate of 15 units/min.

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