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

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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, 12:53
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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?

No. of items in his shop = 2000
after 1 yr total items = 2000 + 200 = 2200
after 2 yrs total items = 2200 + 220 = 2420

He decides to open more 10 shops, so he will have a total of 11 shops.

If he distributes an equal number of items then
2420 / 11 = 220 items per shop.

We will have to a minimum number of items in each shop and max in one shop

We can eliminate option A 200, B 216, C 220 immediately because minimum items are 220.

So we have 2 options 230 and 260.

If we consider,
216 items in each shop: 216 * 10 = 2160
2420 - 2160 = 260

But 210 * 120 % = 259.2
and the shop cannot have more than 20 % so 259 items can be there but not 260.

any number less than 216 will have a lesser value of 120% than 260.
So from remaining options of 230 and 260
We are asked for approx number of items. and as 230 and 260 has a difference of 30 number of items,
we can conclude the answer to be 260.

Answer: E

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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, 13:10
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Total number of items Sam has in the shop= 2000*1.1*1.1=2420

Since we want to maximize the number of products in one shop and total items of no shop are greater than 20% of the total items of any other shop, we need to divide equal number of items in 10 shops and 11th shop has 20% more items than any other shop.

10*x+1.2x=2420
1.2x= (2420*1.2)/11.2
1.2x= 260(approximately)

the approximate maximum possible number of items in a shop= 260

IMO E
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New post 17 Jul 2019, 14:33
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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



First of all, we need to figure out the total number of items after 2 years.
1,1 *1,1 = 1,21.
2000*1,21 = 2420 items

lets assume, that there are 10 shops with identical number of items - X and 1 shop with +20% - 1,2X

so we have 10X+1,2X = 2420
11,2X=2420
X is approximately 216
1,2 X is approximately 260

The answer is E
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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, 18:42
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total items in shop after 2 years; 2000 * 1.1 ; 2200 and 2200*1.1; 2420
now using options we can solve
for max items if avail in shop are 260 ; then for rest 9 shops we have 2420-260 ; 2160 items which when divided amongst 9 shops would yield an integer value ; where as for other options we get a fraction
IMO E ; 260

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, 21:41
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Given:
Present items = 2000
Rate of increase = 10% = *1.1 every year
Number of items after 2 years = 2000 * 1.1 * 1.1 = 2420

Now we need to distribute 2420 items among 10 shops such that total items of no shop are greater than 20% of the total items of any other shop.
We can do this by placing “x” quantities of items in 9 shops and “1.2 x” quantity of the items in the other shop.
Therefore, 9x + 1.2x = 2420 => 10.2x = 2420 => x= 237.25 -> (a)

Now (a) is certainly not the maximum value that can be placed. We can have one shop with lower than x quantity and its difference from the quantity of maximum can still be less than 20%.

However, from (a) we know that the maximum possible number is at least 237.
Thus, from (a) we can eliminate options A, B, C, and D.

Hence, from (a) and the options available we see that the maximum possible number of items with the given constrictions is 260.

Answer E
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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, 22:26
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The number of items in the shop: \(2000\)

If inceased by 10% for 2 years: \(2420\)

The number of shops after 2 years: \(11\)

Average item in every shop: \(\frac{2420}{11}=220\)

Max item is \(20\)% greater than Min item: \(Min*1.2 = Max\)

Note that the difference between Min and Max is 20% of Min. That means that we cannot increase Max however much we want without increasing Min too, becuase the difference may surpass 20%. For this reason we will increase both Max and Min. In order to increase Max we will increase the items of only one shop so that the effect to the average is Max. However, in order to increase the Min we will take equal number of items from the each of the other 10 shops so that the effect to the average is Min.
Note that the average 220 is already higher than the or equal to A, B, and C. Thus they are out.

\(x\) - is the equal number of items we take from each of the 10 shops to give to the 11th shop.

So \((220-x)*1.2=220+10x\) and \(x\approx{4}\)

So Max is \(200+10x\approx{260}\)


Hence E
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New post 17 Jul 2019, 22:26
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Sam by the end of first year will sell 2200 items and by the end of second year 2420 items. He will also have 11 shops in total. \(\frac{2420}{11}\)=220 items on average in each shop. Now, let's maximize one shop's items and minimize others 10. If we subtract from each 10 shops 4*10 and add 40 to the eleventh, we get 216 items in 10 shops and 260 items in 11th shop. Please note that we cannot take all of that 40 from one shop and add to another because then difference will be over 20%. Thus we minimize all 10 shops so that difference becomes negligible and add all to one shop. So, we get

216=100%
x=120%
Or x=259.2, if rounded 260 (E)
Note: question asks for approximate maximum number.
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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, 23:21
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Total items at the start: 2000

each year items will increase by 10%.
Total items after 1 year : 2200
Total items after 2 years : 2420 (overall 21% increase in 2 years).

Total shops after opening 10 new ones : 11

Now, average items per shop = 2420/11 = 220

Condition : difference in total items between any 2 shops can not be more than 20% of any of the shop.

To get maximum items possible in 1 shop, we need to distribute items equally in 10 shops and more in 11th shop. Since average is 220, we can say that maximum possible items in 1 shop will be more than 220. ............A, B, C are eliminated.

Between, D and E:

When 1 shop has 230 items then other shops will have 219 items each. Here, 230 is just 5% more than 219. It doesn't utilize the allowed difference of 20% completely...............................D is eliminated.

So, E must be the answer. Let's check:
10 shops will have 216 items and 1 shop will have 260 items. 260 is approximately 20% more than 216.



ANSWER: E
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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, 00:05
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In two years, Sam will have 2420 items in his shops and 11 shops overall. In average, he will sell 220 items per store. We can eliminate A, B, C. D is too small to be 20% more. E is left, which is our answer
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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, 01:42
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If Sam will own 11 shops and overall 2420 items after 2 years, then what will be the greatest possible number of the items if no shop has more than 20% difference in the number of items?

From the stem we know that 220 is the average number of items. For this reason the greatest number can’t be 200, 216, or 220. Therefore, A, B, and C are incorrect.

If it is 230 as in D, then the difference between 219 and 230 is about 5%. As you see for achieving the greatest number we need raise both the highest and the smallest numbers. Thus D is out.

Basing on the same logic, if it’s 260, then the difference between 216 and 260 is about 20%. Therefore, E is our answer.

IMO E
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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, 02:23
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After 1st year, Number of items = 2000 + 200 = 2200
After 2nd year, Number of items = 2200 + 220 = 2420

Let the items of each shop = x
Number of maximum items available in 1 shop = x + 0.2x = 1.2x

Total items in 11 shops = 10*x + 1.2x = 11.2x
Given, 11.2x = 2420
--> x = 2420/11.2 = 24200/112 = 216.07

Maximum items possible in a shop = 1.2x = 1.2*216.07 = 260 approximately

IMO Option E

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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, 03:05
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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?


Current items in Sam's shop = 2000
1 year after = 2200 (2000+10%)
2 year after = 2420 (2200+10%)

So now Sam has got 2420 items in 1 shop, he wants to open 10 more, 11 in total.
And divide 2420 items among 11 shop, so that difference between items in the shop were not higher than 20 %.

Let's imagine belos situation:

Attachment:
shops.PNG
shops.PNG [ 18.74 KiB | Viewed 210 times ]


Now let's calculate difference:

230 - 219 difference ~ 5%
240 - 218 difference ~ 9.2%
250 - 217 difference ~15%
260 - 216 difference ~ 20% Bingo! :inlove:

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

E. 260

E is the answer. :heart
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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, 04:48
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at present number of items in his shop = 2000
each year item increase by 10%
the total number of items after 2 years = (1.1)*(1.1)*2000 = 2420

he decided to open 10 more shop
so now the total number of shops = 11

he distributed the total item in these shop in such a way -- total items of no shop are greater than 20% of the total items of any other shop

we have to find the maximum possible number of items in a shop
so we will minimize the items of all the other shops

let x be the minimum item distributed to shop

then the total number of items of 10 shops = 10x (since we have to find the maximum item in a shop we will distribute minimum possible item in all the other shop distributing x in each shop)

maximum possible item in the 11th shop will be 20% greater = 1.2x

adding all the items of 11 shops = 10x+1.2x = 2420
11.2x = 2420
x = 216.07

maximum possible item in one shop = 1.2x = 1.2*216.07 =259.28 approxx = 260

E is the answer
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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, 05:20
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No. of items is 2000. (Increasing by 10% every year)
In 1 year: 2000+10%*2000=2200.
In 2 years: 2200+10%2200=2420.

After two years, in total, there will be 2420 items and 11 shops.

The items should be distributed in such way :
The total items of one shop are not greater than 20% of those of any other shop.

In order to find the maximum possible items in a shop,
Total Items of each 10 shops should be the same ('a') and the 11th shop should be '1.2a'.

a+a+...+1.2a=2420
10a+1.2a=2420
11.2a=2420
a=216.07...

(maximum number of items):1.2a=1.2*216.07=259.28≈260

The answer choice is E.
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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, 05:38
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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

Solution:

Sam is increasing the number of accessories per year by 10 % i.e in year 1, it will be 2200, & in the second year, it will be 2420.

So we have the total number of items that is 2420 which shall be divided across a total of 11 stores.

If we divide the number of items by total number of shops, we get \(\frac{2420}{11}\) = 220 items per shop, To find the value of maximum possible value in 1 shop, we have to take the minimum possible values in all the Other shops, Therefore, 200, 216 & 220 are out of the question we can eliminate them all, if we take the maximum value as 260, we get 2420-\(\frac{260}{10}\)= 216 in every other shops, which is around 83 % the value of 260. that is it's less than 20% of the total value. Hence the maximum value can be 260.

Answer is E.
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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, 05:52
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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

Solution:
Sam has shop in which he has 2000 total items in whcih he is incerasing by 10% ever year. So it will be 2000+10% 2000=2200 in first year and 2200+ 10% 2200=2420 in the second year.
Nopw after the sencond year he opens 10 more shops so total number of shops will be 11.
Hence we get 2420/11= 220 items in each shop.
Now to maximise the value in 1 shop we minimise the values in others.
Now as because already the minimum value is 220 so we cannot take into consideration 200,216 and 220.Now if we maximise the number of items to 260 in one shop then the differential of 260-220=40 can be evenly reduced among rest 10 shops.that we reduce 40/10=4 from each of the rest 10.So maximum can be 260 in one and 216 in rest of 10.

Now to see whether its not greater than 20%, we can check as [(260-216)/216]*100=20.3%.we can ignoe .3% as item cant be in .3 value
Hnece E IMO
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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, 06:31
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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, 06:45
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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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Sam owns a football accessory shop. At present, the number of items in  [#permalink]

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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 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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