We have, 4 2 3 + 3 1 4 = 14 3 + 13 4 = 14 × 4 3 × 4 + 13 × 3 4 × 3 [LCM of the denominators 3 and 4 is 12 ] = 56 12 + 39 12 = 56 + 39 12 ∵ a p + b p = a + b p = 97 12 = 8 1 12 . Therefore, the required answer is 8 1 12 .

We have, 16 5 − 7 5 = 16 - 7 5 ∵ a p - b p = a - b p = 9 5 = 1 4 5 . Therefore, the required answer is 1 4 5 .

We have, 4 3 − 1 2 = 4 × 2 3 × 2 − 1 × 3 2 × 3 [LCM of the denominators 3 and 2 is 6 ] = 8 6 - 3 6 = 8 - 3 6 ∵ a p - b p = a - b p = 5 6 . Therefore, the required answer is 5 6 .

Sarita bought 2 5 metre of ribbon and Lalita 3 4 metre of ribbon. What is the total length of the ribbon they bought?

The total length of the ribbon = 2 5 + 3 4 m = 2 × 4 5 × 4 + 3 × 5 4 × 5 m [LCM of the denominators 5 and 4 is 20 ] = 8 20 + 15 20 m = = 8 + 15 20 m ∵ a p + b p = a + b p = 23 20 m = 1 3 20 m . Therefore, the total length of the ribbon they bought is 1 3 20 m .

Naina was given 1 1 2 piece of cake and Najma was given 1 1 3 piece of cake. Find the total amount of cake given to both of them.

The total amount of cake given to both of them = 1 1 2 + 1 1 3 pieces = 3 2 + 4 3 pieces = 3 × 3 2 × 3 + 4 × 2 3 × 2 pieces [LCM of the denominators 2 and 3 is 6 ] = 9 6 + 8 6 pieces = 9 + 8 6 pieces ∵ a p + b p = a + b p = 17 6 pieces = 2 5 6 pieces. Therefore, the total amount of cake given to both of them is 2 5 6 pieces.

Fill in the box. □ − 5 8 = 1 4

The number in the box will be obtained by adding 5 8 and 1 4 . We have, 5 8 + 1 4 = 5 8 + 1 × 2 4 × 2 [LCM of the denominators 8 and 4 is 8 ] = 5 8 + 2 8 = 5 + 2 8 ∵ a p + b p = a + b p = 7 8 . Therefore, the required number in the box is 7 8 .

Fill in the box. □ − 1 5 = 1 2

The number in the box will be obtained by adding 1 5 and 1 2 . We have, 1 5 + 1 2 = 1 × 2 5 × 2 + 1 × 5 2 × 5 [LCM of the denominators 5 and 2 is 10 ] = 2 10 + 5 10 = 2 + 5 10 ∵ a p + b p = a + b p = 7 10 . Therefore, the required number in the box is 7 10 .

Fill in the box. 1 2 − □ = 1 6

The number in the box will be obtained by subtracting 1 6 from 1 2 . We have, 1 2 - 1 6 = 1 × 3 2 × 3 − 1 6 [LCM of the denominators 2 and 6 is 6 ] = 3 6 − 1 6 = 3 - 1 6 ∵ a p - b p = a - b p = 2 6 = 2 ÷ 2 6 ÷ 2 [Divide both the numerator and the denominator by their HCF] = 1 3 . Therefore, the required number in the box is 1 3 .