Write the following in the simplest form : 27 64 , 121 11011 , 5 15 , 42 49 , 33 77

Given fractions are 27 64 , 121 11011 , 5 15 , 42 49 , 33 77 In 27 64 , there is no common factor between numerator and denominator except 1 . So, the simplest form of 27 64 is 27 64 . ⇒ 121 11011 = 121 ÷ 121 11011 ÷ 121 = 1 91 ⇒ 5 15 = 5 ÷ 5 15 ÷ 5 = 1 3 ⇒ 42 49 = 42 ÷ 7 49 ÷ 7 = 6 7 ⇒ 33 77 = 33 ÷ 3 77 ÷ 7 = 3 7 Therefore, the simplest form of given fraction 27 64 , 1 91 , 1 3 , 6 7 , 3 7

Write the following in the simplest form : 27 64 , 121 11011 , 5 15 , 42 49 , 33 77

Given fractions are 27 64 , 121 11011 , 5 15 , 42 49 , 33 77 In 27 64 , there is no common factor between numerator and denominator except 1 . So, the simplest form of 27 64 is 27 64 . ⇒ 121 11011 = 121 ÷ 121 11011 ÷ 121 = 1 91 ⇒ 5 15 = 5 ÷ 5 15 ÷ 5 = 1 3 ⇒ 42 49 = 42 ÷ 7 49 ÷ 7 = 6 7 ⇒ 33 77 = 33 ÷ 3 77 ÷ 7 = 3 7 Therefore, the simplest form of given fraction 27 64 , 1 91 , 1 3 , 6 7 , 3 7

Which is the largest fraction? 5 7 , 2 3 , 4 5

Given: fractions are 5 7 , 2 3 , 4 5 Find the L.C.M. of the denominators of the fractions to make the denominators same. LCM of 7 , 3 , 5 = 105 Now to make the fraction as like fractions divide the L.C.M. by the denominator of fractions, then multiply both the numerator and denominator of fraction with the number get after dividing L.C.M. 5 7 = 5 7 × 15 15 = 75 105 2 3 = 2 3 × 35 35 = 70 105 4 5 = 4 5 × 21 21 = 84 105 Therefore, 75 105 , 70 105 , 84 105 are like fractions. Comparing the numerators, we get : 84 > 75 > 70 So, the largest fraction is 84 105 . Hence, the largest fraction is 4 5 .

Rearrange the following in ascending order : 4 15 , 5 21 , 8 35

Given: fractions are 4 15 , 5 21 , 8 35 Find the L.C.M. of the denominators of the fractions to make the denominators same. LCM of 15 , 21 , 35 = 105 Now to make the fraction as like fractions divide the L.C.M. by the denominator of fractions, then multiply both the numerator and denominator of fraction with the number get after dividing L.C.M. 4 15 = 4 15 × 7 7 = 28 105 5 21 = 5 21 × 5 5 = 25 105 8 35 = 8 35 × 3 3 = 24 105 Therefore, 28 105 , 25 105 , 24 105 are like fractions. Comparing the numerators, we get : 24 < 25 < 28 So, the ascending order of given fractions is 24 105 , 28 105 , 25 105 Hence, the ascending order of given fractions is 8 35 , 5 21 , 4 15

Ram spends 1 3 of his income and saves the rest. Shyam saves 3 5 of his income and spends the rest. Who saves greater part of the income ?

Given, Saving of Ram is 1 3 of his income Saving of Shyam is 3 5 of his income. We have to find who saves greater part of his income. So, we have compare the fraction. Now, 1 3 and 3 5 is not like fraction. So, change the unlike fractions to like fractions. LCM of denominator 3 and 5 is 15 . Each fractions are converted to its equivalent fraction with denominator equal to the LCM 1 3 = 1 × 5 3 × 5 = 5 15 3 5 = 3 × 3 5 × 3 = 9 15 When we compare the like fraction 5 15 and 9 15 , 9 > 5 . Thus, Shyam saves greater part of the income.

Find the sum : 3 5 + 6 7 + 1 4

We have to find the sum of 3 5 + 6 7 + 1 4 We know that when the denominators are different, we find the LCM of denominators and equivalent fraction with the LCM as the denominator then, the sum of the fraction = sum of the numerators common denominator . Now, the LCM of 5 , 7 , and 8 is 280 . ∴ 3 5 = 3 × 56 5 × 56 = 168 280 , 6 7 = 6 × 40 7 × 40 = 240 280 , and 1 4 = 1 × 70 4 × 70 = 70 280 So, the required sum 3 5 + 6 7 + 1 4 = 168 280 + 240 280 + 70 280 = 168 + 240 + 70 280 = 478 280 = 239 140 Hence, 239 140 is the required answer.

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.

E M B I B E

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 3

EASY

6th CBSE

IMPORTANT

Earn 100

Write the following in the simplest form :
$\frac{27}{64}, \frac{121}{11011}, \frac{5}{15}, \frac{42}{49}, \frac{33}{77}$

Important Questions on Fractions

EASY

6th CBSE

IMPORTANT

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 4

Which is the largest fraction?
$\frac{5}{7}, \frac{2}{3}, \frac{4}{5}$

EASY

6th CBSE

IMPORTANT

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 5

Rearrange the following in ascending order :

$\frac{4}{15}, \frac{5}{21}, \frac{8}{35}$

MEDIUM

6th CBSE

IMPORTANT

Excellence in Mathematics>Chapter 7 - Fractions>CHAPTER-END EXERCISES>Q 6

Ram spends $\frac{1}{3}$ of his income and saves the rest. Shyam saves $\frac{3}{5}$ of his income and spends the rest. Who saves greater part of the income ?

MEDIUM

6th CBSE

IMPORTANT

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

Find the sum : $\frac{3}{5} + \frac{6}{7} + \frac{1}{4}$

MEDIUM

6th CBSE

IMPORTANT

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

Simplify: