Ratio Problems - Higher

Convert both quantities to the same unit before simplifying.

1. Convert 0.75\text{ kg} to 750\text{ g}.
2. The ratio is 750:300. Dividing both parts by 150 gives 5:2.

Find the value of one ratio part before finding the larger share.

1. There are 5+7=12 equal parts, so one part is worth £96\div12=8.
2. The larger share is 7\times8= £56.

The difference corresponds to the difference between the ratio parts.

1. The difference is 9-4=5 parts, so one part is 35\div5=7\text{ cm}.
2. The total is 4+9=13 parts.
3. The 35 cm difference spans 5 equal parts, while the total spans 13 parts.

4. The total length is 13\times7=91\text{ cm}.

Scale the first ratio so that both descriptions use the same amount of the middle quantity.

1. Double a:b to get 6:10.
2. The common middle quantity is now 10, so a:b:c=6:10:7.
3. Therefore a:c=6:7.

Use the known blue count to find one part, then use the total number of parts.

1. Three parts represent 18 counters, so one part represents 6.
2. The total is 10 parts, giving 10\times6=60 counters.

Find the number of lemon sweets first, then work out what fraction of the box is lemon or orange.

1. Orange is 5 parts and there are 30 orange sweets, so one part is 30\div5=6 sweets.
2. Lemon is 4 parts, so there are 4\times6=24 lemon sweets.
3. Lemon and orange together are 24+30=54 sweets.
4. The probability of lime is \frac14, so lemon and orange make up 1-\frac14=\frac34 of the box.
5. \frac34 of the box is 54 sweets, so \frac14 is 54\div3=18 and the whole box is 4\times18=72 sweets.

The seeds account for the middle number of ratio parts.

1. The total is 3+4+5=12 parts. Each part has mass 360\div12=30\text{ g}.
2. The seeds are 4 parts, so their mass is 4\times30=120\text{ g}.

The total does not change, so afterwards each of them has half of it.

1. Write Ali's sweets as 5k and Ben's as 3k, so the total is 8k.
2. Giving sweets away does not change the total. Afterwards they have the same number, so each has half of 8k, which is 4k.
3. Each bar has 8 equal parts in total. The transfer changes Ali's 5 parts and Ben's 3 parts into 4 parts each; the one part moved is 12 sweets.

4. Ali goes from 5k to 4k, so he gives away k sweets. He gives 12, so k=12.
5. The total is 8\times12=96 sweets. (Check: Ali 60, Ben 36; after the gift, 48 each.)

Work out what fraction of all the money Finn gets.

1. Dev gets \frac25, so Ella and Finn share 1-\frac25=\frac35 of the money.
2. Finn gets 2 of the 1+2=3 parts of that, which is \frac23\times\frac35=\frac25 of all the money.
3. So \frac25 of the money is £84, and \frac15 is 84\div2= £42.
4. The total is 5\times42= £210.

Find how many tickets of each type were sold before applying the two prices.

1. There are 5 ratio parts, so one part is 25\div5=5 tickets.
2. There are 15 standard and 10 premium tickets.
3. The income is 15\times4+10\times7= £130.

Find the actual numbers in each group rather than adding the ratio parts directly.

1. Group A has 12 boys and 18 girls. Group B has 12 boys and 8 girls.
2. The combined group has 24 boys and 26 girls.
3. The ratio 24:26 simplifies to 12:13.

Convert the total to grams and find each original mass before adding the extra beans.

1. The original mass is 1400\text{ g}, shared into 7 parts of 200\text{ g}.
2. There are 400\text{ g} of A and 1000\text{ g} of B. After the addition, A has mass 500\text{ g}.
3. The new ratio is 500:1000=1:2.

Make the number of silver parts equal in both ratios.

1. Scale the ratios to 9:12 and 12:10, giving gold:silver:bronze =9:12:10.
2. There are 31 parts, so one part represents 155\div31=5 medals.
3. Bronze is 10 parts, giving 50 medals.

Removing half a well-mixed blend halves both components, preserving their ratio.

1. The original blend has 10 litres yellow and 8 litres blue.
2. After half is removed, 5 litres yellow and 4 litres blue remain.
3. The three bars use one common litre scale: the yellow and blue amounts halve together, then 3 litres of blue are added.

4. Adding 3 litres blue gives the final ratio 5:7.

Cube the length scale factor before applying it to the volume.

1. The volume ratio is 3^3:5^3=27:125.
2. The smaller volume represents 27 parts, so one part is 81\div27=3\text{ cm}^3.
3. The larger volume is 125\times3=375\text{ cm}^3.

The amount of concentrate stays the same, so use it to scale the final ratio.

1. The original mixture has 12 litres concentrate and 28 litres water.
2. In the final ratio, 1 concentrate part is still 12 litres, so water must be 4\times12=48 litres.
3. The amount added is 48-28=20 litres.

Find the mass represented by one whole group of the stated ratio.

1. A ratio group of 5 red and 7 blue beads has mass 5\times3+7\times2=29\text{ g}.
2. The total contains 145\div29=5 such groups.
3. There are 5\times5=25 red beads.

Choose numbers of members for the two clubs that fit all three ratios.

1. Choose club sizes that fit all the ratios. Each club splits into 3+2=5 parts, so try 4\times5=20 members in A and 5\times5=25 members in B, which are in the ratio 4:5.
2. Club A: the men are \frac35 of 20, which is 12.
3. Club B: the men are \frac25 of 25, which is 10.
4. The bars show the five equal ratio parts in each club; each part has the member count shown inside it.

5. Altogether there are 12+10=22 men out of 20+25=45 members, so the fraction is \frac{22}{45}.
6. Any other sizes in the ratio 4:5, such as 40 and 50, give the same fraction.

Mass equals density multiplied by volume; the volume ratio comes from cubing the length ratio.

1. Mass = density \times volume, so multiply the density ratio by the volume ratio, part by part.
2. For similar solids, the volume ratio is the cube of the length ratio: 2^3:3^3=8:27.
3. Mass ratio A:B =8\times9:27\times4=72:108.
4. Divide both parts by 36: 2:3. So A is lighter, even though it is made of the denser material.

Combine the ratios, then find the cost of one group with those proportions.

1. Combining the ratios gives A:B:C =9:12:10.
2. One such group costs 9\times2+12\times3+10\times4= £94.
3. The order contains 376\div94=4 groups.
4. There are 12\times4=48 B notebooks.