An investment fund allocates capital between two asset classes: Asset Class P and Asset Cl… — GMAT

An investment fund allocates capital between two asset classes: Asset Class $P$ and Asset Class $Q$ . The annual percentage yields earned by these asset classes remained constant over a two-year period.

- In Year 1, an investment of $\$3,000,000$ in Asset Class $P$ and $\$2,000,000$ in Asset Class $Q$ produced a total return of $\$410,000$ .
- In Year 2, an investment of $\$2,000,000$ in Asset Class $P$ and $\$5,000,000$ in Asset Class $Q$ produced a total return of $\$640,000$ .

Based on the information provided, select the annual percentage yield earned by Asset Class $P$ and the annual percentage yield earned by Asset Class $Q$ .

Annual percentage yield for Asset Class P7%
Annual percentage yield for Asset Class Q10%

Answer

The annual percentage yield for Asset Class P is 7%, and the annual percentage yield for Asset Class Q is 10%.

Solving the system of simultaneous linear equations

3r_P + 2r_Q = 41

and

2r_P + 5r_Q = 64

yields

r_P = 7\%

for Asset Class P and

r_Q = 10\%

for Asset Class Q.

Step-by-Step Solution

Define variables for the unknown annual return rates.

Let

r_P

be the annual percentage yield of Asset Class

P

and

r_Q

be the annual percentage yield of Asset Class

Q

Establishing explicit variables allows translation of the financial statements into a system of linear equations.

Set up the simultaneous equations for Year 1 and Year 2.

Year 1 equation:

3,000,000 \cdot \frac{r_P}{100} + 2,000,000 \cdot \frac{r_Q}{100} = 410,000 \implies 30,000 r_P + 20,000 r_Q = 410,000 \implies 3r_P + 2r_Q = 41

.
Year 2 equation:

2,000,000 \cdot \frac{r_P}{100} + 5,000,000 \cdot \frac{r_Q}{100} = 640,000 \implies 20,000 r_P + 50,000 r_Q = 640,000 \implies 2r_P + 5r_Q = 64

Simplifying by dividing both sides by 10,000 reduces the coefficients to manageable integers.

Solve the system of linear equations using the elimination method.

Multiply the first equation by 5:

15r_P + 10r_Q = 205

.
Multiply the second equation by 2:

4r_P + 10r_Q = 128

.
Subtract the second modified equation from the first:

(15r_P - 4r_P) + (10r_Q - 10r_Q) = 205 - 128 \implies 11r_P = 77 \implies r_P = 7

Eliminating

r_Q

allows direct solution for the value of

r_P

Substitute

r_P = 7

back into the first simplified equation to solve for

r_Q

3(7) + 2r_Q = 41 \implies 21 + 2r_Q = 41 \implies 2r_Q = 20 \implies r_Q = 10

Substituting the known variable yields the remaining unknown.

Key Concept

Simultaneous Linear Equations in Two Variables

Estimated Time:2m 0s

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