Simple Interest

Worked example

1. 1

   $P=2500,R=3.5,T=4$ — Read the three numbers from the question. The 3.5% goes in as 3.5, because the formula divides by 100 for you.

2. 2

   $I=\frac{2500\times 3.5\times 4}{100}$ — Put the numbers in the formula. Write this line down: it earns a mark on its own.

3. 3

   $=\frac{35000}{100}$ — Multiply the top: 2500 × 3.5 = 8750, and 8750 × 4 = 35 000.

4. 4

   $=350$ — Divide by 100. This is the interest: $350 for the 4 years.

5. 5

   $2500+350=\$2850$ — The total (the value at the end) is the money you started with plus the interest, so add them.

Worked example

1. 1

   $\frac{1500\times R\times 3}{100}=270$ — $1500 earns $270 in 3 years. Put P = 1500, T = 3 and I = 270 into the formula. R is the only letter left.

2. 2

   $45R=270$ — Work out the numbers you know: 1500 × 3 ÷ 100 = 45. So the left side is 45 lots of R.

3. 3

   $R=270÷45=6$ — Divide both sides by 45 to get R on its own.

4. 4

   $\text{rate}=6\%$ — R is already the percent, because the formula has divided by 100 for you.

5. 5

   $\frac{1500\times 6\times 3}{100}=270$ — Check: put 6 back in. You get the $270 from the question, so the answer is right.

Worked example

1. 1

   $4560-4000=560$ — $4000 grows to $4560 in 4 years. The interest is only the extra money: total take away start.

2. 2

   $\frac{4000\times R\times 4}{100}=560$ — Now it is the same as before. Put P = 4000, T = 4 and I = 560 into the formula.

3. 3

   $160R=560$ — 4000 × 4 ÷ 100 = 160, so the left side is 160 lots of R.

4. 4

   $R=560÷160=3.5$ — Divide both sides by 160. The rate is 3.5% per year.

Worked example

1. 1

   $1000-800=200$ — $800 at 5% grows to $1000. The interest is the total take away the start: $200.

2. 2

   $\frac{800\times 5\times 1}{100}=40$ — Find the interest for one year: put T = 1 in the formula. Simple interest is the same every year.

3. 3

   $T=200÷40=5$ — Each year adds $40, so count how many lots of $40 make $200: 5 years.

4. 4

   $\frac{800\times 5\times 5}{100}=200$ — Check: 5 years really does give $200 interest.

Worked example

1. 1

   $\frac{P\times 3\times 4}{100}=96$ — $96 interest at 3% for 4 years. Put R = 3, T = 4 and I = 96 into the formula. P is the unknown.

2. 2

   $0.12P=96$ — Work out the numbers you know: 3 × 4 ÷ 100 = 0.12. So the left side is 0.12 lots of P.

3. 3

   $P=96÷0.12=\$800$ — Divide both sides by 0.12 to get P on its own.

4. 4

   $\frac{800\times 3\times 4}{100}=96$ — Check: $800 really does earn $96.

Worked example

1. 1

   $P+\frac{P\times 4\times 3}{100}=1120$ — $1120 after 3 years at 4%. Total = start + interest, and the interest is the formula with P still unknown.

2. 2

   $P+0.12P=1120$ — Work out the interest part: 4 × 3 ÷ 100 = 0.12, so the interest is 0.12P.

3. 3

   $1.12P=1120$ — P is the same as 1P, and 1P + 0.12P = 1.12P: the start plus 12% of it.

4. 4

   $P=1120÷1.12=\$1000$ — Divide both sides by 1.12 to get P on its own.

5. 5

   $1000+120=1120$ — Check: $1000 earns $120 in 3 years at 4%, and the total is $1120.

Worked example

1. 1

   $P=2500,R=3.2,I=560$ — $2500 at 3.2% earns $560. How many years? T is missing, so use the T formula.

2. 2

   $T=\frac{100\times 560}{2500\times 3.2}$ — Put the numbers in: 100 × I on top, P × R underneath.

3. 3

   $=\frac{56000}{8000}$ — Work out the top and the bottom: 100 × 560 = 56 000, and 2500 × 3.2 = 8000.

4. 4

   $=7\text{years}$ — Divide. The time is 7 years.

5. 5

   $\frac{2500\times 3.2\times 7}{100}=560$ — Check in the main formula: 7 years gives the $560 from the question.