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**

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