The process of subtracting units of length is exactly similar to subtracting ordinary numbers.
Learn how the length values for length subtraction are arranged in different columns.
1. Subtract 12 m 36 cm from 48 m 57 cm
Solution:
Case 1:
Minuend and subtrahend are both converted into smaller units.
12 m 36 cm = (12 × 100) cm + 36 cm = (1200 + 36) cm = 1236 cm
48 m 57 cm = (48 × 100) cm + 57 cm = (4800 + 57) cm = 4857 cm
Now subtract
48m 57cm = 4857 cm
– 12 m 36 cm = – 1236 cm
3621 cm
= 3600 cm + 21 cm
= 36 m 21 cm
Therefore 48 m 57 cm – 12 m 36 cm = 36 m 21 cm
Case 2:
Since the minuend is larger than the subtrahend, the minuend is placed above the subtrahend. Then m and cm are placed in different columns.
Now subtract
m cm
48 57
– 12 36
36 21
(i) Deduction of cm, 57 cm – 36 cm = 21 cm.
It is placed under the cm column.
(ii) Subtract m, 48m – 12m = 36m.
It is placed under the column m.
So difference = 36 m 21 cm
2. Subtract 37 m 50 cm from 53 m 30 cm.
Solution:
Case 1:
Minuend and subtrahend are both converted into smaller units.
53 m 30 cm = (53 × 100) cm + 33 cm = (5300 + 30) cm = 5330 cm
37 m 50 cm = (37 × 100) cm + 50 cm = (3700 + 50) cm = 3750 cm
Now subtract
5330 cm
– 3750 cm
1580 cm
= 1500 cm + 80 cm
= 15 m 80 cm
Therefore 53 m 30 cm – 37 m 50 cm = 15 m 80 cm.
Case 2:
Since the minuend is larger than the subtrahend, the minuend is placed above the subtrahend. Then m and cm are placed in different columns. Minuend 53 m 30 cm is placed above the subtrahend 37 m 50 cm.
m cm
1 100
53 30
– 37 50
15 80
(i) 50cm> 30cm. So 50 cm cannot be deducted from 30 cm.
1m or 100cm is borrowed from 53m, leaving 52m, which makes 30cm to 130cm.
Now 130 – 50 = 80. It is placed under the cm column.
(ii) Now in the m column 53 – 1 = 52 m. 52 m – 37 m = 15 m.
It is placed under the column m.
So the difference is 15 m 80 cm.
3. Subtract 56 m 65 cm from 62 m 7 cm.
Solution: Let’s subtract. Step I: Arrange the numbers vertically. Step II: Write the lengths to be subtracted in m and cm as shown. Step III: First subtract inches from the right, then subtract the meters. |
So 62 m 7 cm – 56 m 65 cm = 5 m 42 cm
4th Subtract 37 m 6 cm from 70 m.
Solution:
Case 1:
We have, 70 m – 37 m 6 cm
Minuend and subtrahend are both converted into smaller units.
70 m = (70 × 100) cm = 7000 cm
37 m 6 cm = (37 × 100) cm + 6 cm = (3700 + 6) cm = 3706 cm
Now subtract
7000 cm
– 3706 cm
3294 cm
= 3200 cm + 94 cm
= 32 m 94 cm²
Therefore 70 m – 37 m 6 cm = 32 m 94 cm.
Case 2:
Since the minuend is larger than the subtrahend, the minuend is placed above the subtrahend. Then m and cm are placed in different columns. Minuend 70 m 00 cm is placed above the subtrahend 37 m 06 cm.
m cm
1 100
70 00
– 37 06
32 94
(i) 06cm> 0cm. 6 cm cannot be deducted from 0 cm.
1m or 100cm is borrowed from 70m, leaving 69m in the m column.
(ii) Now 100 cm – 06 cm = 94 cm. It is placed under the cm column.
(iii) Now in the m column 69 m – 37 m = 32 m.
It is placed under the column m.
So the difference is 32 m 94 cm.
5. Subtract 45 km 282 m from 63 km 70 m.
Solution:
Case 1:
Minuend and subtrahend are both converted into smaller units (conversion method).
63 km 70 m = (63 × 1000) m + 70 m = (63000 + 70) m = 63070 m
45 km 282 m = (45 × 1000) m + 282 m = (45000 + 282) m = 45282 m
Now subtract
63070 m
– 45282 m
17788 m
= 17000 m + 788 m
= 17 km 788 m²
So 63 km 70 m – 45 km 282 m = 17 km 788 m.
Case 2:
Since the minuend is larger than the subtrahend, the minuend is placed above the subtrahend. Then km and m are arranged in different columns. Minuend 63 km 70 m is about subtrahend 45 km 282 m (without conversion).
km m
1 1000
63 070
– 45 282
17 788
(i) 282 m> 70 m. 282 m cannot therefore be deducted from 70 m.
1 km or 1000 m is borrowed from 63 km, leaving 62 km, which results in 70 m over 1070 m (since 1 km = 1000 m and 70 m = 1070 m).
Now 1070 m – 282 m = 788 m. It is placed under the column m.
(ii) Now in the km column 63 – 1 = 62 km. 62 km – 45 km = 17 km.
It is placed under the km column.
The difference is 17 km 788 m.
6th Subtract 75 km 345 m from 200 km 20 m.
Solution:
Since the minuend is larger than the subtrahend, the minuend is placed above the subtrahend. Then km and m are arranged in different columns. Minuend 200 km 20 m is above the subtrahend 75 km 345 m.
km m
1 1000
200 020
– 275345
124675
(i) 345 m> 20 m. 345 m cannot therefore be deducted from 20 m.
1 km or 1000 m are borrowed from 200 km with 199 km remaining, which makes 020 m to 1020 m.
Now 1020 m – 345 m = 675 m. It is placed under the column m.
(ii) Now in the km column 200 – 1 = 199 km. 199 km – 75 km = 124 km.
It is placed under the km column.
So the difference is 124 km 675 m.
7th Subtract 8 m 7 dm 5 cm from 26 m 4 dm 8 cm.
Solution:
Since the minuend is larger than the subtrahend, the minuend is placed above the subtrahend. Then m, dm and cm are placed in different columns. Minuend 26 m 4 dm 8 cm is placed over the subtrahend 8 m 7 dm 5 cm.
m dmcm
1 10
26 4 8
– 8 7 5
17 7 3
(i) 8cm – 5cm = 3cm. It is placed under the cm column.
(ii) 7 dm> 4 dm. Therefore 7 dm cannot be subtracted from 4 dm.
1 m or 10 dm is borrowed from 26 m, with 25 m left over, making 4 dm to 14 dm.
Now 14 dm – 7 dm = 7 dm. It is placed under the column dm.
(iii) Now in the m column 26 – 1 = 25 m. 25m – 8m = 17m.
It is placed under the column m.
Thus the difference is 17 m 7 dm 3 cm.
Questions and answers about length subtraction:
1. Subtract the following:
(i) 81m 9cm – 52m 52cm
(ii) 28m 98cm – 16m 20cm
(iii) 352m 917cm – 148m 79cm
(iv) 938 m 33 cm – 619 m 57 cm
(v) 394 m 68 cm – 45 m 79 cm
(vi) 180 m 90 cm – 42 m 33 cm
Reply:
1. (i) 28 m 57 cm
(ii) 12 m 78 cm
(iii) 203 m 38 cm
(iv) 318 m 76 cm
(v) 348 m 89 cm
(vi) 138 m 57 cm
2. Subtract the following:
(i) 81 m 09 cm from 145 m 56 cm
(ii) 14 m 82 cm from 67 m 40 cm
(iii) 174 m 259 cm from 512 m 36 cm
(iv) 79 m 56 cm from 140 m 23 cm
(v) 180 m 90 cm from 287 m 46 cm
Reply:
2. (i) 64 m 47 cm
(ii) 52 m 58 cm
(iii) 338 m 11 cm
(iv) 60 m 67 cm
(v) 106 m 56 cm
● Related concepts
● Standard length unit
● Conversion of the standard length unit
● Add length
Math worksheets for 3rd grade
3rd grade math class
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