Multiply 3 7 ÷ 21 and write the solution in mixed fraction.

Given, 3 7 ÷ 21 Take the reciprocal of the divisor → 1 21 Multiply the dividend and the reciprocal of the divisor. That is, 3 7 × 1 21 = 3 1 147 49 = 1 49 Therefore, the required answer is 1 49 .

The product of two numbers is 15 5 6 . If one of the numbers is 6 1 3 , find the other.

Given, product of the two numbers = 15 5 6 = 95 6 One of the numbers = 6 1 3 = 19 3 Let other number be x ⇒ 19 3 × x = 95 6 On transposing 19 3 to R.H.S, we get, The other number, x = 95 6 ÷ 19 3 To divide two fractions, we need to multiply the first fraction with the reciprocal of second fraction. x = 95 6 × 3 19 Since, we know that 19 × 5 = 95 and 3 × 2 = 6 , then simplifying the fraction to the lowest form, we get x = 5 2 We can convert the improper fraction to mixed fraction by dividing the numerator with denominator. 2 2 5 - 4 1 Therefore, the other number is 2 1 2

Given, 4 5 ÷ 10 1 Take the reciprocal of the divisor → 1 10 Multiply the dividend and the reciprocal of the divisor. That is, 4 5 × 1 10 = 4 2 50 25 = 2 25 Therefore, the required answer is 2 25 .

Given, 3 8 ÷ 8 Take the reciprocal of the divisor → 1 8 Multiply the dividend and the reciprocal of the divisor. That is, 3 8 × 1 8 = 3 64 Therefore, the required answer is 3 64 .

Given, 3 1 2 ÷ 3 14 or 7 2 ÷ 3 14 Take the reciprocal of the divisor → 14 3 Multiply the dividend and the reciprocal of the divisor. That is, 7 2 × 14 3 = 98 49 6 3 = 49 3 = 16 1 3 Therefore, the required answer is 16 1 3 .

Given, 1 3 ÷ 12 Take the reciprocal of the divisor → 1 12 Multiply the dividend and the reciprocal of the divisor. That is, 1 3 × 1 12 = 1 36 Therefore, the required answer is 1 36 .

Vikas can cover a distance of 20 2 3 km in 7 3 4 hours on foot. How many km per hour does he walk?

Given, The distance covered by Vikas in 7 3 4 hours = 20 2 3 km Conversion of mixed fraction to improper fraction is given by Denominator × Quotient + Remainder Denominator Therefore, 20 2 3 = 62 3 and 7 3 4 = 31 4 So, the distance covered by him in 1 hour = 20 2 3 ÷ 7 3 4 km = 62 3 ÷ 31 4 km = 62 3 × 4 31 km = 2 × 4 3 km = 8 3 km Conversion of improper fraction into mixed fraction is given by Quotient Remainder Denominator . If we divide 8 ÷ 3 , we get Quotient = 2 Remainder = 2 Therefore, 8 3 km = 2 2 3 km Therefore, the distance covered by Vikas per hour is 2 2 3 km .

Given, 2 5 ÷ 2 Take the reciprocal of the divisor → 1 2 Multiply the dividend and the reciprocal of the divisor. That is, 2 5 × 1 2 = 2 1 10 5 = 1 5 Therefore, the required answer is 1 5 .

Given to divide 3 6 by 1 3 3 6 ÷ 1 3 Take the reciprocal of the divisor → 3 1 Multiply the dividend and the reciprocal of the divisor. That is, 3 6 × 3 1 = 3 1 6 2 = 1 2 Therefore, the required answer is 1 2 .

By what number should 6 2 9 be divided to obtain 4 2 3 ?

From the given, Required number = 6 2 9 ÷ 4 2 3 = 56 9 ÷ 14 3 = 56 9 × 3 14 [ ∵ Reciprocal of 14 3 = 3 14 ] = 4 3 = 1 1 3 Therefore, we have to divide 6 2 9 by 1 1 3 to get 4 2 3 .

Given, 8 1 2 by 2 5 or 8 1 2 ÷ 2 5 or 17 2 ÷ 2 5 Take the reciprocal of the divisor → 5 2 Multiply the dividend and the reciprocal of the divisor. That is, 17 2 × 5 2 = 85 4 = 21 × 4 + 1 4 = 21 1 4 Therefore, the required answer is 21 1 4 .

Mangoes are sold at ₹ 43 1 2 per kg. What is the weight of mangoes available for ₹ 326 1 4 ?

Given: Selling price of 1 kg mango = ₹ 43 1 2 Also, total amount available = ₹ 326 1 4 Conversion of mixed fraction to improper fraction is given by Denominator × Quotient + Remainder Denominator Therefore, 43 1 2 = 87 2 and 326 1 4 = 1305 4 So, weight of mangoes at ₹ 1305 4 = 1305 4 ÷ 87 2 ⇒ 1305 4 × 2 87 = 435 × 1 2 × 29 = 435 58 = 15 2 Conversion of improper fraction into mixed fraction is given by Quotient Remainder Denominator . If we divide 15 ÷ 2 , we get Quotient = 7 Remainder = 1 Therefore, 15 2 = 7 1 2 Therefore, the weight of mangoes available for ₹ 326 1 4 is 7 1 2 kg .

Given, 1 2 ÷ 7 Take the reciprocal of the divisor → 1 7 Multiply the dividend and the reciprocal of the divisor. That is, 1 2 × 1 7 = 1 14 Therefore, the required answer is 1 14 .

Given, 9 7 ÷ 3 Take the reciprocal of the divisor → 1 3 Multiply the dividend and the reciprocal of the divisor. That is, 9 7 × 1 3 = 9 3 21 7 = 3 7 Therefore, the required answer is 3 7 .

Find the reciprocal of the following fraction 3 7 . Classify the reciprocal as proper fraction, improper fraction or whole number.

Given fraction is 3 7 Reciprocal of 3 7 is 7 3 . Since, Numerator is greater than the denominator, 7 3 is an improper fraction Hence, the reciprocal of the given fraction is an Improper fraction.

18 boxes of nails weigh equally and their total weight is 49 1 2 kg . How much does each box weigh?

Given: Weight of 18 boxes of nails = 49 1 2 kg Conversion of mixed fraction to improper fraction is given by Denominator × Quotient + Remainder Denominator Therefore, 49 1 2 = 99 2 So, Weight of 1 box = 99 2 ÷ 18 kg = 99 2 × 1 18 kg [∵ Reciprocal of 18 = 1 18 ] = 99 × 1 2 × 18 kg = 11 × 1 2 × 2 kg = 11 4 kg Conversion of improper fraction into mixed fraction is given by Quotient Remainder Denominator . If we divide 11 ÷ 4 , we get Quotient = 2 Remainder = 3 Therefore, 11 4 = 2 3 4 Therefore, the weight of each box is 2 3 4 kg .