Ratio · applications

Applications: Ratio

10 question types · Model Method and algebra, side by side

PSLE · O-Level · SAT · GCSE Higher

01

Sharing a Total in a Given Ratio

heuristicTotal Units / Find One Unit from the Whole

Ahmad and Bala share 104 marbles in the ratio 5 : 3. (a) How many marbles does Ahmad get? (b) How many more marbles does Ahmad get than Bala?

Ahmad5 unitsBala3 units
Ahmad has 5 equal units and Bala has 3.
Draw Ahmad's share as 5 equal units and Bala's share as 3 equal units.
step 1 of 5

Draw Ahmad's share as 5 equal units and Bala's share as 3 equal units. All the units together are the 104 marbles.

  1. Draw Ahmad's share as 5 equal units and Bala's share as 3 equal units.
  2. Count all the units: 5 + 3 = 8 units, and 8 units = 104 marbles.
  3. 1 unit = 104 ÷ 8 = 13 marbles.
  4. (a) Ahmad gets 5 × 13 = 65 marbles.
  5. (b) Ahmad has 5 − 3 = 2 units more than Bala, which is 2 × 13 = 26 marbles. Check: 65 + 39 = 104.

answer(a) 65 marbles; (b) 26 marbles

techniqueBar Models for Ratio · The Unitary Method

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing 104 by 5 to find Ahmad's share. The 104 marbles are all 8 units, so divide by 8 to find one unit first.
  • Giving 39 as the answer to part (b). 39 is Bala's share. The question asks how many more Ahmad gets, which is the difference between the two shares.
02

A Part Compared with the Whole Group

heuristicPart-to-Whole Ratio / The Whole Minus One Part Is the Other Part

In a school choir, the ratio of the number of boys to the total number of members is 3 : 8. There are 35 girls in the choir. (a) How many boys are in the choir? (b) How many members does the choir have altogether?

Choir8 units = all the membersboys: 3 units
The whole choir is 8 units, and the boys are 3 of them.
Draw all the members as one bar of 8 equal units. The boys are 3 of these units.
step 1 of 5

Draw the whole choir as one bar of 8 equal units. The boys are 3 of these units, so the girls are the other 5 units.

  1. Draw all the members as one bar of 8 equal units. The boys are 3 of these units.
  2. The rest of the bar is the girls: 8 − 3 = 5 units, so 5 units = 35 girls.
  3. 1 unit = 35 ÷ 5 = 7.
  4. (a) The number of boys is 3 × 7 = 21.
  5. (b) The choir has 8 × 7 = 56 members. Check: 21 + 35 = 56.

answer(a) 21 boys; (b) 56 members

techniquePart-to-Whole Ratios · The Unitary Method

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Reading 3 : 8 as boys to girls and writing 8 units = 35. The 8 units are the whole choir. The girls are 8 − 3 = 5 units.
  • Adding 3 + 8 = 11 units for the total. The boys are already inside the 8 units, so the total is 8 units.
03

Three Quantities in a Ratio with One Difference Known

heuristicThree-Part Ratio / The Difference in Units Equals the Known Difference

A box contains red, blue and green beads in the ratio 2 : 5 : 4. There are 27 more blue beads than red beads. (a) How many green beads are there? (b) How many beads are in the box altogether?

Red2 unitsBlue5 unitsGreen4 units
The three bars are 2 units, 5 units and 4 units.
Draw the red beads as 2 units, the blue beads as 5 units and the green beads as 4 units.
step 1 of 5

Draw three bars of 2, 5 and 4 equal units. The blue bar is longer than the red bar by 3 units, and that piece is the 27 beads.

  1. Draw the red beads as 2 units, the blue beads as 5 units and the green beads as 4 units.
  2. The blue bar is 5 − 2 = 3 units longer than the red bar, so 3 units = 27 beads.
  3. 1 unit = 27 ÷ 3 = 9 beads.
  4. (a) The number of green beads is 4 × 9 = 36.
  5. (b) There are 2 + 5 + 4 = 11 units altogether, which is 11 × 9 = 99 beads. Check: 18 + 45 + 36 = 99.

answer(a) 36 green beads; (b) 99 beads

techniqueThree-Part Ratios · The Unitary Method

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Writing 5 units = 27 or 11 units = 27. The 27 is the difference between the blue beads and the red beads, which is 5 − 2 = 3 units.
  • Finding the total from only two of the colors, 2 + 5 = 7 units. The green beads are in the box as well, so the total is 11 units.
04

Two Ratios That Share One Person

heuristicCombine the Ratios / Make the Shared Term the Same Number of Units

The ratio of Ann's savings to Ben's savings is 2 : 3. The ratio of Ben's savings to Cal's savings is 4 : 5. The three children have saved $210 altogether. (a) How much has Ben saved? (b) How much more has Cal saved than Ann?

Ann : Ben = 2 : 3Ann2 unitsBen3 unitsBen : Cal = 4 : 5Ben4 unitsCal5 units
Ben is 3 units in the first ratio and 4 units in the second. Make him 12 units in both.
Ben is 3 units in the first ratio and 4 units in the second. The lowest common multiple of 3 and 4 is 12, so make Ben 12 units in both.
step 1 of 5

Ben appears in both ratios, but as a different number of units each time. Rewrite both ratios so that Ben is the same number of units, and the three children can then be compared in one ratio.

  1. Ben is 3 units in the first ratio and 4 units in the second. The lowest common multiple of 3 and 4 is 12, so make Ben 12 units in both.
  2. Multiply the first ratio by 4: 2 : 3 = 8 : 12. Multiply the second ratio by 3: 4 : 5 = 12 : 15. So Ann : Ben : Cal = 8 : 12 : 15.
  3. There are 8 + 12 + 15 = 35 units altogether, so 35 units = $210 and 1 unit = 210 ÷ 35 = $6.
  4. (a) Ben has saved 12 × 6 = $72.
  5. (b) Cal has 15 − 8 = 7 units more than Ann, which is 7 × 6 = $42. Check: 48 + 72 + 90 = 210.

answer(a) $72; (b) $42

techniqueCombining Two Ratios · Three-Part Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Joining the two ratios as 2 : 3 : 5 or 2 : 4 : 5. Ben is 3 units in one ratio and 4 units in the other, so the units are different sizes until Ben is made 12 units in both.
  • Multiplying only Ben's term by 4 and leaving Ann's term as 2. Both terms of a ratio must be multiplied by the same number, or the ratio changes.
05

A Mixture Made in Two Batch Sizes

heuristicUnitary Method / Find One Unit from Whichever Quantity Is Known

An orange drink is made by mixing orange concentrate and water in the ratio 2 : 7. (a) How much water is needed to mix with 250 ml of concentrate? (b) A jug holds 3.6 liters of the drink. How much concentrate is in the jug?

Concentrate250 mlWater?
The concentrate is 2 units and the water is 7 units. Here 2 units are 250 ml.
Every batch is 2 units of concentrate and 7 units of water. With 250 ml of concentrate, 2 units = 250 ml.
step 1 of 5

Every batch is 2 units of concentrate and 7 units of water. Use the quantity that is known to find 1 unit, then multiply to find the quantity that is asked for.

  1. Every batch is 2 units of concentrate and 7 units of water. With 250 ml of concentrate, 2 units = 250 ml.
  2. 1 unit = 250 ÷ 2 = 125 ml.
  3. (a) The water is 7 units: 7 × 125 = 875 ml.
  4. In the jug the whole drink is 2 + 7 = 9 units, and 3.6 liters = 3600 ml. So 1 unit = 3600 ÷ 9 = 400 ml.
  5. (b) The concentrate is 2 × 400 = 800 ml. Check: 800 + 7 × 400 = 3600 ml.

answer(a) 875 ml; (b) 800 ml

techniqueThe Unitary Method · Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using 1 unit = 125 ml again in part (b). The jug is a different batch, so one unit has a new size and must be found from the 3600 ml.
  • Dividing 3600 ml by 7 or by 2 in part (b). The 3600 ml is the whole drink, which is 2 + 7 = 9 units.
06

A Scale Model and the Real Object

heuristicScale Factor / Multiply to Enlarge and Divide to Reduce

A model of a ship is built to the scale 1 : 150. The model is 42 cm long. (a) How long is the real ship, in meters? (b) The mast of the real ship is 27 m tall. How tall is the mast of the model, in centimeters?

Scale 1 : 150. Each length on the ship is 150 times the length on the model.Model length42 cmship: ?
Model to ship, multiply by 150. Ship to model, divide by 150.
The scale 1 : 150 means that every length on the real ship is 150 times the matching length on the model.
step 1 of 5

The scale factor is 150. Multiply by it to go from the model to the ship, and divide by it to go from the ship to the model. Keep both lengths in the same unit while doing so.

  1. The scale 1 : 150 means that every length on the real ship is 150 times the matching length on the model.
  2. Multiply the length of the model by 150: 42 × 150 = 6300 cm.
  3. (a) 6300 cm = 63 m, so the real ship is 63 m long.
  4. For the mast, change meters to centimeters first: 27 m = 2700 cm. Going from the ship to the model, divide by 150.
  5. (b) The mast of the model is 2700 ÷ 150 = 18 cm tall. Check: 18 × 150 = 2700 cm.

answer(a) 63 m; (b) 18 cm

techniqueScale Factor

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Calculating 27 ÷ 150 = 0.18 and giving 0.18 cm. The 27 is in meters, so the result is 0.18 m, which is 18 cm. Change to centimeters before dividing.
  • Dividing 42 by 150 in part (a). The real ship is larger than the model, so the length of the model is multiplied by the scale factor.
07

A Ratio Read as a Percentage of the Whole

heuristicRatio to Fraction to Percentage / The Total Units Are the Denominator

The ratio of the number of adults to the number of children in a concert audience is 7 : 13. (a) What percentage of the audience are children? (b) There are 300 people in the audience. How many of them are adults?

Audienceadults: 7 unitschildren: 13 units
The audience is 7 + 13 = 20 units, and the children are 13 of them.
Draw the audience as one bar of 7 + 13 = 20 equal units. The children are 13 of the 20 units, which is 1320 of the audience.
step 1 of 4

Draw the audience as one bar of 7 + 13 = 20 equal units. A part of the bar, written over the 20 units of the whole bar, is a fraction of the audience.

  1. Draw the audience as one bar of 7 + 13 = 20 equal units. The children are 13 of the 20 units, which is 1320 of the audience.
  2. (a) Multiply the numerator and the denominator by 5: 1320 = 65100, so 65% of the audience are children.
  3. The 20 units are the 300 people, so 1 unit = 300 ÷ 20 = 15 people.
  4. (b) The adults are 7 units: 7 × 15 = 105 adults. Check: 105 + 13 × 15 = 300.

answer(a) 65%; (b) 105 adults

techniqueRatio, Fraction, Percent · Part-to-Whole Ratios

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Writing the fraction of children as 137 or 713. Those compare the children with the adults. A fraction of the audience has the total, 20 units, as its denominator.
  • Reading 7 : 13 as 7% and 13%. A percentage is out of 100, so the 20 units must first be scaled to 100: each unit is 5% of the audience.
08

Sides of a Rectangle in a Ratio with the Perimeter Known

heuristicUnits Around the Perimeter / Two Lengths and Two Breadths

The ratio of the length of a rectangle to its breadth is 5 : 3. The perimeter of the rectangle is 96 cm. (a) What is the length of the rectangle? (b) What is the area of the rectangle?

5 units3 units
The length is 5 units and the breadth is 3 units.
Draw the rectangle with a length of 5 units and a breadth of 3 units.
step 1 of 5

Draw the rectangle with a length of 5 units and a breadth of 3 units, and count the units all the way round it.

  1. Draw the rectangle with a length of 5 units and a breadth of 3 units.
  2. The perimeter is two lengths and two breadths: 5 + 3 + 5 + 3 = 16 units, so 16 units = 96 cm.
  3. 1 unit = 96 ÷ 16 = 6 cm.
  4. (a) The length is 5 × 6 = 30 cm.
  5. (b) The breadth is 3 × 6 = 18 cm, so the area is 30 × 18 = 540 cm2. Check: 2 × (30 + 18) = 96.

answer(a) 30 cm; (b) 540 cm2

techniqueThe Unitary Method · Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Writing 5 + 3 = 8 units = 96 cm. Eight units is one length and one breadth, which is only half of the perimeter.
  • Finding the area as 5 × 3 = 15 and multiplying by 6. Each side is multiplied by 6, so the area is 15 × 6 × 6 = 540 cm2.
09

A Ratio from Measurements in Different Units

heuristicSame Unit First / Divide Both Terms by the Highest Common Factor

A ribbon 1.2 m long is cut into two pieces. The shorter piece is 45 cm long. (a) What is the ratio of the length of the shorter piece to the length of the longer piece, in its simplest form? (b) A second ribbon is cut into two pieces in the same ratio, and its shorter piece is 72 cm long. How long is the second ribbon?

Ribbon45 cm75 cm1.2 m = 120 cm
In centimeters the ribbon is 120 cm, so the longer piece is 120 − 45 = 75 cm.
Write both lengths in the same unit: 1.2 m = 120 cm. The longer piece is 120 − 45 = 75 cm.
step 1 of 5

A ratio compares quantities in the same unit, so change the meters to centimeters before comparing. The simplest form of the ratio then gives the units for the second ribbon.

  1. Write both lengths in the same unit: 1.2 m = 120 cm. The longer piece is 120 − 45 = 75 cm.
  2. The ratio is 45 : 75. The highest common factor of 45 and 75 is 15, so divide both terms by 15.
  3. (a) 45 : 75 = 3 : 5.
  4. In the second ribbon the shorter piece is 3 units, so 3 units = 72 cm and 1 unit = 72 ÷ 3 = 24 cm.
  5. (b) The whole ribbon is 3 + 5 = 8 units, which is 8 × 24 = 192 cm.

answer(a) 3 : 5; (b) 192 cm

techniqueTidying a Ratio · The Unitary Method

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Writing the ratio as 45 : 1.2. One length is in centimeters and the other is in meters, and 1.2 m is the whole ribbon, not the longer piece.
  • Stopping at 9 : 15 after dividing by 5. Both terms still divide by 3, so the ratio is not yet in its simplest form.
10

A Transfer That Makes Two Amounts Equal

heuristicThe Total Stays the Same / Half of the Difference Is Moved

The ratio of the number of cards Devi has to the number of cards Farah has is 7 : 3. After Devi gives Farah 12 cards, the two girls have the same number of cards. (a) How many cards did Devi have at first? (b) How many cards does each girl have in the end?

Devi7 unitsFarah3 units
Devi has 7 units and Farah has 3. The total of 10 units does not change.
Draw Devi's cards as 7 units and Farah's cards as 3 units. The cards only move between the two girls, so the total stays at 7 + 3 = 10 units.
step 1 of 5

The cards only move from one girl to the other, so the total number of units does not change. Share the total equally to see how many units Devi must give away.

  1. Draw Devi's cards as 7 units and Farah's cards as 3 units. The cards only move between the two girls, so the total stays at 7 + 3 = 10 units.
  2. In the end the two girls have the same number of cards, so each has 10 ÷ 2 = 5 units.
  3. Devi goes from 7 units to 5 units, so she gives away 2 units. 2 units = 12 cards, and 1 unit = 12 ÷ 2 = 6 cards.
  4. (a) Devi had 7 × 6 = 42 cards at first.
  5. (b) In the end each girl has 5 × 6 = 30 cards. Check: 42 − 12 = 30 and 18 + 12 = 30.

answer(a) 42 cards; (b) 30 cards

techniqueRatios After a Transfer · Bar Models for Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Writing 7 − 3 = 4 units = 12 cards. Devi gives away only half of the difference. If she gave all 4 units, Farah would then have more cards than Devi.
  • Giving 42 as the answer to part (b). 42 is what Devi had at first. In the end she has 12 fewer cards.
Mr. Chalk Drill ratio and proportion