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What is the maximum possible weight for the heaviest car?

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by bhumika.k.shah » Sat Jan 30, 2010 3:47 am
The combined weight of 10 cars is 14000 pounds. No car weighs more than 50% more than any other car. What is the maximum possible weight for the heaviest car?

(A) 1,600
(B) 1,700
(C) 1,800
(D) 1,900
(E) 2,000

HOWWWWWWWWWW TO SOLVEEEEE!!! :(
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by sars72 » Sat Jan 30, 2010 4:32 am
take the mean weight of the 10 cars to be variable M
let H be the weight of the heaviest car
let L be the weight of the leasy heavy (lightest) car

-> 10M= 14000 -> M = 1400

No car weighs more than 50% more than any other car -> H <= 1.5L

lets start of with the higher value choices and eliminate 1 by 1

start off with choice (E) 2000

H = 2000 pounds; since it is no heavier by 50% of any other car
--> 2000< 1.5L --> 1333 <= L, which means that the lightest car should be greater than or equal to (4000/3)

So, to check if this answer is plausible, we assume that all other cars weigh he minimum possible which is (4000/3)

-> (4000/3)*9 = 12000
Add this to the heaviest car which is 2000 -> 12000 + 2000 = 14000; 14000 is the total weight given in the question

So, it is possible for the heaviest car to weight 2000

Since this is the highest answer choice, this is the answer we are looking for and we do not need proceed further.

Answer is (E) 2000
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by ajith » Sat Jan 30, 2010 5:58 am
bhumika.k.shah wrote:The combined weight of 10 cars is 14000 pounds. No car weighs more than 50% more than any other car. What is the maximum possible weight for the heaviest car?

(A) 1,600
(B) 1,700
(C) 1,800
(D) 1,900
(E) 2,000

HOWWWWWWWWWW TO SOLVEEEEE!!! :(

Now if only one car had 1.5 times weight of the other 9 and if that car had 1.5 x weight of the others. and if all the other cars had the same weight x

10.5 x = 14000
x= 4000/3
the heaviest car weighs 4000/3*1.5 = 2000

now since 2000 is the maximum weight that is in the choices select E
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by rahul.s » Sat Jan 30, 2010 6:16 am
sorry. wrong post.
Last edited by rahul.s on Sat Jan 30, 2010 6:35 am, edited 1 time in total.
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by sanju09 » Sat Jan 30, 2010 6:32 am
bhumika.k.shah wrote:The combined weight of 10 cars is 14000 pounds. No car weighs more than 50% more than any other car. What is the maximum possible weight for the heaviest car?

(A) 1,600
(B) 1,700
(C) 1,800
(D) 1,900
(E) 2,000

HOWWWWWWWWWW TO SOLVEEEEE!!! :(
For having the heaviest possible single car, let the remaining 9 cars weigh x pounds each, and the single heaviest car weighs h pounds, such that:

9 x + h = 14000 and h ≤ 1.5 x. For maximum h, we must take h = 1.5 x

Now, 9 x + 1.5 x = 14000

Or x = 14000/10.5 and h = [spoiler]2,000 [/spoiler]pounds.

[spoiler]E[/spoiler]
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by bhumika.k.shah » Sat Jan 30, 2010 8:22 am
How do u expect to solve such sums in less than 2 mins ??? :-P
sars72 wrote:take the mean weight of the 10 cars to be variable M
let H be the weight of the heaviest car
let L be the weight of the leasy heavy (lightest) car

-> 10M= 14000 -> M = 1400

No car weighs more than 50% more than any other car -> H <= 1.5L

lets start of with the higher value choices and eliminate 1 by 1

start off with choice (E) 2000

H = 2000 pounds; since it is no heavier by 50% of any other car
--> 2000< 1.5L --> 1333 <= L, which means that the lightest car should be greater than or equal to (4000/3)

So, to check if this answer is plausible, we assume that all other cars weigh he minimum possible which is (4000/3)

-> (4000/3)*9 = 12000
Add this to the heaviest car which is 2000 -> 12000 + 2000 = 14000; 14000 is the total weight given in the question

So, it is possible for the heaviest car to weight 2000

Since this is the highest answer choice, this is the answer we are looking for and we do not need proceed further.

Answer is (E) 2000
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by bhumika.k.shah » Sat Jan 30, 2010 8:24 am
how did u get this 4000/3 ??

--> 2000< 1.5L --> 1333 <= L, which means that the lightest car should be greater than or equal to (4000/3)

So, to check if this answer is plausible, we assume that all other cars weigh he minimum possible which is (4000/3)

-> (4000/3)*9 = 12000
Add this to the heaviest car which is 2000 -> 12000 + 2000 = 14000; 14000 is the total weight given in the question

So, it is possible for the heaviest car to weight 2000

Since this is the highest answer choice, this is the answer we are looking for and we do not need proceed further.

Answer is (E) 2000[/quote][/quote]
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by bhumika.k.shah » Sat Jan 30, 2010 8:28 am
how did u still get 10.5 ??? :-(
ajith wrote:
bhumika.k.shah wrote:The combined weight of 10 cars is 14000 pounds. No car weighs more than 50% more than any other car. What is the maximum possible weight for the heaviest car?

(A) 1,600
(B) 1,700
(C) 1,800
(D) 1,900
(E) 2,000

HOWWWWWWWWWW TO SOLVEEEEE!!! :(

Now if only one car had 1.5 times weight of the other 9 and if that car had 1.5 x weight of the others. and if all the other cars had the same weight x

10.5 x = 14000
x= 4000/3
the heaviest car weighs 4000/3*1.5 = 2000

now since 2000 is the maximum weight that is in the choices select E
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by MFCR » Sat Jan 30, 2010 8:47 am
I think the correct answer is 1600 because if the 10 cars weight 14000 in total, the average weight is 1400. If there is one car heavier, there is another one ligther, and the other 8 cars have the average weight of 1400. I mean:

8*1400=11200 pounds
14000-11200=2800 pounds (the total possible weight of other 2 cars, called A and B). And if the weight of one of them (i.e, A) is 1.5 the another one (B), we have the following:

A+B=2800 , if B=1,5A => A+1,5A=2800 => 2,5A=2800 => A=1120 ; B= 1120+1,5= 1680

And now: (1400*8)+1680 (Car B) +1120 (Car A)=14000

If we choose the answer 1700, we have the following: (1400*8)+1700+(1700/1.5)= 14033 pounds (and the total sum of the weight of the 10 cars is 14000). And if we choose another cuantity higher, the tota sum of the 10 cars increse more and more.

(Sorry for my english because I'm not english speaker)
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by sars72 » Sat Jan 30, 2010 9:56 am
bhumika.k.shah wrote:How do u expect to solve such sums in less than 2 mins ??? :-P
It's just the explanation that looks long. It will take two mins max to solve this. In any case, Sanju's is explanation is much more straightforward and you can see how simple he has made the problem. I went about solving the problem in a round about fashion. :(
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by sars72 » Sat Jan 30, 2010 9:58 am
bhumika.k.shah wrote:how did u get this 4000/3 ??

--> 2000< 1.5L --> 1333 <= L, which means that the lightest car should be greater than or equal to (4000/3)
sorry, didnt explain that.. I divided both sides of the inequality 2000< 1.5L by 1.5 i.e. 3/2
-> 2000 * 2/3 < L --> 4000/3 < L

and that's how i got 4000/3
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by sars72 » Sat Jan 30, 2010 10:06 am
MFCR wrote:I think the correct answer is 1600 because if the 10 cars weight 14000 in total, the average weight is 1400. If there is one car heavier, there is another one ligther, and the other 8 cars have the average weight of 1400. I mean:

8*1400=11200 pounds
14000-11200=2800 pounds (the total possible weight of other 2 cars, called A and B). And if the weight of one of them (i.e, A) is 1.5 the another one (B), we have the following:

A+B=2800 , if B=1,5A => A+1,5A=2800 => 2,5A=2800 => A=1120 ; B= 1120+1,5= 1680

And now: (1400*8)+1680 (Car B) +1120 (Car A)=14000
Classic silly mistake. The mean value of 1400 is for that 10 cars combined. By doing the calculation involving (1400*8), you are assuming that the mean value of the 8 cars is also 1400. You cannot take out 2 cars and assume that the mean of the remaining cars will remain the same.

take the following example. I have 6 numbers: 0,2,2,2,2,40. The total is 48 and hence the mean is 48/6= 8.

If i follow what you did, I will get the total by adding the smallest and largest and then adding the multiple of the remaining items (4) by the mean (*) --> Total = 0+40+(8*4) = 72

Hope this helps MFCR.
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by sumanr84 » Sat Jan 30, 2010 10:40 am
Ans: E

Avg wt = 1400. Now to make one car heaviest than the rest of the cars we will shift the wt of 9 cars and add the diff from mean to the other car. Let me put it in the form of eqn.
Im going to take off wt from 9 cars and add that extra to 10th car so that it becomes heavy.

(1400 - x)*9 + (1400 + 9x ) = 1400 * 10

Now we are given with a condition that max car weight should not exceed more than 50% of other car weights

(1400 - x ) 1.5 >= 1400 + 9x
2100 - 1.5x >= 1400 + 9x
700 >= 10.5 x
66.66 >= x ( as soon as I reach this step I will mark Ans - E)

So, heaviest car would be 1400 + 66.667 * 9 = 2000
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by bhumika.k.shah » Sat Jan 30, 2010 7:51 pm
Hey,
thanks for making it so simple :D
sanju09 wrote:
bhumika.k.shah wrote:The combined weight of 10 cars is 14000 pounds. No car weighs more than 50% more than any other car. What is the maximum possible weight for the heaviest car?

(A) 1,600
(B) 1,700
(C) 1,800
(D) 1,900
(E) 2,000

HOWWWWWWWWWW TO SOLVEEEEE!!! :(
For having the heaviest possible single car, let the remaining 9 cars weigh x pounds each, and the single heaviest car weighs h pounds, such that:

9 x + h = 14000 and h ≤ 1.5 x. For maximum h, we must take h = 1.5 x

Now, 9 x + 1.5 x = 14000

Or x = 14000/10.5 and h = [spoiler]2,000 [/spoiler]pounds.

[spoiler]E[/spoiler]
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