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Manager  Joined: 05 Sep 2007
Posts: 124
Location: New York
A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 74% (01:57) correct 26% (02:13) wrong based on 367 sessions

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A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?

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

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Director  Joined: 10 Sep 2007
Posts: 708
Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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Lets assume there are 2X people.
Half of them have average weight of 180 and other half has 215.
Maximum Weight is = 2000
So 180*X + 215*X = 2000
=> 395X = 2000
=> X is approximately equal to 5.
So total people is 2*5 = 10

We are not taking 11 as answer because say 11th person has minimum of 180 weight then
180*6 + 215*5 = 2155 (Which is more than 2000)

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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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A certain elevator has a safety weight limit of 2000 pounds. Whats is the greatest possible number of people who can safely ride on the elevator at one time with the average (aritmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?

A. 7
B. 8
C. 9
D. 10
E. 11
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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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.5X(180)+.5X(215)<=2000
X<=~10.125
Greatest possible integer is 10

So--D
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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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1
x/2(180) + x/2(215) = 2000

That gives x = (2000 * 2)/395, x ~ 10.1

So it has to be D
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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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el1981 wrote:
A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11

2000 pounds total

180*x/2 + 215*x/2 = 2000

Max is 10

Hence D

Hope it helps
Cheers!
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GMAT 1: 620 Q42 V33 Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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3
Since it is 50%-50% it has to be an even number.
Only 8 and 10 are even.

Average weight = (180+215)/2 = 197.5

We can then easily see that 197.5*10 = 1,975 pounds, so answer = 10.
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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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Enael wrote:
Since it is 50%-50% it has to be an even number.
Only 8 and 10 are even.

Average weight = (180+215)/2 = 197.5

We can then easily see that 197.5*10 = 1,975 pounds, so answer = 10.

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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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10 people.

Equation to use: --> (#/2)(180)+(#/2)(215) = CANNOT EXCEED 2,000

5(180)+5(215) = 1975 [We've maxed out on this option, thus this is correct]
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GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33 Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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Alternative Approach :

People can not be divided in half..... so we remove odd options . Remaining with 8 & 10.
Start with the largest :10: so 5*180 +5 * 215=900+1075=1975<2000.
Hence, Ans is 10 (D)
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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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walkerme13 wrote:
x/2(180) + x/2(215) = 2000

That gives x = (2000 * 2)/395, x ~ 10.1

So it has to be D

how did we get this ? $$\frac{180x}{2}$$and this $$\frac{215x}{2}$$

I understand that average weight of half group is 180 , and average weight of another one is 215

but why do we divide by 2 ; we dont know number of persons in each group right thanks! Senior PS Moderator D
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GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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1
dave13, So let n folks have an average weight of 180 pounds ... total weight will be 180n.

The other n will have average wt of 215, so total weight will be 215n

Now, the total weight of the entire group will be 215n + 180n = 395n

whereas total no. of folks will be 2n... so the new average will be = 395n/2n = 395/2 = 197.5

That's how we get the divide by 2

Hope it is clear.

best,

dave13 wrote:
walkerme13 wrote:
x/2(180) + x/2(215) = 2000

That gives x = (2000 * 2)/395, x ~ 10.1

So it has to be D

how did we get this ? $$\frac{180x}{2}$$and this $$\frac{215x}{2}$$

I understand that average weight of half group is 180 , and average weight of another one is 215

but why do we divide by 2 ; we dont know number of persons in each group right thanks! _________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
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Re: A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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el1981 wrote:
A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?

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

Now as we see that average weight of two groups is divided in half, then only even values will work.

Said that out of 8 and 10,only 10 will work

(180 + 215)/2 = 197.5

Since we will get 1975 as the sum, which when divided by 10, gives us the average as 197.5
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GMAT 1: 800 Q51 V51 A certain elevator has a safe weight limit of 2,000 pounds.  [#permalink]

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el1981 wrote:
A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?

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

Since the correct answer will include two equal numbers of people, and since people do not come in fractional quantities and, thus, may be represented only by integer values, the correct answer will be the sum of two equal integer values, and thus, must be an even number.

So, the only two choices that work are (B) 8 and (D) 10.

While the riders may have various weights, we can ignore the variation in the weights because we know that the average weight of half of the riders is 180 pounds and the average weight of the other half of the riders is 215 pounds. So, we can proceed as we would if half the riders weighed exactly 180 pounds each and the other half weighed exactly 215 pounds each.

Since the average rider weight of half of the riders is 180 pounds, and the average rider weight of other half of the riders is 215 pounds, the average rider weight of all the riders is the average of these two numbers.

We can see that, because 180 is 20 less than 200, 215 is 15 greater than 200, and 20 > 15, the average rider weight is under 200 pounds.

Thus, 10 riders can ride, as 10 x (a number < 200) < 2000.

The correct answer is (D).
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