Z is the set of the first n positive odd numbers, where n is

Question

Z is the set of the first n positive odd numbers, where n is a positive integer. Given that n > k, where k is also a positive integer, x is the maximum value of the sum of k distinct members of Z, and y is the minimum value of the sum of k distinct members of Z, what is x + y?

(A) kn
(B) kn + k^2
(C) kn + 2k^2
(D) 2kn – k^2
(E) 2kn

(A) kn
(B) kn + k^2
(C) kn + 2k^2
(D) 2kn – k^2
(E) 2kn

Bunuel · Accepted Answer

Probably the easiest way to solve this question would be to assume some values for n and k.

Say n=3, so Z, the set of the first n positive odd numbers would be: Z={1, 3, 5};
Say k=1, so X, the maximum value of the sum of K distinct members of Z would simply be 5. Similarly, Y, the minimum value of the sum of K distinct members of Z would simply be 1.

X+Y=5+1=6.

Now, substitute n=3 and k=1 in the options provided to see which one yields 6. Only asnwer choice E fits: 2kn=2*3*1=6.

Answer: E.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Hope it helps.

akashganga · Answer

In the given case, sum of all n odd numbers is n^2.

If the numbers are arranged in increasing order, the max sum is when k is taken from end. So the sum of k odd numbers taken from the end
 = n^2 - (n-k)^2
 (Sum of all numbers) - (Sum of remaining initial numbers)
which is equal to 2nk - k^2 -- (1)

Now for minimum sum, take k numbers from the starting and their sum is k^2 -- (2)

Adding (1) and (2) = 2nk.

DJ

sagarsir · Answer

for different values of n and k.. the options are invalid.

sambam · Answer

Just use hypothetical numbers. Say n = 5 (1,3,5,7,9) and k = 3. This would mean that x = 21 and y = 9. 
x+y = 30

Only answer E gives this result.

fdecazaux · Answer

Let us name Zi the general term of the set Z.
z(i) = 2*i - 1 for i = 1 to n.
For instance z(1) = 2*1 - 1 = 1
 z(2) = 2*2 - 1 = 3

z(n) = 2*n - 1

y is the sum of the first k terms of the set Z.
x is the sum of the terms from order (n-k+1) to n.

Let us express the sums x and y. Each of the sums contains k terms.
y = Sum (2*i - 1) for i = 1, ... k
x = Sum (2*j - 1) for j = (n-k+1), ,n

By developing y, we obtain 
y = 2*Sum (i) - k for i=1,....k
y = 2*(1/2)  /2 - k = k^2 + k - k = k^2
(we can also remember that the sum of the first k odd numbers equals k^2).

By developing x, we obtain
x = 2*Sum (j) - k for j=n-k+1, n
We can notice that x is the difference of 
 - the sum of the first "n" odd numbers
 - and the sum of the first (n-k) odd numbers.
We deduce 
x = n^2 - (n-k)^2=2*n*k - k^2

Then x + y = 2*n*k - k^2 + k^2 = 2*n*k

The answer is E.

Tanvi94 · Answer

You mentioned that  Say k=1, so X, the maximum value of the sum of K distinct members of Z would simply be 5.

but for the example that you have used - Z={1, 3, 5} - shouldn't the maximum value of the sum of K distinct members of Z be = 1+3+5 = 9 ?

Please help were am I going wrong ?

blizzardAres · Answer

hi, He assumed K = One distinct member ; so max sum of One distinct member of Z going to be 5 and minimum is going to be 1.

ScottTargetTestPrep · Answer

We can let n = 5 and k = 3. Thus Z = {1, 3, 5, 7, 9}, x = 5 + 7 + 9 = 21, y = 1 + 3 + 5 = 9 and hence x + y = 21 + 9 = 30.

Now let’s check the given answer choices (note: we will be looking for the one that is equal to 30):

A) kn = 5(3) = 15 → This is not 30.

B) kn + k^2 = 5(3) + 3^2 = 15 + 9 = 24 → This is not 30.

C) kn + 2k^2 = 5(3) + 2(3)^2 = 15 + 18 = 33 → This is not 30.

D) 2kn - k^2 = 2(5)(3) - 3^2 = 30 - 9 = 21 → This is not 30.

E) 2kn = 2(5)(3) = 30 → This is 30

Answer: E

jsam98 · Answer

idk what's wrong in my approach : min sum = sum of first k odd integer is k^2 max sum = sum of last k odd integer is n + /2 * k n + /2 * K n-k+1 * k x+y = K^2 +nK - K^2 +K = nk + k no option

Krunaal · Answer

Z = (1, 3, ......., 2n-1) 
Largest terms in  Z = (2n-1, 2n-3,........, 2n-2k+1) 
Max sum (x) =  k/2 * (2n-1 + 2n-2k+1) = k/2 * (4n-2k) = 2nk-k^{2} 
Min sum (y) = Sum of first k odd integers =  k^{2} 
 x+y = 2nk

totamofficiis · Answer

So just to make it clear. first n consecutive positive odd number sum is always n^2. 
so it becomes y=k^2 x= n^2-(n-k)^2
x+y = k^2+n^2-n^2-k^2+2kn = 2kn

Z is the set of the first n positive odd numbers, where n is

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Z is the set of the first n positive odd numbers, where n is

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