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Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 07:59
Question Stats:
45% (01:59) correct 55% (02:35) wrong based on 213 sessions
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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
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Sam owns a football accessory shop. At present, the number of items in
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Updated on: 17 Jul 2019, 11:14
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 At present = 2000 items 10 % increase every year so, after year 1 = 2200 items after year 2 = 24200 items Now he opens 10 more shops i.e. 11 shops. therefore each shop will have \(\frac{24200}{11}\) = 220 Now as we are asked maximum possible items in the shop with no shop having more than 20 % items greater than any other shop, possible answer has to be greater than avg (220). Only option D and E possible. Lets try option E, as max 260 which is 40 greater than average SO for if there are 11 shops with max as 260, then rest have to be 216. And 260 is 20% more than 216 so thats out. Hence go with option D. 230 is 10 more than avg. So apart from 230 rest all will be 219 and they are within 20 % of each other. Hence option 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, 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
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17 Jul 2019, 08:23
IMO answe is E:
total number of items after 2 years = 2420 so, at this point i looked into the answer choices and checked if the total items can exceed 2420, given if 260 is the max, then min can be 26020*260/100 = 208. if 208 in all 9 shops then 208*9+260<2420, so 260 is possible max



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17 Jul 2019, 08:34
Answer is D
after year 2 the total # of items = 2000*(110%)*(110%)=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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17 Jul 2019, 08:38
Total no of items after 2 years = 2000X(1.1)^2 = 2420
open 10 more shops. Total shops = 11
Condition >>no shop can have more than 20% items than any other shop
To find max items in one shop.
To max items in one shop. let other 10 shops has equal number of items.
let 100x be the items in 10 store and 120x items in the store with max items. 10(100x) + 120x = 2420 ==> 120x =~ 260
Hence, 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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17 Jul 2019, 08:38
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?
This is an easy question. A total number of items in the shop after 2 years is 2420. Now just in case, he has to divide equally then each shop will have 2420/11 = 220
But we are looking for the maximum increase he can do so that none of the shops have greater than 20% By now you should eliminate options A, B, C.
Take 230, which means that he is going 10 extra on 1 shop 10 pieces will be subtracted from each of the remaining stores.
If you take 260, which means that he is going for 40 extra. which means that 40/10 = 4 will be deducted equally for each of the remaining stores. So the maximum possible answer is 260.
A. 200 B. 216 C. 220 D. 230 E. 260
Hence the 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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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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Re: Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 08:42
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 the first year  2000*1.1=2200 After the second year  2200*1.1=2420
Now, he opens 10 more shops and the total number of shops is 11. Let's make an equation 10x+1,2x=2420 11,2x=2420 x=216 but we need 1.2x =259.2
IMO E



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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, 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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Re: Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 08:44
Total items at end of two years = 2000*1.1*1.1 = 2420
Total shops now= 10+1 = 11
Let's say we distribute all the first 10 shops with x and last shop with 1.2x. this is the way to maximise the 1.2x because this keeps all the other shops at low "X" number
Then 11.2x = 2420
X= 216
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17 Jul 2019, 08:49
E.
Currently owner has total 2,000 items, and she will increase it by 10% for two consecutive years, so total no of items after two years = 2000*(1.1)^2 = 2420.
Now, owner opens 10 more stores, so total no of stores are now 11. If owner were to distribute items equally, each store will have 2420/11 = 220 items each.
Now, suppose maximum items one store can have is 260, which is 40 more than average 220 items. So, lets say that these 40 items in one store come from remaining 10 stores, so these 10 store will now have 220  40/10 = 216 items each.
Compare the difference we get: (260216)/216 ~0.2037 or 20.37%



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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, 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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Re: Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 08:54
After 2 years total items  2000*1.1*1.1=2420 For having maximum number of items in one shop the other 10 shops must have the same value.
With the above logic substitute answer choices. Option C  220 (maximum) means others have 220/1.2=183.33  total is short. Option E  260 (maximum) means others have 260/1.2=216  2160+260=2420  Matches to Total Hence Option E  260
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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, 09:02
After two years, the total amount of items will be 2420 & if 10 new shops were to receive the equal amount of items then each one will have around 242 pieces & maximum variance of one shop can have 20% more items than any other than this shop is supposed to receive items above the equal amount, only one option is there that is 260
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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, 09:05
IMO : E
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
Sol:
2000*110/100*110/110=2420
That means 2420 is the number of items after two years.
242 per shop,
and even if one shop has 260 then others will have 240 and then it would be 8 percent only.
So E is correct.



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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, 01:29
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 Given: 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. Asked: What is the approximate maximum possible number of items in a shop? At present, the number of items in his shop = 2000 he plans to increase it by 10% every year. Next year, the number of items in his shop will be = 2000 * 1.1 = 2200 Second year, the number of items in his shop will be = 2200 * 1.1 = 2420 After 2 years, the number of shops he will have = 1 + 10 = 11 shops He 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. If all other shops have number of items = x => Total number of items in 10 shops = 10x Then the maximum number of items in 11th shop < 1.2x Total no of items in 11 shops approximately = 10x + 1.2x = 11.2x But total no of items in 11 shops = 2420 11.2x = 2420 x = 2420/11.2 1.2x = 2420/11.2*1.2 = 260 approx Maximum no of items a shop may have = 260 since no of items is a positive integer. IMO E
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Originally posted by Kinshook on 17 Jul 2019, 09:08.
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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, 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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17 Jul 2019, 09:31
Total items will be 2420
To if we have 260 in one
Then on average each can't have more than 240
20% of 240 is 48 and difference is 20
E is the right answer
A short cut is 2420/10 = 242 value of.maximum can't be less than it .
E it is
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Sam owns a football accessory shop. At present, the number of items in
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17 Jul 2019, 09:31
 let's calculate the number of items after two years: \(2000*1.1^2=2420\)  After two years sam will have 11 shops The average items in a shop will be 2420/11=220 items (instantly dismiss A, B and C as they are equal or less than the average and 20% difference will give us some room to go above the average) Let's test the values and begin with the maximum number 260 260 is 40 more than the average (220) We don't want to drop the lowest number, so we will distribute those 40 items burden evenly on the remaining 10 shops (40/10=4) so each shop can have 2204=216 items (216*10+260=2420 we are not missing anything) Now we have got one shop with 260 items and the rest ten shops with 216 items. Do we maintain 20% restriction? ({260216}/216)*100~20.3% As we could approximate, lets round to the nearest integer to get 20%, Fits IMO ANS: E




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