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Question:

How many units of currency C did the outlet sell on that day?

Explanation:

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

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Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold 

$=\frac{88000}{110 m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132 m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B = 

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114 m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

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If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

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The outlet sold 19000 units of currency C on that day.

Hence, option (c).