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.