Subtraction of Length

Learn how the values of length are arranged in different columns for the subtraction of length.

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

48 m 57 cm       =          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:

As minuend is greater than subtrahend, the minuend is placed above the subtrahend. Then m and cm are arranged in different columns.

Now subtract

m   cm
48   57

- 12   36
36   21

(i) Subtracting cm, 57 cm - 36 cm = 21 cm.
It is placed under cm column.

(ii) Subtracting m, 48 m - 12 m = 36 m.
It is placed under m column.

Hence, 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:

As minuend is greater than subtrahend, the minuend is placed above the subtrahend. Then m and cm are arranged in different columns. Minuend 53 m 30 cm is placed above subtrahend 37 m 50 cm.

m     cm
1     100
53     30

-  37     50
15     80

(i) 50 cm > 30 cm. So, 50 cm cannot be subtracted from 30 cm.
1 m or 100 cm is borrowed from 53 m leaving 52 m making 30 cm to 130 cm.
Now 130 - 50 = 80. It is placed under cm column.

(ii) Now in the m column, 53 - 1 = 52 m. 52 m - 37 m = 15 m.
It is placed under m column.

Hence, the difference is 15 m 80 cm.

3. 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 cm

Therefore, 70 m - 37 m 6 cm = 32 m 94 cm.

Case 2:

As minuend is greater than subtrahend, the minuend is placed above the subtrahend. Then m and cm are arranged in different columns. Minuend 70 m 00 cm is placed above subtrahend 37 m 06 cm.

m     cm
1     100
70     00

-  37     06

32     94

(i) 06  cm > 0 cm. So, 6 cm cannot be subtracted from 0 cm.
1 m or 100 cm is borrowed from 70 m leaving 69 m in m column.

(ii) Now 100 cm - 06 cm = 94 cm. It is placed under cm column.

(iii) Now in the m column, 69 m - 37 m = 32 m.
It is placed under m column.

Hence, the difference is 32 m 94 cm.

4. 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 m

Therefore, 63 km 70 m - 45 km 282 m = 17 km 788 m.

Case 2:

As minuend is greater than subtrahend, the minuend is placed above the subtrahend. Then km and m are arranged in different columns. Minuend 63 km 70 m is placed above subtrahend 45 km 282 m (without conversion).

km      m
1     1000
63     070

-  45     282
17     788

(i) 282 m > 70 m. So, 282 m cannot be subtracted from 70 m.
1 km or 1000 m is borrowed from 63 km leaving 62 km making 70 m to 1070 m (Since, 1 km = 1000 m and 70 m = 1070 m).
Now 1070 m - 282 m = 788 m. It is placed under m column.

(ii) Now in the km column, 63 - 1 = 62 km. 62 km - 45 km = 17 km.
It is placed under km column.

Hence, the difference is 17 km 788 m.

5. Subtract 75 km 345 m from 200 km 20 m.

Solution:

As minuend is greater than subtrahend, the minuend is placed above the subtrahend. Then km and m are arranged in different columns. Minuend 200 km 20 m is placed above subtrahend 75 km 345 m.

km      m
1    1000
200    020

-  275    345
124    675

(i) 345 m > 20 m. So, 345 m cannot be subtracted from 20 m.
1 km or 1000 m is borrowed from 200 km leaving 199 km making 020 m to 1020 m.
Now 1020 m - 345 m = 675 m. It is placed under m column.

(ii) Now in the km column, 200 - 1 = 199 km. 199 km - 75 km = 124 km.
It is placed under km column.

Hence, the difference is 124 km 675 m.

6. Subtract 8 m 7 dm 5 cm from 26 m 4 dm 8 cm.

Solution:

As minuend is greater than subtrahend, the minuend is placed above the subtrahend. Then m, dm and cm are arranged in different columns. Minuend 26 m 4 dm 8 cm is placed above subtrahend 8 m 7 dm 5 cm.

m    dm    cm
1      10
26     4      8

-    8     7      5
17     7      3

(i) 8 cm - 5 cm = 3 cm. It is placed under cm column.

(ii) 7 dm > 4 dm. So, 7 dm cannot be subtracted from 4 dm.
1 m or 10 dm is borrowed from 26 m leaving 25 m making 4 dm to 14 dm.
Now 14 dm - 7 dm = 7 dm. It is placed under dm column.

(iii) Now in the m column, 26 - 1 = 25 m. 25 m - 8 m = 17 m.
It is placed under m column.

Hence, the difference is 17 m 7 dm 3 cm.

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