Question

A and b set out to meet each other from two places 165 km apart. a travels 15 km on first day, 14 km the second day , 13 km the third day and so on. b travels 10 km the first day, 12 km the second day, 14 km the third day and so on. after how many days will they meet?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

CONCEPT

Here we can use the concept of arithmetic progressions.

GIVEN

We are given that A travels 15,14,13.... on consecutive days; and B travels 10,12,14.... on those days. The total distance they need to cover collectively is 165 km.

FIND

We need to find after how many days will A and B meet.

SOLUTION

Suppose they meet after n days.

As we can see, A and B are covering distances per day in arithmetic progression or AP.

So first lets focus on A:     AP made by A is 15,14,13....

                                          so,  first term a=15

                                                 common difference d= 14-15= -1

Distance covered by A in n days = sum of distances covered on 1st,2nd,.......,nth days.

                                                       = n/2 [2a+(n-1)d]

                                                       = n/2 [2*15+(n-1)(-1)]

                                                       = n/2 [31-n]

So likewise now lets focus on B: 10,12,14.........

                                               first term a=10

                                               common difference d=12-10=2

                                         sum to n terms= n/2 [2*10+(n-1)(2)]

                                                                  = n/2 [18+2n]

Now, sum of distance covered by A + sum of distance covered by B=165

n/2 [31-n] + n/2[18+2n] = 165

n/2 [31-n+18+2n] =165

n*[49+n]= 165*2

n²+49n=330      

n² +49n -330=0

by solving this quadratic equation, we get n=6 as the answer.

So, A and B will meet after 6 days

SPJ#3