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# If p and n are positive integers, with p > n, what is the average of n

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Math Expert
Joined: 02 Sep 2009
Posts: 58335
If p and n are positive integers, with p > n, what is the average of n  [#permalink]

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14 Sep 2015, 22:53
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If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?

A. (p + n)/2
B. (p – n)/2
C. p + n
D. p – n
E. p^2 – n^2

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If p and n are positive integers, with p > n, what is the average of n  [#permalink]

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14 Sep 2015, 23:27
Bunuel wrote:
If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?

A. (p + n)/2
B. (p – n)/2
C. p + n
D. p – n
E. p^2 – n^2

Let, n = 2
and p = 3

i.e. Average of 2 (n) consecutive multiples of 3 (p) starting from 3 (p) = (3+6)/2 = 9/2
and Average of 3 (p) consecutive multiples of 2 (n) starting from 2 (n) = (2+4+6)/3 = 12/3 = 4

The average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n) = (9/2)-4 = (1/2)

A. (p + n)/2 = (3+2)/2 = 7/2 INCORRECT
B. (p – n)/2 = (3-2)/2 = 1/2 CORRECT
C. p + n = (3+2) = 7 INCORRECT
D. p – n = (3-2) = 1 INCORRECT
E. p^2 – n^2 = 3^2 - 2^2 = 5 INCORRECT

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Re: If p and n are positive integers, with p > n, what is the average of n  [#permalink]

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14 Sep 2015, 23:44
4
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?

A. (p + n)/2
B. (p – n)/2
C. p + n
D. p – n
E. p^2 – n^2

The n consecutive multiples of p (starting with p) : p, 2p, 3p,...., np
So the average is [n(n+1)/2]*p/n = p(n+1)/2.
Similarly the p consecutive multiples of n (starting with n) : n, 2n,..., pn
--> the average is [p(p+1)/2]*n/p = n(p+1)/2.
So the difference is p(n+1)/2 – n(p+1)/2 = (p-n)/2.
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Re: If p and n are positive integers, with p > n, what is the average of n  [#permalink]

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15 Sep 2015, 21:15
Hello bunuel
Please post the concept behind this question when you post the answer! I got the answer as b by picking numbers.
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Re: If p and n are positive integers, with p > n, what is the average of n  [#permalink]

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17 Sep 2015, 14:09
3
Bunuel wrote:
If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?

A. (p + n)/2
B. (p – n)/2
C. p + n
D. p – n
E. p^2 – n^2

This question contains 2 evenly spaced sets.
1) p, 2p, 3p, ..... np
2) n, 2n, 3n, ..... pn

Now,
Average of any evenly spaced set is given by ( sum of first term + sum of last term)/2
Average of first set will be

$$\frac{p+np}{2}$$

Average of 2nd set will be

$$\frac{n+np}{2}$$

The difference between averages will be

$$\frac{p+np}{2}$$-$$\frac{n+np}{2}$$ = $$\frac{p-n}{2}$$

Math Expert
Joined: 02 Sep 2009
Posts: 58335
Re: If p and n are positive integers, with p > n, what is the average of n  [#permalink]

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20 Sep 2015, 21:07
1
Bunuel wrote:
If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?

A. (p + n)/2
B. (p – n)/2
C. p + n
D. p – n
E. p^2 – n^2

MANHATTAN GMAT OFFICIAL SOLUTION:

The easiest way to solve this problem is to pick convenient numbers for p and n. Say that p = 3 and n = 2, to follow the constraint that p > n.

Then we need to take the difference of two quantities:

1. The average of 2 consecutive multiples of 3 (starting with 3)
2. The average of 3 consecutive multiples of 2 (starting with 2).

The first quantity is the average of 3 and 6. That number is 4.5.

The second quantity is the average of 2, 4, and 6. That number is 4.

The difference between these two numbers is 0.5. We now look for the answer choice that yields 0.5 when p = 3 and n = 2. The only answer choice that does so is (p – n)/2.

Proving that this answer is correct in the abstract is significantly harder, but not outlandish.

The sum of n consecutive multiples of p = p + 2p + 3p + … + np.

Factor out a p to get p(1 + 2 + 3 + … + n). Now, the sum of n consecutive integers starting at 1 is the average of those integers times the number of integers.

The average of the integers is (n+ 1)/2. The number of integers is n.

So the sum of 1 through n is n(n + 1)/2. This is a formula that you shouldn’t memorize, but you should be distantly familiar with it, as passing acquaintances.

Thus, the sum of n consecutive multiples of p is pn(n + 1)/2, and the average of those n multiples is that sum divided by n, or p(n + 1)/2.

A similar argument gives you the average of p consecutive multiples of n. You can just switch p and n in the formula above: this average is n(p + 1)/2.

Subtracting these two gives you (p – n)/2.

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Re: If p and n are positive integers, with p > n, what is the average of n  [#permalink]

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25 Jan 2019, 23:18
Bunuel wrote:
If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?

A. (p + n)/2
B. (p – n)/2
C. p + n
D. p – n
E. p^2 – n^2

They are +ive Integers, 3 > 2

average of n consecutive multiples of p (starting with p) = (3+6)/2 = 4.5

average of p consecutive multiples of n (starting with n) => 2+4+6 / 3 = 4

Less than ?? => 4.5 - 4 = 0.5, we have to look for this now

3-2 / 2

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Re: If p and n are positive integers, with p > n, what is the average of n   [#permalink] 25 Jan 2019, 23:18
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