Question

A particular cattle is fed 7 kg, 8 kg and 9 kg of special food on Day 1, 2 and 3 respectively by a farmer.

On the Day 4 onwards, the amount of food fed to the cattle follows the following rule:

(i) if the total amount of food fed to the cattle in the past 3 days is 24 kg or more, the cattle will be fed 2 kg less than the day before.

(ii) if the total amount of food fed to the cattle in the past 3 days is less than 24 kg, the cattle will be fed 1 kg more than the day before.

Calculate the minimum number of days for the total amount of food fed to the cattle to exceed 2020 kg.


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Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

Solution:

Day # Amount of food fed on Day # Amount of food fed in the past three days (at the end of Day #)

Day 1 7 -

Day 2 8 -

Day 3 9 7 + 8 + 9 = 24

Day 4 9 – 2 = 7 8 + 9 + 7 = 24

Day 5 7 – 2 = 5 21

Day 6 5 + 1 = 6 18

Day 7 7 18

Day 8 8 21

Day 9 9 24

Day 10 7 24

So, the amount of food fed to the cattle follows the pattern:

7 8 9 7 5 6   7 8 9 7 5 6   7 8 9 7 5 6   7 8 9 7 5 6 ....

Total amount of food fed in the first 6 days = 42 kg

2020 = 48 × 42 + 4

So, in 48 × 6 = 288 days, the total amount of food fed to the cattle is 48 × 42 = 2016 kg.

The total amount of food fed will exceed 2020 on the next day (that is on the 289th day)

The total amount of food fed will exceed 2020 kg in 289 days.