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When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

Re: When a person aged 39 is added to a group of n people, the average age [#permalink]

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19 Aug 2015, 02:34

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Answer according to me: A

Let's assume that the total no. of people is n and the initial average is x.

So we can assume that the total sum of ages would be nx initially.

When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:

(nx+39)/(n+1) = x+2 solving this equation: nx+39=(x+2)(n+1) nx+39=nx +x +2n+2

we get,

2n+x=37

When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:

(nx+15)/(n+1) = x-1

we get

x-n=16

solving the two equation simultaneously, we get n's value as 7.

Re: When a person aged 39 is added to a group of n people, the average age [#permalink]

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19 Aug 2015, 02:55

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Bunuel wrote:

When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Ans: A

Solution: lets say avg is x so total is (xn): now with the first condition (nx+39)/n+1 = x+2 2n+x = 37 Eq.1

with second condition (nx+15)/n+1 = x-1 x-n = 16 Eq.2

by solving Eq1 and Eq2 we can find out n=7
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]

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19 Aug 2015, 12:56

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A simple and elegant solution.

As addition of 39, shifts mean by 2, and addition of 15, shifts mean by 1 to the other side, we have the mean lying between 39 & 15, and in a ratio of 2:1

39-15 = 24 24 divide by 3 is 8.

Meaning mean of the n terms is 15+8 = 39-16 = 23

Now, from first statement, When a person aged 39 is added to a group of n people, the average age increases by 2.

When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.
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When a person aged 39 is added to a group of n people, the average age [#permalink]

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08 Dec 2016, 21:40

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Here is my solution to this one --> Let the original Mean = \(p\) Thus original sum = \(p*n\) Where \(n\) is the number of people in the original set. we need to get

As per question => \(\frac{p*n+39}{n+1} = p+2\) \(p+2n=37\)-> Equation 1

Re: When a person aged 39 is added to a group of n people, the average age [#permalink]

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11 Feb 2018, 12:44

Bunuel wrote:

Bunuel wrote:

When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.

Would it be possible to figure out the initial average in a problem like this?

Re: When a person aged 39 is added to a group of n people, the average age [#permalink]

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11 Feb 2018, 20:32

Bunuel wrote:

When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Kudos for a correct solution.

let t=total age of group (t+39)/(n+1)-2=t/n (t+15)/(n+1)+1=t/n subtracting equation 2 from 1, 24/(n+1)=3 n=7 A

When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.

Would it be possible to figure out the initial average in a problem like this?

Yes. Initial average is 23. From: 7x = 8*(x-1)-15 --> x = 23.
_________________

When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Kudos for a correct solution.

Let m = mean of ORIGINAL n people This means the SUM of the ages of the ORIGINAL n people = nm

When a person aged 39 is added to a group of n people, the average age increases by 2 In other words: new mean (with extra person) of n+1 people = original mean + 2 Rewrite as: (nm + 39)/(n+1) = m + 2 Cross multiply to get: nm + 39 = (n + 1)(m + 2) Simplify: nm + 39 = nm + 2n + m + 2 Simplify: 39 = 2n + m + 2 Rearrange to get: 2n+ m = 37

When a person aged 15 is added instead, the average age decreases by 1 In other words: new mean (with extra person) of n+1 people = original mean - 1 Rewrite as: (nm + 15)/(n+1) = m - 1 Cross multiply to get: nm + 15 = (n + 1)(m - 1) Simplify: nm + 15 = nm - n + m - 1 Simplify: 15 = -n + m - 1 Rearrange to get: -n + m = 16

We now have the following system: 2n+ m = 37 -n + m = 16

Subtract the bottom equation from the top equation to get: 3n = 21 Solve: n = 7

When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

We can let the original average = x, and thus, the original sum is nx; thus:

x + 2 = (39 + nx)/(n + 1)

(n + 1)(x + 2) = 39 + nx

nx + x + 2n + 2 = 39 + nx

x + 2n = 37 (Eq 1)

and

x - 1 = (15 + nx)/(n + 1)

(n + 1)(x - 1) = 15 + nx

nx + x - n - 1 = 15 + nx

x - n = 16

x = 16 + n (Eq 2)

Substituting Eq 2 into Eq 1, we have:

16 + n + 2n = 37

3n = 21

n = 7

Answer: A
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