Fractions

We have:

$$\begin{array}{ll} \phantom{=} \frac{5}{6} \times \frac{3}{7} \\\\ = \frac{5 \times 3}{6 \times 7} \\\\ = \frac{5 \times 1}{2 \times 7} & \text{[Simplifying 3 and 6 ⇒ Divide by 3 ]} \\\\ = \frac{5}{14} \\\\ \end{array}$$

∴ The answer is $\frac{5}{14}$

We have:

$$\begin{array}{ll} \phantom{=} \frac{7}{18} \times \frac{9}{14} \\\\ = \frac{7 \times 9}{18 \times 14} \\\\ = \frac{7 \times 1}{2 \times 14} & \text{[Simplifying 9 and 18 ⇒ Divide by 9]} \\\\ = \frac{1 \times 1}{2 \times 2} & \text{[Simplifying 7 and 14 ⇒ Divide by 7]} \\\\ = \frac{1}{4} & \end{array}$$

∴ The answer is $\frac{1}{4}$

We have:

$$\begin{array}{ll} \phantom{=} 28 \times \frac{7}{8} \\\\ = \frac{28}{1} \times \frac{7}{8} \\\\ = \frac{7}{1} \times \frac{7}{2} & \text{[Simplifying 28 and 8 ⇒ Divide by 4]} \\\\ = \frac{7 \times 7}{1 \times 2} \\\\ = \frac{49}{2} & \\\\ = 24 \frac{1}{2} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The answer is 24 $\frac{1}{2}$

We have:

$$\begin{array}{ll} \phantom{=} 7 \times \frac{1}{7} \\\\ = \frac{7}{1} \times \frac{1}{7} \\\\ = \frac{7 \times 1}{1 \times 7} \\\\ = \frac{7}{7} \\\\ = 1 & \text{[Dividing both by 7]} \end{array}$$

∴ The answer is 1

We have:

$$\begin{array}{ll} \phantom{=} 2\frac{1}{25} \times \frac{5}{17} \\\\ = \frac{51}{25} \times \frac{5}{17} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \frac{51}{5} \times \frac{1}{17} & \text{[Simplifying 5 and 25 ⇒ Divide by 5]} \\\\ = \frac{3}{5} \times \frac{1}{1} & \text{[Simplifying 51 and 17 ⇒ Divide by 17]} \\\\ = \frac{3 \times 1}{5 \times 1} & \\\\ = \frac{3}{5} \end{array}$$

∴ The answer is $\frac{3}{5}$

We have:

$$\begin{array}{ll} \phantom{=} 1\frac{1}{13} \times 7\frac{3}{7} \\\\ = \frac{14}{13} \times \frac{52}{7} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \frac{2}{13} \times \frac{52}{1} & \text{[Simplifying 14 and 7 ⇒ Divide by 7]} \\\\ = \frac{2}{1} \times \frac{4}{1} & \text{[Simplifying 52 and 13 ⇒ Divide by 13]} \\\\ = \frac{2 \times 4}{1 \times 1} \\\\ = \frac{8}{1} \\\\ = 8 \end{array}$$

∴ The answer is 8

We have:

$$\begin{array}{ll} \phantom{=} \frac{4}{17} \times 7\frac{1}{12} \\\\ = \frac{4}{17} \times \frac{85}{12} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \frac{4}{1} \times \frac{5}{12} & \text{[Simplifying 85 and 17 ⇒ Divide by 17 ]} \\\\ = \frac{1}{1} \times \frac{5}{3} & \text{[Simplifying 4 and 12 ⇒ Divide by 4 ]} \\\\ = \frac{1 \times 5}{1 \times 3} \\\\ = \frac{5}{3} \end{array}$$

∴ The answer is $\frac{5}{3}$

We have:

$$\begin{array}{ll} \phantom{=} 7\frac{1}{4}\times\frac{7}{58}\times1\frac{11}{21} \\\\ = \frac{29}{4} \times \frac{7}{58} \times \frac{32}{21} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \frac{1}{4} \times \frac{7}{2} \times \frac{32}{21} & \text{[Simplifying 29 and 58 ⇒ Divide by 29 ]} \\\\ = \frac{1}{4} \times \frac{1}{2} \times \frac{32}{3} & \text{[Simplifying 7 and 21 ⇒ Divide by 7 ]} \\\\ = \frac{1}{1} \times \frac{1}{2} \times \frac{8}{3} & \text{[Simplifying 32 and 4 ⇒ Divide by 4 ]} \\\\ = \frac{1 \times 1 \times 8}{1 \times 2 \times 3} \\\\ = \frac{8}{6} \\\\ = \frac{4}{3} & \text{[Dividing both by 2]} \\\\ = 1\frac{1}{3} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The answer is $1\frac{1}{3}$

We have:

$$\begin{array}{ll} \phantom{=} \frac{1}{19}\times91\times5\frac{11}{13} \\\\ = \frac{1}{19} \times \frac{91}{1} \times \frac{76}{13} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \frac{1}{19} \times \frac{7}{1} \times \frac{76}{1} & \text{[Simplifying 91 and 13 ⇒ Divide by 13 ]} \\\\ = \frac{1}{1} \times \frac{7}{1} \times \frac{4}{1} & \text{[Simplifying 76 and 19 ⇒ Divide by 19 ]} \\\\ = \frac{1 \times 7 \times 4}{1 \times 1 \times 1} \\\\ = \frac{28}{1} \\\\ = 28 \end{array}$$

∴ The answer is 28

We have:

$$\begin{array}{ll} \phantom{=} 7\frac{1}{4}\times2\frac{3}{16}\times2\frac{2}{7} \\\\ = \frac{29}{4} \times \frac{35}{16} \times \frac{16}{7} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \frac{29}{4} \times \frac{35}{1} \times \frac{1}{7} & \text{[Simplifying 16 and 16 ⇒ Divide by 16 ]} \\\\ = \frac{29}{4} \times \frac{5}{1} \times \frac{1}{1} & \text{[Simplifying 35 and 7 ⇒ Divide by 7 ]} \\\\ = \frac{29 \times 5 \times 1}{4 \times 1 \times 1} \\\\ = \frac{145}{4} \\\\ = 36\frac{1}{4} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The answer is $36\frac{1}{4}$

We have:

$\frac{3}{4}$ of $\frac{8}{9}$

$$\begin{array}{ll} = \frac{8}{9} \times \frac{3}{4} \\\\ = \frac{2}{9} \times \frac{3}{1} & \text{[Simplifying 8 and 4 ⇒ Divide by 4 ]} \\\\ = \frac{2}{3} \times \frac{1}{1} & \text{[Simplifying 3 and 9 ⇒ Divide by 3 ]} \\\\ = \frac{2 \times 1}{3 \times 1} \\\\ = \frac{2}{3} \end{array}$$

∴ The answer is $\frac{2}{3}$

We have:

$\frac{1}{2}$ of $2\frac{2}{3}$

$$\begin{array}{ll} = \frac{8}{3} \times \frac{1}{2} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \frac{4}{3} \times \frac{1}{1} & \text{[Simplifying 8 and 2 ⇒ Divide by 2 ]} \\\\ = \frac{4}{3} \\\\ = 1\frac{1}{3} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The answer is 1 $\frac{1}{3}$

We have:

$\frac{4}{5}$ of 1 hour

$$\begin{array}{ll} 1 \text{ hour} = 60 \text{ minutes} & \text{[Converting hour into minutes]} \\\\ = \Big(60 \times \frac{4}{5}\Big) \text{ minutes} \\\\ = \Big(\frac{60}{1} \times \frac{4}{5}\Big) \text{ minutes} \\\\ = \Big(\frac{12}{1} \times \frac{4}{1}\Big) \text{ minutes} & \text{[Simplifying 60 and 5 ⇒ Divide by 5 ]} \\\\ = \frac{48}{1} \text{ minutes} \\\\ = 48 \text{ minutes} \end{array}$$

∴ The answer is 48 minutes

We have:

$\frac{3}{5}$ of ₹1

₹1 = 100 paise $\hspace{1.5cm}$[Converting rupees into paise]

$$\begin{array}{ll} = \Big(100 \times \frac{3}{5}\Big) \text{ paise} \\\\ = \Big(\frac{100}{1} \times \frac{3}{5}\Big) \text{ paise} \\\\ = \Big(\frac{20}{1} \times \frac{3}{1}\Big)\text{ paise} & \text{[Simplifying 100 and 5 ⇒ Divide by 5 ]} \\\\ = \Big(\frac{20 \times 3}{1 \times 1}\Big)\text{ paise} \\\\ = \frac{60}{1}\text{ paise} \\\\ = 60\text{ paise} \end{array}$$

∴ The answer is 60 paise

We have:

$\frac{8}{15}$ of $1\frac{1}{2}$ metres

$1\frac{1}{2}$ metres = $\frac{3}{2}$ metres $\hspace{1.5cm}$[Converting mixed to improper fraction]

1 meter = 100 cm

∴$\frac{3}{2}$ metres = 150 cm $\hspace{1.5cm}$[Converting meter to centimeter]

Now we have:

$\frac{8}{15}$ of 150 centimetres

$$\begin{array}{ll} = \Big(150 \times \frac{8}{15}\Big) \text{ cm} \\\\ = \Big(\frac{150}{1} \times \frac{8}{15}\Big) \text{ cm} \\\\ = \Big(\frac{10}{1} \times \frac{8}{1}\Big) \text{ cm} & \text{[Simplifying 150 and 15 ⇒ Divide by 15 ]} \\\\ = \Big(\frac{10 \times 8}{1 \times 1}\Big) \text{ cm} \\\\ = \frac{80}{1} \text{ cm} \\\\ = 80 \text{ cm} \end{array}$$

∴ The answer is 80 cm

We have:

$\frac{5}{7}$ of $2\frac{1}{3}$kg

$2\frac{1}{3}\text{ kg} = \frac{7}{3}\text{ kg} \hspace{1.5cm}$[Converting mixed to improper fraction]

Now we have:

$\frac{5}{7}$ of $\frac{7}{3}$ kg

$$\begin{array}{ll} = \Big(\frac{7}{3} \times \frac{5}{7}\Big) \text{ kg} \\\\ = \Big(\frac{1}{3} \times \frac{5}{1}\Big) \text{ kg} & \text{[Simplifying 7 and 7 ⇒ Divide by 7 ]} \\\\ = \Big(\frac{1 \times 5}{3 \times 1}\Big) \text{ kg} \\\\ = \frac{5}{3} \text{ kg} \\\\ = 1\frac{2}{3} \text{ kg} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The answer is $1\frac{2}{3}$ kg

Given:

Distance travelled in 1 litre of petrol = $12\frac{1}{2}$ km

= $\frac{25}{2}$ km $\hspace{1cm}$ [Converting mixed to improper fraction]

Distance travelled in $42\frac{3}{5}$ litres of petrol = ?

Number of litres = $42\frac{3}{5}$ = $\frac{213}{5}$ litres

Distance travelled in $\frac{213}{5}$ litres of petrol = (Distance travelled in 1 litre) x (Number of litres)

Substituting the values in above, we get:

$$\begin{array}{ll} = \Big(\frac{25}{2} \times \frac{213}{5}\Big)\text{ km} \\\\ = \Big(\frac{5}{2} \times \frac{213}{1}\Big)\text{ km} & \text{[Simplifying 25 and 5 ⇒ Divide by 5 ]} \\\\ = \Big(\frac{5 \times 213}{2 \times 1}\Big) \text{ km} \\\\ = \frac{1065}{2}\text{ km} \\\\ = 532 \frac{1}{2}\text{ km} & \text{[Converting improper fraction to mixed fraction]} \end{array}$$

∴ The distance travelled in $42\frac{3}{5}$ litres of petrol = $532\frac{1}{2}$ km.

Given:

Charge of 1 diagram = ₹$27\frac{3}{5} = ₹\frac{138}{5}$

Charge for 186 diagrams = ?

Number of diagrams = 186

Charge for 186 diagrams = (Charge of 1 diagram) x (Number of diagrams)

Substituting the values in above, we get:

$$\begin{array}{ll} = ₹\Big(\frac{138}{5} \times 186\Big) \\\\ = ₹\Big(\frac{138 \times 186}{5 \times 1}\Big) \\\\ = ₹\frac{25668}{5} \\\\ = ₹5133\frac{3}{5} & \text{[Converting improper fraction to mixed fraction]} \end{array}$$

∴ The charge for 186 diagrams = ₹ $5133\frac{3}{5}$.

Cost of 1 meter of cloth = ₹$715\frac{1}{4} = ₹\frac{2861}{4}$

Cost of $3\frac{2}{5}$ meters of cloth = ?

Length of cloth = $3\frac{2}{5} \text{ meters} = \frac{17}{5}$ meters

Cost of $3\frac{2}{5}$ meters of cloth = (Cost of 1 meter of cloth) x (Length of cloth)

Substituting the values in above, we get:

$$\begin{array}{ll} = ₹\Big(\frac{2861}{4} \times \frac{17}{5}\Big) \\\\ = ₹\Big(\frac{2861 \times 17}{4 \times 5}\Big) \\\\ = ₹\frac{48637}{20} \\\\ = ₹2431\frac{17}{20} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The cost of $3\frac{2}{5}$ meters of cloth = ₹ $2431\frac{17}{20}$.

Cost of an advertisement of 1 square inch = ₹$1472\frac{2}{5} = ₹\frac{7362}{5}$

Cost of an advertisement of $17\frac{6}{7}$ square inch = ?

Area of an advertisement = $17\frac{6}{7}$ square inch

Cost of an advertisement of $17\frac{6}{7}$ square inch = (Cost of an advertisement of 1 square inch) x (Area of an advertisement)

Substituting the values in above, we get:

$$\begin{array}{ll} = ₹\Big(\frac{7362}{5} \times 17\frac{6}{7}\Big) \\\\ = ₹\Big(\frac{7362}{5} \times \frac{125}{7}\Big) & \text{[Converting mixed to improper fraction]} \\\\ = ₹\Big(\frac{7362}{1} \times \frac{25}{7}\Big) \\\\ = ₹\Big(\frac{7362 \times 25}{1 \times 7}\Big) \\\\ = ₹\frac{184050}{7} \\\\ = ₹26292 \frac{6}{7} & \text{[Converting improper fraction to mixed fraction]} \end{array}$$

∴ The cost of an advertisement of $17\frac{6}{7}$ square inch = ₹ $26292\frac{6}{7}$.

Given:

Speed = $52\frac{2}{7}$ km/hour

Time = $4\frac{3}{8}$ hours

Distance = ?

We know the formula,

Distance = Speed x Time

By substituting the values we get,

$$\begin{array}{ll} \text{Distance} = \Big(52\frac{2}{7} \times 4\frac{3}{8}\Big) \text{ km} & \space \\\\ = \Big(\frac{366}{7} \times \frac{35}{8}\Big) \text{ km} & \text{[Converting mixed to improper fraction]} \\\\ = \Big(\frac{366}{1} \times \frac{5}{8}\Big)\text{ km} & \text{[Simplifying 35 and 7 ⇒ Divide by 7 ]} \\\\ = \frac{366 \times 5}{1 \times 8}\text{ km} \\\\ = \frac{1830}{8} \text{ km} \\\\ = 228\frac{3}{4}\text{ km} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The distance between two cities is $228\frac{3}{4}$ km.

Given:

Length of a rectangular plot of land = $29\frac{3}{7}$ m

Breadth of a rectangular plot of land = $12\frac{8}{11}$ m

Area of a rectangular plot of land = ?

We know the formula,

Area of rectangle = Length x Breadth

By substituting the values we get,

Area of rectangle = Length x Breadth

$$\begin{array}{ll} = \Big(29\frac{3}{7} \times 12\frac{8}{11}\Big) \text{m}^2 \\\\ = \Big(\frac{206}{7} \times \frac{140}{11}\Big) \text{m}^2 & \text{ [Converting mixed to improper fraction]} \\\\ = \Big(\frac{206}{1} \times \frac{20}{11}\Big)\text{m}^2 & \text{[Simplifying 140 and 7 ⇒ Divide by 7 ]} \\\\ = \frac{206 \times 20}{1 \times 11} \text{m}^2 \\\\ = \frac{4120}{11}\text{m}^2 \\\\ = 374\frac{6}{11}\text{m}^2 & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The area of a rectangular plot of land is $374\frac{6}{11}$ m²

We have:

$\frac{13}{17}$

To find the reciprocal of a number, we interchange its numerator and denominator.

$\frac{13}{17}$ = $\frac{17}{13}$

= $1\frac{4}{13} \hspace{2cm}\text{[Converting improper to mixed fraction]}$

∴ The reciprocal of $\frac{13}{17}$ is $1\frac{4}{13}$

We have:

$\frac{3}{313}$

Interchanging its numerator and denominator we get,

$\frac{3}{313}$ = $\frac{313}{3}$

= $104\frac{1}{3} \hspace{2cm}\text{[Converting improper to mixed fraction]}$

∴ The reciprocal of $\frac{3}{313}$ is $104\frac{1}{3}$

We have:

217

This can be written as $\frac{217}{1}$

Interchanging its numerator and denominator we get,

$\frac{217}{1}$ = $\frac{1}{217}$

∴ The reciprocal of $\frac{217}{1}$ is $\frac{1}{217}$

We have:

$\frac{1}{1024}$

Interchanging its numerator and denominator we get,

$\frac{1}{1024}$ = $\frac{1024}{1}$ = 1024

∴ The reciprocal of $\frac{1}{1024}$ is 1024

We have:

$3\frac{1}{3} = \frac{10}{3} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}$

Interchanging its numerator and denominator we get,

$\frac{10}{3}$ = $\frac{3}{10}$

∴ The reciprocal of $\frac{10}{3}$ is $\frac{3}{10}$

We have:

$11\frac{1}{11}$ = $\frac{122}{11} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}$

Interchanging its numerator and denominator we get,

$\frac{122}{11} = \frac{11}{122}$

∴ The reciprocal of $\frac{122}{11}$ is $\frac{11}{122}$

We have:

$1\frac{1}{2} = \frac{3}{2} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}$

Interchanging its numerator and denominator we get,

$\frac{3}{2}$ = $\frac{2}{3}$

∴ The reciprocal of $\frac{3}{2}$ is $\frac{2}{3}$

We have:

$125\frac{1}{8} = \frac{1001}{8} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}$

Interchanging its numerator and denominator we get,

$\frac{1001}{8}$ = $\frac{8}{1001}$

∴ The reciprocal of $\frac{1001}{8}$ is $\frac{8}{1001}$

We have:

$$\begin{array}{ll} \phantom{=} \frac{1}{2} ÷ \frac{1}{3} \\\\ = \frac{1}{2} \times \frac{3}{1} & [\text{Reciprocal of } \frac{1}{3} \text{ is } \frac{3}{1}] \\\\ = \frac{1 \times 3}{2 \times 1} \\\\ = \frac{3}{2} \\\\ = 1\frac{1}{2} \end{array}$$

∴ The answer is $1\frac{1}{2}$

We have:

$$\begin{array}{ll} \phantom{=} \frac{3}{14} ÷ \frac{11}{42} \\\\ = \frac{3}{14} \times \frac{42}{11} & [\text{Reciprocal of } \frac{11}{42} \text{ is } \frac{42}{11}] \\\\ = \frac{3 \times 42}{14 \times 11} \\\\ = \frac{3 \times 3}{1 \times 11} & \text{[Simplifying 42 and 14 ⇒ Divide by 14]} \\\\ = \frac{9}{11} \\\\ \end{array}$$

∴ The answer is $\frac{9}{11}$

We have:

$$\begin{array}{ll} \phantom{=} 1 ÷ 6\frac{2}{3} \\\\ = 1 ÷ \frac{20}{3} & \text{[Converting mixed to improper fraction]} \\\\ = 1 \times \frac{3}{20} & [\text{Reciprocal of } \frac{20}{3} \text{ is } \frac{3}{20}] \\\\ = \frac{1}{1} \times \frac{3}{20} \\\\ = \frac{1 \times 3}{1 \times 20} \\\\ = \frac{3}{20} \\\\ \end{array}$$

∴ The answer is $\frac{3}{20}$

We have:

$$\begin{array}{ll} \phantom{=} 13\frac{1}{3} ÷ 10 \\\\ = \frac{40}{3} ÷ 10 & \text{[Converting mixed to improper fraction]} \\\\ = \frac{40}{3} \times \frac{1}{10} & [\text{Reciprocal of } 10 \text{ is } \frac{1}{10}] \\\\ = \frac{4}{3} \times \frac{1}{1} & \text{[Simplifying 40 and 10 ⇒ Divide by 10]} \\\\ = \frac{4 \times 1}{3 \times 1} \\\\ = \frac{4}{3} \\\\ = 1\frac{1}{3} \end{array}$$

∴ The answer is $1\frac{1}{3}$

We have:

$$\begin{array}{ll} \phantom{=} 12\frac{1}{12} ÷ \frac{5}{36} \\\\ = \frac{145}{12} ÷ \frac{5}{36} & \text{[Converting mixed to improper fraction]} \\\\ = \frac{145}{12} \times \frac{36}{5} & [\text{Reciprocal of } \frac{5}{36} \text{ is } \frac{36}{5}] \\\\ = \frac{29}{12} \times \frac{36}{1} & \text{[Simplifying 145 and 5 ⇒ Divide by 5]} \\\\ = \frac{29}{1} \times \frac{3}{1} & \text{[Simplifying 36 and 12 ⇒ Divide by 12]} \\\\ = \frac{29 \times 3}{1 \times 1} \\\\ = \frac{87}{1} \\\\ = 87 \end{array}$$

∴ The answer is 87

We have:

$$\begin{array}{ll} \phantom{=} \frac{13}{46} ÷ 2\frac{1}{11} \\\\ = \frac{13}{46} ÷ \frac{23}{11} & \text{[Converting mixed to improper fraction]} \\\\ = \frac{13}{46} \times \frac{11}{23} & [\text{Reciprocal of } \frac{23}{11} \text{ is } \frac{11}{23}] \\\\ = \frac{13 \times 11}{46 \times 23} \\\\ = \frac{143}{1058} \end{array}$$

∴ The answer is $\frac{143}{1058}$

We have:

$$\begin{array}{ll} \phantom{=} \frac{20}{31} ÷ 54 \\\\ = \frac{20}{31} \times \frac{1}{54} & [\text{Reciprocal of } 54 \text{ is } \frac{1}{54}] \\\\ = \frac{10}{31} \times \frac{1}{27} & \text{[Simplifying 20 and 54 ⇒ Divide by 2]} \\\\ = \frac{10 \times 1}{31 \times 27} \\\\ = \frac{10}{837} \end{array}$$

∴ The answer is $\frac{10}{837}$

We have:

$$\begin{array}{ll} \phantom{=} 9\frac{4}{5} ÷ 3\frac{23}{25} \\\\ = \frac{49}{5} ÷ \frac{98}{25} & \text{[Converting mixed to improper fraction]} \\\\ = \frac{49}{5} \times \frac{25}{98} & [\text{Reciprocal of } \frac{98}{25} \text{ is } \frac{25}{98}] \\\\ = \frac{1}{5} \times \frac{25}{2} & \text{[Simplifying 49 and 98 ⇒ Divide by 49]} \\\\ = \frac{1}{1} \times \frac{5}{2} & \text{[Simplifying 25 and 5 ⇒ Divide by 5]} \\\\ = \frac{1 \times 5}{1 \times 2} \\\\ = \frac{5}{2} \\\\ = 2\frac{1}{2} \end{array}$$

∴ The answer is $2\frac{1}{2}$

We have:

$$\begin{array}{ll} \phantom{=} 32\frac{1}{2} ÷ 8\frac{3}{4} \\\\ = \frac{65}{2} ÷ \frac{35}{4} & \text{[Converting mixed to improper fraction]} \\\\ = \frac{65}{2} \times \frac{4}{35} & [\text{Reciprocal of } \frac{35}{4} \text{ is } \frac{4}{35}] \\\\ = \frac{13}{2} \times \frac{4}{7} & \text{[Simplifying 65 and 35 ⇒ Divide by 5]} \\\\ =\frac{13}{1} \times \frac{2}{7} & \text{[Simplifying 4 and 2 ⇒ Divide by 2]} \\\\ = \frac{13 \times 2}{1 \times 7} \\\\ = \frac{26}{7} \\\\ = 3\frac{5}{7} \end{array}$$

∴ The answer is $3\frac{5}{7}$

We have:

$$\begin{array}{ll} \phantom{=} 8\frac{1}{21} ÷ 1\frac{6}{7} \\\\ = \frac{169}{21} ÷ \frac{13}{7} & \text{[Converting mixed to improper fraction]} \\\\ = \frac{169}{21} \times \frac{7}{13} & [\text{Reciprocal of } \frac{13}{7} \text{ is} \frac{7}{13}] \\\\ = \frac{13}{21} \times \frac{7}{1} & \text{[Simplifying 169 and 13 ⇒ Divide by 13]} \\\\ = \frac{13}{3} \times \frac{1}{1} & \text{[Simplifying 7 and 21 ⇒ Divide by 7]} \\\\ = \frac{13 \times 1}{3 \times 1} \\\\ = \frac{13}{3} \\\\ = 4\frac{1}{3} \end{array}$$

∴ The answer is $4\frac{1}{3}$

We have:

$$\begin{array}{ll} \phantom{=} 6\frac{3}{16} ÷ 3\frac{1}{7} \\\\ = \frac{99}{16} ÷ \frac{22}{7} & \text{[Converting mixed to improper fraction]} \\\\ = \frac{99}{16} \times \frac{7}{22} & [\text{Reciprocal of } \frac{22}{7} \text{ is } \frac{7}{22}] \\\\ = \frac{9}{16} \times \frac{7}{2} & \text{[Simplifying 99 and 22 ⇒ Divide by 11]} \\\\ = \frac{9 \times 7}{16 \times 2} \\\\ = \frac{63}{32} \\\\ = 1\frac{31}{32} \end{array}$$

∴ The answer is $1\frac{31}{32}$

We have:

$$\begin{array}{ll} \phantom{=} 3\frac{11}{15} ÷ 19\frac{3}{5} \\\\ = \frac{56}{15} ÷ \frac{98}{5} & \text{[Converting mixed to improper fraction]} \\\\ = \frac{56}{15} \times \frac{5}{98} & [\text{Reciprocal of } \frac{98}{5} \text{ is } \frac{5}{98}] \\\\ = \frac{56}{3} \times \frac{1}{98} & \text{[Simplifying 5 and 15 ⇒ Divide by 5]} \\\\ = \frac{4}{3} \times \frac{1}{7} & \text{[Simplifying 56 and 98 ⇒ Divide by 14]} \\\\ = \frac{4 \times 1}{3 \times 7} \\\\ = \frac{4}{21} \end{array}$$

∴ The answer is $\frac{4}{21}$

Given:

The fraction by which $\frac{9}{35}$ must be multiplied to get $\frac{7}{15}$ = ?

Let the required fraction be x

Now we have,

$\frac{9}{35} \times x = \frac{7}{15}$

x = $\frac{7}{15} ÷ \frac{9}{35}$

$$\begin{array}{ll} = \frac{7}{15} \times \frac{35}{9} & [\text{Reciprocal of } \frac{9}{35} \text{ is } \frac{35}{9}] \\\\ = \frac{7}{3} \times \frac{7}{9} & \text{[Simplifying 35 and 15 ⇒ Divide by 5]} \\\\ = \frac{7 \times 7}{3 \times 9} \\\\ = \frac{49}{27} \\\\ = 1\frac{22}{27} \end{array}$$

∴ The answer is $1\frac{22}{27}$

Given:

Total cost = ₹$1421\frac{7}{8} = ₹\frac{11375}{8}$

Total length = $6\frac{1}{4}\text{ m} = \frac{25}{4}\text{ m}$

Cost of 1 metre of cloth = ?

Cost of 1 metre of cloth = (Total cost) ÷ (Total length)

Substituting the values in above, we get:

Cost of 1 metre of cloth = ₹ $\frac{11375}{8} ÷ \frac{25}{4}$ m

$$\begin{array}{ll} = ₹\Big(\frac{11375}{8} \times \frac{4}{25}\Big) & [\text{Reciprocal of } \frac{25}{4} \text{ is } \frac{4}{25}] \\\\ = ₹\Big(\frac{455}{8} \times \frac{4}{1}\Big) & \text{[Simplifying 11375 and 25 ⇒ Divide by 25]} \\\\ = ₹\Big(\frac{455}{2} \times \frac{1}{1}\Big) & \text{[Simplifying 4 and 8 ⇒ Divide by 4]} \\\\ = ₹\frac{455 \times 1}{2 \times 1} \\\\ = ₹\frac{455}{2} \\\\ = ₹227\frac{1}{2} \end{array}$$

∴ The cost of 1 metre of cloth = ₹$227\frac{1}{2}$

Given:

Piece length = $3\frac{5}{7}$ m = $\frac{26}{7}$ m $\hspace{1.5cm}\text{[Converting mixed to improper fraction]}$

Total length = 52 m

Number of pieces = ?

Number of pieces = (Total length) ÷ (Piece length)

Substituting the values in above, we get:

Number of pieces = 52 m ÷ $\frac{26}{7}$ m

$$\begin{array}{ll} = \Big(\frac{52}{1} \times \frac{7}{26}\Big) & [\text{Reciprocal of } \frac{26}{7} \text{ is } \frac{7}{26}] \\\\ = \Big(\frac{2}{1} \times \frac{7}{1}\Big) & \text{[Simplifying 52 and 26 ⇒ Divide by 26]} \\\\ = \Big(\frac{2 \times 7}{1 \times 1}\Big) \\\\ = \frac{14}{1} \\\\ = 14 \end{array}$$

∴ Number of pieces = 14

Given :

Total length of a log of wood = $6\frac{2}{7}$ m = $\frac{44}{7}$ m

Number of pieces = 11

Length of each piece = ?

Length of each piece = (Total length of a log of wood) ÷ (Number of pieces)

Substituting the values in above, we get:

Length of each piece = $\frac{44}{7}$ m ÷ 11

$$\begin{array}{ll} = \Big(\frac{44}{7} \times \frac{1}{11}\Big)\text{ m} & [\text{Reciprocal of } 11 \text{ is } \frac{1}{11}] \\\\ = \Big(\frac{4}{7} \times \frac{1}{1}\Big)\text{ m} & \text{[Simplifying 44 and 11 ⇒ Divide by 11]} \\\\ = \frac{4}{7}\text{ m} \\\\ \end{array}$$

∴ The length of each piece = $\frac{4}{7}$ m

Given:

Total distance = $641\frac{1}{4}\text{ km} = \frac{2565}{4}$ km

Total fuel = $43\frac{1}{8}\text{ litres} = \frac{345}{8}\text{ litres}$

Distance per litre = ?

Distance per litre = (Total distance) ÷ (Total fuel)

Substituting the values in above, we get:

Distance per litre = $\frac{2565}{4}\text{ km} ÷ \frac{345}{8}\text{ litres}$

$$\begin{array}{ll} = \Big(\frac{2565}{4} \times \frac{8}{345}\Big)\text{ km} & [\text{Reciprocal of } \frac{345}{8} \text{ is } \frac{8}{345}] \\\\ = \Big(\frac{2565}{1} \times \frac{2}{345}\Big)\text{ km} & \text{[Simplifying 8 and 4 ⇒ Divide by 4]} \\\\ = \Big(\frac{171}{1} \times \frac{2}{23}\Big)\text{ km} & \text{[Simplifying 2565 and 345 ⇒ Divide by 15]} \\\\ = \frac{171 \times 2}{1 \times 23} \text{ km} \\\\ = \frac{342}{23} \text{ km} \\\\ = 14\frac{20}{23} \text{ km} \end{array}$$

∴ The distance per litre of petrol = $14\frac{20}{23}$ km

Given:

Area of rectangular plot of land = $817\frac{4}{5}\text{ m}^2 = \frac{4089}{5}\text{ m}^2$

Breadth = $21\frac{3}{4}\text{ m} = \frac{87}{4} \text{ m}$

Length of rectangular plot of land = ?

Length of rectangular plot of land = (Area of rectangular plot of land) ÷ (Breadth)

Substituting the values in above, we get:

Length of rectangular plot of land = $\frac{4089}{5}\text{ m}^2 ÷ \frac{87}{4} \text{ m}$

$$\begin{array}{ll} = \Big(\frac{4089}{5} \times \frac{4}{87}\Big)\text{ m} & [\text{Reciprocal of } \frac{87}{4} \text{ is } \frac{4}{87}] \\\\ = \Big(\frac{47}{5} \times \frac{4}{1}\Big)\text{ m} & \text{[Simplifying 2565 and 345 ⇒ Divide by 15]} \\\\ = \frac{47 \times 4}{5 \times 1} \text{ m} \\\\ = \frac{188}{5} \text{ m} \\\\ = 37\frac{3}{5}\text{ m} \end{array}$$

∴ The breadth of rectangular plot of land = $37\frac{3}{5}$ m

Given:

Area of a sheet of paper = $623\frac{7}{10}\text{ cm}^2 = \frac{6237}{10}\text{ cm}^2$

Length = $29\frac{7}{10}\text{ cm} = \frac{297}{10}\text{ cm}$

Width = ?

The sheet of paper will be in rectangular shape,

∴ Area = Length x Width

Width = Area ÷ Length

Substituting the values in above, we get:

Width = $\frac{6237}{10}\text{ cm}^2$ ÷ $\frac{297}{10}\text{ cm}$

$$\begin{array}{ll} = \Big(\frac{6237}{10} \times \frac{10}{297}\Big)\text{ cm} & [\text{Reciprocal of } \frac{297}{10} \text{ is } \frac{10}{297}] \\\\ = \Big(\frac{6237}{1} \times \frac{1}{297}\Big)\text{ cm} & \text{[Simplifying 10 and 10 ⇒ Divide by 1]} \\\\ = \frac{6237 \times 1}{1 \times 297} \text{ cm} \\\\ = \frac{6237}{297}\text{ cm} \\\\ = 21 \text{ cm} \end{array}$$

∴ The width of a sheet of paper = 21 cm

Given:

Total weigth = $2\frac{3}{10} \text{ kg} = \frac{23}{10} \text{ kg}$

Number of sheets = 500

Weigth of 1 sheet of paper = (Total weigth) ÷ (Number of sheets)

Substituting the values in above, we get:

Weigth of 1 sheet of paper = $\frac{23}{10}$ kg ÷ 500

$$\begin{array}{ll} = \Big(\frac{23}{10} \times \frac{1}{500}\Big)\text{ kg} & [\text{Reciprocal of } 500 \text{ is } \frac{1}{500}] \\\\ = \frac{23 \times 1}{10 \times 500} \text{ kg} \\\\ = \frac{23}{5000} \text{ kg} \end{array}$$

∴ The weigth of 1 sheet of paper = $\frac{23}{5000}$ kg

Given:

Total distance = $283\frac{1}{2}\text{ km} = \frac{567}{2}\text{ km}$

Total time = $4\frac{2}{3}\text{ hours} = \frac{14}{3}\text{ hours}$

Distance covered in 1 hour = ?

Distance covered in 1 hour = (Total distance) ÷ (Total time)

Substituting the values in above, we get:

Distance covered in 1 hour = $\frac{567}{2}\text{ km} ÷ \frac{14}{3}\text{ hours}$

$$\begin{array}{ll} = \Big(\frac{567}{2} \times \frac{3}{14}\Big) \text{ km} & [\text{Reciprocal of } \frac{14}{3} \text{ is } \frac{3}{14}] \\\\ = \Big(\frac{81}{2} \times \frac{3}{2}\Big) \text{ km} & \text{[Simplifying 567 and 14 ⇒ Divide by 7]} \\\\ = \frac{81 \times 3}{2 \times 2}\text{ km} \\\\ = \frac{243}{4}\text{ km} \\\\ = 60\frac{3}{4}\text{ km} \end{array}$$

∴ The distance covered in 1 hour = $60\frac{3}{4}$ km

Given:

Let the two fractions be p and q.

p = $3\frac{1}{15} = \frac{46}{15}$

Let the other fraction be q.

p x q = $8\frac{7}{25} = \frac{207}{25}$

q = $\frac{207}{25}$ ÷ p $\hspace{2cm}\text{[Solving for q]}$

Substituting the value of p, we get:

q = $\frac{207}{25} ÷ \frac{46}{15}$

$$\begin{array}{ll} = \frac{207}{25} \times \frac{15}{46} & [\text{Reciprocal of } \frac{46}{15} \text{ is } \frac{15}{46}] \\\\ = \frac{207}{5} \times \frac{3}{46} & \text{[Simplifying 15 and 25 ⇒ Divide by 5]} \\\\ = \frac{9}{5} \times \frac{3}{2} & \text{[Simplifying 207 and 46 ⇒ Divide by 23]} \\\\ = \frac{9 \times 3}{5 \times 2} \\\\ = \frac{27}{10} \\\\ = 2\frac{7}{10} \end{array}$$

∴ The other fraction = $2\frac{7}{10}$

We have:

$1\frac{2}{3} + \frac{5}{6}$ of $\frac{24}{25}$

= $\frac{5}{3}$ + $\frac{5}{6}$ of $\frac{24}{25}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$$\begin{array}{ll} = \frac{5}{3} + \frac{5}{6} \times \frac{24}{25} \\\\ = \frac{5}{3} + \frac{1}{6} \times \frac{24}{5} \\\\ = \frac{5}{3} + \frac{24}{30} \\\\ = \frac{5}{3} + \frac{4}{5} & \text{[Of simplified]} \\\\ = \frac{25 + 12}{15} = \frac{37}{15} & \text{[Addition simplified]} \\\\ = 2\frac{7}{15} & \text{[Converting improper to mixed fraction]} \end{array}$$

∴ The answer is $2\frac{7}{15}$

We have:

$\frac{1}{3}$ of $4\frac{2}{3}$ ÷ $2\frac{1}{3}\times 1\frac{1}{2}$

= $\frac{1}{3}$ of $\frac{14}{3}$ ÷ $\frac{7}{3}\times \frac{3}{2}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$$\begin{array}{ll} = \frac{1}{3} \times \frac{14}{3} ÷ \frac{7}{3}\times \frac{3}{2} \\\\ = \frac{1 \times 14}{3 \times 3} ÷ \frac{7}{3}\times \frac{3}{2} \\\\ = \frac{14}{9} ÷ \frac{7}{3}\times \frac{3}{2} & \text{[Of simplified]} \\\\ = \frac{14}{9} \times \frac{3}{7}\times \frac{3}{2} & [\text{Reciprocal of } \frac{7}{3} \text{ is } \frac{3}{7}] \\\\ = \frac{14}{3} \times \frac{1}{7} \times \frac{3}{2} \\\\ = \frac{2}{3} \times \frac{1}{1} \times \frac{3}{2} \\\\ = \frac{2 \times1}{3 \times 1} \times \frac{3}{2} = \frac{2}{3} \times \frac{3}{2} & \text{[Division simplified]} \\\\ = \frac{2 \times 3}{3 \times 2} = \frac{6}{6}\\\\ = \frac{1}{1} = 1 & \text{[Multiplication simplified]} \end{array}$$

∴ The answer is 1

We have:

$2\frac{1}{4} + 1\frac{1}{6} - 1\frac{2}{3} ÷ 2\frac{2}{3}$ of $3\frac{3}{4}$

= $\frac{9}{4}$ + $\frac{7}{6}$ - $\frac{5}{3}$ ÷ $\frac{8}{3}$ of $\frac{15}{4}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$$\begin{array}{ll} = \frac{9}{4} + \frac{7}{6} - \frac{5}{3} ÷ \frac{8}{3}\times \frac{15}{4} \\\\ = \frac{9}{4} + \frac{7}{6} - \frac{5}{3} ÷ \frac{8 \times 15}{3 \times 4} \\\\ = \frac{9}{4} + \frac{7}{6} - \frac{5}{3} ÷ \frac{120}{12} \\\\ = \frac{9}{4} + \frac{7}{6} - \frac{5}{3} ÷ \frac{10}{1} & \text{[Of simplified]} \\\\ = \frac{9}{4} + \frac{7}{6} - \frac{5}{3} \times \frac{1}{10} & [\text{Reciprocal of } \frac{10}{1} \text{ is } \frac{1}{10}] \\\\ = \frac{9}{4} + \frac{7}{6} - \frac{5 \times 1}{3 \times 10} \\\\ = \frac{9}{4} + \frac{7}{6} - \frac{5}{30} \\\\ = \frac{9}{4} + \frac{7}{6} - \frac{1}{6} \\\\ = \frac{27 + 14}{12} - \frac{1}{6} \\\\ = \frac{41}{12} - \frac{1}{6} & \text{[Addition simplified]} \\\\ = \frac{41 - 2}{12} \\\\ = \frac{39}{12} \\\\ = \frac{13}{4} & \text{[Subtraction simplified]} \\\\ = 3\frac{1}{4} & \text{[Converting improper to mixed fraction]} \\\\ \end{array}$$

∴ The answer is $3\frac{1}{4}$

We have:

$1\frac{1}{2}\times2\frac{3}{4} ÷ 1\frac{4}{7}$ of $2\frac{5}{8}$

= $\frac{3}{2}\times \frac{11}{4}$ ÷ $\frac{11}{7}$ of $\frac{21}{8}$ [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

$$\begin{array}{ll} = \frac{3}{2}\times \frac{11}{4} ÷ \frac{11}{7} \times \frac{21}{8} \\\\ = \frac{3}{2}\times \frac{11}{4} ÷ \frac{11}{1} \times \frac{3}{8} \\\\ = \frac{3}{2}\times \frac{11}{4} ÷ \frac{11 \times 3}{1 \times 8} \\\\ = \frac{3}{2}\times \frac{11}{4} ÷ \frac{33}{8} & \text{[Of simplified]} \\\\ = \frac{3}{2}\times \frac{11}{4} \times \frac{8}{33} & [\text{Reciprocal of } \frac{33}{8} \text{ is } \frac{8}{33}] \\\\ = \frac{3}{2}\times \frac{1}{4} \times \frac{8}{3} \\\\ = \frac{3}{2}\times \frac{1}{1} \times \frac{2}{3}\\\\ = \frac{3}{2}\times \frac{1 \times 2}{1 \times 3} \\\\ = \frac{3}{2}\times \frac{2}{3} & \text{[Division simplified]} \\\\ = \frac{6}{6} = 1 & \text{[Multiplication simplified]} \\\\ \end{array}$$

∴ The answer is 1

We have:

$\Big(2\frac{3}{4} + 1\frac{5}{6}\Big)$ ÷ $2\frac{1}{5}$ of $3\frac{1}{3}$

= $\Big(\frac{11}{4} + \frac{11}{6}\Big)$ ÷ $\frac{11}{5}$ of $\frac{10}{3}$ [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

$$\begin{array}{ll} = \Big(\frac{33 + 22}{12}\Big) ÷ \frac{11}{5} \text{ of } \frac{10}{3} \\\\ = \frac{55}{12} ÷ \frac{11}{5} \text{ of } \frac{10}{3} & \text{[Brackets simplified]} \\\\ = \frac{55}{12} ÷ \frac{11}{5} \times \frac{10}{3} \\\\ = \frac{55}{12} ÷ \frac{110}{15} \\\\ = \frac{55}{12} ÷ \frac{22}{3} & \text{[Of simplified]} \\\\ = \frac{55}{12} \times \frac{3}{22} & [\text{Reciprocal of } \frac{22}{3} \text{ is } \frac{3}{22}] \\\\ = \frac{5}{12} \times \frac{3}{2} & \text{[Dividing 55 and 22 by 11]} \\\\ = \frac{15}{24} = \frac{5}{8} & \text{[Division simplified]} \end{array}$$

∴ The answer is $\frac{5}{8}$

We have:

$\frac{7}{15}$ of $\Big(\frac{2}{3} + \frac{7}{12}\Big)$ ÷ $\Big(\frac{5}{6}-\frac{3}{5}\Big)$

According to BODMAS rule, we simplify brackets first

$$\begin{array}{ll} = \frac{7}{15}\text{ of }\Big(\frac{8 + 7}{12}\Big) ÷ \Big(\frac{5}{6}-\frac{3}{5}\Big) \\\\ = \frac{7}{15}\text{ of } \Big(\frac{15}{12}\Big) ÷ \Big(\frac{5}{6}-\frac{3}{5}\Big) \\\\ = \frac{7}{15}\text{ of } \frac{15}{12} ÷ \Big(\frac{5}{6}-\frac{3}{5}\Big) & \text{[First bracket simplified]} \\\\ = \frac{7}{15}\text{ of } \frac{15}{12} ÷ \Big(\frac{25 - 18}{30}\Big) \\\\ = \frac{7}{15}\text{ of } \frac{15}{12} ÷ \frac{7}{30} & \text{[Second bracket simplified]} \\\\ = \frac{7}{15} \times \frac{15}{12} ÷ \frac{7}{30} \\\\ = \frac{7}{12} ÷ \frac{7}{30} & \text{[Of simplified]} \\\\ = \frac{7}{12} \times \frac{30}{7} & [\text{Reciprocal of } \frac{7}{30} \text{ is } \frac{30}{7}] \\\\ = \frac{1}{12} \times \frac{30}{1} & \text{[Dividing 7 and 7 by 7]} \\\\ = \frac{1}{2} \times \frac{5}{1} & \text{[Dividing 30 and 12 by 6]} \\\\ = \frac{5}{2} & \text{[Division simplified]} \\\\ = 2\frac{1}{2} \end{array}$$

∴ The answer is $2\frac{1}{2}$

We have:

$\Big(22÷5\frac{1}{2}\Big)$ ÷ $2\frac{1}{5}$ of $3\frac{1}{3} + 1\frac{5}{11}$

= $\Big(22÷\frac{11}{2}\Big)$ ÷ $\frac{11}{5}$ of $\frac{10}{3}$ + $\frac{16}{11}$ [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

$$\begin{array}{ll} = \Big(22 \times \frac{2}{11}\Big) ÷ \frac{11}{5}\text{ of }\frac{10}{3} + \frac{16}{11} & [\text{Reciprocal of } \frac{11}{2} \text{ is } \frac{2}{11}] \\\\ = \Big(\frac{44}{11}\Big) ÷ \frac{11}{5}\text{ of }\frac{10}{3} + \frac{16}{11} & \text{[Divide 44 and 11 by 11]} \\\\ = \frac{4}{1} ÷ \frac{11}{5}\text{ of }\frac{10}{3} + \frac{16}{11} & \text{[Bracket simplified]} \\\\ = \frac{4}{1} ÷ \frac{11}{5} \times \frac{10}{3} + \frac{16}{11} \\\\ = \frac{4}{1} ÷ \frac{110}{15} + \frac{16}{11} & \text{[Divide 110 and 15 by 5]} \\\\ = \frac{4}{1} ÷ \frac{22}{3} + \frac{16}{11} & \text{[Of simplified]} \\\\ = \frac{4}{1} \times \frac{3}{22} + \frac{16}{11} & [\text{Reciprocal of } \frac{22}{3} \text{ is } \frac{3}{22}] \\\\ = \frac{12}{22} + \frac{16}{11} & \text{[Divide 12 and 22 by 2]} \\\\ = \frac{6}{11} + \frac{16}{11} & \text{[Division simplified]} \\\\ = \frac{6 + 16}{11} = \frac{22}{11} & \text{[Divide 22 and 11 by 11]} \\\\ = 2 & \text{[Division simplified]} \end{array}$$

∴ The answer is 2

We have:

$6\frac{1}{3}$ ÷ $\Big(2\frac{1}{5} + 3\frac{1}{2}\Big)$ of $3\frac{1}{3}$

= $\frac{19}{3}$ ÷ $\Big(\frac{11}{5} + \frac{7}{2}\Big)$ of $\frac{10}{3}$ [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

$$\begin{array}{ll} = \frac{19}{3} ÷ \Big(\frac{22 + 35}{10}\Big)\text{ of }\frac{10}{3} \\\\ = \frac{19}{3} ÷ \frac{57}{10}\text{ of }\frac{10}{3} & \text{[Bracket simplified]} \\\\ = \frac{19}{3} ÷ \frac{57}{10} \times \frac{10}{3} \\\\ = \frac{19}{3} ÷ \frac{570}{30} \\\\ = \frac{19}{3} ÷ \frac{19}{1} & \text{[Of simplified]} \\\\ = \frac{19}{3} \times \frac{1}{19} & [\text{Reciprocal of } \frac{19}{1} \text{ is } \frac{1}{19}] \\\\ = \frac{1}{3} & \text{[Division simplified]} \end{array}$$

∴ The answer is $\frac{1}{3}$

A fraction can be expressed as a mixed fraction only if it is an improper fraction (numerator > denominator).

Hence, option 3 is the correct option.

An improper fraction has numerator greater than denominator.

Hence, option 2 is the correct option.

Given:

Three-fourths of 3

= $\frac{3}{4}$ of 3

= $\frac{3}{4} \times$ 3 = $\frac{9}{4}$

where numerator > denominator, therefore it is an improper fraction.

Hence, option 3 is the correct option.

Given:

Cost of 1 pen = ₹$16\frac{2}{3}$

Number of pens = 60

Cost of 60 pens = ?

Cost of 60 pens = (Cost of 1 pen) x (Number of pens)

Substituting the values in above, we get:

Cost of 60 pens = ₹$16\frac{2}{3}$ x 60

= ₹$\frac{50}{3}$ x 60 = ₹$\frac{3000}{3}$ = ₹1000

Hence, option 4 is the correct option.

Given:

Area of sheet = $42\frac{1}{4}$ cm²

Number of strips = 13

Area of each strip = ?

Area of each strip = (Area of sheet) ÷ (Number of strips)

Substituting the values in above, we get:

Area of each strip = $42\frac{1}{4}$ cm² ÷ 13

$$\begin{array}{ll} = \frac{169}{4} \text{ cm}^2 ÷ 13 \\\\ = \Big(\frac{169}{4} \times \frac{1}{13}\Big) \text{ cm}^2 \\\\ = \Big(\frac{13}{4}\Big) \text{ cm}^2 \\\\ = 3\frac{1}{4} \text{ cm}^2 \end{array}$$

Hence, option 2 is the correct option.

(i) If $\frac{a}{b}$ is a fraction, then a and b are natural numbers.

(ii) In an improper fraction, the numerator is greater than or equal to the denominator.

(iii) The product of two improper fractions is a/an improper fraction.

(iv) $\frac{a}{b}$ ÷ $\frac{c}{d}$ equal to the product of $\frac{a}{b}$ and the reciprocal of $\frac{c}{d}$.

(v) The product of two proper fractions is smaller than each of the given fractions.

(vi) The reciprocal of a proper fraction is a/an improper fraction.

(vii) The fraction whose reciprocal is equal to the fraction itself is 1 .

(i) False
Reason — In an improper fraction, the numerator may be equal to or greater than the denominator.

(ii) False
Reason — A proper fraction is less than 1. The product of two numbers less than 1 is always less than 1. Hence, it remains a proper fraction.

(iii) True
Reason — An improper fraction is greater than or equal to 1. A proper fraction is always less than 1. Therefore, any improper fraction is greater than any proper fraction.

(iv) False
Reason — A unit fraction is any fraction with a numerator of 1, such as $\frac{1}{5}, \frac{1}{9}$. Only the fraction $\frac{1}{1}$ is equal to 1.

(v) True
Reason — In a proper fraction, the denominator is larger than the numerator. When you take the reciprocal, the larger number becomes the numerator, making it an improper fraction.

(vi) False
Reason — The product of two proper fractions is always less than each of them.

(vii) True
Reason — The number 1 can be written as the fraction $\frac{1}{1}$. Its multiplicative inverse (reciprocal) is also $\frac{1}{1}$, which is equal to the original fraction. (Similarly, $-1$ or $\frac{-1}{1}$ also shares this property).

(viii) True
Reason — A proper fraction is less than 1. When a number is multiplied by a fraction less than 1, the product becomes smaller than the original number.

(1) Given:

Total Money = ₹56,00,000

Total Grand-children = 3 Boys + 5 Girls = 8 children

Each child receives equal share.

∴ Share of each child = Total Money ÷ Total Grand-children

Substituting the values in above, we get:

Share of each child = ₹56,00,000 ÷ 8

= ₹$\frac{5600000}{8}$ = ₹7,00,000

Hence, option 4 is the correct option.

(2) Given:

Number of girls = 5

Total children = 8

Fraction of money received by girls = Number of girls ÷ Total children

Substituting the values in above, we get:

Fraction of money received by girls = 5 ÷ 8

= $\frac{5}{8}$

Hence, option 3 is the correct option.

(3) Number of boys = 3

Given,

Amount received by each boy = ₹7,00,000 [From step 1]

Total amount received by boys = (Number of boys) x (Amount received by each boy)

Substituting the values in above, we get:

Total amount received by boys = 3 x ₹7,00,000

= ₹21,00,000

Hence, option 1 is the correct option.

(4) Given:

If one girl did not take her share,

Remaining children = 7 (3 boys + 4 girls)

Now total money is divided among 7 children equally.

Fraction received by boys = Number of boys ÷ Total children

Substituting the values in above, we get:

Fraction received by boys = 3 ÷ 7 = $\frac{3}{7}$

Hence, option 4 is the correct option.

(1) Given:

Total bundles = $7 \frac{1}{2} = \frac{15}{2}$

Length per bundle = $202 \frac{4}{5} = \frac{1014}{5}$ m

Total length of the cable purchased = (Total bundles) x (Length per bundle)

Substituting the values in above, we get:

Total length of the cable purchased = $\frac{15}{2}$ x $\frac{1014}{5}$ m

= $\frac{3}{2}$ x $\frac{1014}{1}$ m

= $\frac{3}{1}$ x $\frac{507}{1}$ m

= 1521 m

Hence, option 2 is the correct option.

(2) The cost of cable per meter = ₹$6\frac{2}{3} = ₹\frac{20}{3}$ $\hspace{2cm}$

Given

Total length of the cable = 1521 m $\hspace{2.5cm}$ [From previous step]

The amount paid by Amar = (Cost of cable per meter) x (Total length of the cable)

Substituting the values in above, we get:

The amount paid by Amar = ₹$\frac{20}{3}$ x 1521 m

= ₹20 x 507

= ₹10140

Hence, option 3 is the correct option.

(3) Given:

Bundles of cable used for the top floor = $2\frac{1}{2} = \frac{5}{2}$

Length per bundle = $\frac{1014}{5}$ m

Length of cable used for the top floor = (Bundles of cable used for the top floor) x (Length per bundle)

Substituting the values in above, we get:

Length of cable used for the top floor = $\frac{5}{2}$ x $\frac{1014}{5}$ m = 507 m

Hence, option 2 is the correct option.

(4) Given:

Length of one bundle = $\frac{1014}{5}$ m

Cut length = $13\frac{4}{5}$ m = $\frac{69}{5}$ m

Remaining length = (Length of one bundle - Cut length)

Substituting the values in above, we get:

Remaining length = $\Big(\frac{1014}{5} - \frac{69}{5}\Big)$ m

= $\frac{1014 - 69}{5}$ m = $\frac{945}{5}$ m = 189 m

Number of pieces of 21 m length = Remaining length ÷ 21

Substituting the values in above, we get:

Number of pieces of 21 m length = 189 m ÷ 21 = $\frac{189}{21}$

= 9

Hence, option 1 is the correct option.

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

An improper fraction is a fraction where the numerator is greater than or equal to the denominator (e.g., $\frac{5}{3}$).

Its reciprocal is obtained by interchanging numerator and denominator:

$\frac{5}{3} \rightarrow \frac{3}{5}$

Since $\frac{3}{5}$ is less than 1, it is a proper fraction.

So, the Assertion is true.

The Reason is also true because reciprocal is indeed called the multiplicative inverse.

However, the reason does not explain why the reciprocal of an improper fraction becomes a proper fraction. It only defines what a reciprocal is.

Hence, option 2 is the correct option.

Assertion (A) is true but Reason (R) is false.

Explanation

A proper fraction is less than 1.

When we multiply two numbers less than 1, the result becomes even smaller.

Example:

$\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

$\frac{1}{6}$ is smaller than both $\frac{1}{2}$ and $\frac{1}{3}$

So, the Assertion is true.

The given formula in the Reason is incorrect.
Correct multiplication rule is:

$\frac{a}{b} \times \frac{c}{b} = \frac{ac}{bd}$

But the reason states the denominator remains b, which is wrong.

So, the Reason is false.

Hence, option 3 is the correct option.

Given:

The sum of numbers in each row and each column is the same.

First, we find the common sum by adding the numbers in the first row.

Sum of first row = $7\frac{3}{10} + 1\frac{1}{2} + \frac{4}{5}$

$$\begin{aligned}= \frac{73}{10} + \frac{3}{2} + \frac{4}{5} \\[1em] = \frac{73}{10} + \frac{15}{10} + \frac{8}{10} \hspace{1.5cm}\text{[Taking LCM = 10]} \\[1em] = \frac{73 + 15 + 8}{10} \\[1em] = \frac{96}{10}\end{aligned}$$

So, the common sum of each row and each column = $\frac{96}{10}$

Finding P (in the second column):

Sum of second column = $1\frac{1}{2}$ + P + $2\frac{4}{5}$ = $\frac{96}{10}$

$$\begin{aligned}\frac{3}{2} + P + \frac{14}{5} = \frac{96}{10} \\[1em] \frac{15}{10} + \frac{28}{10} + P = \frac{96}{10} \\[1em] \frac{43}{10} + P = \frac{96}{10} \\[1em] P = \frac{96}{10} - \frac{43}{10} \\[1em] P = \frac{53}{10}\end{aligned}$$

Finding Q (in the third row):

Sum of third row = $1\frac{7}{10}$ + $2\frac{4}{5}$ + Q = $\frac{96}{10}$

$$\begin{aligned}\frac{17}{10} + \frac{14}{5} + Q = \frac{96}{10} \\[1em] \frac{17}{10} + \frac{28}{10} + Q = \frac{96}{10} \\[1em] \frac{45}{10} + Q = \frac{96}{10} \\[1em] Q = \frac{96}{10} - \frac{45}{10} \\[1em] Q = \frac{51}{10}\end{aligned}$$

Finding the product P × Q:

$$\begin{aligned}P \times Q = \frac{53}{10} \times \frac{51}{10} \\[1em] = \frac{53 \times 51}{10 \times 10} \\[1em] = \frac{2703}{100} \\[1em] = 27\frac{3}{100}\end{aligned}$$

Hence, option 2 is the correct option.

We have:

$\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)............\left(1 - \frac{1}{10}\right)$

Simplifying each bracket:

$$\begin{aligned}= \frac{2-1}{2} \times \frac{3-1}{3} \times \frac{4-1}{4} \times \frac{5-1}{5} \times \frac{6-1}{6} \times \frac{7-1}{7} \times \frac{8-1}{8} \times \frac{9-1}{9} \times \frac{10-1}{10} \\[1em] = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} \times \frac{7}{8} \times \frac{8}{9} \times \frac{9}{10} \\[1em] = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9}{2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10} \hspace{1.5cm}\text{[Cancelling common terms]} \\[1em] = \frac{1}{10}\end{aligned}$$

Hence, option 3 is the correct option.

From the figure:

Total number of unit squares = 18

Number of squares already shaded = 6

Required fraction of shaded squares = $\frac{7}{9}$

Number of squares that must be shaded = $\frac{7}{9}$ × 18 = 14

Number of more squares to be shaded = (Required shaded squares) − (Already shaded squares)

Substituting the values, we get:

Number of more squares to be shaded = 14 − 6 = 8

Hence, option 4 is the correct option.

Given:

Fraction of red roses = $\frac{2}{10}$

Fraction of pink roses = $\frac{5}{10}$

Total fraction of all roses = 1

Fraction of blue roses = 1 − (Fraction of red roses + Fraction of pink roses)

Substituting the values, we get:

$$\begin{aligned}\text{Fraction of blue roses} = 1 - \Big(\frac{2}{10} + \frac{5}{10}\Big) \\[1em] = 1 - \frac{2 + 5}{10} \\[1em] = 1 - \frac{7}{10} \\[1em] = \frac{10 - 7}{10} \\[1em] = \frac{3}{10}\end{aligned}$$

Hence, option 3 is the correct option.

From the pie chart:

Reema occupies about half of the circle, i.e., $\frac{1}{2}$ of the total votes.

The remaining half is divided between Kumar and Manish, with Kumar's share being much larger than Manish's share.

Kumar's share is approximately $\frac{1}{3}$ of the total votes.

Total votes = 150

Number of votes received by Kumar = $\frac{1}{3}$ × 150

$$\begin{aligned}= \frac{1}{3} \times \frac{150}{1} \\[1em] = \frac{150}{3} \\[1em] = 50\end{aligned}$$

Hence, option 3 is the correct option.

From the figure, the fractions shown are:

$\frac{2}{3}, \frac{2}{4}, \frac{2}{5}, \frac{2}{6}$

Observation:

The numerator remains the same (2) in all the fractions, while the denominator keeps increasing (3, 4, 5, 6).

As the denominator increases, the whole is divided into more parts, so each part becomes smaller. Hence, the value of the fraction decreases.

This is shown by the shaded regions in the circles, which keep getting smaller as we move from $\frac{2}{3}$ to $\frac{2}{6}$.

Hence, option 4 is the correct option.

Given:

Total length of rope = 30 m

Length of each piece = $3\frac{3}{4}$ m = $\frac{15}{4}$ m

Number of pieces = (Total length of rope) ÷ (Length of each piece)

Substituting the values, we get:

Number of pieces = 30 ÷ $\frac{15}{4}$

$$\begin{aligned}= \frac{30}{1} \times \frac{4}{15} \\[1em] = \frac{2}{1} \times \frac{4}{1} \\[1em] = \frac{2 \times 4}{1 \times 1} \\[1em] = 8\end{aligned}$$

Hence, option 4 is the correct option.

We have:

$3 - \frac{1}{1 + \frac{2}{2 - \frac{3}{5}}}$

We simplify from the innermost fraction outwards.

Step 1: Simplify $2 - \frac{3}{5}$

$2 - \frac{3}{5} = \frac{10 - 3}{5} = \frac{7}{5}$

Step 2: Simplify $\frac{2}{2 - \frac{3}{5}} = \frac{2}{\frac{7}{5}}$

$\frac{2}{\frac{7}{5}} = 2 \times \frac{5}{7} = \frac{10}{7}$

Step 3: Simplify $1 + \frac{10}{7}$

$1 + \frac{10}{7} = \frac{7 + 10}{7} = \frac{17}{7}$

Step 4: Simplify $\frac{1}{1 + \frac{2}{2 - \frac{3}{5}}} = \frac{1}{\frac{17}{7}}$

$\frac{1}{\frac{17}{7}} = 1 \times \frac{7}{17} = \frac{7}{17}$

Step 5: Simplify $3 - \frac{7}{17}$

$3 - \frac{7}{17} = \frac{51 - 7}{17} = \frac{44}{17}$

Hence, option 4 is the correct option.

From the figure:

First triangle:

The triangle is divided into 4 equal smaller triangles. Out of these, only 1 (the central inverted) triangle is shaded.

So, fraction of shaded portion = $\frac{1}{4}$

Second triangle:

The triangle is divided into 4 equal smaller triangles. Out of these, 3 triangles (the corners) are shaded.

So, fraction of shaded portion = $\frac{3}{4}$

Total shaded fraction = $\frac{1}{4} + \frac{3}{4}$

Hence, option 4 is the correct option.

We have:

$\left(1\frac{1}{3} \times 1\frac{1}{2}\right) \div 1\frac{1}{9}$

$$\begin{aligned}= \Big(\frac{4}{3} \times \frac{3}{2}\Big) \div \frac{10}{9} \\[1em] = \frac{4 \times 3}{3 \times 2} \div \frac{10}{9} \\[1em] = \frac{12}{6} \div \frac{10}{9} \\[1em] = 2 \div \frac{10}{9} \hspace{1.5cm}\text{[Bracket simplified]} \\[1em] = 2 \times \frac{9}{10} \hspace{1.5cm}\text{[Reciprocal of } \frac{10}{9} \text{ is } \frac{9}{10}] \\[1em] = \frac{18}{10} \\[1em] = \frac{9}{5}\end{aligned}$$

To make the result a natural number, we must add the least fraction that takes $\frac{9}{5}$ to the next natural number.

The next natural number greater than $\frac{9}{5}$ is 2.

Required least fraction = 2 − $\frac{9}{5}$

$$\begin{aligned}= \frac{10 - 9}{5} \\[1em] = \frac{1}{5}\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Let the total number of pages in the book = x

Day 1:

Pages read on day 1 = $\frac{4}{9}$ of x = $\frac{4x}{9}$

Remaining pages after day 1 = x − $\frac{4x}{9}$ = $\frac{9x - 4x}{9}$ = $\frac{5x}{9}$

Day 2:

Pages read on day 2 = $\frac{3}{5}$ of remaining = $\frac{3}{5} \times \frac{5x}{9}$

$$\begin{aligned}= \frac{3 \times 5x}{5 \times 9} \\[1em] = \frac{15x}{45} \\[1em] = \frac{x}{3}\end{aligned}$$

Pages remaining after day 2 = $\frac{5x}{9} - \frac{x}{3}$

$$\begin{aligned}= \frac{5x}{9} - \frac{3x}{9} \hspace{1.5cm}\text{[Taking LCM = 9]} \\[1em] = \frac{5x - 3x}{9} \\[1em] = \frac{2x}{9}\end{aligned}$$

According to the question, pages still left unread = 100

$$\begin{aligned}\frac{2x}{9} = 100 \\[1em] 2x = 100 \times 9 \\[1em] 2x = 900 \\[1em] x = \frac{900}{2} \\[1em] x = 450\end{aligned}$$

Hence, option 1 is the correct option.

Given:

Let the capacity of the drum = x litres

Initial petrol in drum = $\frac{3}{4}$ of x = $\frac{3x}{4}$ litres

Petrol drawn out = 30 litres

Final petrol in drum = $\frac{7}{12}$ of x = $\frac{7x}{12}$ litres

According to the question:

Initial petrol − Petrol drawn out = Final petrol

Substituting the values, we get:

$$\begin{aligned}\frac{3x}{4} - 30 = \frac{7x}{12} \\[1em] \frac{3x}{4} - \frac{7x}{12} = 30 \\[1em] \frac{9x - 7x}{12} = 30 \hspace{1.5cm}\text{[Taking LCM = 12]} \\[1em] \frac{2x}{12} = 30 \\[1em] \frac{x}{6} = 30 \\[1em] x = 30 \times 6 \\[1em] x = 180\end{aligned}$$

So, the capacity of the drum = 180 litres

Hence, option 2 is the correct option.