RS Aggarwal Solutions for Class 6 Math Chapter 5 - Fractions

Answer:

(i) The shaded portion is 3 parts of the whole figure.
     $\therefore$ $\frac{3}{4}$            
(ii) The shaded portion is 1 parts of the whole figure.
     $\therefore$ $\frac{1}{4}$      
(iii) The shaded portion is 2 parts of the whole figure.
     $\therefore$ $\frac{2}{3}$        
(iv) The shaded portion is 3 parts of the whole figure.
     $\therefore$$\frac{3}{10}$              
(v)The shaded portion is 4 parts of the whole figure.
     $\therefore$$\frac{4}{9}$          
(vi) The shaded portion is 3 parts of the whole figure.
     $\therefore$ $\frac{3}{8}$

Answer:

Answer:

The given rectangle is not divided into four equal parts.

Thus, the shaded region is not equal to $\frac{1}{4}$ of the whole.

Answer:

(i) $\frac{3}{4}$        (ii) $\frac{4}{7}$             (iii) $\frac{2}{5}$           (iv) $\frac{3}{10}$           (v) $\frac{1}{8}$
(vi) $\frac{5}{6}$             (vii)$\frac{8}{9}$              (viii) $\frac{7}{12}$

Answer:

     Numerator        Denominator
(i) 4                         9
(ii) 6                       11
(iii) 8                      15
(iv) 12                     17
(v) 5                        1

Answer:

(i)$\frac{3}{8}$        (ii) $\frac{5}{12}$          (iii)$\frac{7}{16}$             (iv) $\frac{8}{15}$

Answer:

(i) two-thirds
(ii) four$-$ninths
(iii) two$-$fifths
(iv) seven$-$tenths
(v) one$-$thirds
(vi) three$-$fourths
(vii) three$-$eighths
(viii) nine$-$fourteenths 
(ix) five$-$elevenths
(x) six$-$fifteenths

Answer:

We know: 1 hour = 60 minutes
∴ The required fraction = $\frac{24}{60}=\frac{2}{5}$  

Answer:

There are total 9 natural numbers from 2 to 10. They are 2, 3, 4, 5, 6, 7, 8, 9, 10.
Out of these natural numbers, 2, 3, 5, 7 are the prime numbers.
∴ The required fraction = $\frac{4}{9}$.

Answer:

(i) $\frac{2}{3}$ of 15 pens = $\left(\frac{2}{\cancel{3_1}}\times \frac{\cancel{{15}^5}}{1}\right) = 10 \mathrm{pens}$
(ii) $\frac{2}{3}$ of 27 balls = $\left(\frac{2}{\cancel{3_1}}\times \frac{\cancel{{27}^9}}{1}\right) = 18 \mathrm{balls}$
(iii) $\frac{2}{3}$ of 36 balloons = ​$\left(\frac{2}{\cancel{3_1}}\times \frac{\cancel{{36}^{12}}}{1}\right) = 24 \mathrm{balloons}$

Answer:

(i) $\frac{3}{4}$ of 16 cups = $\left(\frac{3}{\cancel{4_1}} \times \frac{\cancel{{16}^4}}{1}\right) = 12 \mathrm{cups}$
(ii) $\frac{3}{4}$ of 28 rackets = $\left(\frac{3}{\cancel{4_1}} \times \frac{\cancel{{28}^7}}{1}\right) = 21 \mathrm{rackets}$
(iii) $\frac{3}{4}$ of 32 books = $\left(\frac{3}{\cancel{4_1}} \times \frac{\cancel{{32}^8}}{1}\right) = 24 \mathrm{books}$

Answer:

Neelam gives $\frac{4}{5}$ of 25 pencils to Meena.
$\left(\frac{4}{\cancel{5_1}} \times \frac{\cancel{{25}^5}}{1}\right) = 20 \mathrm{Pencils}$
Thus, Meena gets 20 pencils.
∴ Number of pencils left with Neelam = 25 $-$ 20 = 5 pencils
Thus, 5 pencils are left with Neelam.

Answer:

Draw a 0 to 1 on a number line. Label point 1 as A and mark the starting point as 0.

(i) Divide the number line from 0 to 1 into 8 equal parts and take out 3 parts from it to reach point P.



(ii) Divide the number line from 0 to 1 into 9 equal parts and take out 5 parts from it to reach point P

.

(iii) Divide the number line from 0 to 1 into 7 equal parts and take out 4 parts from it to reach point P.



(Iv) Divide the number line from 0 to 1 into 5 equal parts and take out 2 parts from it to reach point P.



(v) Divide the number line from 0 to 1 into 4 equal parts and take out 1 part from it to reach point P.

Answer:

$\frac{1}{2}, \frac{3}{5}, \frac{10}{11}$

Answer:

A fraction whose numerator is greater than or equal to its denominator is called an improper fraction. Hence, $\frac{3}{2}, \frac{9}{4}, \frac{8}{8}, \frac{27}{16}, \frac{19}{18} \mathrm{and} \frac{26}{26}$ are improper fractions.

Answer:

Clearly, $\frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, \frac{11}{5}\mathrm{and} \frac{12}{5}$ are improper fractions, each with 5 as the denominator.

Answer:

Clearly, $\frac{13}{2}, \frac{13}{3}, \frac{13}{4}, \frac{13}{5}, \frac{13}{6}, \frac{13}{7}$ are improper fractions, each with 13 as the numerator.

Answer:

We have:
(i) $5\frac{5}{7} = \frac{(5 \times 7) + 5}{7} = \frac{40}{7}$

(ii) $9\frac{3}{8} = \frac{(9 \times 8) + 3}{8} = \frac{75}{8}$

(iii) $6\frac{3}{10} = \frac{(6 \times 10) + 3}{10} = \frac{63}{10}$

(iv) $3\frac{5}{11} = \frac{(3 \times 11) + 5}{11} = \frac{38}{11}$

(v) $10\frac{9}{14} = \frac{(10 \times 14) + 9}{14} = \frac{149}{14}$

(vi) $12\frac{7}{15} = \frac{(12 \times 15) + 7}{15} = \frac{187}{15}$

(vii) $8\frac{8}{13} = \frac{(8 \times 13) + 8}{13} = \frac{112}{13}$

(viii) $51\frac{2}{3} = \frac{(51 \times 3) + 2}{3} = \frac{155}{3}$

Answer:

(i) On dividing 17 by 5, we get:
    Quotient = 3
    Remainder = 2
   ∴ $\frac{17}{5} = 3 +\frac{2}{5} = 3\frac{2}{5}$

(ii) On dividing 62 by 7, we get:
    Quotient = 8
    Remainder = 6
   ∴ $\frac{62}{7} = 8 +\frac{6}{7} = 8\frac{6}{7}$

(iii) On dividing 101 by 8, we get:
    Quotient = 12
    Remainder = 5
   ∴ $\frac{101}{8} = 12 +\frac{5}{8} = 12\frac{5}{8}$

(iv) On dividing 95 by 13, we get:
    Quotient = 7
    Remainder = 4
   ∴ $\frac{95}{13} = 7 +\frac{4}{13} = 7\frac{4}{13}$

(v) On dividing 81 by 11, we get:
    Quotient = 7
    Remainder = 4
   ∴ $\frac{81}{11} = 7 +\frac{4}{11} = 7\frac{4}{11}$

(vi) On dividing 87 by 16, we get:
    Quotient = 5
    Remainder = 7
   ∴ $\frac{87}{16} = 5 +\frac{7}{16} = 5\frac{7}{16}$

(vii) On dividing 103 by 12, we get:
    Quotient = 8
    Remainder = 7
   ∴ $\frac{103}{12} = 8 +\frac{7}{12} = 8\frac{7}{12}$

(viii) On dividing 117 by 20, we get:
    Quotient = 5
    Remainder = 17
   ∴ $\frac{117}{20} = 5 +\frac{17}{20} = 5\frac{17}{20}$

Answer:

An improper fraction is greater than 1. Hence, it is always greater than a proper fraction, which is less than 1.
(i) $\frac{1}{2} \boxed{ < } 1$

(ii) $\frac{3}{4} \boxed{ < } 1$

(iii) $1 \boxed{ > } \frac{6}{7}$

(iv) $\frac{6}{6} \boxed{ = } 1$

(v) $\frac{3016}{3016} \boxed{ = } 1$

(vi) $\frac{11}{5} \boxed{ > } 1$

Answer:

(i) Draw a number line. Mark 0 as the starting point and 1 as the ending point.
Then, divide 0 to 1 in four equal parts, where each part is equal to 1/4.
Show the consecutive parts as 1/4, 1/2, 3/4 and at 1 show 4/4 = 1.



(ii) Draw 0 to 1 on a number line. Divide the segment into 8 equal parts, each part corresponds to 1/8. Show the consecutive parts as 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8 and 8/8. Highlight the required ones only.



(iii) Draw 0 to 2 on a number line. Divide the segment between 0 and 1 into 5 equal parts, where each part is equal to 1/5.
Show 2/5, 3/5, 4/5 and 8/5 3 parts away from 1 towards 2. (1 < 8/5 < 2)

Answer:

(i) $\frac{2}{3} =\frac{2\times 2}{3\times 2} = \frac{2\times 3}{3\times 3}= \frac{2\times 4}{3\times 4}= \frac{2\times 5}{3\times 5} = \frac{2\times 6}{3\times 6}$

   ∴ $\frac{2}{3} = \frac{4}{6} = \frac{6}{9} = \frac{8}{12} = \frac{10}{15} = \frac{12}{18}$

Hence, the five fractions equivalent to $\frac{2}{3}$ are $\frac{4}{6}, \frac{6}{9}, \frac{8}{12}, \frac{10}{15} \mathrm{and} \frac{12}{18}$.


(ii) ​ $\frac{4}{5} =\frac{4\times 2}{5\times 2} = \frac{4\times 3}{5\times 3}= \frac{4\times 4}{5\times 4}= \frac{4\times 5}{5\times 5} = \frac{4\times 6}{5\times 6}$

   ∴ $\frac{4}{5} = \frac{8}{10} = \frac{12}{15} = \frac{16}{20} = \frac{20}{25} = \frac{24}{30}$

Hence, the five fractions equivalent to $\frac{4}{5}$ are $\frac{8}{10}, \frac{12}{15}, \frac{16}{20}, \frac{20}{25} \mathrm{and} \frac{24}{30}$.


(iii) ​ $\frac{5}{8} =\frac{5\times 2}{8\times 2} = \frac{5\times 3}{8\times 3}= \frac{5\times 4}{8\times 4}= \frac{5\times 5}{8\times 5} = \frac{5\times 6}{8\times 6}$

   ∴ $\frac{5}{8} = \frac{10}{16} = \frac{15}{24} = \frac{20}{32} = \frac{25}{40} = \frac{30}{48}$

Hence, the five fractions equivalent to $\frac{5}{8}$ are $\frac{10}{16}, \frac{15}{24}, \frac{20}{32}, \frac{25}{40} \mathrm{and} \frac{30}{48}$.



(iv) ​ $\frac{7}{10} =\frac{7\times 2}{10\times 2} = \frac{7\times 3}{10\times 3}= \frac{7\times 4}{10\times 4}= \frac{7\times 5}{10\times 5} = \frac{7\times 6}{10\times 6}$

   ∴ $\frac{7}{10} = \frac{14}{20} = \frac{21}{30} = \frac{28}{40} = \frac{35}{50}= \frac{42}{60}$

Hence, the five fractions equivalent to $\frac{7}{10}$ are $\frac{14}{20}, \frac{21}{30}, \frac{28}{40}, \frac{35}{50} \mathrm{and} \frac{42}{60}$.


(v) ​​ $\frac{3}{7} =\frac{3\times 2}{7\times 2} = \frac{3\times 3}{7\times 3}= \frac{3\times 4}{7\times 4}= \frac{3\times 5}{7\times 5} = \frac{3\times 6}{7\times 6}$

   ∴ $\frac{3}{7} = \frac{6}{14} = \frac{9}{21} = \frac{12}{28} = \frac{15}{35}= \frac{18}{42}$

Hence, the five fractions equivalent to $\frac{3}{7}$ are $\frac{6}{14}, \frac{9}{21}, \frac{12}{28},\frac{15}{35} \mathrm{and} \frac{18}{42}$.


(vi)  ​ $\frac{6}{11} =\frac{6\times 2}{11\times 2} = \frac{6\times 3}{11\times 3}= \frac{6\times 4}{11\times 4}= \frac{6\times 5}{11\times 5} = \frac{6\times 6}{11\times 6}$

   ∴ $\frac{6}{11} = \frac{12}{22} = \frac{18}{33} = \frac{24}{44} = \frac{30}{55}= \frac{36}{66}$

Hence, the five fractions equivalent to $\frac{6}{11}$ are $\frac{12}{22}, \frac{18}{33}, \frac{24}{44}, \frac{30}{55} \mathrm{and} \frac{36}{66}$.


(vii)  $\frac{7}{9} =\frac{7\times 2}{9\times 2} = \frac{7\times 3}{9\times 3}= \frac{7\times 4}{9\times 4}= \frac{7\times 5}{9\times 5} = \frac{7\times 6}{9\times 6}$

   ∴ $\frac{7}{9} = \frac{14}{18} = \frac{21}{27} = \frac{28}{36} = \frac{35}{45}= \frac{42}{54}$

Hence, the five fractions equivalent to $\frac{7}{9}$ are $\frac{14}{18}, \frac{21}{27}, \frac{28}{36}, \frac{35}{45} \mathrm{and} \frac{42}{54}$.


(viii)  $\frac{5}{12} =\frac{5\times 2}{12\times 2} = \frac{5\times 3}{12\times 3}= \frac{5\times 4}{12\times 4}= \frac{5\times 5}{12\times 5} = \frac{5\times 6}{12\times 6}$

   ∴ $\frac{5}{12} = \frac{10}{24} = \frac{15}{36} = \frac{20}{48} = \frac{25}{60}= \frac{30}{72}$

Hence, the five fractions equivalent to $\frac{5}{12}$ are $\frac{10}{24}, \frac{15}{36}, \frac{20}{48},\frac{25}{60} \mathrm{and} \frac{30}{72}$.

Answer:

The pairs of equivalent fractions are as follows:
(i) $\frac{5}{6} \mathrm{and} \frac{20}{24} \left(\frac{20}{24} = \frac{5\times 4}{6\times 4}\right)$
(ii) $\frac{3}{8} \mathrm{and} \frac{15}{40} \left(\frac{15}{40} = \frac{3\times 5}{8\times 5}\right)$
(iv) $\frac{2}{9} \mathrm{and} \frac{14}{63} \left(\frac{14}{63} = \frac{2\times 7}{9\times 7}\right)$

Answer:

(i) Let $\frac{3}{5} = \frac{\boxed{}}{30}$
Clearly, 30 = 5 $\times$ 6
So, we multiply the numerator by 6.

∴ ​$\frac{3}{5} = \frac{3\times 6}{5\times 6}= \frac{18}{30}$
Hence, the required fraction is $\frac{18}{30}$.
(ii)  ​Let $\frac{3}{5} = \frac{24}{\boxed{}}$
   Clearly, 24 = 3 $\times$ 8
   So, we multiply the denominator by 8.

∴ ​$\frac{3}{5} = \frac{3\times 8}{5\times 8}= \frac{24}{40}$
Hence, the required fraction is $\frac{24}{40}$.

Answer:

(i) Let $\frac{5}{9} = \frac{\boxed{}}{54}$
Clearly, 54 = 9 $\times$ 6
So, we multiply the numerator by 6.
∴ ​$\frac{5}{9} = \frac{5\times 6}{9\times 6}= \frac{30}{54}$
Hence, the required fraction is $\frac{30}{54}$.
(ii)  ​Let $\frac{5}{9} = \frac{35}{\boxed{}}$
   Clearly, 35 = 5 $\times$ 7
   So, we multiply the denominator by 7.
∴ ​$\frac{5}{9} = \frac{5\times 7}{9\times 7}= \frac{35}{63}$
Hence, the required fraction is $\frac{35}{63}$.

Answer:

(i) Let $\frac{6}{11} = \frac{\boxed{}}{77}$
   Clearly, 77 = 11 $\times$ 7
   So, we multiply the numerator by 7.

∴ ​$\frac{6}{11} = \frac{6\times 7}{11\times 7}= \frac{42}{77}$
Hence, the required fraction is $\frac{42}{77}$.
(ii)  ​Let $\frac{6}{11} = \frac{60}{\boxed{}}$
   Clearly, 60 = 6 $\times$ 10
   So, we multiply the denominator by 10.

∴ ​$\frac{6}{11} = \frac{6\times 10}{11\times 10}= \frac{60}{110}$
Hence, the required fraction is $\frac{60}{110}$.

Answer:

   Let $\frac{24}{30} = \frac{4}{\boxed{}}$
   Clearly, 4 = 24 $÷$ 6
   So, we divide the denominator by 6.
 ∴ ​$\frac{24}{30} = \frac{24÷6}{30÷6}= \frac{4}{5}$
  Hence, the required fraction is $\frac{4}{5}$.

Answer:

(i) Let $\frac{36}{48} = \frac{9}{\boxed{}}$
   Clearly, 9 = 36 $÷$ 4
   So, we divide the denominator by 4.
∴ ​$\frac{36}{48} = \frac{36÷4}{48÷4}= \frac{9}{12}$
Hence, the required fraction is $\frac{9}{12}$.
(ii)  ​Let $\frac{36}{48} = \frac{\boxed{}}{4}$
   Clearly, 4 = 48 $÷$ 12
   So, we divide the numerator by 12.
∴ ​$\frac{36}{48} = \frac{36÷12}{48÷12}= \frac{3}{4}$
Hence, the required fraction is $\frac{3}{4}$.

Answer:

(i) Let $\frac{56}{70} = \frac{4}{\boxed{}}$
   Clearly, 4 = 56 $÷$ 14
   So, we divide the denominator by 14.
  ∴ ​$\frac{56}{70} = \frac{56÷14}{70÷14}= \frac{4}{5}$
  Hence, the required fraction is $\frac{4}{5}$.
(ii)  ​Let $\frac{56}{70} = \frac{\boxed{}}{10}$
     Clearly, 10 = 70 $÷$ 7
     So, we divide the numerator by 7.
   ∴ ​$\frac{56}{70} = \frac{56\times 7}{70\times 7}= \frac{8}{10}$
   Hence, the required fraction is $\frac{8}{10}$.

Answer:

(i) Here, numerator = 9 and denominator = 15
Factors of 9 are 1, 3 and 9.
Factors of 15 are 1, 3, 5 and 15.
Common factors of 9 and 15 are 1 and 3.
H.C.F. of 9 and 15 is 3.
∴ $\frac{9}{15} =\frac{9÷3}{15÷3} = \frac{3}{5}$
Hence, the simplest form of $\frac{9}{15} \mathrm{is} \frac{3}{5}$.

(ii) Here, numerator = 48 and denominator = 60
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 48 and 60 are 1, 2, 3, 4, 6 and 12.
H.C.F. of 48 and 60 is 12.
∴ $\frac{48}{60} =\frac{48÷12}{60÷12} = \frac{4}{5}$
Hence, the simplest form of $\frac{48}{60} \mathrm{is} \frac{4}{5}$.

(iii) Here, numerator = 84 and denominator = 98
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 42 and 84.
Factors of 98 are 1, 2, 7, 14, 49 and 98.
Common factors of 84 and 98 are 1, 2, 7 and 14.
H.C.F. of 84 and 98 is 14.
∴ $\frac{84}{98} =\frac{84÷14}{98÷14} = \frac{6}{7}$
Hence, the simplest form of $\frac{84}{98} \mathrm{is} \frac{6}{7}$.

(iv) Here, numerator = 150 and denominator = 60
Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75 and 150.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 150 and 60 are 1, 2, 3, 5, 6, 10, 15 and 30.
H.C.F. of 150 and 60 is 30.
∴ $\frac{150}{60} =\frac{150÷30}{60÷30} = \frac{5}{2}$
Hence, the simplest form of $\frac{150}{60} \mathrm{is} \frac{5}{2}$.

(v) ​Here, numerator = 72 and denominator = 90
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90.
Common factors of 72 and 90 are 1, 2, 3, 6, 9 and 18.
H.C.F. of 72 and 90 is 18.
∴ $\frac{72}{90} =\frac{72÷18}{90÷18} = \frac{4}{5}$
Hence, the simplest form of $\frac{72}{90} \mathrm{is} \frac{4}{5}$.

Answer:

(i) Here, numerator = 8 and denominator = 11
    Factors of 8 are 1, 2, 4 and 8.
    Factors of 11 are 1 and 11.

Answer:

(i) 28            $\left(\frac{2}{7} = \frac{2\times 4}{7\times 4} = \frac{8}{28}\right)$
(ii) 21           $\left(\frac{3}{5} = \frac{3\times 7}{5\times 7} = \frac{21}{35}\right)$
(iii) 32          $\left(\frac{5}{8} = \frac{5\times 4}{8\times 4} = \frac{20}{32}\right)$
(iv) 12          $\left(\frac{45}{60} = \frac{45÷5}{60÷5} = \frac{9}{12}\right)$
(v) 5             $\left(\frac{40}{56} = \frac{40÷8}{56÷8} = \frac{5}{7}\right)$ 
(vi) 9              $\left(\frac{42}{54} = \frac{42÷6}{54÷6} = \frac{7}{9}\right)$

Answer:

Like fractions:
Fractions having the same denominator are called like fractions.
Examples: $\frac{3}{11}, \frac{5}{11}, \frac{7}{11}, \frac{9}{11}, \frac{10}{11}$

Unlike fractions:
Fractions having different denominators are called unlike fractions.
Examples: $\frac{3}{4}, \frac{4}{5}, \frac{6}{7}, \frac{9}{11}, \frac{2}{13}$

Answer:

The given fractions are $\frac{3}{5}, \frac{7}{10}, \frac{8}{15} \mathrm{and} \frac{11}{30}.$


L.C.M. of 5, 10, 15 and 30 = (5 $\times$ 2 $\times$ 3) = 30
So, we convert the given fractions into equivalent fractions with 30 as the denominator.
(But, one of the fractions already has 30 as its denominator. So, there is no need to convert it into an equivalent fraction.)
Thus, we have:
$\frac{3}{5} = \frac{3\times 6}{5\times 6} = \frac{18}{30}; \frac{7}{10} = \frac{7\times 3}{10\times 3} = \frac{21}{30} ; \frac{8}{15} = \frac{8\times 2}{15\times 2} = \frac{16}{30}$

Hence, the required like fractions are $\frac{18}{30}, \frac{21}{30}, \frac{16}{30} \mathrm{and} \frac{11}{30}.$

Answer:

The given fractions are $\frac{1}{4}, \frac{5}{8}, \frac{7}{12} \mathrm{and} \frac{13}{24} .$
L.C.M. of 4, 8, 12 and 24 = (4 $\times$ 2 $\times$ 3) = 24
So, we convert the given fractions into equivalent fractions with 24 as the denominator.
(But one of the fractions already has 24 as the denominator. So, there is no need to convert it into an equivalent fraction.)
Thus, we have:
$\frac{1}{4} = \frac{1\times 6}{4\times 6} = \frac{6}{24}; \frac{5}{8} = \frac{5\times 3}{8\times 3} = \frac{15}{24} ; \frac{7}{12} = \frac{7\times 2}{12\times 2} = \frac{14}{24}$

Hence, the required like fractions are $\frac{6}{24}, \frac{15}{24}, \frac{14}{24} \mathrm{and} \frac{13}{24}.$

Answer:

Between two fractions with the same denominator, the one with the greater numerator is the greater of the two.

(i) >
(ii) >
(iii) <
(iv) >
(v) >
(vi) <

Answer:

Between two fractions with the same numerator, the one with the smaller denominator is the greater of the two.

(i) >
(ii) >
(iii)<
(iv) >
(v) <
(vi) >

Answer:

$\frac{4}{5}, \frac{5}{7}$
By cross multiplying:
5 $\times$ 5 = 25 and 4 $\times$ 7 = 28     
Clearly, 28 > 25
$\therefore$ $\frac{4}{5} > \frac{5}{7}$

Answer:

$\frac{3}{8}, \frac{5}{6}$
By cross multiplying:
3 $\times$ 6 = 18 and 5 $\times$ 8 = 40        
Clearly, 18 < 40
$\therefore$ $\frac{3}{8} < \frac{5}{6}$

Answer:

$\frac{7}{11} , \frac{6}{7}$

By cross multiplying:
7 $\times$ 7 = 49 and 11 $\times$ 6 = 66        
Clearly, 49 < 66
$\therefore$ $\frac{7}{11} < \frac{6}{7}$

Answer:

$\frac{7}{11} , \frac{6}{7}$
By cross multiplying:
5 $\times$ 11 = 55 and 9 $\times$ 6 = 54         
Clearly, 55 > 54
$\therefore$ $\frac{5}{6} > \frac{9}{11}$

Answer:

$\frac{7}{11} , \frac{6}{7}$
By cross multiplying:
2 $\times$ 9 = 18 and 4 $\times$ 3 = 12     
Clearly, 18 > 12
$\therefore$ $\frac{2}{3} > \frac{4}{9}$

Answer:

$\frac{6}{13} , \frac{3}{4}$
By cross multiplying:
6 $\times$ 4 = 24 and 13 $\times$ 3 = 39      
Clearly, 24 < 39
$\therefore$ $\frac{6}{13} < \frac{3}{4}$

Answer:

$\frac{6}{13}, \frac{3}{4}$
By cross multiplying:
3 $\times$ 6 = 18 and 4 $\times$ 5 = 20     
Clearly, 18 < 20
$\therefore$ $\frac{3}{4} < \frac{5}{6}$

Answer:

$\frac{5}{8} ,\frac{7}{12}$
By cross multiplying:
5 $\times$ 12 = 60 and 8 $\times$ 7 = 56     
Clearly, 60 > 56
$\therefore$ $\frac{5}{8} > \frac{7}{12}$

Answer:

L.C.M. of 9 and 6 = (3 $\times$ 3 $\times$ 2) = 18
Now, we convert $\frac{4}{9} \mathrm{and} \frac{5}{6}$ into equivalent fractions having 18 as the denominator. 
∴​ $\frac{4}{9} = \frac{4\times 2}{9\times 2} = \frac{8}{18} \mathrm{and} \frac{5}{6} = \frac{5\times 3}{6\times 3} = \frac{15}{18}$4949
         
Clearly, $\frac{8}{18} < \frac{15}{18}$
$\therefore$ $\frac{4}{9} < \frac{5}{6}$

Answer:

L.C.M. of 5 and 10 = (5 $\times$ 2) = 10
Now, we convert $\frac{4}{5}$ into an equivalent fraction having 10 as the denominator as the other fraction has already 10 as its denominator.
∴​ $\frac{4}{5} = \frac{4\times 2}{5\times 2} = \frac{8}{10}$4949
         
Clearly, $\frac{8}{10} > \frac{7}{10}$
$\therefore$ $\frac{4}{5} > \frac{7}{10}$

Answer:

L.C.M. of 8 and 10 = (2 $\times$ 5 $\times$ 2 $\times$ 2) = 40
Now, we convert $\frac{7}{8} \mathrm{and} \frac{9}{10}$ into equivalent fractions having 40 as the denominator.
∴​ $\frac{7}{8} = \frac{7\times 5}{8\times 5} = \frac{35}{40} \mathrm{and} \frac{9}{10} = \frac{9\times 4}{10\times 4} = \frac{36}{40}$4949
         
Clearly, $\frac{35}{40} < \frac{36}{40}$
$\therefore$ $\frac{7}{8} < \frac{9}{10}$

Answer:

L.C.M. of 12 and 15 = (2 $\times$ 2 $\times$ 3 $\times$ 5) = 60
Now, we convert $\frac{11}{12} \mathrm{and} \frac{13}{15}$ into equivalent fractions having 60 as the denominator.
∴​ $\frac{11}{12} = \frac{11\times 5}{12\times 5} = \frac{55}{60} \mathrm{and} \frac{13}{15} = \frac{13\times 4}{15\times 4} = \frac{52}{60}$4949
         
Clearly, $\frac{55}{60} > \frac{52}{60}$
$\therefore$ $\frac{11}{12} > \frac{13}{15}$

Answer:

The given fractions are $\frac{1}{2}, \frac{3}{4}, \frac{5}{6} \mathrm{and} \frac{7}{8}$.
L.C.M. of 2, 4, 6 and 8 = (2 $\times$ 2 $\times$ 2 $\times$ 3) = 24
We convert each of the given fractions into an equivalent fraction with denominator 24.
Now, we have:
 $$\begin{gathered} \frac{1}{2} = \frac{1\times 12}{2\times 12} = \frac{12}{24}; \frac{3}{4} = \frac{3\times 6}{4\times 6} = \frac{18}{24} \\ \frac{5}{6} = \frac{5\times 4}{6\times 4} = \frac{20}{24}; \frac{7}{8} = \frac{7\times 3}{8\times 3} = \frac{21}{24} \end{gathered}$$

Clearly, $\frac{12}{24} <\frac{18}{24} <\frac{20}{24} <\frac{21}{24}$

∴ ​$\frac{1}{2} <\frac{3}{4} <\frac{5}{6} <\frac{7}{8}$
Hence, the given fractions can be arranged in the ascending order as follows:
$\frac{1}{2}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}$​

Answer:

The given fractions are $\frac{2}{3}, \frac{5}{6}, \frac{7}{9} \mathrm{and} \frac{11}{18}.$


L.C.M. of 3, 6, 9 and 18 = (3 $\times$ 2  $\times$ 3) = 18
So, we convert each of the fractions whose denominator is not equal to 18 into an equivalent fraction with denominator 18.
Now, we have:
$\frac{2}{3} = \frac{2\times 6}{3\times 6} = \frac{12}{18}; \frac{5}{6} = \frac{5\times 3}{6\times 3} = \frac{15}{18}; \frac{7}{9} = \frac{7\times 2}{9\times 2} = \frac{14}{18}$
Clearly, $\frac{11}{18} <\frac{12}{18} <\frac{14}{18} <\frac{15}{18}$
∴ ​$\frac{11}{18} <\frac{2}{3} <\frac{7}{9} <\frac{5}{6}$

Hence, the given fractions can be arranged in the ascending order as follows:
$\frac{11}{18} ,\frac{2}{3} ,\frac{7}{9} ,\frac{5}{6}$

Answer:

The given fractions are $\frac{2}{5},\frac{7}{10}, \frac{11}{15} \mathrm{and} \frac{17}{30}.$
L.C.M. of 5, 10, 15 and 30 = (2 $\times$ 5 $\times$ 3) = 30  

                                        
So, we convert each of the fractions whose denominator is not equal to 30 into an equivalent fraction with denominator 30.
Now, we have:
$\frac{2}{5} = \frac{2\times 6}{5\times 6} = \frac{12}{30}; \frac{7}{10} = \frac{7\times 3}{10\times 3} = \frac{21}{30}; \frac{11}{15} = \frac{11\times 2}{15\times 2} = \frac{22}{30}$
Clearly, $\frac{12}{30} <\frac{17}{30} <\frac{21}{30} <\frac{22}{30}$
∴ ​$\frac{2}{5} <\frac{17}{30} <\frac{7}{10} <\frac{11}{15}$

Hence, the given fractions can be arranged in the ascending order as follows:
$\frac{2}{5}, \frac{17}{30}, \frac{7}{10}, \frac{11}{15}$ 

Answer:

The given fractions are $\frac{3}{4}, \frac{7}{8}, \frac{11}{16} \mathrm{and} \frac{23}{32}.$
L.C.M. of 4, 8, 16 and 32 = (2 ⨯ 2 ⨯ 2 ⨯ 2 ⨯ 2) = 32    

                                      
So, we convert each of the fractions whose denominator is not equal to 32 into an equivalent fraction with denominator 32.
Now, we have:
$\frac{3}{4} = \frac{3\times 8}{4\times 8} = \frac{24}{32}; \frac{7}{8} = \frac{7\times 4}{8\times 4} = \frac{28}{32}; \frac{11}{16} = \frac{11\times 2}{16\times 2} = \frac{22}{32}$
Clearly, $\frac{22}{32} <\frac{23}{32} <\frac{24}{32} <\frac{28}{32}$
∴ ​$\frac{11}{16} <\frac{23}{32} <\frac{3}{4} <\frac{7}{8}$

Hence, the given fractions can be arranged in the ascending order as follows:
$\frac{11}{16}, \frac{23}{32}, \frac{3}{4}, \frac{7}{8}$

Answer:

The given fractions are $\frac{3}{4}, \frac{5}{8}, \frac{11}{12} \mathrm{and} \frac{17}{24}.$
L.C.M. of 4, 8, 12 and 24 = (2 ⨯ 2 ⨯ 2 ⨯ 3) = 24            

                             
So, we convert each of the fractions whose denominator is not equal to 24 into an equivalent fraction with denominator 24.
Thus, we have;
$\frac{3}{4} = \frac{3\times 6}{4\times 6} = \frac{18}{24}; \frac{5}{8} = \frac{5\times 3}{8\times 3} = \frac{15}{24}; \frac{11}{12} = \frac{11\times 2}{12\times 2} = \frac{22}{24}$
Clearly, $\frac{22}{24} >\frac{18}{24} >\frac{17}{24} >\frac{15}{24}$

∴ ​$\frac{11}{12} >\frac{3}{4} >\frac{17}{24} >\frac{5}{8}$

Hence, the given fractions can be arranged in the descending order as follows:
$\frac{11}{12}, \frac{3}{4}, \frac{17}{24}, \frac{5}{8}$

Answer:

The given fractions are $\frac{7}{9}, \frac{5}{12}, \frac{11}{18} \mathrm{and} \frac{17}{36}.$
L.C.M. of 9, 12, 18 and 36 = (3 ⨯ 3 ⨯ 2 ⨯ 2) = 36      

                                   
We convert each of the fractions whose denominator is not equal to 36 into an equivalent fraction with denominator 36.
Thus, we have:
$\frac{7}{9} = \frac{7\times 4}{9\times 4} = \frac{28}{36}; \frac{5}{12} = \frac{5\times 3}{12\times 3} = \frac{15}{36}; \frac{11}{18} = \frac{11\times 2}{18\times 2} = \frac{22}{36}$
Clearly, $\frac{28}{36} >\frac{22}{36} >\frac{17}{36} >\frac{15}{36}$

∴ ​$\frac{7}{9} >\frac{11}{18} >\frac{17}{36} >\frac{5}{12}$

Hence, the given fractions can be arranged in the descending order as follows:
$\frac{7}{9} ,\frac{11}{18},\frac{17}{36},\frac{5}{12}$

Answer:

The given fractions are $\frac{2}{3}, \frac{3}{5}, \frac{7}{10} \mathrm{and} \frac{8}{15}.$
L.C.M. of 3, 5,10 and 15 = (2 ⨯ 3 ⨯ 5) = 30  

                                       
So, we convert each of the fractions into an equivalent fraction with denominator 30.
Thus, we have:
$$\begin{gathered} \frac{2}{3} = \frac{2\times 10}{3\times 10} = \frac{20}{30}; \frac{3}{5} = \frac{3\times 6}{5\times 6} = \frac{18}{30}; \\ \frac{7}{10} = \frac{7\times 3}{10\times 3} = \frac{21}{30}; \frac{8}{15} = \frac{8\times 2}{15\times 2} = \frac{16}{30} \end{gathered}$$
Clearly, $\frac{21}{30} >\frac{20}{30} >\frac{18}{30} >\frac{16}{30}$
∴ ​$\frac{7}{10} >\frac{2}{3} >\frac{3}{5} >\frac{8}{15}$

Hence, the given fractions can be arranged in the descending order as follows:
$\frac{7}{10} ,\frac{2}{3} ,\frac{3}{5} ,\frac{8}{15}$

Answer:

The given fractions are $\frac{5}{7}, \frac{9}{14}, \frac{17}{21} \mathrm{and} \frac{31}{42}.$
L.C.M. of 7, 14, 21 and 42 = (2 ⨯ 3 ⨯ 7) = 42                                          


We convert each one of the fractions whose denominator is not equal to 42 into an equivalent fraction with denominator 42.
Thus, we have:
$\frac{5}{7} = \frac{5\times 6}{7\times 6} = \frac{30}{42}; \frac{9}{14} = \frac{9\times 3}{14\times 3} = \frac{27}{42}; \frac{17}{21} = \frac{17\times 2}{21\times 2} = \frac{34}{42}$
Clearly, $\frac{34}{42} >\frac{31}{42} >\frac{30}{42} >\frac{27}{42}$
∴ ​$\frac{17}{21} >\frac{31}{42} >\frac{5}{7} >\frac{9}{14}$
Hence, the given fractions can be arranged in the descending order as follows:
$\frac{17}{21},\frac{31}{42},\frac{5}{7}, \frac{9}{14}$

Answer:

The given fractions are $\frac{1}{12},\frac{1}{23}, \frac{1}{7}, \frac{1}{9} , \frac{1}{17} \mathrm{and} \frac{1}{50}.$
As the fractions have the same numerator, we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.
Clearly, $\frac{1}{7} >\frac{1}{9} >\frac{1}{12} >\frac{1}{17}>\frac{1}{23}>\frac{1}{50}$
Hence, the given fractions can be arranged in the descending order as follows:
$\frac{1}{7}, \frac{1}{9}, \frac{1}{12}, \frac{1}{17}, \frac{1}{23}, \frac{1}{50}$

Answer:

The given fractions are $\frac{3}{7}, \frac{3}{11}, \frac{3}{5}, \frac{3}{13}, \frac{3}{4} \mathrm{and} \frac{3}{17}.$
As the fractions have the same numerator, so we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.

Clearly, $\frac{3}{4} >\frac{3}{5} >\frac{3}{7} >\frac{3}{11}>\frac{3}{13}>\frac{3}{17}$
Hence, the given fractions can be arranged in the descending order as follows:
$\frac{3}{4}, \frac{3}{5}, \frac{3}{7}, \frac{3}{11}, \frac{3}{13}, \frac{3}{17}$ 

Answer:

Lalita read 30 pages of a book having 100 pages.
Sarita read $\frac{2}{5}$ of the same book.
 $\frac{2}{5}$ of 100 pages = ​$\frac{2}{5} \times 100 = \frac{200}{5} = 40 \mathrm{pages}$
Hence, Sarita read more pages than Lalita as 40 is greater than 30.

Answer:

To know who exercised for a longer time, we have to compare $\frac{2}{3} \mathrm{hour} \mathrm{with} \frac{3}{4} \mathrm{hour}$.
On cross multiplying:
4 $\times$ 2 = 8 and 3 $\times$ 3 = 9
Clearly, 8 < 9
$\therefore$ $\frac{2}{3} \mathrm{hour} < \frac{3}{4} \mathrm{hour}$
Hence, Rohit exercised for a longer time.

Answer:

Fraction of students who passed in VI A = $\frac{20}{25} = \frac{20÷5}{25÷5} = \frac{4}{5}$

Fraction of students who passed in VI B = $\frac{24}{30} = \frac{24÷6}{30÷6} = \frac{4}{5}$
In both the sections, the fraction of students who passed is the same, so both the sections have the same result.

Answer:

The given fractions are like fractions.
We know:
Sum of like fractions  = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{numerators}}{\mathrm{C}\mathrm{ommon} \mathrm{d}\mathrm{enominator}}$
Thus, we have:
$\frac{5}{8} + \frac{1}{8} = \frac{\left(5+1\right) }{8} = \frac{{\cancel{6 }}^3}{{\cancel{8}}_4} = \frac{3}{4}$

Answer:

The given fractions are like fractions.
We know:
Sum of like fractions  = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{numeratos}}{\mathrm{Common} \mathrm{d}\mathrm{enominator}}$
Thus, we have:
$\frac{4}{9} + \frac{8}{9} = \frac{\left(4+8\right) }{9} = \frac{{\cancel{12}}^4}{{\cancel{9}}_3} = \frac{4}{3} = 1\frac{1}{3}$

Answer:

The given fractions are like fractions.
We know:
Sum of like fractions  = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{numerators}}{\mathrm{Common} \mathrm{d}\mathrm{enominator}}$
Thus, we have:
$1\frac{3}{5} + 2\frac{4}{5} = \frac{8}{5} + \frac{14}{5} = \frac{\left(8+14\right) }{5} = \frac{22}{5} = 4\frac{2}{5}$

Answer:

L.C.M. of 9 and 6 = (2 $\times$ 3 $\times$ 3) = 18                                   


Now, we have:

$$\begin{gathered} \frac{2}{9} = \frac{2 \times 2}{9 \times 2} = \frac{4}{18}; \frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18} \\ \therefore \frac{2}{9} + \frac{5}{6} = \frac{4}{18} + \frac{15}{18} = \frac{\left(4 + 15\right)}{18} = \frac{19}{18} = 1\frac{1}{18} \end{gathered}$$