RD Sharma solution class 7 chapter 2 Fractions Exercise Objective Type Questions

Objective Type Questions

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Question 1:

Mark the correct alternative in each of the following:

If a fraction $\frac{a}{b}$ is a lowest terms, then HCF of a and b is

(a) a                                       (b) b                                       (c) 1                                       (d) ab                                 

Answer 1:

We know that a fraction is in its lowest terms if its numerator and denominator have no common factor other than 1.

Thus, if the fraction $\frac{a}{b}$ is in its lowest terms, then the HCF of a and b is 1.

Hence, the correct answer is option (c). 

Question 2:

Mark the correct alternative in each of the following:

The fraction $\frac{84}{98}$ in its lowest terms is

(a) $\frac{42}{49}$                                       (b) $\frac{12}{14}$                                       (c) $\frac{6}{7}$                                       (d) $\frac{3}{7}$                                 

Answer 2:

Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

Factors of 98: 1, 2, 7, 14, 49, 98

Common factors of 84 and 98: 1, 2, 14

∴ HCF of 84 and 98 = 14

Now,

$\frac{84}{98}=\frac{84÷14}{98÷14}=\frac{6}{7}$             (Dividing numerator and denominator by the HCF of 84 and 98 i.e. 14)

Hence, the correct answer is option (c).

Question 3:

Mark the correct alternative in each of the following:

Which of the following is a vulgar fraction?

(a) $\frac{7}{10}$                                       (b) $\frac{13}{1000}$                                       (c) $2\frac{9}{10}$                                       (d) $\frac{7}{9}$                                                                 

Answer 3:

The fractions with denominator not equal to 10, 100, 1000 etc are called vulgar fractions.

Thus, the fraction $\frac{7}{9}$ is a vulgar fraction.

Hence, the correct answer is option (d).

Question 4:

Mark the correct alternative in each of the following:

Which of the following fraction is an irreducible (or in its lowest terms)?

(a) $\frac{91}{104}$                                       (b) $\frac{105}{112}$                                       (c) $\frac{51}{85}$                                       (d) $\frac{43}{83}$                                                    

Answer 4:

We know that a fraction is irreducible (or is in its lowest terms) if the HCF of its numerator and denominator is 1.

Consider the fraction $\frac{91}{104}$.

HCF of 91 and 104 = 13 ≠ 1

So, the fraction $\frac{91}{104}$ is reducible.

Consider the fraction $\frac{105}{112}$.

HCF of 105 and 112 = 7 ≠ 1

So, the fraction $\frac{105}{112}$ is reducible.

Consider the fraction $\frac{51}{85}$.

HCF of 51 and 85 = 17 ≠ 1

So, the fraction $\frac{51}{85}$ is reducible.

Now,

Consider the fraction $\frac{43}{83}$.

HCF of 43 and 83 = 1

So, the fraction $\frac{43}{83}$ is irreducible (or is in its lowest terms).

Hence, the correct answer is option (d).

Question 5:

Mark the correct alternative in each of the following:

Which of the following is a proper fraction?

(a) $\frac{13}{17}$                                       (b) $\frac{17}{13}$                                       (c) $\frac{12}{5}$                                       (d) $1\frac{3}{4}$                                                                  

Answer 5:

A fraction whose numerator is less than the denominator is called a proper fraction.

The numerator in each of the fractions $\frac{17}{13}$, $\frac{12}{5}$, $1\frac{3}{4}=\frac{7}{4}$ is more than the denominator, so these fractions are improper fractions.

The numerator of the fraction $\frac{13}{17}$ is less than the denominator, so this fraction is a proper fraction.

Hence, the correct answer is option (a).

Question 6:

Mark the correct alternative in each of the following:

The reciprocal of the fraction $2\frac{3}{5}$ is

(a) $2\frac{5}{3}$                                       (b) $\frac{13}{5}$                                       (c) $\frac{5}{13}$                                       (d) $2\frac{2}{5}$                                 

Answer 6:

The reciprocal of a non-zero fraction $\frac{a}{b}$ is the fraction $\frac{b}{a}$.

$2\frac{3}{5}=\frac{2\times 5+3}{5}=\frac{13}{5}$

Now,

Reciprocal of the fraction $\frac{13}{5}$ = $\frac{5}{13}$

∴ Reciprocal of the fraction $2\frac{3}{5}$ = $\frac{5}{13}$

Hence, the correct answer is option (c).

Question 7:

Mark the correct alternative in each of the following:

$4\frac{1}{3}-2\frac{1}{3}=$

(a) $2\frac{1}{3}$                                      (b) 2                                       (c) $3\frac{1}{3}$                                      (d) $\frac{1}{2}$                                 

Answer 7:

$$\begin{gathered} 4\frac{1}{3}-2\frac{1}{3} \\ =\frac{13}{3}-\frac{7}{3} \\ =\frac{13-7}{3} \\ =\frac{6}{3} \\ =2 \end{gathered}$$

Hence, the correct answer is option (b).

Question 8:

Mark the correct alternative in each of the following:

$2\frac{3}{5}÷\frac{5}{7}=$

(a) $\frac{13}{7}$                                      (b) $\frac{13}{25}$                                       (c) $\frac{91}{25}$                                      (d) $\frac{25}{91}$                                 

Answer 8:

$$\begin{gathered} 2\frac{3}{5}÷\frac{5}{7} \\ =\frac{13}{5}÷\frac{5}{7} \\ =\frac{13}{5}\times \frac{7}{5} \\ =\frac{13\times 7}{5\times 5} \\ =\frac{91}{25} \end{gathered}$$

Hence, the correct answer is option (c).

Question 9:

Mark the correct alternative in each of the following:

By what number should $1\frac{3}{4}$ be divided to get $2\frac{1}{2}$?

(a) $\frac{3}{7}$                                      (b) $1\frac{2}{5}$                                       (c) $\frac{7}{10}$                                      (d) $1\frac{3}{7}$

Answer 9:

Let the required number be x.

Now,

$$\begin{gathered} 1\frac{3}{4}÷x=2\frac{1}{2} \\ \Rightarrow \frac{7}{4}\times \frac{1}{x}=\frac{5}{2} \\ \Rightarrow x=\frac{7}{4}\times \frac{2}{5} \\ \Rightarrow x=\frac{7\times 1}{2\times 5} \\ \Rightarrow x=\frac{7}{10} \end{gathered}$$

Thus, the required number is $\frac{7}{10}$.

Hence, the correct answer is option (c).

Question 10:

Mark the correct alternative in each of the following:

By what number $4\frac{3}{5}$ be multiplied to get $2\frac{3}{7}$?

(a) $\frac{391}{35}$                                      (b) $\frac{85}{91}$                                       (c) $\frac{91}{85}$                                      (d) None of these                                

Answer 10:

Product of two numbers = $2\frac{3}{7}=\frac{17}{7}$

One of the numbers = $4\frac{3}{5}=\frac{23}{5}$

∴ Other number = Product of two numbers ÷ One of the numbers

                          $$\begin{gathered} =\frac{17}{7}÷\frac{23}{5} \\ =\frac{17}{7}\times \frac{5}{23} \\ =\frac{17\times 5}{7\times 23} \\ =\frac{85}{161} \end{gathered}$$

Hence, the correct answer is option (d).

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Question 11:

Mark the correct alternative in each of the following:

$\left(5\frac{1}{4}-3\frac{1}{3}\right)=$

(a) $\frac{12}{23}$                                      (b) 2                                       (c) $1\frac{11}{12}$                                      (d) $\frac{11}{12}$   

Answer 11:

$$\begin{gathered} 5\frac{1}{4}-3\frac{1}{3} \\ =\frac{21}{4}-\frac{10}{3} \\ =\frac{21\times 3}{4\times 3}-\frac{10\times 4}{3\times 4} \left(\mathrm{LCM} \mathrm{of} 3 \mathrm{and} 4 \mathrm{is} 12\right) \\ =\frac{63}{12}-\frac{40}{12} \end{gathered}$$
$$\begin{gathered} =\frac{63-40}{12} \\ =\frac{23}{12} \\ =1\frac{11}{12} \end{gathered}$$

Hence, the correct answer is option (c).

Question 12:

Mark the correct alternative in each of the following:

The fraction equivalent to $1\frac{2}{3}$ is

(a) $\frac{10}{3}$                                      (b) $\frac{3}{5}$                                       (c) $\frac{10}{6}$                                      (d) $\frac{6}{10}$  

Answer 12:

The given fraction is $1\frac{2}{3}=\frac{5}{3}$.

We know that if $\frac{a}{b}$ and $\frac{c}{d}$ are two equivalent fractions, then

$a\times d=b\times c$ 

Now,

$5\times 6=3\times 10$

So, the fractions $\frac{5}{3}$ and $\frac{10}{6}$ are equivalent fractions.

Thus, the fraction equivalent to $1\frac{2}{3}$ is $\frac{10}{6}$.

Hence, the correct answer is option (c).

Question 13:

Mark the correct alternative in each of the following:

By what number $9\frac{4}{5}$ be multiplied to get 42?

(a) $\frac{30}{7}$                                      (b) $\frac{7}{30}$                                       (c) $4\frac{1}{7}$                                      (d) $4\frac{3}{7}$   

Answer 13:

Product of two numbers = 42

One of the numbers = $9\frac{4}{5}=\frac{49}{5}$

∴ Other number = Product of two numbers ÷ One of the numbers

                          $$\begin{gathered} =42÷\frac{49}{5} \\ =\frac{42}{1}\times \frac{5}{49} \\ =\frac{6\times 5}{1\times 7} \\ =\frac{30}{7} \end{gathered}$$

Hence, the correct answer is option (a).

Question 14:

Mark the correct alternative in each of the following:

Which of the following statements is true?

(a) $\frac{7}{12}<\frac{4}{21}$                                      (b) $\frac{7}{12}=\frac{4}{21}$                                       (c) $\frac{7}{12}>\frac{4}{21}$                                      (d) None of these   

Answer 14:

Consider the fractions $\frac{7}{12}$ and $\frac{4}{21}$.

Prime factorisation of 12 = 2 × 2 × 3

Prime factorisation of 21 = 3 × 7

∴ LCM of 12 and 21 = 2 × 2 × 3 × 7 = 84

Firstly, convert the fractions to equivalent fractions with denominator 84.

$$\begin{gathered} \frac{7}{12}=\frac{7\times 7}{12\times 7}=\frac{49}{84} \\ \frac{4}{21}=\frac{4\times 4}{21\times 4}=\frac{16}{84} \end{gathered}$$

Now,

49 > 16

$$\begin{gathered} \therefore \frac{49}{84}>\frac{16}{84} \\ \Rightarrow \frac{7}{12}>\frac{4}{21} \end{gathered}$$

Hence, the correct answer is option (c).

Question 15:

Mark the correct alternative in each of the following:

Which one of the following is the correct statement?

(a) $\frac{3}{4}<\frac{2}{3}<\frac{12}{15}$                  (b) $\frac{2}{3}<\frac{3}{4}<\frac{12}{15}$                  (c) $\frac{2}{3}<\frac{12}{15}<\frac{3}{4}$                  (d) $\frac{12}{15}<\frac{2}{3}<\frac{3}{4}$  

Answer 15:

Consider the fractions $\frac{3}{4}$, $\frac{2}{3}$ and $\frac{12}{15}$.

LCM of 4, 3 and 15 = 60

Firstly, convert the fractions into equivalent fractions with denominator 60.

$$\begin{gathered} \frac{3}{4}=\frac{3\times 15}{4\times 15}=\frac{45}{60} \\ \frac{2}{3}=\frac{2\times 20}{3\times 20}=\frac{40}{60} \\ \frac{12}{15}=\frac{12\times 4}{15\times 4}=\frac{48}{60} \end{gathered}$$

Now,

40 < 45 <  48

$$\begin{gathered} \therefore \frac{40}{60}<\frac{45}{60}<\frac{48}{60} \\ \Rightarrow \frac{2}{3}<\frac{3}{4}<\frac{12}{15} \end{gathered}$$

Hence, the correct answer is option (b).

Question 16:

Mark the correct alternative in each of the following:

Which of the following fractions lies between $\frac{2}{3}$ and $\frac{5}{7}$?

(a) $\frac{3}{4}$                                      (b) $\frac{4}{5}$                                       (c) $\frac{5}{6}$                                      (d) None of these

Answer 16:

Consider the fractions $\frac{2}{3}$, $\frac{5}{7}$, $\frac{3}{4}$, $\frac{4}{5}$ and $\frac{5}{6}$.

LCM of 3, 4, 5, 6 and 7 = 420

Firstly, convert the fractions into equivalent fractions with denominator 420.

$$\begin{gathered} \frac{2}{3}=\frac{2\times 140}{3\times 140}=\frac{280}{420} \\ \frac{5}{7}=\frac{5\times 60}{7\times 60}=\frac{300}{420} \\ \frac{3}{4}=\frac{3\times 105}{4\times 105}=\frac{315}{420} \\ \frac{4}{5}=\frac{4\times 84}{5\times 84}=\frac{336}{420} \\ \frac{5}{6}=\frac{5\times 70}{6\times 70}=\frac{350}{420} \end{gathered}$$

Now,

280 < 300 < 315 < 336 < 350

$$\begin{gathered} \therefore \frac{280}{420}<\frac{300}{420}<\frac{315}{420}<\frac{336}{420}<\frac{350}{420} \\ \Rightarrow \frac{2}{3}<\frac{5}{7}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6} \end{gathered}$$

Thus, none of the fractions $\frac{3}{4}$, $\frac{4}{5}$, $\frac{5}{6}$ lies between the fractions $\frac{2}{3}$ and $\frac{5}{7}$.

Hence, the correct answer is option (d).

Question 17:

Mark the correct alternative in each of the following:

Which one of the following is true?

(a) $\frac{1}{2}<\frac{9}{13}<\frac{3}{4}<\frac{12}{17}$                                                     (b) $\frac{3}{4}<\frac{9}{13}<\frac{1}{2}<\frac{12}{17}$                                      

(c) $\frac{1}{2}<\frac{3}{4}<\frac{9}{13}<\frac{12}{17}$                                                     (d) $\frac{1}{2}<\frac{9}{13}<\frac{12}{17}<\frac{3}{4}$   

Answer 17:

Consider the fractions $\frac{1}{2}$, $\frac{9}{13}$, $\frac{3}{4}$ and $\frac{12}{17}$.

LCM of 2, 4, 13 and 17 = 884

Firstly, convert the fractions into equivalent fractions with denominator 884.

$$\begin{gathered} \frac{1}{2}=\frac{1\times 442}{2\times 442}=\frac{442}{884} \\ \frac{9}{13}=\frac{9\times 68}{13\times 68}=\frac{612}{884} \\ \frac{3}{4}=\frac{3\times 221}{4\times 221}=\frac{663}{884} \\ \frac{12}{17}=\frac{12\times 52}{17\times 52}=\frac{624}{884} \end{gathered}$$

Now,

442 < 612 < 624 < 663

$$\begin{gathered} \therefore \frac{442}{884}<\frac{612}{884}<\frac{624}{884}<\frac{663}{884} \\ \Rightarrow \frac{1}{2}<\frac{9}{13}<\frac{12}{17}<\frac{3}{4} \end{gathered}$$

Hence, the correct answer is option (d).

Question 18:

Mark the correct alternative in each of the following:

The smallest of the fractions $\frac{2}{3},\frac{4}{7},\frac{8}{11}$ and $\frac{5}{9}$ is

(a) $\frac{2}{3}$                                      (b) $\frac{4}{7}$                                       (c) $\frac{8}{11}$                                      (d) $\frac{5}{9}$   

Answer 18:

The given fractions are $\frac{2}{3},\frac{4}{7},\frac{8}{11}$ and $\frac{5}{9}$.

LCM of 3, 7, 9 and 11 = 693

Firstly, convert the fractions into equivalent fractions with denominator 693.

$$\begin{gathered} \frac{2}{3}=\frac{2\times 231}{3\times 231}=\frac{462}{693} \\ \frac{4}{7}=\frac{4\times 99}{7\times 99}=\frac{396}{693} \\ \frac{8}{11}=\frac{8\times 63}{11\times 63}=\frac{504}{693} \\ \frac{5}{9}=\frac{5\times 77}{9\times 77}=\frac{385}{693} \end{gathered}$$

Now,

385 < 396 < 462 < 504

$$\begin{gathered} \therefore \frac{385}{693}<\frac{396}{693}<\frac{462}{693}<\frac{504}{693} \\ \Rightarrow \frac{5}{9}<\frac{4}{7}<\frac{2}{3}<\frac{8}{11} \end{gathered}$$

Thus, the smallest of the given fractions is $\frac{5}{9}$.

Hence, the correct answer is option (d).

Question 19:

Mark the correct alternative in each of the following:

$9\times \left(-\frac{1}{3}\right)\times \left(-3\right)\times \left(-\frac{1}{9}\right)=$

(a) 1                                      (b) −1                                       (c) −3                                      (d) 3  

Answer 19:

Since the number of negative terms in the product is odd. Therefore, their product is negative.

$$\begin{gathered} 9\times \left(-\frac{1}{3}\right)\times \left(-3\right)\times \left(-\frac{1}{9}\right) \\ =9\times \left(-\frac{1}{9}\right)\times \left(-\frac{1}{3}\right)\times \left(-3\right) \\ =-\left(9\times \frac{1}{9}\times \frac{1}{3}\times 3\right) \\ =-\left(1\times 1\right) \\ =-1 \end{gathered}$$

Hence, the correct answer is option (b).

Question 20:

Mark the correct alternative in each of the following:

Which of the following is correct?

(a) $\frac{2}{3}<\frac{3}{5}<\frac{11}{15}$                  (b) $\frac{3}{5}<\frac{2}{3}<\frac{11}{15}$                  (c) $\frac{11}{15}<\frac{3}{5}<\frac{2}{3}$                  (d) $\frac{3}{5}<\frac{11}{15}<\frac{2}{3}$  

Answer 20:

Consider the fractions $\frac{2}{3}$, $\frac{3}{5}$ and $\frac{11}{15}$.

LCM of 3, 5 and 15 = 15

Firstly, convert the fractions into equivalent fractions with denominator 15.

$$\begin{gathered} \frac{2}{3}=\frac{2\times 5}{3\times 5}=\frac{10}{15} \\ \frac{3}{5}=\frac{3\times 3}{5\times 3}=\frac{9}{15} \end{gathered}$$

Now,

9 < 10 < 11

$$\begin{gathered} \therefore \frac{9}{15}<\frac{10}{15}<\frac{11}{15} \\ \Rightarrow \frac{3}{5}<\frac{2}{3}<\frac{11}{15} \end{gathered}$$

Hence, the correct answer is option (b).

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Question 21:

Mark the correct alternative in each of the following:

Which is the smallest of the following fractions?

(a) $\frac{4}{9}$                                 (b) $\frac{2}{5}$                                 (c) $\frac{3}{7}$                                  (d) $\frac{1}{4}$  

Answer 21:

Consider the fractions $\frac{4}{9}$, $\frac{2}{5}$, $\frac{3}{7}$ and $\frac{1}{4}$.

LCM of 4, 5, 7 and 9 = 1260

Firstly, convert the fractions into equivalent fractions with denominator 1260.

$$\begin{gathered} \frac{4}{9}=\frac{4\times 140}{9\times 140}=\frac{560}{1260} \\ \frac{2}{5}=\frac{2\times 252}{5\times 252}=\frac{504}{1260} \\ \frac{3}{7}=\frac{3\times 180}{7\times 180}=\frac{540}{1260} \\ \frac{1}{4}=\frac{1\times 315}{4\times 315}=\frac{315}{1260} \end{gathered}$$

Now,

315 < 504 < 540 < 560

$$\begin{gathered} \therefore \frac{315}{1260}<\frac{504}{1260}<\frac{540}{1260}<\frac{560}{1260} \\ \Rightarrow \frac{1}{4}<\frac{2}{5}<\frac{3}{7}<\frac{4}{9} \end{gathered}$$

Thus, the smallest fraction is $\frac{1}{4}$.

Hence, the correct answer is option (d).

Question 22:

Mark the correct alternative in each of the following:

The difference between the greatest and the least fractions out of $\frac{6}{7},\frac{7}{8},\frac{8}{9}$ and $\frac{9}{10}$ is

(a) $\frac{3}{10}$                            (b) $\frac{1}{56}$                             (c) $\frac{1}{40}$                            (d) $\frac{1}{72}$  

Answer 22:

Consider the fractions $\frac{6}{7},\frac{7}{8},\frac{8}{9}$  and $\frac{9}{10}$.

LCM of 7, 8, 9 and 10 = 2520

Firstly, convert the fractions into equivalent fractions with denominator 2520.

$$\begin{gathered} \frac{6}{7}=\frac{6\times 360}{7\times 360}=\frac{2160}{2520} \\ \frac{7}{8}=\frac{7\times 315}{8\times 315}=\frac{2205}{2520} \\ \frac{8}{9}=\frac{8\times 280}{9\times 280}=\frac{2240}{2520} \\ \frac{9}{10}=\frac{9\times 252}{10\times 252}=\frac{2268}{2520} \end{gathered}$$

Now,

2160 < 2205 < 2240 < 2268

$$\begin{gathered} \therefore \frac{2160}{2520}<\frac{2205}{2520}<\frac{2240}{2520}<\frac{2268}{2520} \\ \Rightarrow \frac{6}{7}<\frac{7}{8}<\frac{8}{9}<\frac{9}{10} \end{gathered}$$

So,

Greatest fraction = $\frac{9}{10}$

Least fraction = $\frac{6}{7}$

∴ Required difference

$$\begin{gathered} =\frac{9}{10}-\frac{6}{7} \\ =\frac{2268}{2520}-\frac{2160}{2520} \\ =\frac{2268-2160}{2520} \\ =\frac{108}{2520} \end{gathered}$$

$$\begin{gathered} =\frac{108÷36}{2520÷36} \left(\mathrm{HCF} \mathrm{of} 108 \mathrm{and} 2520=36\right) \\ =\frac{3}{70} \end{gathered}$$


Disclaimer: None of the options given in the question matches with the answer.

Question 23:

Mark the correct alternative in each of the following:

Which of the following fractions is greater than $\frac{3}{4}$ and less than $\frac{5}{6}$?

(a) $\frac{2}{3}$                               (b) $\frac{1}{2}$                               (c) $\frac{4}{5}$                               (d) $\frac{9}{10}$  

Answer 23:

Consider the fractions $\frac{3}{4},\frac{5}{6},\frac{2}{3},\frac{1}{2},\frac{4}{5}$ and $\frac{9}{10}$.

LCM of 2, 3, 4, 5, 6 and 10 = 60

Firstly, convert the fractions into equivalent fractions with denominator 60.

$$\begin{gathered} \frac{3}{4}=\frac{3\times 15}{4\times 15}=\frac{45}{60} \\ \frac{5}{6}=\frac{5\times 10}{6\times 10}=\frac{50}{60} \\ \frac{2}{3}=\frac{2\times 20}{3\times 20}=\frac{40}{60} \\ \frac{1}{2}=\frac{1\times 30}{2\times 30}=\frac{30}{60} \\ \frac{4}{5}=\frac{4\times 12}{5\times 12}=\frac{48}{60} \\ \frac{9}{10}=\frac{9\times 6}{10\times 6}=\frac{54}{60} \end{gathered}$$

Now,

30 < 40 < 45 < 48 < 50 < 54

$$\begin{gathered} \therefore \frac{30}{60}<\frac{40}{60}<\frac{45}{60}<\frac{48}{60}<\frac{50}{60}<\frac{54}{60} \\ \Rightarrow \frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{9}{10} \end{gathered}$$

Thus, the fraction $\frac{4}{5}$ is greater than $\frac{3}{4}$ and less than $\frac{5}{6}$.

Hence, the correct answer is option (c).

Question 24:

Mark the correct alternative in each of the following:

Which of the following fractions is more than one-third?

(a) $\frac{23}{70}$                               (b) $\frac{205}{819}$                               (c) $\frac{26}{75}$                               (d) $\frac{118}{335}$  

Answer 24:

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two fractions. Then, $\frac{a}{b}>\frac{c}{d}$ if $a\times d>b\times c$.

Consider the fractions $\frac{23}{70}$ and $\frac{1}{3}$.
$$\begin{gathered} 23\times 3=69 \\ 1\times 70=70 \\ \therefore 23\times 3<1\times 70 \\ \Rightarrow \frac{23}{70}<\frac{1}{3} \end{gathered}$$

Consider the fractions $\frac{205}{819}$ and $\frac{1}{3}$.
$$\begin{gathered} 205\times 3=615 \\ 1\times 819=819 \\ \therefore 205\times 3<1\times 819 \\ \Rightarrow \frac{205}{819}<\frac{1}{3} \end{gathered}$$

Consider the fractions $\frac{26}{75}$ and $\frac{1}{3}$.
$$\begin{gathered} 26\times 3=78 \\ 1\times 75=75 \\ \therefore 26\times 3>1\times 75 \\ \Rightarrow \frac{26}{75}>\frac{1}{3} \end{gathered}$$

Consider the fractions $\frac{118}{335}$ and $\frac{1}{3}$.
$$\begin{gathered} 118\times 3=354 \\ 1\times 335=335 \\ \therefore 118\times 3>1\times 335 \\ \Rightarrow \frac{118}{335}>\frac{1}{3} \end{gathered}$$

Thus, the fractions $\frac{26}{75}$ and $\frac{118}{335}$ are more than the fraction $\frac{1}{3}$.

Hence, the correct answers are options (c) and (d).


Disclaimer: There are two correct options in the question. One of the two options among (c) and (d) must be changed accordingly to get only one correct answer.