What should be added to 4 3 5 to get 15 ?

To find the number to be added to 4 3 5 to get 15 Let the number that needs to be added be x Then, according to question, 4 3 5 + x = 15 ⇒ 20 + 3 5 + x = 15 1 ⇒ 23 5 + x = 15 1 Rearranging we get ⇒ x = 15 1 - 23 5 ⇒ x = ( 15 × 5 ) - 23 5 ⇒ x = 75 - 23 5 = 52 5 Hence, the number to be added to 4 3 5 to get 15 is 52 5 .

Simplify: 2 3 + 5 6 - 1 9

To Simplify: 2 3 + 5 6 - 1 9 Taking the LCM of denominators, we get 3 = 3 × 1 6 = 2 × 3 9 = 3 × 3 Since the maximum number of times the prime factor 3 occurs is two; this happens for 9 and the maximum number of times the prime factor 2 occurs is one. ∴ LCM 3 , 6 , 9 = 3 × 3 × 2 = 18 Now changing the fractions to the like fractions and then adding or subtracting the numerator values we get 2 3 + 5 6 - 1 9 = 2 × 6 3 × 6 + 5 × 3 6 × 3 - 1 × 2 9 × 2 = 12 18 + 15 18 - 2 18 = 12 + 15 - 2 18 = 25 18 Hence, 2 3 + 5 6 - 1 9 = 25 18 .

Simplify 8 5 6 - 3 3 8 + 1 7 12

Given, to simplify, 8 5 6 - 3 3 8 + 1 7 12 Converting the mixed fractions to improper fractions, we get 8 5 6 - 3 3 8 + 1 7 12 = 6 × 8 + 5 6 - 8 × 3 + 3 8 + 1 × 12 + 7 12 = 53 6 - 27 8 + 19 12 Now, taking the LCM of denominators and following the process of changing the unlike fractions to the like fractions, we get 6 = 2 × 3 8 = 2 × 2 × 2 12 = 2 × 2 × 3 Since the maximum number of times the prime factor 2 occurs is three; this happens for 8 and prime factor 3 occurs is one. ∴ LCM ( 6 , 8 , 12 ) = 2 × 2 × 2 × 3 = 24 Now changing the fractions to the like fractions and then adding or subtracting the numerator values we get = 53 6 - 27 8 + 19 12 = 53 × 4 6 × 4 - 27 × 3 8 × 3 + 19 × 2 12 × 2 = 212 24 - 81 24 + 38 24 = 212 - 81 + 38 24 = 169 24 Hence, 8 5 6 - 3 3 8 + 1 7 12 = 169 24 .

Find the difference 5 6 - 3 4

Given, 5 6 - 3 4 Since the fractions are unlike fractions. So, firstly, we need to take the LCM of the denominators, We get 6 = 2 × 3 4 = 2 × 2 Since the maximum number of times the prime factor 2 occurs is two; this happens for 4 and the prime factor 3 occurs is one. ∴ LCM ( 6 , 4 ) = 2 × 2 × 3 = 12 Now changing the fractions to the like fractions, and then subtracting the numerator values keeping the denominator same, we have 5 6 - 3 4 = 5 × 2 6 × 2 - 3 × 3 4 × 3 = 10 12 - 9 12 = 1 12 Hence, 5 6 - 3 4 = 1 12 .

A piece of wire 3 3 4 m long broke into two pieces. One piece is 1 1 2 m long. How long is the other piece ?

Given, a piece of wire 3 3 4 m long broke into two pieces and one piece is 1 1 2 m long. Let, the length of the other piece of wire be x m. According to question, 1 1 2 + x = 3 3 4 ⇒ 3 2 + x = 15 4 Rearranging and taking LCM, we get x = 15 4 - 3 2 ⇒ x = 15 - ( 3 × 2 ) 4 = 15 - 6 4 ⇒ x = 9 4 m Hence, the length of the other piece is 9 4 m .

Of 3 4 and 5 7 , which is greater and by how much?

Let us compare 3 4 and 5 7 . By cross multiplying: 3 × 7 = 21 and 4 × 5 = 20 Clearly, 21 > 20 ∴ 3 4 > 5 7 Their difference: 3 4 - 5 7 = 21 - 20 28 [ 28 ÷ 4 = 7 , 7 × 3 = 21 ] and [ 28 ÷ 7 = 4 , 4 × 5 = 20 ] = 1 28 Hence, 3 4 is greater than 5 7 by 1 28 .

The sum of two rational numbers is 11 10 . If one of the rational number is 3 4 , find the other number.

Given, the sum of two rational numbers is 11 10 . If one of the rational number is 3 4 , we have to find the other number. Let the other number be x According to the question, 3 4 + x = 11 10 Rearranging, we get ⇒ x = 11 10 - 3 4 Taking LCM of denominators and simplifying we get 10 = 2 × 5 4 = 2 × 2 ∴ LCM ( 10 , 4 ) = 2 × 2 × 5 = 20 ⇒ x = ( 11 × 2 ) - ( 3 × 5 ) 20 = 22 - 15 20 = 7 20 Hence, the required number is 7 20 .

Find the difference 3 3 10 - 1 7 15

Given, 3 3 10 - 1 7 15 Converting the mixed fractions into improper fractions we get 3 3 10 - 1 7 15 = 3 × 10 + 3 10 - 1 × 15 + 7 15 = 33 10 - 22 15 Now, we need to take the LCM of the denominators, we get 10 = 2 × 5 15 = 3 × 5 Since the maximum number of times the prime factor 2 , 3 , and 5 occurs is one. ∴ LCM ( 10 , 15 ) = 2 × 3 × 5 = 30 Now changing the fractions to the like fractions, and then subtracting the numerator values keeping the denominator same, we have 33 10 - 22 15 = 33 × 3 30 - 22 × 2 30 = 99 30 - 44 30 = 99 - 44 30 = 55 30 = 11 6 Hence, 3 3 10 - 1 7 15 = 11 6

Complete the addition-subtraction box. Addition → Subtraction ↓ 2 3 4 3 ___ 1 3 2 3 ____ ____ ____ ____

From the given figure, 2 3 + 4 3 = 2 + 4 3 = 6 3 = 2 1 3 + 2 3 = 1 + 2 3 = 3 3 = 1 Addition → Subtraction ↓ 2 3 4 3 2 1 3 2 3 1 2 3 - 1 3 = 2 - 1 3 = 1 3 4 3 - 2 3 = 4 - 2 3 = 2 3 Addition → Subtraction ↓ 2 3 4 3 2 1 3 2 3 1 1 3 2 3 Now, 2 - 1 = 1 Also, 1 3 + 2 3 = 1 + 2 3 = 1 Addition → Subtraction ↓ 2 3 4 3 2 1 3 2 3 1 1 3 2 3 1 Hence, the completed box is given as above.

Given, 4 7 - 2 7 Since the denominators are same, we can directly subtract the numerators keeping the denominator common. We get ⇒ 4 7 - 2 7 = 4 - 2 7 = 2 7 Hence, 4 7 - 2 7 = 2 7

Find the difference: 7 - 4 2 3

Given, to find the difference: 7 - 4 2 3 First convert the mixed fraction 4 2 3 to improper fraction and simplify it, 7 - 4 2 3 = 7 - 3 × 4 + 2 3 = 7 1 - 14 3 On simplifying, we get 7 1 - 14 3 = 7 × 3 - 14 3 = 21 - 14 3 = 7 3 Hence, 7 - 4 2 3 = 7 3 .