CAT 2018 Slot 2 — DILR Question 23

Mixed PracticeEasy

Passage / Data

Answer the following question based on the information given below.

The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.

A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly withanother. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates.

The following facts are known about the outlet on a particular day:

The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
The amounts of L used by the outlet to buy A and B are in the ratio 5 : 3.
The amounts of L the outlet received from the sales of A and B are in the ratio 5 : 9.
The outlet received 88000 units of L by selling A during the day.
The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.

What was the base exchange rate of currency B with respect to currency L on that day?

Answer & solution

Answer: 240

Solution

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 $(800 + \frac{800}{m})$ = 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 base exchange rate of B w.r.t. L on that day was 240

Therefore the required answer is 240.

Answer: 240