a is the sum of x consecutive positive integers. b is the

For part a)
a=10+11=b=1+2+3+4+5+6=21
for part b)
a= 6+7+8=1+2+3+4+5+6=21
for part c)
a=6+7+8+9+10+11=3+4+5+6+7+8+9+10+11=63
for part d) impossible
for part e)
a=6+7+8+9+10+11+12+13+14+15=12+13+14+15+16+17+18=105
answer is d)

further explanation in support for d
a= n+(n+1)+(n+2)+(n+3)=4n+6
b=m+ (m+1)+ (m+2)+……(m+9)= 10m+45

4n+6=10m+45
4n=10m+39
4n is even but 10m+39 is odd , so it is impossible

Although (D) is correct the assertion below is not.
5*a(x)(decimal) = 2*a(y) Integer => DECIMAL = Integer

if a(x) = 0.4, a(y)=1 then 5*0.4 = 2 * 1, there are other examples too like a(x) = 7.2 and a(y) = 18

I solved it a slightly different way.

Sum of X cons integers = Kx + x(x+1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y+1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*11/2 = 10K + 55
SUM2 = 4K1 + 4*5/2 = 4K1 + 10

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.

the sum of x consecutive positive integers. b is the sum of y consecutive positive integers

i think your example is not right.

Saha, you are close. But i believe the formula would be

Sum of X cons integers = Kx + x(x-1)/2 (where K is lowest) - 1
Sum of Y cons integers = K1y + y(y-1)/2 (where K1 is lowest) - 2

If x = 10 and y = 4

SUM1 = 10K + 10*9/2 = 10K + 45
SUM2 = 4M + 4*3/2 = 4M + 6

SUM1 = ODD and SUM2 = EVEN, hence D is correct. The others do not provide such an inconsistency.

a(x) is the average of X consecutive integers, a(y) is the averge of Y consecutive integers according to the example given above ? What's wrong ?

Yes, thanks for the correction. My formula would be correct if K+1 and K1+1 were the lowest, not K and K1.

(a) take any 6 consecutive integers (+ve) for y = 10, 11, 12, 13, 14 and 15. their sum is 75. divide it by 2, we get 37.5. so take immidiate integers above and below 37.5. they are 37 and 38 and their sum is75.

(b) take any 6 consecutive integers for y = 10, 11, 12, 13, 14 and 15. their sum is 75. divide it by 3, we get 25. so put 25 in the middle. the 3 consecutive integers are 24, 25 and 26. their sum is 75.

(c) take any 9 consecutive integers for y = 3, 4, 5, 6, 7, 8, 9, 10, and 11. their sum is 63. divide it by 7, we get 9. so put 9 in the middle of 7 consecutive integers. they are 6, 7, 8, 9, 10, 11, and 12. their sum is 63.

(d) take any 10 consecutive integers for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. their sum is 55. similarly take any 4 consecutive integers and add them up, the sum is even number/integer.

here we have interesting pattren with 10 and 4 consecutive integers. any 10 such integers have sum ending at 5 where as sum of 4 consecutive integers end at even integer. so these two consecutive integers never be equal.

(e) take any 10 consecutive integers for x = 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15. their sum is 105. divide it by 7, we get 15. so put 15 in the middle of 7 consecutive integers. they are 12, 13, 14, 15, 16, 17, and 18. their sum is 105.

it is D, however it is lengthy.

The decimal average can never be 0.4 or anything different from
.5 in tenths. And it will never give an integer when multiplied by 5.
So my assertion is perfectly correct.

I realised that later but it wasn't mentioned in that form earlier. its a good method i think.

I did the same way, but i dont agree with the first solution of sum=average*number

Instead of going into averages etc. Here is shorter method:

(A) x = 2; y = 6=> a is always ODD : b is always ODD
(B) x = 3; y = 6 => a may be ODD or EVEN : b is always ODD
(C) x = 7; y = 9 => a may be ODD or EVEN : b may be ODD or EVEN
(D) x = 10; y = 4=> a is always ODD : b is always EVEN
(E) x = 10; y = 7 => a may be ODD or EVEN : b is always ODD

Great shortcut! However, i think you made a typo in (A) --> a should be odd.

Yep I solved the same way!

Yes that was a typo. Thanks buddy. I edited that.

Shouldn't (E) be: a is always odd, b maybe odd or even?