A fruiterer had the same number of apples, pears and oranges at first. After 98 oranges, some apples and pears were sold, there were 392 fruits...

For the stated word problem, we need to use variable to represent the unknown counts per each fruit. We may let:

o = original number of oranges

a= original number of apples

p = original number of pears.

 To set-up an equation, we translate the given conditions in the problem.

Condition 1: A fruiterer had the same number of apples, pears and oranges at first.

 This implies that we can equate them as `o =a=p` .

Condition 2:After 98 oranges, some apples and pears were sold, there were 392 fruits left. 

We may let:

unsold oranges = o'

unsold apples = a'

unsold pears = p'

 It indicates that the  sum of the remaining number of fruits = 392 such that: sold oranges =98

`o' +a' +p' = 392`

Condition 3: There were thrice as many apples as pears left

This means that `a' =3p'` or `p' =(a')/3`

Condition 4: The number of oranges left was 35 fewer than the number of apples left.

`o' = a' -35 `

Using ` a'=3p'` , we get: `o'=3p'-35`

Applying condition 3: `a' = 3p'` and condition 4: `o' = 3p'-35` on condition 2:

`3p'-35 +3p' +p' = 392`

`7p'-35=392`

`7p'=392+35`

`7p'=427`

`p' =427/7`

p'=61 as the number of "unsold pears".   

Plug-in `p' =61` on `o'=3p'-35` , we get:

`o' = 3(61)-35`

`o' =148` as number of unsold oranges

With sold oranges = 98 and unsold oranges=148 then

original number of oranges: `o= 246` .

Applying ` o=a=p` , we can determine that we also have:

246 original number of pears and 61 unsold pears.

Then,

sold pears: `246-61 =185`   [FINAL ANSWER]

In addition, the other unsold apples and oranges are:

Plug-in `p' =61` on `a'=3p'` , we get:

`a'=3*61=183` as number of unsold apples

then sold apples: `246-83=63`

Here is the tally.

Number of sold fruits: 63 apples, 185 pears, and 98 oranges

Number of unsold fruits: 183 apples, 61 pears, and 148 oranges