Simplify: 3 - 1 1 4 + 2 3

Given, to simplify 3 - 1 1 4 + 2 3 We should convert the mixed fraction to improper fraction, we get = 3 - 1 1 4 + 2 3 = 3 - 5 4 + 2 3 = 3 1 - 5 4 + 2 3 = 36 - 15 + 8 12 = 29 12 Therefore, the solution is 29 12 .

Find the product of 2 1 7 and 3 1 5 .

Given, to find the product of 2 1 7 and 3 1 5 , The given fractions are in mixed fractions, we should convert them into improper fractions. When multiplying fractions is to multiply the two numerators. Then multiply the two denominators. Finally, simplify the new fractions. The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator. = 2 1 7 × 3 1 5 = 15 7 × 16 5 = 3 7 × 16 = 48 7 The obtained result is in improper fraction, we should convert into mixed fraction, we get 48 7 = 6 6 7 Therefore, the resultant value is 6 6 7 .

Divide 5 1 2 by 3 1 4 .

Given, to divide 5 1 2 by 3 1 4 , To dividing fractions is to find the reciprocal (reverse the numerator and denominator) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. Finally, simplify the fractions if needed. The given fractions are in mixed fractions, we should convert them into improper fraction in order to divide, = 5 1 2 by 3 1 4 = 11 2 ÷ 13 4 = 11 2 × 4 13 = 22 13 By converting the resultant into mixed fraction, we get 22 13 = 1 9 13 Therefore, 1 9 13 is the product of given fractions.

Shalini bought two pieces of ribbon measuring 7 1 6 m and 6 3 4 m respectively. She used 11 2 3 m of it on her doll. How much ribbon is left with her?

Given, Two pieces of ribbon bought by Shalini which measures 7 1 6 m and 6 3 4 m , Length of total ribbon shalini had = 7 1 6 m + 6 3 4 m = 43 6 + 27 4 = 86 + 81 12 = 167 12 She used 11 2 3 part for her doll, i.e. 35 3 = 167 12 - 35 3 = 167 - 140 12 = 27 12 After simplifying, we get, 9 4 Therefore, 9 4 th part of ribbon is left.

Arrange the following fractions in descending order : 7 11 , 2 11 , 5 11 , 1 11 , 9 11

Given fractions are 7 11 , 2 11 , 5 11 , 1 11 , 9 11 As we know if two fractions have the same denominator, then the fraction with the smallest numerator is the smallest fraction. Here the smallest numerator is 1 , then 1 11 is the smallest fraction . After that, 2 is the smallest one , then 2 11 is the second smallest fraction . After that 5 is the smallest one , then 5 11 is the third smallest fraction. Hence, the given fractions can be written in ascending order as: 7 11 , 2 11 , 5 11 , 1 11 , 9 11

Arrange the following fractions in descending order : 3 7 , 1 7 , 5 7 , 2 7 , 6 7

Given fractions are 3 7 , 1 7 , 5 7 , 2 7 , 6 7 As we know if two fractions have the same denominator, then the fraction with the smallest numerator is the smallest fraction. Here the smallest numerator is 1 , then 1 7 is the smallest fraction . After that, 2 is the smallest one , then 2 7 is the second smallest fraction . After that 3 is the smallest one , then 3 7 is the third smallest fraction. Hence, the given fractions can be written in ascending order as: 3 7 , 1 7 , 5 7 , 2 7 , 6 7

Arrange the following fractions in descending order : 1 6 , 5 6 , 2 6 , 3 6 , 4 6

Given fractions are 1 6 , 5 6 , 2 6 , 3 6 , 4 6 As we know if two fractions have the same denominator, then the fraction with the smallest numerator is the smallest fraction. Here the smallest numerator is 1 , then 1 6 is the smallest fraction . After that, 2 is the smallest one , then 2 6 is the second smallest fraction . After that 3 is the smallest one , then 3 6 is the third smallest fraction. Hence, the given fractions can be written in ascending order as: 1 6 , 5 6 , 2 6 , 3 6 , 4 6

E M B I B E

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES

Dr. Sanjeev Verma and R.K. Tyagi Solutions for Chapter: Fractions, Exercise 13: CHAPTER-END EXERCISES

Author:Dr. Sanjeev Verma & R.K. Tyagi

Dr. Sanjeev Verma Mathematics Solutions for Exercise - Dr. Sanjeev Verma and R.K. Tyagi Solutions for Chapter: Fractions, Exercise 13: CHAPTER-END EXERCISES

Attempt the practice questions on Chapter 7: Fractions, Exercise 13: CHAPTER-END EXERCISES with hints and solutions to strengthen your understanding. Excellence in Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Dr. Sanjeev Verma and R.K. Tyagi Solutions for Chapter: Fractions, Exercise 13: CHAPTER-END EXERCISES with Hints & Solutions

MEDIUM

6th CBSE

IMPORTANT

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 8

Simplify: $3 - 1 \frac{1}{4} + \frac{2}{3}$

MEDIUM

6th CBSE

IMPORTANT

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 9

Find the product of $2 \frac{1}{7} and 3 \frac{1}{5}$ .

HARD

6th CBSE

IMPORTANT

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 10

Divide $5 \frac{1}{2} by 3 \frac{1}{4} .$

MEDIUM

6th CBSE

IMPORTANT

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 11

Shalini bought two pieces of ribbon measuring $7 \frac{1}{6} m and 6 \frac{3}{4} m$ respectively. She used $11 \frac{2}{3} m$ of it on her doll. How much ribbon is left with her?