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

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New post 17 Jul 2019, 08:23
D has to be the right answer.


After 2 years total items will be 2420

It can be divided into 220x11

But if we make 260 for any one then we can reduce 4 by other shops. so new combination could be 260, 216x10
difference between 260 and 216 is greater than 20% of 216.
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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 17 Jul 2019, 08:34
Answer is D

after year 2 the total # of items = 2000*(1-10%)*(1-10%)=2420

2420/10=242 shops if we divide equally. 242/2420=10%. this is the minimum he can distribute the items and its no greater than 10%.

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
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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 17 Jul 2019, 08:41
After 2 years total items in shop = 2420

Equation b/w two shops => x+ 1.2x=484 (for 2 shops)
x=220

Hence C
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New post Updated on: 18 Jul 2019, 04:46
2000*1.1*1.1 = 2420 /11 = 220

(C) it is the answer.

Originally posted by Mizar18 on 17 Jul 2019, 08:44.
Last edited by Mizar18 on 18 Jul 2019, 04:46, edited 1 time in total.
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New post 17 Jul 2019, 08:50
the total number of items have to be in such a way that total items of no shop are greater than 20% of the total items of any other shop, then
percentage difference between the shop with least number of items and the shop with max number of item cannot be greater than 20%
so if 2200 items have to be divided among 11 shops
if divided equally 200 each,
so the least +max sum should be 400
so lets take option a)
200 (this is not the maximum)
Lets take B
if max is 216 least would be 184
percentage difference less than 20%. hence could be the answer
Lets take C
if max is 220 least would be 180
Percentage difference greater than 20. hence not possible
Same goes for D and E
Hence the answer is B
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New post Updated on: 18 Jul 2019, 05:39
The total number of items after 2 years = \(2000*(\frac{11}{10})^2 = 2420\)
If the shop with the smallest number of items has a items, then the one with the greatest number of items is \(\frac{6a}{5}\)
To maximize the number of items in the largest shop, we will assume that all the rest shops have each \(a\) items.
so \(10a + \frac{6a}{5} = 2420\)
then \(a\approx{216}\) which is choice B
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Originally posted by Mahmoudfawzy83 on 17 Jul 2019, 09:27.
Last edited by Mahmoudfawzy83 on 18 Jul 2019, 05:39, edited 2 times in total.
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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, 09:09
Quote:
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


yr0: 2000
yr1: 2000*11/10
yr2: 2000*121/100=2420 items and 11 stores

condition that no store has more than 20% items of any other store;
so lets make 10 stores with the same items and 1 store with 20% more;
10x+1.2x=2420
56x/5=2420
x=2420*5/56
now the store with no more than 20% is 1.2x
1.2x=2420*5/56,… 1.2x=(aprox.) 260

Answer (E).

Originally posted by exc4libur on 17 Jul 2019, 09:37.
Last edited by exc4libur on 18 Jul 2019, 09:09, edited 1 time in total.
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New post 17 Jul 2019, 09:46
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.
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New post 17 Jul 2019, 10:04
so total books considering 20% each year increase in 2 years will be 2000+ 200 + 220=2420.Total no of shops=11
so if consider all equal its 2420/11 i.e. 220. Now a shop can have not more than 20% from any other.When we consider 260 it goes more than 20% as 20% of 220 is 44, if 1 hs 260 other will be less than 220 by more than 44.

Therefore D is correct.
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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 17 Jul 2019, 10:07
Total items = 2000

Increase of 10% per year

After 1st year 1.10 x 2000 = 2200
After 2nd year 1.10 x 2200 = 2420

Total shops after 2 years = 1 + 10 = 11
Average item per shop = 2420/11 = 220

2420 - 220 = 2200

20% of 2200 = 440

220 < 440

Thus each shop will have a maximum item of 220

Hence answer choice C.
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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 17 Jul 2019, 11:15
Given,

1. Total number of items in Sam's shop = 2000
2. Percentage increase in number of items in Sam's shop every year = 10%
=> Total number of items in Sam's shop after 2 years = 2420
3. Total number of shops after 2 years = 11
4. Number of items of no shop is greater than 20% of the total number of items of any other shop.

To find,

Approximate maximum possible number of items in a shop.

Now if we see then we find that 2420 can be equally divided into 11 shops by giving 220 items to each shop.
=> At 220 items in each shop, all conditions are met. But note that this is not the maximum.
=> Hence maximum will be any number greater than 220.

Let us check the options -

A. 200 --> As the maximum needs to be greater than 220, this is wrong.

B. 216 --> As the maximum needs to be greater than 220, this is wrong.

C. 220 --> As the maximum needs to be greater than 220, this is wrong.

D. 230 --> This is a possibility, let us check this further.

Now 230 is 10 more than the average number of items per shop which is 220. Now, if we balance the increase of 10 by reducing the same number of items in another shop and making it to be 210 and keeping every other shop constant at 220, all our conditions are met. Hence this is a possible answer but note that it is still not a maximum as for 210, the maximum can go up to 252.

E. 260 --> This is a possibility, let us check this further.

Now 260 is 40 more than the average number of items per shop which is 220. Now, if we balance the increase of 40 by reducing the same number of items in anoher shop and making it to be 180 and keeping every other shop constant at 220, then our main condition of 260 not being more by 20% of other shops fails. Hence this is not the correct answer.

If we try to find the maximum then we can find it by using the following equation -->

220 + x = 220 - x + 0.2 * (220 - x)
=> 2.2 *x = 44
=> x = 20

Hence the maximum is at 240.

Given our options we have,

Answer: D
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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, 03:56
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?

Given:
# number of items after 2 years = 2000 * 1.1 * 1.1 = 2000 * 1.21 = 2420
# number of shops after 2 years = 10 + 1 =11
total items of no shop are greater than 20% of the total items of any other shop means Nmax <= 1.2*Nmin, where Nmax = maximum number of items in a shop & Nmin = minimum number of items in a shop
Need to find out the maximum number of items in a shop.


A. 200 --> 230> 200
B. 216 --> 230> 216
C. 220 --> 230> 220
D. 230 --> correct: let say Nmax = 230, other 10 shops has Nmin number of items each, so Nmin = (2420 - 230)/10 = 219, & 230 < 1.2 * 219 = 262.8
E. 260 --> wrong: let say Nmax = 260, other 10 shops has Nmin number of items each, so Nmin = (2420 - 260)/10 = 216, but 260 > 1.2 * 216 = 259.2

Originally posted by hiranmay on 17 Jul 2019, 11:29.
Last edited by hiranmay on 18 Jul 2019, 03:56, edited 1 time in total.
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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 17 Jul 2019, 12:07
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?

After two years, total number of items = 2420
Total shops = 11
maximum possible number of items in a shop = 2420 / 11 = 220

So, the correct answer choice is (C)
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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 17 Jul 2019, 12:08
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?

Total no. of items in shop = 2000.
Total no. of items in shop after 1 year = 1.1*2000 = 2200 (10%increase)
Total no. of items in shop after 2nd year = 1.1*2200 = 2420 (10%increase)
Now Sam opens 10 more shops, so total shops = 11.

Now Sam distributes the items(2420) in 11 shops in such a manner that total items of no shop are greater than 20% of the total items of any other shop.
Lets say a shop has 200 items. 20% of 200 = 40. Therefore no other shop should have items more than 240.

If items are equally distributed among 11 shops then each shop will have 220 items.

Consider one shop has 230 and one has 210. Rest other being 220. In this case the difference between max. items and min. items is 230 - 210 = 20. If we take 20% of 210 = 42. This scenario is possible as the the shop having max. items is less than 20% of the items with the shop having min. items. Therefore, a shop can have 230 items.

Consider one shop has 260 and one has 180. Rest others being 220. In this case the difference between max. items and min. items is 260-180=80. If we take 20%of 180 = 36. This scenario is not possible as the shop with max. items is more than 20% of the items with shop with min. items. Therefore a shop cannot have 260 items.

Hence from the answer choices, the approximate maximum possible number of items in a shop = 230.


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

Answer Choice: D
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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, 06:09
Total items in shop currently =2000
Yearly projection for increment =10%
Hence after two years new stock level = 1.1x1.1x2000 =2420 items

Ten additional shops were opened after two years, implying total shops now is 11.
A requirement of sharing the total items into the eleven shops is that no shop must contain more than 20% more items than any other shop.
With this constraint, the maximum possible items in a shop can only occur if ten shops have the same number of items while one has 20% more items.
If x is the minimum items in the ten shops, then 10x+1.2x=2420
11.2x=2420
x=216
Maximum possible =1.2x216=259.2
This is approximately 260.

Hence E.

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Originally posted by eakabuah on 17 Jul 2019, 14:15.
Last edited by eakabuah on 18 Jul 2019, 06:09, edited 1 time in total.
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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 17 Jul 2019, 14:53
Answer D
Stock at end of 2 years: 2420
Now distribution of stock in 11 shops, 1 current shop and 10 new:
2420/11 : 220 if question ask equal distribution, But question is asking maximum no of items
So test 230 and 260 :
Consider 230 given to 8 shops : 230* 8 = 1840
Left : 580 /3 : 193,
193+20% = 231.5, So clearly other shops has less than 20% of shops with 193 items

260 : will not give such solution,
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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 17 Jul 2019, 15:23
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

2000 items in the shop at 10% increase per year, there will be 2200 after 1 year and 2420 after 2 years.
Now there are 10 more shops, so total of 11 shops to split the 2420 items.

The split is such that no shop carries more than 20% of 2420. So, the maximum possible number of items in any shop would be 20% of 2420.
Since there are 11 shops, each shop should carry a maximum of 220 items in order to sell all 2420 items.

Correct answer choice C
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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 17 Jul 2019, 18:17
1st year: 2000 * 1.1 = 200*11
2nd year: 200*11*1.1 = 20*11*11

Dividing by number of total shop = 20*11 = 220 (equally distributed)
Now divide take the last answer choice E (260-220 = 40) and divide it among 10 other shops to push all other item to one remaining shop.
Each has to supply 40/10 = 4 to one shop. 220 - 4 = 216.
216*1.2 < 260. Therefore answer is D.
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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, 07:44
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?

Given : Sam Owns 1 shop with 2000 items
After first year = 1.1*2000 = 2200
After second year = 1.1*2200 = 2420

After 2 years he opens 10 more shops , Total no. of shops now = 10+1(current shop) = 11

Maximum possible number of items in shop =1.1x
== >10*x + 1.2*x =2420
=> 11.2x =2420
=> x = 216

Max number of items in 1 shop = 1.2*216 = 260 approximately = Option E is the Correct Answer

Originally posted by baru on 17 Jul 2019, 19:18.
Last edited by baru on 18 Jul 2019, 07:44, edited 1 time in total.
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Sam owns a football accessory shop. At present, the number of items in  [#permalink]

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New post 17 Jul 2019, 20:50
No of items = 2000
no of shop = 1
After 10% increase in year 1 = 2200
After 10% increase in year 2 = 2420
No of shops after opening new ones = 1 + 10 = 11

Average no of items that can be divided evenly = \(\frac{2420}{11}\) = 220
So the maximum no will be >= 220
We have margin of 20% higher. there two numbers in answer choice greater than 220. Let's try bigger number 260
Max no for one shop = 260
Rest all shops to divide the items = 2420 - 260 = \(\frac{2160}{10}\) = 216
Is 260 within 20% of 216 , apparently it is close enough but higher than 20% -> 20% of 216 = 43.2 .
216+43.2 = 259.2 < 260 , 260 is higher than 20% of 216..
As we know 260 was close enough, we pick other choice Max = 230 is the answer

D is the answer
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