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.

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

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

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