Comparing Two Quantities by Percentage

Draw a bar model. $40 and $50 are in the ratio $40 : 50 = 4 : 5$, so Siti's bar is 4 units and Ravi's bar is 5 units, and 1 unit is $10.

(a) "As a percentage of Siti's savings": Siti's $40 is the base, so it goes in the denominator.

Ravi's savings as a percentage of Siti's $= \frac{50}{40} \times 100\% = 125\%$

This is more than 100% because Ravi has saved more than Siti. In units, 5 units out of 4 units is $\frac{5}{4} \times 100\% = 125\%$.

(b) Now Ravi's $50 is the base: Siti's savings as a percentage of Ravi's $= \frac{40}{50} \times 100\% = 80\%$

This is less than 100% because Siti has saved less than Ravi. In units, $\frac{4}{5} \times 100\% = 80\%$.

The ages are not given, but they are not needed. Draw the son's age as 1 unit. The father is 4 times as old, so his age is 4 of the same units.

(a) The son's age is the base: the father's age as a percentage of the son's $= \frac{4 \text{ units}}{1 \text{ unit}} \times 100\% = 4 \times 100\% = 400\%$

(b) The father's age is the base: the son's age as a percentage of the father's $= \frac{1 \text{ unit}}{4 \text{ units}} \times 100\% = \frac{1}{4} \times 100\% = 25\%$

Check with ages: if the son is 10, the father is 40. Then $\frac{40}{10} \times 100\% = 400\%$ and $\frac{10}{40} \times 100\% = 25\%$. Any pair of ages with the father 4 times as old gives the same two answers.

Draw the bar model of Example 1. The difference between the two bars is 1 unit, which is $10, since $50 - 40 = 10$. It is drawn in its own colour at the end of Ravi's bar.

(a) "More than Siti's": the base is Siti's $40.

Percentage more $= \frac{\text{difference}}{\text{Siti's savings}} \times 100\% = \frac{10}{40} \times 100\% = 25\%$

(b) "Less than Ravi's": the base is Ravi's $50.

Percentage less $= \frac{\text{difference}}{\text{Ravi's savings}} \times 100\% = \frac{10}{50} \times 100\% = 20\%$

Check with Example 1. Ravi's savings are 125% of Siti's, which is $125\% - 100\% = 25\%$ more than Siti's. Siti's savings are 80% of Ravi's, which is $100\% - 80\% = 20\%$ less than Ravi's.

The son's age is 1 unit and the father's age is 4 units. The difference is $4 - 1 = 3$ units, drawn in its own colour at the end of the father's bar.

(a) "More than the son's age": the base is the son's 1 unit. Percentage more $= \frac{3}{1} \times 100\% = 300\%$

Check: the father's age is 400% of the son's (Example 2), and $400\% - 100\% = 300\%$.

(b) "Less than the father's age": the base is the father's 4 units. Percentage less $= \frac{3}{4} \times 100\% = 75\%$

Check: the son's age is 25% of the father's (Example 2), and $100\% - 25\% = 75\%$.

The difference: $560 - 525 = 35$ marks.

(a) "More than Mei Ling's": the base is Mei Ling's 525 marks.

Percentage more $= \frac{35}{525} \times 100\% = \frac{1}{15} \times 100\% = 6\frac{2}{3}\% \approx 6.67\%$ (to 3 s.f.)

(b) "Fewer than Hui Min": the base is Hui Min's 560 marks.

Percentage fewer $= \frac{35}{560} \times 100\% = \frac{1}{16} \times 100\% = 6.25\%$, which is exact.

Check (a) the second way: Hui Min's marks as a percentage of Mei Ling's $= \frac{560}{525} \times 100\% = 106\frac{2}{3}\%$, which is $6\frac{2}{3}\%$ more than 100%.

Check (b) the second way: Mei Ling's marks as a percentage of Hui Min's $= \frac{525}{560} \times 100\% = 93.75\%$, which is $100\% - 93.75\% = 6.25\%$ less than 100%.

"More than the number of girls": the girls are the base, 100%. 50% is $\frac{1}{2}$ of 100%, so draw the girls as 2 units: 1 unit is 50% of the girls.

The boys are 50% more, which is 1 unit more: $2 + 1 = 3$ units, or 150% of the girls.

Now turn it round. "Fewer girls than boys": the base is the boys' 3 units. The difference is 1 unit.

Percentage fewer $= \frac{1}{3} \times 100\% = 33\frac{1}{3}\% \approx 33.3\%$ (to 3 s.f.)

Check with numbers: if there are 20 girls, there are 30 boys (50% more, as 10 is 50% of 20). The girls are 10 fewer, and $\frac{10}{30} \times 100\% = 33\frac{1}{3}\%$. Any number of girls gives the same answer.

Mathematics, out of its own total of 20: $\frac{16}{20} \times 100\% = 80\%$

Science, out of its own total of 25: $\frac{19}{25} \times 100\% = 76\%$

Compare: $80\% > 76\%$.

He scored 3 marks more in Science, but out of a total that is 5 marks larger, so his Science result is the lower one.

Each player had a different number of attempts, so express each player's goals as a percentage of her own attempts.

Siti: $\frac{36}{40} \times 100\% = 90\%$

Mei Ling: $\frac{40}{50} \times 100\% = 80\%$

Hui Min: $\frac{36}{45} \times 100\% = 80\%$

Siti's 90% is the highest. Mei Ling and Hui Min have the same record, 80%, even though Mei Ling scored 4 more goals: she also had 5 more attempts.

Shop P: 25% of 400 g is $\frac{25}{100} \times 400 = 100$, so 100 g extra. The packet holds $400 + 100 = 500$ g and costs $4.00.

Shop Q: 20% off means paying $100\% - 20\% = 80\%$ of the usual price: $\frac{80}{100} \times 4 = 3.2$, so $3.20 for 400 g.

Compare the price of the same amount, 100 g. Shop P: 500 g is 5 lots of 100 g, so 100 g costs $4.00 \div 5 = 0.80$, or $0.80. Shop Q: 400 g is 4 lots of 100 g, so 100 g costs $3.20 \div 4 = 0.80$, or $0.80.

They are the same. This is the pair of this lesson again: 500 g is 25% more than 400 g, and 400 g is 20% less than 500 g. Getting 25% more for the same money is the same as paying 20% less for the same amount.

(a) Difference in mass: $10 - 5 = 5$ kg. The base is the small bag's 5 kg: $\frac{5}{5} \times 100\% = 100\%$ more rice.

(b) Difference in price: $22.80 - 12.00 = 10.80$, so $10.80. The base is the small bag's $12.00: $\frac{10.80}{12} \times 100\% = 90\%$ more money.

(c) The large bag gives 100% more rice for only 90% more money, so it is the better buy.

Check by the price of 1 kg. Small bag: $12.00 \div 5 = 2.40$, so $2.40. Large bag: $22.80 \div 10 = 2.28$, so $2.28. The large bag's price per kilogram is less by $\frac{0.12}{2.40} \times 100\% = 5\%$.