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Sam owns a football accessory shop. At present, the number of items in

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Sam owns a football accessory shop. At present, the number of items in  [#permalink]

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New post Updated on: 18 Jul 2019, 06:34
Sam owns a football accessory shop. At present, the number of items in his shop is 2000 and every year, he plans to increase it by 10% every year. After 2 years, he decides to open 10 more shops and distributes the total number of items he has in such a way that total items of no shop are greater than 20% of the total items of any other shop. What is the approximate maximum possible number of items in a shop?

Intital items : 2000
After 1st year : 2200 (10%+)
After 2nd year : 2420 (10%+)

10 more shops, so total shops will be 11.

Even if we evenly distribute, items per shop will be 2420/11 = 220

But with the condition given, max items in a shop will be more than 220.

A. 200
B. 216
C. 220
D. 230
E. 260

So, option D & E are our contenders.

distribute 2420 among 11 shops in such a way that,
total items of no shop are greater than 20% of the total items of any other shop

I am confused with this condition. Unable to solve further :|

Originally posted by sj24 on 18 Jul 2019, 06:29.
Last edited by sj24 on 18 Jul 2019, 06:34, edited 1 time in total.
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New post 18 Jul 2019, 06:31
1
Sam owns a football accessory shop. At present, the number of items in his shop is 2000 and every year, he plans to increase it by 10% every year. After 2 years, he decides to open 10 more shops and distributes the total number of items he has in such a way that total items of no shop are greater than 20% of the total items of any other shop. What is the approximate maximum possible number of items in a shop?

1. NumberOfShops = 1 + 10 = 11

2. Avearge # of items per shop = 2000*1,1*1,1 / 11 = 220

3. "distributes the total number of items he has in such a way that total items of no shop are greater than 20% of the total items of any other shop"
That means: TotalItemsOfAnyShop * 1,2 > TotalItemsOfMaxPossibleItemsInOtherShop

4. "What is the approximate maximum possible number of items in a shop?"
If Average = 220, than let's try several variants:
219 in 10 shops and 230 in 1 shop: 219*1,2 = 262,8
218 in 10 shops and 240 in 1 shop: 218*1,2 = 261,6
217 in 10 shops and 250 in 1 shop: 217*1,2 = 260,4
216 in 10 shops and 260 in 1 shop: 216*1,2 = 259,2

So, the only answer that fits is E

A. 200
B. 216
C. 220
D. 230
E. 260
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New post 18 Jul 2019, 06:38
couldn't understand what is meant here:
"the total number of items he has in such a way that total items of no shop are greater than..." what does "no shop" mean?
Thanks!
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New post 18 Jul 2019, 06:45
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Sam sells 220 items in every shop (11 shops in total 200*1.1*1.1). Maximum number of items sold in one store should not be greater than 20% of any other stores. If one store sells maximum number possible, then rest should sell minimum and that difference should not exceed 20%. 216 items sell 10 shops and 1 shop sells 260, which is 20% more.
Answer - E
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New post 18 Jul 2019, 07:19
IMO C

the number of items in his shop is 2000 and every year, he plans to increase it by 10% every year.

therefore in 2 years he will have -> 2000x1.1x1.1 items = 2420


to distribute (max) items acc to the given condition, each store will have 2420/11= 220 items !!
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New post 18 Jul 2019, 07:22
Answer C

total item after 2 yr=2000x1.1x1.1=2420
item in each shop so that fulfill the condition=2420/11=220
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Re: Sam owns a football accessory shop. At present, the number of items in  [#permalink]

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New post 18 Jul 2019, 07:43
Presently, no of items = 2000
After 1 year, no of items = 110% of 2000 = 2200
After 2 years, no of items = 110% of 2200 = 2420

He opens 10 more shops. So, total no. of shops = 11
Avg. no. of items in a shop = 2420/11 = 220

Max. no. of items in any shop will be greater than the average (220). Hence options A, B and C are eliminated.

Let's consider option E: max. no of items in any shop is 260
In this case lets consider the highest possible no. of items in all other shops so as to arrive at the least difference between the maximum value and other values (we are trying to bring the value below 20% as per the question).
So, the other 10 shops should have (2420-260)/10 = 216 items

So, the % by which the largest shop (260 items) is greater than the smallest (216 items) is
\(\frac{260-216}{216}=20.37%\)
So, option E is eliminated.

Option D is the answer
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New post 18 Jul 2019, 10:09
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I don't understand how 260 is the answer.

If one is 260 then others have to be 216.

20% of 216 is 43.2 = 259.2

The question says no store should have items greater than 20% of any other store.

But 260 is greater than 259.2

How is 260 possible??

It has to be 230.

The question said APPROX MAXIMUM POSSIBLE.

BUT 260 IS NOT POSSIBLE.


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New post 18 Jul 2019, 20:51
ccheryn wrote:
Sam owns a football accessory shop. At present, the number of items in his shop is 2000 and every year, he plans to increase it by 10% every year. After 2 years, he decides to open 10 more shops and distributes the total number of items he has in such a way that total items of no shop are greater than 20% of the total items of any other shop. What is the approximate maximum possible number of items in a shop?

A. 200
B. 216
C. 220
D. 230
E. 260

no of items in the shop after 2 years = 2000 x 1.1 x 1.1 = 20 x 11 x 11

he adds 10 more shops so total is 11 shops so average items per shop = 20 x 11 x 11 / 11 = 20 x 11 = 220 items

so options A, B, are ruled out.

for c, yeah ans can be easily greater than 220 ie 221 and which is easily greater than 219 for one of the shops ie ( 221, 219 , 220 x 9)
so better leave that as of now without checking.

remaining we can check for 260 now for upper limit

if one shop is 260 then remaining 10 shops has to have 216 items to maintain the higher level. but 1.2 x 216 is 259.2 which is less than 260. ( but it cant be greater than 259.2 as per the question).

so e is ruled out. even 259 is possible. so between d and c, definitely d is greater. so D is the answer.




Bunuel ,

i vary in my opinion against this answer, for everybodys calculation, ans comes to 259.2 , how it can be approximated to 260 , ( against the condition it cannot be greater than 20%)

so next best answer( the greatest) has to be selected right?

i calculated but how can i approximate it to 260. hence i marked d,

Do i deserve kudos for this??( along with the option E guys might be, inspite of me believing their ans are wrong) you have to decide...

am i wrong???

moreover few people approximated the other way 260-216/216= for those ans whould be below 20 as the upper limit is 20% max given. so what do you say , am i wrong?
please enlighten me if i am wrong
thanks

PS: i posted the same in game of timers page also, its been removed. actually for conflict in kudos i have to put over there right
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Re: Sam owns a football accessory shop. At present, the number of items in  [#permalink]

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New post 19 Jul 2019, 02:44
Bunuel wrote:
Sam owns a football accessory shop. At present, the number of items in his shop is 2000 and every year, he plans to increase it by 10% every year. After 2 years, he decides to open 10 more shops and distributes the total number of items he has in such a way that total items of no shop are greater than 20% of the total items of any other shop. What is the approximate maximum possible number of items in a shop?

A. 200
B. 216
C. 220
D. 230
E. 260

 

This question was provided by e-GMAT
for the Game of Timers Competition

 



OFFICIAL EXPLANATION FROM eGMAT:



Given:
Sam owns a football accessory shop.
The number of items in his shop is 2000.
Every year, he is increasing the number of items in his shop 10%
After 2 years, he opens 10 more shops.
He distributes the total number of items in all the shops in such a way that total items of no shop are greater than 20% of the total items of any other shop.

To Find:
Approximate maximum possible number of items in a shop.

Approach & Working Out:
After 2 years, number of items in the shop of Sam= 2000 * 1.1 * 1.1 = 2420
Now, Sam distributes these items among 1 old shop and 10 new shops.
Let Minimum number of items in a shop is x.
Then, maximum number of items in a shop can be = 1.2 x

Let the number of shops having minimum number of items =a
a × x + (11 - a) × 1.2 x = 2420
Now, we want the maximum possible number of items in a shop, maximum possible number of shops should have minimum number of items.
This will ensure at least one shop has maximum number of items.
So, 11 - a = 1
Thus, a = 10
Hence, 10x + 1.2x = 2420
11.2x = 2420

x = 2420/11.2
Hence, maximum possible number of items = 1.2 × 2420/11.2 = 259.28 ≈ 260.

Therefore, the correct answer is option E.
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New post 19 Jul 2019, 02:54
Bunuel

thanks for official explanation from egmat.. but being the best in brain and analysis.. i want your answer.. i dont want their ans.

if you feel the answer is correct then its fine.. but as the last line states how can 259.2 can be approximated to 260 ( against the condition it cannot be greater than 20%). this is my question..

you are the final authority..

if you feel.. that it can be approximated.. against the condition of (should not be greater than 20%). then fine..

i want your answer

thanks . really thanks for posting official answer..

( an analogy i am trying if i have $10 , and if an apple costs $0.93, how many approximate no of apples i can buy). ( ans cannot be 11, ans can be only 10. even though i will have $.7 remaining out of which i can buy around 3/4th of apple). ) it cannot be the higher no. btw my analogy can be wrong.. but for the above question.. those are whole items.. cannot be bought in parts.. so it cannot be as easily approximated to the number on the higher side.). this is for normal situation..

in this question.. against this logic, there is one more condition it cannot be greater than 20%.Also ans is 259.3 not even higher than 259.5 . so it cannot be approximated to 260. this is my humble but strong opinion. At the end. here u r the authority. whatever you decide. Its fine for me.

thanks again
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New post 19 Jul 2019, 05:01
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Hi Bunuel,

I do agree with ccheryn on the solution.

In fact if we have the max no of items as 260, then the average no of items per shop in the remaining 10 would turn out to be (2420-260)/10 = 2160/10 = 216.

Now, 260 is 20.37% greater than 216. This violates the criteria provided in the question.

Had the answer been 259, that would have been acceptable but 260 does not seem right.
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Sam owns a football accessory shop. At present, the number of items in   [#permalink] 19 Jul 2019, 05:01

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