Algebra
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Q.

The currencies in countries $X$ and $Y$ are denoted by $X_s$. and $Y_s$. respectively. The exchange rate in 1990 was $1$ $X_s. = 0.6 Y_s$. The price level in 2006 in $X$ and $Y$ are 150 and 400 respectively with 1990 as a base of 100. The exchange rate in 2006, based solely on this purchasing power parity consideration, is 1 $X_s.=$

 A.

0.225 $Y_s$

 B.

0.625 $Y_s$

 C.

1.6 $Y_s$

 D.

3.6 $Y_s$.

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Solution:
Option(C) is correct

In 1990, $1$ $X_s=0.6Y_s$

Price levels in 2006 change by 150 and 400 for $X$ and $Y$ respectively with 1990 as base 100.

In 2006, $150×1X_s=400×0.6Y_s$

\(X_s=\left(\dfrac{240}{150}\right)Y_s\)

\(1X_s=1.6Y_s\)


(4) Comment(s)


ABHIJEET
 ()

the most understandable solution would be interpreted as this

if X is 0.6 times of Y

then 150 and 400 currency shares a relation along with the previous exchange rate

so 150 of X i.e., 150 x X and 400 of Y i.e. 400 x 0.6 Y (as a exchange rate in 1990) should give u the right answer



Chinmay Pathak
 ()

I think the answer is wrong, it must be A.

Reasoning:

Let in 1990, $1Y_s = 100$, so $1X_s = 60$.

Now, in 2006, there is $50\%$ increase in value of $X_s$ while there is $300\%$ increase in value of $Y_s$.

So in 2006, the values of $1X_s$ and $1 Y_s$ are as follows:

$1Y_s = 100 + 3(100) = 400$ and

$1X_s = 60 + 0.5(60) = 90$

$= k*400$

Here, $k$ is our ans which is $0.225$


Ritika
 ()

You don't go that far while solving such questions. Short and right way is just to scale them up as below,

In 2006,

$1 X_s=0.6 \times \dfrac{450}{150}$

$1 X_s=1.6 Y_s$

Which is the right answer.


Salman
 ()

didn't understand d solution...any alternate

way 2 do this question..