OG Problem no 70

[inertia2010]

Can someone please explain this?

List S consists of 10 consecutive odd integers, and list T consists of 5 even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average of the integers in S than the average of the integers in T?

Thanks!!!

[pankajks2010]

I think the information about the second list T is incomplete. Shouldn't it be consecutive even integers? If its not the case, then the information provided is insufficient.

In case the elements in the list T are consecutive, the question can be solved as below:

Let the first element of list S be 2x+1 and that of T be 2y.

Now, as per question, 2x+1=7+2y; 2y=2x-6

The average of elements in S = average of 5th & 6th element = average of (2x+9) and (2x+11) = 2x+10

The average of elements in T = third term of T = 2x-2

Required difference= 2x+10-(2x-2)=12

[inertia2010]

I am extremely sorry. Yes, it is consecutive even integers. Here is the question again:

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average of the integers in S than the average of the integers in T?

This is how i worked:
Let S = 1,3,5,7,9,11,13,15,17,19 - Avg is 100/10 = 10
Let T = 2,4,6,8,10 - Avg is 30/6 =

Least integer in S = (Leasst integer in T ) + 7
1 = 7 + 2

I am struck after this. Can someone help? I dont know if this is the right approach.

Thanks!!

[Anurag@Gurome]

I am extremely sorry. Yes, it is consecutive even integers. Here is the question again:

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average of the integers in S than the average of the integers in T?

This is how i worked:
Let S = 1,3,5,7,9,11,13,15,17,19 - Avg is 100/10 = 10
Let T = 2,4,6,8,10 - Avg is 30/6 =

Least integer in S = (Leasst integer in T ) + 7
1 = 7 + 2

I am struck after this. Can someone help? I dont know if this is the right approach.

Thanks!!

You are taking the list wrongly.
If T is starting with 2, S should start with 2+7 = 9.
Anyway, here is another approach.
Solution:
Let the list S be 2n+1, 2n+3, 2n+5,.....,2n+19 where n is an integer.
Let list T be 2k, 2k+2, 2k+4,...2k+8.
Now 2n+1 - 2k = 7. Or 2n-2k = 6.
Also, average of the integers in S is (4n+20)/2 = 2n+10.
Average of integers in T is 2k+4.
So, difference is (2n+10) - (2k+4) = 2n-2k+6 = 12

[pankajks2010]

If you are fitting in the numbers, then choose them as per the information provided in the question:
As, least integer in S is 7 more than the first integer in T, choose the numbers accordingly.

Let least integer in T be 2, then the least integer in S is 9.

digits in T=2,4,6,8,10; Average=6
digits in S=9,11,13,15,17,19,21,23,25,27; Average=18

Required Difference=18-6=12

[linhshining]

Call the least integer in S: L(s), the most M(s)
the least integer in T: L(t), the most M(t)

Average(s) = [ L(s) + M(s) ] / 2 = [ L(s) + L(s) + 9*2 ] / 2 = L(s) + 9
Average(t) = [ L(t) + M(t) ] / 2 = [ L(t) + L(t) + 4*2 ] / 2 = L(t) + 4

Average(s) - average(t) = L(s) + 9 - [ L(t) + 4 ] = 7 + 5 = 12

I love to solve it this way