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...

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?

Asked on by varshlok

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

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

`x+y =310`

2) If `c_1` is the cost of a phone and `c_2` the cost of a tab then

`c_1*x =c_2*y+2484`

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`

`c_2 =475.2`


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

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