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There were 310 phones and tabs for sale in a shop.All the phones in the shop cost $2484 more than all the tabs. After 1/3 of the phones and 10/11 of the tabs were sold, there were thrice as many phones as tabs left.If each phone cost $714 more than each tab,how much did each tab cost?

Let P be the number of phones and T be the number of tabs.

Let x be the price of a phone and y be the price of a tab.

There were 310 phones and tabs for sale in a shop, so P + T = 310

and P and T are both integers.

All the phones in the shop cost $2484 more than all the tabs, so Px - Ty = 2484.

After 1/3 of the phones and 10/11 of the tabs were sold, there were thrice as many phones as tabs left. With the amount sold, that leave 2P/3 and T/11,

so this says that 2P/3 = 3(T/11).

If each phone cost $714 more than each tab sounds a little fishy,

but what it says is that x = y + 714. The reason it sounds fishy is that the price of most phones is far under $714.

The equations are, then,

P + T = 310,

Px - Ty = 2484,

2P/3 - 3T/11 = 0, and

x - y = 714.

Using the 1st equation, it can be rewritten as P = 310 - T.

Putting this into the 3rd equation gives 2(310-T)/3 - 3T/11 = 0.

Multiplying this out gives 620/3 – 2T/3 - 3T/11 = 0.

This is the same as 2T/3 + 3T/11 = 620/3.

Multiplying by 33 gives 22T + 9T = 6820.

That is the same as 31T = 6820.

That gives T=220.

Using the 1st equation again, this says that P = 90.

Putting these values into the 2nd equation gives 90x – 220y = 2484.

The 4th equation can be converted to x = 714 – y.

Putting this back in the prior equation gives 90(714-y) – 220y = 2484.

That multiplies out to 64260 – 90y – 220y = 2484.

This says that 61776 = 310y, so y = $2059.20, and that's what the question is asking for.

If we look a little farther, however, putting that into the 4th equation gives

x – 2059.2 = 714, so x = $2773.20. That seems a little expensive for a phone,

but that’s what I get from what’s given.

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