Earnings, Simple and Compound Interest Worked ExamplesIGCSE Maths 0580

(a) Find the time between clocking in and clocking out for each day, then take off the $45$-minute lunch break. Monday: $08{:}15$ to $16{:}45$ is $8$ h $30$ min.

Do the same for the other days. Tuesday is $8$ h $30$ min, Wednesday $8$ h $45$ min, Thursday $8$ h $15$ min and Friday $5$ h $45$ min before the break is removed.

Add the paid hours for all five days.

(b) Multiply the hours by the hourly rate. The pay is $496.80.

(c) Find how much more he needs to reach $600.

Divide by the hourly rate. The answer is not a whole number, and $7$ hours would leave him short, so round up.

(a) Commission is paid only on the part of the sales above $10,000.

Find $4\%$ without a calculator: $1\%$ of $21\,250$ is $212.5$, so $4\%$ is four times that.

Add the basic salary. Her April pay is $2650.

(b) Work backwards. First remove the basic salary to find the commission.

$540 is $4\%$ of the sales above $10,000. Since $1\%$ is $540 \div 4 = 135$, $100\%$ is $13\,500$.

This is only the amount above $10,000, so add the $10,000 back on. Her May sales were $23,500.

(a) Simple interest is the same every year, because it is always worked out on the original amount. Find one year's interest: $1\%$ of $2400$ is $24$.

Divide the total interest by the interest for one year.

(b) Amir is right that $5$ years at $4\%$ adds $20\%$ in total. His mistake is that the $20\%$ is a percentage of the amount invested, not of the final total $3360.

So $3360 is $120\%$ of the amount invested. Divide by $1.2$, or find $1\%$ first: $3360 \div 120 = 28$.

Check: the interest on $2800 for $5$ years at $4\%$ is $5 \times 112$.

(a) The deposit is $15\%$ of the cash price. $10\%$ of $1250$ is $125$ and $5\%$ is $62.5$.

Find the total of the monthly payments, then add the deposit. The easy payment plan costs $1312.50.

(b) Find the extra cost, then write it as a percentage of the cash price.

(c) This is a reverse percentage: the cash price is $125\%$ of the price the shop paid.

(d) A $12\%$ reduction leaves $88\%$ of the cash price.

Profit is the sale price minus the price the shop paid. Write it as a percentage of the price the shop paid, not of the selling price.

(a) The pension is $6\%$ of the gross pay, so $A$ is $231.00.

Taxable income is the gross pay minus the tax-free allowance, so $B$ is $2850.00.

Tax is $20\%$ of the taxable income only, so $C$ is $570.00.

Net pay is the gross pay minus both deductions. The allowance is not a deduction, so it is not subtracted here. $D$ is $3049.00.

(b) Work through the payslip again with the new gross pay.

New pension: $0.06 \times 4004 = 240.24$. New tax: $0.2 \times (4004 - 1000) = 600.80$. New net pay is $3162.96.

Percentage increase is the change divided by the original net pay.

(a) Read two points from line $A$: it starts at $3000 and reaches $4800 after $12$ years. The interest is the same every year, so share it equally.

Write one year's interest as a percentage of the amount invested.

(b) Each year the value is multiplied by $1.07$, so after $12$ years the multiplier is $1.07^{12}$. The value is $5405.26, which agrees with the end of curve $B$ on the graph.

(c) The accounts have the same value where the line and the curve cross. Read down to the time axis.

(d) The graph suggests the answer is $8$ or $9$ years, but the lines are very close there, so calculate. After $8$ years, $A$ is $3000 + 8 \times 150 = 4200$.

After $9$ years, $A$ is $3000 + 9 \times 150 = 4350$.

(a) Losing $20\%$ means multiplying by $0.8$. Losing $12\%$ in each of the next $3$ years means multiplying by $0.88$ three times. Combine them into one multiplier.

The value after $4$ years is the new price times this multiplier, so divide to go backwards. The new price was $22,000.

(b) The first car keeps $0.5451776$ of its value over $4$ years. For the second car, the same yearly multiplier $m$ is used $4$ times.

Take the fourth root to find the yearly multiplier.

The yearly multiplier is $1 - \frac{x}{100}$, so find the percentage lost.

(c) The car must keep less than $0.5$ of its value, so we need $\left(1 - \frac{y}{100}\right)^{4} < 0.5$. Try whole numbers near the answer to (b).

$y = 15$ is not enough, so try $y = 16$.

(a) Add $6\%$ interest by multiplying by $1.06$, then subtract the repayment.

(b) Repeat the two steps each year. You cannot use $12\,000 \times 1.06^{3}$, because the amount owed changes with each repayment. Year $2$:

Year $3$, rounding to the nearest cent. She owes $4741.39.

(c) Year $4$:

Year $5$: add the interest. This is less than $3000, so it is the final payment: $2147.42.

(d) Find everything she paid back, then subtract the amount she borrowed.

The interest is $2147.42. As a check, add the interest from each year: $720 + 583.20 + 438.19 + 284.48 + 121.55$ gives the same total.

(a) Basic pay for the first $37.5$ hours.

The overtime rate is $25\%$ more than $14.40, so multiply by $1.25$. The overtime rate is $18.00 per hour.

She works $44 - 37.5 = 6.5$ hours of overtime.

Add the basic pay and the overtime pay. Her pay is $657.

(b) Work backwards. Take off the basic pay for the first $37.5$ hours to find the overtime pay.

Divide by the overtime rate to find the overtime hours.

Add the basic hours back on.

(a) Simple interest is worked out on the original $3500 every year. Find $1\%$, then $2.5\%$.

The interest is the same each year, so multiply by $4$.

The question asks what the bond is worth, so add the interest to the amount invested. The bond is worth $3850.

(b) The interest must be more than $4000 - 3500 = 500$ dollars. Each year adds $87.50.

After $5$ years the value is $3937.50; after $6$ years it is $4025. So it takes $6$ complete years.

(a) From 2019 to 2025 is $6$ years. Let the yearly multiplier be $k$; the balance has been multiplied by $k$ six times.

Divide by $2800$, then take the sixth root — not divide by $6$.

Subtract $1$ and multiply by $100$ to get the rate.

(b) From 2019 to 2035 is $16$ years. Use the unrounded multiplier, kept in the calculator memory (or [Ans]).

Round to the nearest cent. (Starting from the 2025 balance gives the same answer: $3500 \times k^{10}$. Using $1.038$ instead gives $5085.29, which is more than $8 out.)

(a) The starting amount does not matter: call it $P$, and call the yearly multiplier $k$. After $12$ years the money is $2P$.

Take the twelfth root of both sides.

The rate is the part of the multiplier above $1$, as a percentage. Show more figures than the value you are asked to show, then round.

(b) Use the unrounded multiplier $\sqrt[12]{2}$. Tripling needs a little more than one and a half doubling times — about $19$ or $20$ years — so test both.

$19$ years is not quite enough, so try $20$ years.

So the least number of complete years is $20$. Note how close $19$ years is: using a rounded rate of $6\%$ would give $1000 \times 1.06^{19} = 3025.60$, and the wrong answer.