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 first set up the equations for the given data in the problem and then solve them.
1) If `x` and `y` are the number of phones and tabs then
2) If `c_1` is the cost of a phone and `c_2` the cost of a tab then
3) After a part of the phones and another part of the tabs were sold there were 3 times as many phones as tabs:
`(x-1/3*x) =3*(y-10/11*y) `
4) Each phone cost 714 more than a tab
`c_1 =c_2 +714`
Now let us solve the above equations. From 3) we get
`2/3*x =3*1/11*y rArr x =9/22*y`
Which replaced is 1) gives
`9/22*y +y =310 rArr 31/22*y =310 rArr y=220 and x =90`
Now we work on the price. From 2) we have
`90*c_1 =220*c_2 +2484`
This combined with 4) gives
`90*(c_2+714) =220*c_2 +2484`
`90*c_2 +64260 =220*c_2 +2484`
` ` `130*c_2 =61776`
Check: one phone price is
`c_1 =475.2+714 =1189.2`
The cost of all phones is 2484 more than the cost of all tabs
`1189.2*90 -475.2*220 =2484` check OK!
Answer: A tab cost 475.2 USD