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Re: Sam owns a football accessory shop. At present, the number of items in
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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 (260220 = 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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Re: Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 18:42
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 2420260 ; 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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Sam owns a football accessory shop. At present, the number of items in
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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
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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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Re: Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 21:41
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
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17 Jul 2019, 22:25
No. of items by the end of 2 years=2000*1.1*1.1=2420. If items in one of the shops need to be maximized, then rest of the shops must have equal number of items(lease possible). Now, lets check options in descending order of their value. E) If max. items in a shop are 260, remaining items is 2420260=2160. Rest of the 10 shops get 216 items(if divided equally). 216*1.2<260. Not possible D)If max. items in a shop are 230, remaining items is 2420230=2190. Rest of the 10 shops get 219 items(if divided equally). 219*1.2>230. Possible Since remaining choices are <230, ans is D.
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Re: Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 22:26
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 \((220x)*1.2=220+10x\) and \(x\approx{4}\) So Max is \(200+10x\approx{260}\) Hence E
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Re: Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 22:26
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
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17 Jul 2019, 23:21
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
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18 Jul 2019, 00:05
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
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18 Jul 2019, 00:57
Sam owns a football accessory shop. At present, the number of items =2000
increase it by 10% every year. Year 1 = 2000+200=2200 year 2 = 2200+220 = 2420
After 2 years, Total number of Shops = 11 Total Items to distribute = 2420
Total items of no shop are greater than 20% of the total items of any other shop.
We would have to take the max number of items to be x and make all the other shops to be minimum to get a maximum number of items x
10x + 0.8x=2420 10.8x=2420 x=24200/108 x=224.07 items in each shop Approx 220 Answer choice C



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Re: Sam owns a football accessory shop. At present, the number of items in
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18 Jul 2019, 01:01
At present items = 2000 After two years the items = 2000*1.1*1.1 = 2420 When he opens 10 more shops, then total shops = 11 As per question: 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, if one shop has the maximum possible number of items, then it should not be greater than other 10 shop items. Only 216 satisfies the condition.
IMO the answer is B.
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Re: Sam owns a football accessory shop. At present, the number of items in
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18 Jul 2019, 01:42
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
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18 Jul 2019, 02:23
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
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18 Jul 2019, 03:05
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 [ 18.74 KiB  Viewed 191 times ]
Now let's calculate difference: 230  219 difference ~ 5% 240  218 difference ~ 9.2% 250  217 difference ~15% 260  216 difference ~ 20% Bingo! A. 200 B. 216 C. 220 D. 230E. 260 E is the answer.
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Re: Sam owns a football accessory shop. At present, the number of items in
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18 Jul 2019, 03:32
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 At the end of second year, total items he has = 2000*1.1*1.1 = 2420 2420 items distributed across 11 shops. Average = 2420/11 = 220, so the max value should be > 220. Eliminate A, B and C. If one shop has 260, then among the remaining 10 shops the average would be (2420260)/10 = 216 260 is greater than 20% greater than 216. Hence eliminate E Hence answer is 230. Option D.
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Re: Sam owns a football accessory shop. At present, the number of items in
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18 Jul 2019, 04:48
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
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18 Jul 2019, 05:20
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
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18 Jul 2019, 05:38
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
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18 Jul 2019, 05:52
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 260220=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 [(260216)/216]*100=20.3%.we can ignoe .3% as item cant be in .3 value Hnece E IMO




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