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# If a racehorse runs an average(arithmetic mean) of m miles per race fo

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Manager
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If a racehorse runs an average(arithmetic mean) of m miles per race fo  [#permalink]

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18 Oct 2016, 22:24
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If a racehorse runs an average(arithmetic mean) of m miles per race for r races and then runs n miles in its next race, what is the average number of miles the horse has run for the r + 1 races?

a) (rm + n)/r+1
b) (m + n)/r+1
c) (m + n)/r
d) r(m + n)/r+1
e) (m + rn)/r+1

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Re: If a racehorse runs an average(arithmetic mean) of m miles per race fo  [#permalink]

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18 Oct 2016, 23:17
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Average miles ran per race = m for r no of races. So total miles = m*r

Miles ran in next race = n

Total Miles ran = m*r + n in r+1 races

Average miles = Total miles/Total no of races = (m*r + n)/(r + 1)

Option A

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Re: If a racehorse runs an average(arithmetic mean) of m miles per race fo  [#permalink]

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19 Oct 2016, 08:24
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alanforde800Maximus wrote:
If a racehorse runs an average(arithmetic mean) of m miles per race for r races and then runs n miles in its next race, what is the average number of miles the horse has run for the r + 1 races?

a) (rm + n)/r+1
b) (m + n)/r+1
c) (m + n)/r
d) r(m + n)/r+1
e) (m + rn)/r+1

Total distance run in $$r$$ races = $$rm$$
Distance run in last race is $$n$$

So, Total Distance Run = $$rm$$ $$+$$ $$n$$
Total run = $$r + 1$$

Hence average will be $$\frac{rm + n}{r+1}$$

So, Correct answer will be (A)

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Re: If a racehorse runs an average(arithmetic mean) of m miles per race fo  [#permalink]

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05 Dec 2017, 18:52
1
alanforde800Maximus wrote:
If a racehorse runs an average(arithmetic mean) of m miles per race for r races and then runs n miles in its next race, what is the average number of miles the horse has run for the r + 1 races?

a) (rm + n)/r+1
b) (m + n)/r+1
c) (m + n)/r
d) r(m + n)/r+1
e) (m + rn)/r+1

We are given that a racehorse runs an average (arithmetic mean) of m miles per race for r races and then runs n miles in its next race. We need to determine the average number of miles the horse has run for r + 1 races. Let’s plug the given information into the average formula.

average = sum/quantity

Since the racehorse runs an average (arithmetic mean) of m miles per race for r races, the initial sum = mr and the initial quantity is r.

Since the racehorse then runs n miles in its next race, the new sum is mr + n and the new quantity is r + 1.

Thus, the average number of miles that the horse runs for r + 1 races is:

(mr + n)/(r + 1)

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Re: If a racehorse runs an average(arithmetic mean) of m miles per race fo  [#permalink]

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24 Jan 2018, 21:02
Hi All,

This question can be solved Algebraically or by TESTing VALUES. Here's how you can use the second method:

IF...
M = 2
R = 3
Then the racecourse ran (2 miles/race)(3 races) = 6 miles

N = 4
So the horse ran 4 miles in that 1 race.

Thus, the average for the 4 races was (10 miles)/(4 races) = 2.5 miles/race

There's only one answer that matches...

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Re: If a racehorse runs an average(arithmetic mean) of m miles per race fo  [#permalink]

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25 Jan 2018, 08:30
alanforde800Maximus wrote:
If a racehorse runs an average(arithmetic mean) of m miles per race for r races and then runs n miles in its next race, what is the average number of miles the horse has run for the r + 1 races?

a) (rm + n)/r+1
b) (m + n)/r+1
c) (m + n)/r
d) r(m + n)/r+1
e) (m + rn)/r+1

Total distance run in r races is $$mr$$
Total distance run after n miles is $$mr + n$$

So, Average Distance run is $$\frac{mr + n}{r + 1}$$ , Answer must be (A)
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Thanks and Regards

Abhishek....

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Re: If a racehorse runs an average(arithmetic mean) of m miles per race fo &nbs [#permalink] 25 Jan 2018, 08:30
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