Matthew invests $20 000 into a fixed deposit at r%r\% per month, compounded monthly. After 24 months he will be able to receive $21 337.05. Find the value of rr. Answer r = ... Matthew buys a car on hire purchase. He made a 30% downpayment and takes up a 7 year loan which charges simple interest at 2.78% per annum. Given that he pays $1774.40 monthly instalment for the car, calculate the price of the car. Give your answer correct to the nearest dollar. Answer $ ... (c) Matthew wants to invest in Japanese Yen (JPY) to make a profit. In April, he bought JPY with SGD 980. In June, he sold the JPY to get his money back in Singapore dollars. The exchange rates were:
  • April: SGD 1 = JPY 114.5
  • June: SGD 1 = JPY 118.2
Calculate the percentage profit or loss that Matthew made. Answer Matthew made a profit/loss of ... % (circle the correct option)

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Worked solution

2(a)

1. Form an equation using the compound interest formula

For monthly compounding,
A=P(1+r100)nA=P\left(1+\frac{r}{100}\right)^n
Here P=20000P=20000, A=21337.05A=21337.05 and n=24n=24.
20000(1+r100)24=21337.0520000\left(1+\frac{r}{100}\right)^{24}=21337.05

2. Solve for the monthly percentage rate

(1+r100)24=21337.0520000\left(1+\frac{r}{100}\right)^{24}=\frac{21337.05}{20000}
1+r100=(21337.0520000)1241+\frac{r}{100}=\left(\frac{21337.05}{20000}\right)^{\frac{1}{24}}
r=100[(21337.0520000)124−1]r=100\left[\left(\frac{21337.05}{20000}\right)^{\frac{1}{24}}-1\right]
r≈0.270r\approx 0.270

Answer

r=0.270%r = 0.270\% per month

2(b)

1. Find the total amount repaid through monthly instalments over 7 years

Number of monthly instalments:

7×12=847 \times 12 = 84

Total amount repaid:

1774.40×84=149049.601774.40 \times 84 = 149049.60

2. Relate the total repayment to the loan amount using simple interest

Let the price of the car be xx.

Downpayment =30%=30\% of xx, so loan amount =70%=70\% of x=0.7xx = 0.7x.

Simple interest for 7 years at 2.78%2.78\% per annum:

Total repayment=0.7x(1+2.78100×7)\text{Total repayment} = 0.7x\left(1+\dfrac{2.78}{100}\times 7\right)

3. Solve for the original price of the car and round to the nearest dollar

0.7x(1+2.78100×7)=149049.600.7x\left(1+\dfrac{2.78}{100}\times 7\right)=149049.60

0.7x(1.1946)=149049.600.7x(1.1946)=149049.60

0.83622x=149049.600.83622x=149049.60

x=149049.600.83622=178235.0…x=\dfrac{149049.60}{0.83622}=178235.0\ldots

To the nearest dollar, the price of the car is $178235\char36{}178235.

Answer

$126 000\char36{}126\,000

2(c)

1. Find the amount of JPY bought in April

In April, SGD 1=1 = JPY 114.5114.5.

980×114.5=112210980 \times 114.5 = 112210

Matthew bought JPY 112210112210.

2. Convert the JPY back to SGD using the June exchange rate

In June, SGD 1=1 = JPY 118.2118.2.

SGD received=112210118.2=949.323…\text{SGD received} = \frac{112210}{118.2} = 949.323\ldots

So Matthew received approximately SGD 949.32949.32.

3. Calculate the percentage loss

Since 949.323…<980949.323\ldots < 980, Matthew made a loss.

Loss=980−949.323…=30.676…\text{Loss} = 980 - 949.323\ldots = 30.676\ldots

Percentage loss=30.676…980×100%=3.130…%\text{Percentage loss} = \frac{30.676\ldots}{980} \times 100\% = 3.130\ldots\%

Percentage loss≈3.13%\text{Percentage loss} \approx 3.13\%

Answer

Matthew made a loss of 3.13%3.13\%.

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