Question

At 8: 00 am there are 8 pepole waiting in a queue of ration shop . 4 people jpins them after 1 min the no. Of persons joining the queue per min increases at a rate of two persons per minute that is in second min 6 people more joins the queue , i third min 8 people and so on the shop opens at 8 : 20 am and distributes ration available for 300 people only find the no. Of people who have to return empty handed if more people stops comimg at 8 : 20 am sharp

Answer 1

Answer:

Step-by-step explanation:

There are 176 people who have to return empty handed if more people stops coming at 8:20 am sharp

Answer 2

Given,

At 8: 00 am number of people standing in a queue = 8

At 8:01 am, a number of people join them = 4

The rate of people joining them = 2 person/minutes

Total time = 20 minutes

The total ration is available for 300 people

To FInd,

The no. of people who have to return empty-handed if more people stop coming at 8:20 am sharp =?

Solution,

At 8:01 am , people joined = 4

At 8:02 am , people joined = 6

At 8:03 am , people joined = 8

This is an AP with a = 4 and common difference(d) = 6 - 4 = 2

According tot the question, time taken = 20 minutes

Therefore, Last term of AP = 20 = nth term

From the formula upto nth term in AP,

$S_n = n / 2(2a + (n-1)d)\\S_{20} = 20/ 2(2*4 + (20-1)2)\\S_{20} = 10 * (8 + (19)*2)\\S_{20} = 10 * (8 + 38)\\S_{20} = 10 * 46\\S_{20} = 460$

Total people joined  = 460

People were already standing  = 8

Total people = 468

The ration is available for 300

The number of people who have to return empty-handed = 468 - 300

The number of people who have to return empty-handed = 168

Hence, if the number of people who have to return empty-handed if more people stop coming at 8:20 am sharp are 168.