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the velocity of money (v) is defined as v = y / l(r, y), where l(r, y) …

Question

the velocity of money (v) is defined as v = y / l(r, y), where l(r, y) = m^s / p. suppose real gdp (y) is $1,000 billion, the money supply (m^s) is $500 billion, and the price level (p) is 2. if the interest rate (r) increases, causing l(r, y) to fall by 10%, what is the new velocity of money? 5 4.4 5.5 4

Explanation:

Step 1: Calculate the initial \( L(R, Y) \)

Given \( L(R, Y)=\frac{M^s}{P} \), substitute \( M^s = 500 \) billion and \( P = 2 \).
\( L(R, Y)=\frac{500}{2}=250 \) billion.

Step 2: Find the new \( L(R, Y) \) after a 10% fall

A 10% fall means the new \( L(R, Y) \) is \( 250\times(1 - 0.1)=250\times0.9 = 225 \) billion.

Step 3: Calculate the new velocity of money \( V \)

Using \( V=\frac{Y}{L(R, Y)} \), substitute \( Y = 1000 \) billion and the new \( L(R, Y)=225 \) billion.
\( V=\frac{1000}{225}=\frac{40}{9}\approx4.44\approx4.4 \) (rounded to one decimal place). Wait, wait, let's recalculate: Wait, no, wait, 1000 divided by 225: 2254 = 900, 1000 - 900 = 100, 100/225 = 4/9 ≈ 0.444, so total is 4.444..., which is approximately 4.4? Wait, but let's check again. Wait, initial L is 250. After 10% fall, 2500.9=225. Then Y is 1000. So 1000/225 = 40/9 ≈ 4.444..., which is approximately 4.4 (if we take one decimal place) or 4.44. But the options have 4.4, 5, etc. Wait, maybe I made a mistake? Wait, no, let's check the steps again.

Wait, initial L: \( M^s / P = 500 / 2 = 250 \). Correct. Then 10% fall: 250 - 25 = 225. Correct. Then V = Y / L = 1000 / 225 = 4.444... So the closest option is 4.4.

Answer:

4.4 (corresponding to the option with 4.4, e.g., if the options are as given, the answer is the option with 4.4, like the second option: 4.4)