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

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.

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

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

- 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 220-4=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?
({260-216}/216)*100~20.3%

As we could approximate, lets round to the nearest integer to get 20%, Fits

IMO
ANS: E

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

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:



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. 230
E. 260

E is the answer.

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.

Confusing stem:

"in such a way that total items of no shop are greater than 20% of the total items of any other shop"

Taken by the wording, it is impossible. It should be something like "are more than 20 % greater than".