Contents
- A fraction is a number
- A fraction of a whole: divide, then multiply
- A fraction of a fraction: a selection within a selection
- Recovering the whole from a fraction
- Before the problems: choose the operation thoughtfully
- 25 problems: medium to difficult
A fraction is a number
A fraction is not merely a slice of cake: 3/8 is the number obtained by dividing 3 by 8. Applied to a quantity, it acts as a multiplier. The denominator tells us how many equal parts to divide the unit into; the numerator tells us how many to take. For example, 3/2 is greater than 1: a fraction need not be smaller than one.
Parts must be equal, and the reference whole must be clear. The number 1/2 does not change, but half of 20 and half of 80 are different quantities. Equivalent fractions name the same number: 1/2 = 2/4 = 3/6. We use positive denominators below.
A fraction of a whole: divide, then multiply
What is 3/8 of 48? Divide 48 into 8 equal parts: each is 6. Take 3 of them to get 18. Do not divide by 3: the denominator 8 defines the partition.
48 ÷ 8 = 6; 6 × 3 = 18
| ● 6 | ● 6 | ● 6 | ○ 6 | ○ 6 | ○ 6 | ○ 6 | ○ 6 |
The three filled symbols mark the selected parts; the five empty symbols mark the others. All cells together add to 48.
In general, a/b of N is N × a/b. You may divide first or multiply first without changing the result, provided you do not round intermediate values.
A fraction of a fraction: a selection within a selection
Take 2/3 of a rectangle, then 3/4 of that part. The rectangle has 3 columns and 4 rows: 12 equal cells. The first two columns are the initial 2/3, or 8 cells. Three quarters of those 8 cells is 6 cells.
| ● | ● | — |
| ● | ● | — |
| ● | ● | — |
| ○ | ○ | — |
(3/4) × (2/3) = 6/12 = 1/2
● = final selected cells; ○ = cells in the first selection not retained; — = cells outside the first selection. The final region contains 6 of the whole rectangle’s 12 cells.
Multiply numerators and denominators: (a/b) × (c/d) = (a × c)/(b × d). Watch the word “remaining”: calculate that remainder first. Removing 2/3 of 3/4 leaves (1/3) × (3/4) = 1/4, not 1 − 2/3 − 3/4.
Recovering the whole from a fraction
Three fifths of a quantity is 27. Three parts equal 27, so one part is 27/3 = 9. The whole contains 5 parts: 9 × 5 = 45. This time divide by the numerator because the known value covers three parts.
| ● 9 | ● 9 | ● 9 | ○ 9 | ○ 9 |
27 ÷ 3 = 9; 9 × 5 = 45
The first three parts total 27; all five total 45. Check by recalculating 3/5 of 45 = 27.
In general, if a/b of N equals P, then N = P ÷ (a/b) = P × b/a, with a nonzero. Multiplying P by a/b again would reduce the known part instead of recovering the whole.
Before the problems: choose the operation thoughtfully
Fractions of the same whole can be added using a common denominator: 1/3 + 1/4 = 4/12 + 3/12 = 7/12. When a fraction is applied to another fractional quantity, multiply. When a part is known and the whole is required, divide by the fraction.
For each problem, identify the whole, note the quantity each fraction refers to, write down what remains, and check the result against the original wording. Some exercises consolidate the basics; others combine several steps or a changing whole. Hints and solutions start closed: open them only when needed.
25 problems: medium to difficult
- The library
- The changing tank level
- Two successive expenses
- A book in three days
- The class and music
- The final stage
- The vegetable garden
- Two quantities, one total
- The equalising transfer
- Saving part of a salary
- Absences and early departures
- Resizing a recipe
- Two cuts from one roll
- Two discounts, not a sum
- Equal increase and decrease?
- Diluting twice
- Two people working
- The tap and the leak
- The hidden number
- A ratio and a difference
- The party pizzas
- Three colours in an urn
- Adding changes the whole too
- Working backwards through donations
- Four successive shares
1. The library
In a library, 3/8 of the books are novels. Essays account for 2/5 of the remaining books. The other 84 books are manuals. How many books are there altogether and in each group?
Hint
The 2/5 refers to books that are not novels, not to all the books.
Step-by-step solution
After the novels, 5/8 of the total remains. Essays occupy (2/5) × (5/8) = 1/4 of the total. Manuals occupy 1 − 3/8 − 1/4 = 3/8. Since 3/8 represents 84, the total is 84 × 8/3 = 224 books: 84 novels, 56 essays and 84 manuals. Check: 84 + 56 + 84 = 224.
2. The changing tank level
A tank is 2/3 full. One quarter of its water is removed, then 18 litres are added. It is now 4/5 full. What is its capacity?
Hint
After removal, 3/4 of the original 2/3 remains.
Step-by-step solution
The remaining water occupies (3/4) × (2/3) = 1/2 of the tank. The added 18 litres correspond to 4/5 − 1/2 = 3/10 of capacity. Capacity is 18 × 10/3 = 60 litres. Check: start with 40 litres, remove 10 and add 18: 48 litres, which is 4/5 of 60.
3. Two successive expenses
Anna spends 1/3 of her money and then 2/5 of what remains. She has 18 euros left. How much did she start with?
Hint
Multiply the fractions remaining, or work backwards.
Step-by-step solution
After the first expense, 2/3 remains; after the second, 3/5 of that: (3/5) × (2/3) = 2/5 of the original money. The whole is 18 × 5/2 = 45 euros. Backwards: 18 euros is 3/5 of 30; 30 is 2/3 of 45. The expenses are 15 and 12 euros: 45 − 15 − 12 = 18.
4. A book in three days
On day one, Luca reads 2/7 of a book; on day two, he reads 3/5 of the remaining pages. On day three, he reads the last 48 pages. How long is the book, and how many pages does he read each day?
Hint
On day two he does not read 3/5 of the whole book.
Step-by-step solution
After day one, 5/7 remains. On day two he reads (3/5) × (5/7) = 3/7 of the book. Thus 2/7 remains, representing 48 pages. The book has 48 × 7/2 = 168 pages. Daily totals are 48, 72 and 48 pages, adding to 168.
5. The class and music
In a class, 3/5 of the pupils are girls. Exactly 2/3 of the girls study music: 12 girls. How many pupils and girls are there?
Hint
First reconstruct the group of girls, then the class.
Step-by-step solution
The 12 musicians are 2/3 of the girls: there are 12 × 3/2 = 18 girls. These 18 are 3/5 of the class, so there are 18 × 5/3 = 30 pupils. Directly, the girls studying music are (2/3) × (3/5) = 2/5 of the class; 12 × 5/2 = 30.
6. The final stage
A hiker covers 1/4 of a route, then 2/3 of the remaining distance. There are still 18 km to go. How long is the whole route?
Hint
After stage two, 1/3 of the previously remaining 3/4 is left.
Step-by-step solution
The fraction still to cover is (1/3) × (3/4) = 1/4. The route is 18 × 4 = 72 km. Stage one covers 18 km, leaving 54; stage two covers 36 km. Check: 18 + 36 + 18 = 72.
7. The vegetable garden
Three quarters of a plot are cultivated. Two fifths of the cultivated part form a 180 m² vegetable garden. Find the areas of the whole plot and the cultivated part.
Hint
The garden is a fraction of a fraction.
Step-by-step solution
The garden occupies (2/5) × (3/4) = 3/10 of the whole plot. The plot measures 180 × 10/3 = 600 m²; the cultivated area measures (3/4) × 600 = 450 m². Check: 2/5 of 450 is 180.
8. Two quantities, one total
Crate A contains 3/7 as many apples as crate B. There are 100 apples altogether. How many are in each crate?
Hint
Represent A by 3 parts and B by 7 parts of the same size.
Step-by-step solution
Together they contain 3 + 7 = 10 equal parts. Each part is 100/10 = 10 apples. A contains 30 and B contains 70. Check: 30/70 = 3/7. A holds 3/10 of the total, not 3/7 of the total.
9. The equalising transfer
Sack A contains 2/3 as much flour as sack B. Moving 8 kg from B to A makes their contents equal. How much was originally in each sack?
Hint
The transfer reduces the difference by 16 kg, not 8.
Step-by-step solution
A gains 8 kg while B loses 8, so the original difference was 16 kg. As a fraction of B's contents, the difference is 1 − 2/3 = 1/3. Thus B contained 16 × 3 = 48 kg and A 32 kg. After transfer, both contain 40 kg.
10. Saving part of a salary
Someone spends 3/8 of their salary on rent and 1/4 of the salary on food. They save 1/5 of what remains, which is 90 euros. Find the salary and the amount left after saving.
Hint
Rent and food refer to the same whole; savings do not.
Step-by-step solution
The first expenses use 3/8 + 1/4 = 5/8. Thus 3/8 remains. Savings represent (1/5) × (3/8) = 3/40 of salary, so salary is 90 × 40/3 = 1,200 euros. After rent and food, 450 euros remain; after saving 90, the available amount is 360.
11. Absences and early departures
One sixth of a class's enrolled pupils is absent. Then 1/4 of those present leave early. There are 30 pupils left. Find enrolment, initial absences and early departures.
Hint
The second fraction applies to those present: 5/6 of enrolment.
Step-by-step solution
The remaining pupils represent (3/4) × (5/6) = 5/8 of enrolment. Enrolment is 30 × 8/5 = 48. There were 8 absences, 40 pupils present and 10 early departures. Check: 48 − 8 − 10 = 30.
12. Resizing a recipe
Six servings require 3/4 kg of flour at a constant ratio. How much is needed for 10 servings? How many whole servings can be made with 9/8 kg?
Hint
First calculate flour for one serving.
Step-by-step solution
One serving needs (3/4)/6 = 1/8 kg. Ten servings need 10/8 = 5/4 kg. With 9/8 kg, you can make (9/8)/(1/8) = 9 whole servings. One more serving needs another 1/8 kg; do not round 9/8 up to 5/4.
13. Two cuts from one roll
From a roll, first 3/5 of its original length is cut off, then 2/7 of its original length. Twelve metres remain. How long was the roll?
Hint
Both fractions refer to the original length, so add them.
Step-by-step solution
The fraction removed is 3/5 + 2/7 = 21/35 + 10/35 = 31/35. The remaining 4/35 is 12 metres. Original length is 12 × 35/4 = 105 metres. The cuts are 63 and 30 metres; 105 − 63 − 30 = 12. Taking 2/7 of the remainder would be a different problem.
14. Two discounts, not a sum
A product is reduced by 1/5 of its original price. A further reduction of 1/4 is applied to the new price. You pay 72 euros. Find the original price and the total fractional discount.
Hint
Multiply the fractions of price retained.
Step-by-step solution
The final price is (4/5) × (3/4) = 3/5 of the original. The original price was 72 × 5/3 = 120 euros. Total discount is 1 − 3/5 = 2/5, or 48 euros. Check: 120 − 24 = 96; then 96 − 24 = 72. Do not add 1/5 and 1/4: they use different bases.
15. Equal increase and decrease?
A price increases by 1/4, then decreases by 1/4 of the increased price and becomes 90 euros. What was the original price? Has it returned to its starting value?
Hint
The multiplying factors are 5/4 and 3/4.
Step-by-step solution
The overall factor is (5/4) × (3/4) = 15/16. The original price was 90 × 16/15 = 96 euros. It rises to 120, then falls by 30 to 90. The loss is 6 euros, or 1/16 of 96. An increase and decrease by the same fraction do not cancel when the base changes.
16. Diluting twice
A container holds 30 litres of pure juice. One third of the contents is removed and replaced with water to restore 30 litres. Mix thoroughly and repeat once. How many litres of original juice remain? Assume volumes add.
Hint
Each time, 2/3 of the juice currently present remains.
Step-by-step solution
After the first operation, 30 × 2/3 = 20 litres of juice remain. Because the mixture is uniform, removing 1/3 also removes 1/3 of the juice present. This leaves 20 × 2/3 = 40/3 litres, or 13 and 1/3. The retained fraction is (2/3)² = 4/9. Water occupies 30 − 40/3 = 50/3 litres.
17. Two people working
A completes a job in 6 hours and B in 8 hours. They work together at constant, additive rates without interfering. How long does the job take?
Hint
Add the fractions of work completed in one hour, not the times.
Step-by-step solution
In one hour, A completes 1/6 and B 1/8 of the job. Together they complete 1/6 + 1/8 = 7/24 per hour. One whole job takes 1/(7/24) = 24/7 hours, or 3 and 3/7 hours. This is less than 6 hours, as required. Check: (24/7) × (7/24) = 1 job.
18. The tap and the leak
A tap would fill an empty tank in 6 hours. A leak would empty a full tank in 15 hours. Both flow rates are constant and independent of water level. How long does filling an initially empty tank take with both active?
Hint
Subtract the leak's contribution.
Step-by-step solution
Each hour, 1/6 of capacity enters and 1/15 leaves. Net increase is 1/6 − 1/15 = 1/10 of capacity per hour, so filling takes 10 hours. Check: over 10 hours, 5/3 of capacity enters and 2/3 leaves, retaining one full capacity. Constant flow rates are essential to this model.
19. The hidden number
Half of a positive number plus one third of the same number is 35. Find the number without setting up an equation with an unknown.
Hint
Both parts are fractions of the same whole.
Step-by-step solution
The sum is 1/2 + 1/3 = 5/6 of the number. If five sixths equals 35, one sixth equals 7 and six sixths equals 42. Check: half of 42 is 21, one third is 14 and 21 + 14 = 35.
20. A ratio and a difference
Two positive quantities satisfy A = (2/5) × B. Their difference is 27. Find them without using an equation.
Hint
The difference between 5 parts and 2 parts is 3 parts.
Step-by-step solution
B corresponds to 5 equal parts and A to 2. Three parts equal 27, so one part equals 9. A is 18 and B is 45. Check: 18/45 = 2/5 and 45 − 18 = 27.
21. The party pizzas
At a party, 3/8 of the guests eat half a pizza each. All the others eat 2/5 of a pizza each. The pizzas are identical and exactly 21 are eaten. How many guests are there?
Hint
Consider a representative group of 8 guests.
Step-by-step solution
Of every 8 guests, 3 eat 3/2 pizzas altogether and the other 5 eat 5 × 2/5 = 2 pizzas. Each group of 8 consumes 7/2 pizzas. Twenty-one pizzas represent 21/(7/2) = 6 groups, hence 48 guests. Check: 18 guests eat 9 pizzas and 30 eat 12; 9 + 12 = 21.
22. Three colours in an urn
In an urn, 2/5 of the balls are red, 1/3 are blue and all the others are green. There are 16 green balls. How many balls of each colour are there?
Hint
The fractions for the three colours must sum to 1.
Step-by-step solution
Red and blue make up 2/5 + 1/3 = 11/15 of the total. Green is 4/15, so the total is 16 × 15/4 = 60. There are 24 red, 20 blue and 16 green balls. Check: 24 + 20 + 16 = 60, with the required ratios to 60.
23. Adding changes the whole too
In an urn, 2/5 of the balls are red. Twelve red balls are added without removing any, and red balls now make up half the total. How many balls were there initially?
Hint
After adding, red and non-red counts are equal.
Step-by-step solution
Initially, non-red balls make up 3/5 and red balls 2/5: the difference is 1/5 of the original total. Adding 12 reds exactly fills this gap, so 1/5 equals 12 and the original total is 60. There were 24 red and 36 non-red balls; afterwards there are 36 of each, totalling 72.
24. Working backwards through donations
Marta gives away half her money plus 2 euros. Then she gives away one third of the remaining money plus 3 euros. She ends with 7 euros. How much did she have initially?
Hint
Reverse the final operation: first undo the extra 3 euros, then the removal of one third.
Step-by-step solution
Before the extra 3 euros, 7 + 3 = 10 remained: this was 2/3 of the money after the first donation. That amount was 10 × 3/2 = 15 euros. Before the extra 2 euros in the first donation, 17 remained, representing half the initial amount. She started with 34 euros. Check: donate 17 + 2 = 19, leaving 15; donate 5 + 3 = 8, leaving 7.
25. Four successive shares
A fund is shared as follows: A receives 1/3 of the total; B receives 1/4 of the remainder; C receives 2/5 of the new remainder; D receives the last 90 euros. Find the fund and each share.
Hint
D's share is the product of three remaining fractions.
Step-by-step solution
After A, 2/3 remains; after B, (3/4) × (2/3) = 1/2 of the fund remains; after C, (3/5) × (1/2) = 3/10 remains. Thus the fund is 90 × 10/3 = 300 euros. A receives 100; B receives 1/4 of 200, or 50; C receives 2/5 of 150, or 60; D receives 90. Check: 100 + 50 + 60 + 90 = 300.
Continue with a surprising property of fractions: Telescoping sums and products of fractions.