Wholes with wholes, parts with parts
A mixed number is a whole number and a fraction added together: means , and means . To add two mixed numbers, add the wholes together and add the parts together.
For , the wholes are 2 + 1 = 3 and the parts are . Put them back together: .
The 2 wholes and the 1 whole make 3 whole bars. The 1 fifth and the 3 fifths make 4 fifths of a fourth bar: .
When the parts make more than a whole
Add . The wholes are 1 + 2 = 3. The parts are , and is more than a whole.
Trade the full whole into the whole-number part. 4 of the 6 quarters make 1 whole, and 2 quarters are left, so . The 3 wholes gain that 1 and become 4: . In simplest form that is .
The 6 quarters from the two parts: quarters 1 to 4 fill a whole bar, and quarters 5 and 6 are left over.
When the parts will not subtract
Subtract . The wholes would be 3 − 1, but the parts would be , and is too small to take away from.
Break one of the 3 wholes into quarters, just as a ten is broken into ones in column subtraction. That leaves 2 wholes, and the broken whole is , which joins the to make . So , the same amount written another way.
1 1/4 is a whole and 1 pieces of 1/4: the whole is still one piece, a different size, so the pieces cannot be counted together yet
Break the whole into pieces of 1/4: 1 1/4 = ?/4
Drag the handle to break the whole into quarters. At 4 pieces it matches the below, and . The other 2 wholes stay as they are, so .
Then subtract as before
Now the parts subtract: has wholes 2 − 1 = 1 and parts . The answer is , which is in simplest form.
A second route avoids breaking a whole: write both numbers as improper fractions first. and , so , and . It is the same answer. This route is safer when the numbers are awkward.
Different denominators
When the fractions have different denominators, give them a common denominator first, exactly as for fractions on their own. For , write . The wholes are 2 + 1 = 3 and the parts are , so the answer is .
The usual mistakes
Swapping the parts round. In , it is tempting to work out and answer . But is being taken away from , not the other way round, and is too small. Break a whole instead.
Forgetting the traded whole. In the parts make , and the 1 whole from them must join the 3 in front: the answer is , not . When subtracting, the broken whole leaves the front, so 3 becomes 2.
Adding the denominators. is , not . The pieces are fifths before and after.
Worked example: Resin in Three Containers Linked by More and Less
Question A workshop keeps epoxy resin in three containers. Container A holds 316 liters of resin. Container B holds 134 liters more than Container A. Container C holds 223 liters less than Containers A and B together. During a project, a technician uses 458 liters of resin from Container B and 156 liters from Container C. (a) How much resin do Containers A and B hold together at first? (b) How much resin is left in the three containers altogether at the end of the project? Give each answer as a mixed number in its simplest form.
1.12 is a multiple of 6 and of 4, so write the parts in twelfths: 316 = 3212 and 134 = 1912. Container B holds 3212 + 1912 = 41112 liters.
In twelfths, 316 = 3212 and 134 = 1912, so B holds 3212 + 1912 = 41112 liters. 2.(a) Containers A and B hold 3212 + 41112 = 71312 liters together. Twelve twelfths make one liter, so this is 8112 liters.
(a) A and B hold 3212 + 41112 = 71312 = 8112 liters together. 3.Container C holds 223 = 2812 liters less. 112 is too small to take 812 from, so change one of the 8 liters into twelfths: 8112 = 71312, and C holds 71312 − 2812 = 5512 liters.
C is 2812 liters short of A and B together: 71312 − 2812 = 5512 liters. 4.At first the three containers hold 8112 + 5512 = 13612 = 1312 liters.
At first the three containers hold 8112 + 5512 = 1312 liters. 5.24 is a multiple of 8 and of 6, so the resin used is 41524 + 12024 = 53524 = 61124 liters.
In twenty-fourths, the resin used from B and C is 41524 + 12024 = 61124 liters. 6.(b) Write 1312 as 131224: 131224 − 61124 = 7124 liters are left. Check, container by container: A still holds 3424, B keeps 42224 − 41524 = 724 and C keeps 51024 − 12024 = 31424, and 3424 + 724 + 31424 = 62524 = 7124.
(b) 131224 − 61124 = 7124 liters are left: 3424 in A, 724 in B and 31424 in C.
Answer: (a) 8112 liters; (b) 7124 liters
Common mistakes
- Working out 8112 − 2812 as 6712 by taking the smaller numerator from the larger one. The first fraction is the smaller, so one whole liter is changed into 1212 first: 8112 = 71312, which gives 5512.
- Taking Container C as 223 liters less than Container B alone, which gives 214 liters. The question compares C with Containers A and B together, which is 8112 liters.
More adding and subtracting fractions problems, worked step by step →