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Chapter 2

Fractions - Exercise 2(A)

Class - 7 RS Aggarwal Mathematics Solutions



Exercise 2(A)

Question 1

Find the product :

56×37\dfrac{5}{6} \times \dfrac{3}{7}

Answer

We have:

=56×37=5×36×7=5×12×7[Simplifying 3 and 6 ⇒ Divide by 3 ]=514\begin{array}{ll} \phantom{=} \dfrac{5}{6} \times \dfrac{3}{7} \\\\ = \dfrac{5 \times 3}{6 \times 7} \\\\ = \dfrac{5 \times 1}{2 \times 7} & \text{[Simplifying 3 and 6 ⇒ Divide by 3 ]} \\\\ = \dfrac{5}{14} \\\\ \end{array}

∴ The answer is 514\dfrac{5}{14}

Question 2

Find the product :

718×914\dfrac{7}{18} \times \dfrac{9}{14}

Answer

We have:

=718×914=7×918×14=7×12×14[Simplifying 9 and 18 ⇒ Divide by 9]=1×12×2[Simplifying 7 and 14 ⇒ Divide by 7]=14\begin{array}{ll} \phantom{=} \dfrac{7}{18} \times \dfrac{9}{14} \\\\ = \dfrac{7 \times 9}{18 \times 14} \\\\ = \dfrac{7 \times 1}{2 \times 14} & \text{[Simplifying 9 and 18 ⇒ Divide by 9]} \\\\ = \dfrac{1 \times 1}{2 \times 2} & \text{[Simplifying 7 and 14 ⇒ Divide by 7]} \\\\ = \dfrac{1}{4} & \end{array}

∴ The answer is 14\dfrac{1}{4}

Question 3

Find the product :

28×7828 \times \dfrac{7}{8}

Answer

We have:

=28×78=281×78=71×72[Simplifying 28 and 8 ⇒ Divide by 4]=7×71×2=492=2412[Converting improper to mixed fraction]\begin{array}{ll} \phantom{=} 28 \times \dfrac{7}{8} \\\\ = \dfrac{28}{1} \times \dfrac{7}{8} \\\\ = \dfrac{7}{1} \times \dfrac{7}{2} & \text{[Simplifying 28 and 8 ⇒ Divide by 4]} \\\\ = \dfrac{7 \times 7}{1 \times 2} \\\\ = \dfrac{49}{2} & \\\\ = 24 \dfrac{1}{2} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The answer is 24 12\dfrac{1}{2}

Question 4

Find the product :

7×177 \times \dfrac{1}{7}

Answer

We have:

=7×17=71×17=7×11×7=77=1[Dividing both by 7]\begin{array}{ll} \phantom{=} 7 \times \dfrac{1}{7} \\\\ = \dfrac{7}{1} \times \dfrac{1}{7} \\\\ = \dfrac{7 \times 1}{1 \times 7} \\\\ = \dfrac{7}{7} \\\\ = 1 & \text{[Dividing both by 7]} \end{array}

∴ The answer is 1

Question 5

Find the product :

2125×5172\dfrac{1}{25} \times \dfrac{5}{17}

Answer

We have:

=2125×517=5125×517[Converting mixed fraction to improper fraction]=515×117[Simplifying 5 and 25 ⇒ Divide by 5]=35×11[Simplifying 51 and 17 ⇒ Divide by 17]=3×15×1=35\begin{array}{ll} \phantom{=} 2\dfrac{1}{25} \times \dfrac{5}{17} \\\\ = \dfrac{51}{25} \times \dfrac{5}{17} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \dfrac{51}{5} \times \dfrac{1}{17} & \text{[Simplifying 5 and 25 ⇒ Divide by 5]} \\\\ = \dfrac{3}{5} \times \dfrac{1}{1} & \text{[Simplifying 51 and 17 ⇒ Divide by 17]} \\\\ = \dfrac{3 \times 1}{5 \times 1} & \\\\ = \dfrac{3}{5} \end{array}

∴ The answer is 35\dfrac{3}{5}

Question 6

Find the product :

1113×7371\dfrac{1}{13} \times 7\dfrac{3}{7}

Answer

We have:

=1113×737=1413×527[Converting mixed fraction to improper fraction]=213×521[Simplifying 14 and 7 ⇒ Divide by 7]=21×41[Simplifying 52 and 13 ⇒ Divide by 13]=2×41×1=81=8\begin{array}{ll} \phantom{=} 1\dfrac{1}{13} \times 7\dfrac{3}{7} \\\\ = \dfrac{14}{13} \times \dfrac{52}{7} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \dfrac{2}{13} \times \dfrac{52}{1} & \text{[Simplifying 14 and 7 ⇒ Divide by 7]} \\\\ = \dfrac{2}{1} \times \dfrac{4}{1} & \text{[Simplifying 52 and 13 ⇒ Divide by 13]} \\\\ = \dfrac{2 \times 4}{1 \times 1} \\\\ = \dfrac{8}{1} \\\\ = 8 \end{array}

∴ The answer is 8

Question 7

Find the product :

417×7112\dfrac{4}{17} \times 7\dfrac{1}{12}

Answer

We have:

=417×7112=417×8512[Converting mixed fraction to improper fraction]=41×512[Simplifying 85 and 17 ⇒ Divide by 17 ]=11×53[Simplifying 4 and 12 ⇒ Divide by 4 ]=1×51×3=53\begin{array}{ll} \phantom{=} \dfrac{4}{17} \times 7\dfrac{1}{12} \\\\ = \dfrac{4}{17} \times \dfrac{85}{12} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \dfrac{4}{1} \times \dfrac{5}{12} & \text{[Simplifying 85 and 17 ⇒ Divide by 17 ]} \\\\ = \dfrac{1}{1} \times \dfrac{5}{3} & \text{[Simplifying 4 and 12 ⇒ Divide by 4 ]} \\\\ = \dfrac{1 \times 5}{1 \times 3} \\\\ = \dfrac{5}{3} \end{array}

∴ The answer is 53\dfrac{5}{3}

Question 8

Find the product :

714×758×111217\dfrac{1}{4}\times\dfrac{7}{58}\times1\dfrac{11}{21}

Answer

We have:

=714×758×11121=294×758×3221[Converting mixed fraction to improper fraction]=14×72×3221[Simplifying 29 and 58 ⇒ Divide by 29 ]=14×12×323[Simplifying 7 and 21 ⇒ Divide by 7 ]=11×12×83[Simplifying 32 and 4 ⇒ Divide by 4 ]=1×1×81×2×3=86=43[Dividing both by 2]=113[Converting improper to mixed fraction]\begin{array}{ll} \phantom{=} 7\dfrac{1}{4}\times\dfrac{7}{58}\times1\dfrac{11}{21} \\\\ = \dfrac{29}{4} \times \dfrac{7}{58} \times \dfrac{32}{21} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \dfrac{1}{4} \times \dfrac{7}{2} \times \dfrac{32}{21} & \text{[Simplifying 29 and 58 ⇒ Divide by 29 ]} \\\\ = \dfrac{1}{4} \times \dfrac{1}{2} \times \dfrac{32}{3} & \text{[Simplifying 7 and 21 ⇒ Divide by 7 ]} \\\\ = \dfrac{1}{1} \times \dfrac{1}{2} \times \dfrac{8}{3} & \text{[Simplifying 32 and 4 ⇒ Divide by 4 ]} \\\\ = \dfrac{1 \times 1 \times 8}{1 \times 2 \times 3} \\\\ = \dfrac{8}{6} \\\\ = \dfrac{4}{3} & \text{[Dividing both by 2]} \\\\ = 1\dfrac{1}{3} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The answer is 1131\dfrac{1}{3}

Question 9

Find the product :

119×91×51113\dfrac{1}{19}\times91\times5\dfrac{11}{13}

Answer

We have:

=119×91×51113=119×911×7613[Converting mixed fraction to improper fraction]=119×71×761[Simplifying 91 and 13 ⇒ Divide by 13 ]=11×71×41[Simplifying 76 and 19 ⇒ Divide by 19 ]=1×7×41×1×1=281=28\begin{array}{ll} \phantom{=} \dfrac{1}{19}\times91\times5\dfrac{11}{13} \\\\ = \dfrac{1}{19} \times \dfrac{91}{1} \times \dfrac{76}{13} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \dfrac{1}{19} \times \dfrac{7}{1} \times \dfrac{76}{1} & \text{[Simplifying 91 and 13 ⇒ Divide by 13 ]} \\\\ = \dfrac{1}{1} \times \dfrac{7}{1} \times \dfrac{4}{1} & \text{[Simplifying 76 and 19 ⇒ Divide by 19 ]} \\\\ = \dfrac{1 \times 7 \times 4}{1 \times 1 \times 1} \\\\ = \dfrac{28}{1} \\\\ = 28 \end{array}

∴ The answer is 28

Question 10

Find the product :

714×2316×2277\dfrac{1}{4}\times2\dfrac{3}{16}\times2\dfrac{2}{7}

Answer

We have:

=714×2316×227=294×3516×167[Converting mixed fraction to improper fraction]=294×351×17[Simplifying 16 and 16 ⇒ Divide by 16 ]=294×51×11[Simplifying 35 and 7 ⇒ Divide by 7 ]=29×5×14×1×1=1454=3614[Converting improper to mixed fraction]\begin{array}{ll} \phantom{=} 7\dfrac{1}{4}\times2\dfrac{3}{16}\times2\dfrac{2}{7} \\\\ = \dfrac{29}{4} \times \dfrac{35}{16} \times \dfrac{16}{7} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \dfrac{29}{4} \times \dfrac{35}{1} \times \dfrac{1}{7} & \text{[Simplifying 16 and 16 ⇒ Divide by 16 ]} \\\\ = \dfrac{29}{4} \times \dfrac{5}{1} \times \dfrac{1}{1} & \text{[Simplifying 35 and 7 ⇒ Divide by 7 ]} \\\\ = \dfrac{29 \times 5 \times 1}{4 \times 1 \times 1} \\\\ = \dfrac{145}{4} \\\\ = 36\dfrac{1}{4} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The answer is 361436\dfrac{1}{4}

Question 11(i)

Find the value of :

34\dfrac{3}{4} of 89\dfrac{8}{9}

Answer

We have:

34\dfrac{3}{4} of 89\dfrac{8}{9}

=89×34=29×31[Simplifying 8 and 4 ⇒ Divide by 4 ]=23×11[Simplifying 3 and 9 ⇒ Divide by 3 ]=2×13×1=23\begin{array}{ll} = \dfrac{8}{9} \times \dfrac{3}{4} \\\\ = \dfrac{2}{9} \times \dfrac{3}{1} & \text{[Simplifying 8 and 4 ⇒ Divide by 4 ]} \\\\ = \dfrac{2}{3} \times \dfrac{1}{1} & \text{[Simplifying 3 and 9 ⇒ Divide by 3 ]} \\\\ = \dfrac{2 \times 1}{3 \times 1} \\\\ = \dfrac{2}{3} \end{array}

∴ The answer is 23\dfrac{2}{3}

Question 11(ii)

Find the value of :

12\dfrac{1}{2} of 2232\dfrac{2}{3}

Answer

We have:

12\dfrac{1}{2} of 2232\dfrac{2}{3}

=83×12[Converting mixed fraction to improper fraction]=43×11[Simplifying 8 and 2 ⇒ Divide by 2 ]=43=113[Converting improper to mixed fraction]\begin{array}{ll} = \dfrac{8}{3} \times \dfrac{1}{2} & \text{[Converting mixed fraction to improper fraction]} \\\\ = \dfrac{4}{3} \times \dfrac{1}{1} & \text{[Simplifying 8 and 2 ⇒ Divide by 2 ]} \\\\ = \dfrac{4}{3} \\\\ = 1\dfrac{1}{3} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The answer is 1 13\dfrac{1}{3}

Question 11(iii)

Find the value of :

45\dfrac{4}{5} of 1 hour

Answer

We have:

45\dfrac{4}{5} of 1 hour

1 hour=60 minutes[Converting hour into minutes]=(60×45) minutes=(601×45) minutes=(121×41) minutes[Simplifying 60 and 5 ⇒ Divide by 5 ]=481 minutes=48 minutes\begin{array}{ll} 1 \text{ hour} = 60 \text{ minutes} & \text{[Converting hour into minutes]} \\\\ = \Big(60 \times \dfrac{4}{5}\Big) \text{ minutes} \\\\ = \Big(\dfrac{60}{1} \times \dfrac{4}{5}\Big) \text{ minutes} \\\\ = \Big(\dfrac{12}{1} \times \dfrac{4}{1}\Big) \text{ minutes} & \text{[Simplifying 60 and 5 ⇒ Divide by 5 ]} \\\\ = \dfrac{48}{1} \text{ minutes} \\\\ = 48 \text{ minutes} \end{array}

∴ The answer is 48 minutes

Question 11(iv)

Find the value of :

35\dfrac{3}{5} of ₹1

Answer

We have:

35\dfrac{3}{5} of ₹1

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

=(100×35) paise=(1001×35) paise=(201×31) paise[Simplifying 100 and 5 ⇒ Divide by 5 ]=(20×31×1) paise=601 paise=60 paise\begin{array}{ll} = \Big(100 \times \dfrac{3}{5}\Big) \text{ paise} \\\\ = \Big(\dfrac{100}{1} \times \dfrac{3}{5}\Big) \text{ paise} \\\\ = \Big(\dfrac{20}{1} \times \dfrac{3}{1}\Big)\text{ paise} & \text{[Simplifying 100 and 5 ⇒ Divide by 5 ]} \\\\ = \Big(\dfrac{20 \times 3}{1 \times 1}\Big)\text{ paise} \\\\ = \dfrac{60}{1}\text{ paise} \\\\ = 60\text{ paise} \end{array}

∴ The answer is 60 paise

Question 11(v)

Find the value of :

815\dfrac{8}{15} of 1121\dfrac{1}{2} metres

Answer

We have:

815\dfrac{8}{15} of 1121\dfrac{1}{2} metres

1121\dfrac{1}{2} metres = 32\dfrac{3}{2} metres \hspace{1.5cm}[Converting mixed to improper fraction]

1 meter = 100 cm

∴32\dfrac{3}{2} metres = 150 cm \hspace{1.5cm}[Converting meter to centimeter]

Now we have:

815\dfrac{8}{15} of 150 centimetres

=(150×815) cm=(1501×815) cm=(101×81) cm[Simplifying 150 and 15 ⇒ Divide by 15 ]=(10×81×1) cm=801 cm=80 cm\begin{array}{ll} = \Big(150 \times \dfrac{8}{15}\Big) \text{ cm} \\\\ = \Big(\dfrac{150}{1} \times \dfrac{8}{15}\Big) \text{ cm} \\\\ = \Big(\dfrac{10}{1} \times \dfrac{8}{1}\Big) \text{ cm} & \text{[Simplifying 150 and 15 ⇒ Divide by 15 ]} \\\\ = \Big(\dfrac{10 \times 8}{1 \times 1}\Big) \text{ cm} \\\\ = \dfrac{80}{1} \text{ cm} \\\\ = 80 \text{ cm} \end{array}

∴ The answer is 80 cm

Question 11(vi)

Find the value of :

57\dfrac{5}{7} of 2132\dfrac{1}{3}kg

Answer

We have:

57\dfrac{5}{7} of 2132\dfrac{1}{3}kg

213 kg=73 kg2\dfrac{1}{3}\text{ kg} = \dfrac{7}{3}\text{ kg} \hspace{1.5cm}[Converting mixed to improper fraction]

Now we have:

57\dfrac{5}{7} of 73\dfrac{7}{3} kg

=(73×57) kg=(13×51) kg[Simplifying 7 and 7 ⇒ Divide by 7 ]=(1×53×1) kg=53 kg=123 kg[Converting improper to mixed fraction]\begin{array}{ll} = \Big(\dfrac{7}{3} \times \dfrac{5}{7}\Big) \text{ kg} \\\\ = \Big(\dfrac{1}{3} \times \dfrac{5}{1}\Big) \text{ kg} & \text{[Simplifying 7 and 7 ⇒ Divide by 7 ]} \\\\ = \Big(\dfrac{1 \times 5}{3 \times 1}\Big) \text{ kg} \\\\ = \dfrac{5}{3} \text{ kg} \\\\ = 1\dfrac{2}{3} \text{ kg} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The answer is 1231\dfrac{2}{3} kg

Question 12

A car can travel 121212\dfrac{1}{2}km in 1 litre of petrol. How much distance can it travel in 423542\dfrac{3}{5} litres of petrol?

Answer

Given:

Distance travelled in 1 litre of petrol = 121212\dfrac{1}{2} km

= 252\dfrac{25}{2} km \hspace{1cm} [Converting mixed to improper fraction]

Distance travelled in 423542\dfrac{3}{5} litres of petrol = ?

Number of litres = 423542\dfrac{3}{5} = 2135\dfrac{213}{5} litres

Distance travelled in 2135\dfrac{213}{5} litres of petrol = (Distance travelled in 1 litre) x (Number of litres)

Substituting the values in above, we get:

=(252×2135) km=(52×2131) km[Simplifying 25 and 5 ⇒ Divide by 5 ]=(5×2132×1) km=10652 km=53212 km[Converting improper fraction to mixed fraction]\begin{array}{ll} = \Big(\dfrac{25}{2} \times \dfrac{213}{5}\Big)\text{ km} \\\\ = \Big(\dfrac{5}{2} \times \dfrac{213}{1}\Big)\text{ km} & \text{[Simplifying 25 and 5 ⇒ Divide by 5 ]} \\\\ = \Big(\dfrac{5 \times 213}{2 \times 1}\Big) \text{ km} \\\\ = \dfrac{1065}{2}\text{ km} \\\\ = 532 \dfrac{1}{2}\text{ km} & \text{[Converting improper fraction to mixed fraction]} \end{array}

∴ The distance travelled in 423542\dfrac{3}{5} litres of petrol = 53212532\dfrac{1}{2} km.

Question 13

A graphic designer charges ₹273527\dfrac{3}{5} for each diagram. Find the amount he will charge if he designs 186 diagrams for a book.

Answer

Given:

Charge of 1 diagram = ₹2735=₹138527\dfrac{3}{5} = ₹\dfrac{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:

=₹(1385×186)=₹(138×1865×1)=₹256685=₹513335[Converting improper fraction to mixed fraction]\begin{array}{ll} = ₹\Big(\dfrac{138}{5} \times 186\Big) \\\\ = ₹\Big(\dfrac{138 \times 186}{5 \times 1}\Big) \\\\ = ₹\dfrac{25668}{5} \\\\ = ₹5133\dfrac{3}{5} & \text{[Converting improper fraction to mixed fraction]} \end{array}

∴ The charge for 186 diagrams = ₹ 5133355133\dfrac{3}{5}.

Question 14

If a cloth costs ₹71514715\dfrac{1}{4} per metre, find the cost of 3253\dfrac{2}{5} metres of this cloth.

Answer

Cost of 1 meter of cloth = ₹71514=₹28614715\dfrac{1}{4} = ₹\dfrac{2861}{4}

Cost of 3253\dfrac{2}{5} meters of cloth = ?

Length of cloth = 325 meters=1753\dfrac{2}{5} \text{ meters} = \dfrac{17}{5} meters

Cost of 3253\dfrac{2}{5} meters of cloth = (Cost of 1 meter of cloth) x (Length of cloth)

Substituting the values in above, we get:

=₹(28614×175)=₹(2861×174×5)=₹4863720=₹24311720[Converting improper to mixed fraction]\begin{array}{ll} = ₹\Big(\dfrac{2861}{4} \times \dfrac{17}{5}\Big) \\\\ = ₹\Big(\dfrac{2861 \times 17}{4 \times 5}\Big) \\\\ = ₹\dfrac{48637}{20} \\\\ = ₹2431\dfrac{17}{20} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The cost of 3253\dfrac{2}{5} meters of cloth = ₹ 243117202431\dfrac{17}{20}.

Question 15

Advertising in a magazine costs ₹1472251472\dfrac{2}{5} per square inch. Find the cost of an advertisement of 176717\dfrac{6}{7} square inch.

Answer

Cost of an advertisement of 1 square inch = ₹147225=₹736251472\dfrac{2}{5} = ₹\dfrac{7362}{5}

Cost of an advertisement of 176717\dfrac{6}{7} square inch = ?

Area of an advertisement = 176717\dfrac{6}{7} square inch

Cost of an advertisement of 176717\dfrac{6}{7} square inch = (Cost of an advertisement of 1 square inch) x (Area of an advertisement)

Substituting the values in above, we get:

=₹(73625×1767)=₹(73625×1257)[Converting mixed to improper fraction]=₹(73621×257)=₹(7362×251×7)=₹1840507=₹2629267[Converting improper fraction to mixed fraction]\begin{array}{ll} = ₹\Big(\dfrac{7362}{5} \times 17\dfrac{6}{7}\Big) \\\\ = ₹\Big(\dfrac{7362}{5} \times \dfrac{125}{7}\Big) & \text{[Converting mixed to improper fraction]} \\\\ = ₹\Big(\dfrac{7362}{1} \times \dfrac{25}{7}\Big) \\\\ = ₹\Big(\dfrac{7362 \times 25}{1 \times 7}\Big) \\\\ = ₹\dfrac{184050}{7} \\\\ = ₹26292 \dfrac{6}{7} & \text{[Converting improper fraction to mixed fraction]} \end{array}

∴ The cost of an advertisement of 176717\dfrac{6}{7} square inch = ₹ 262926726292\dfrac{6}{7}.

Question 16

A car travelled from city A to city B with a uniform speed of 522752\dfrac{2}{7} km per hour. Find the distance between the two cities, if it took 4384\dfrac{3}{8} hours for the car to reach city B from city A.

Answer

Given:

Speed = 522752\dfrac{2}{7} km/hour

Time = 4384\dfrac{3}{8} hours

Distance = ?

We know the formula,

Distance = Speed x Time

By substituting the values we get,

Distance=(5227×438) km =(3667×358) km[Converting mixed to improper fraction]=(3661×58) km[Simplifying 35 and 7 ⇒ Divide by 7 ]=366×51×8 km=18308 km=22834 km[Converting improper to mixed fraction]\begin{array}{ll} \text{Distance} = \Big(52\dfrac{2}{7} \times 4\dfrac{3}{8}\Big) \text{ km} & \space \\\\ = \Big(\dfrac{366}{7} \times \dfrac{35}{8}\Big) \text{ km} & \text{[Converting mixed to improper fraction]} \\\\ = \Big(\dfrac{366}{1} \times \dfrac{5}{8}\Big)\text{ km} & \text{[Simplifying 35 and 7 ⇒ Divide by 7 ]} \\\\ = \dfrac{366 \times 5}{1 \times 8}\text{ km} \\\\ = \dfrac{1830}{8} \text{ km} \\\\ = 228\dfrac{3}{4}\text{ km} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The distance between two cities is 22834228\dfrac{3}{4} km.

Question 17

The length of a rectangular plot of land is 293729\dfrac{3}{7}m. If its breadth is 1281112\dfrac{8}{11}m, find its area.

Answer

Given:

Length of a rectangular plot of land = 293729\dfrac{3}{7} m

Breadth of a rectangular plot of land = 1281112\dfrac{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

=(2937×12811)m2=(2067×14011)m2 [Converting mixed to improper fraction]=(2061×2011)m2[Simplifying 140 and 7 ⇒ Divide by 7 ]=206×201×11m2=412011m2=374611m2[Converting improper to mixed fraction]\begin{array}{ll} = \Big(29\dfrac{3}{7} \times 12\dfrac{8}{11}\Big) \text{m}^2 \\\\ = \Big(\dfrac{206}{7} \times \dfrac{140}{11}\Big) \text{m}^2 & \text{ [Converting mixed to improper fraction]} \\\\ = \Big(\dfrac{206}{1} \times \dfrac{20}{11}\Big)\text{m}^2 & \text{[Simplifying 140 and 7 ⇒ Divide by 7 ]} \\\\ = \dfrac{206 \times 20}{1 \times 11} \text{m}^2 \\\\ = \dfrac{4120}{11}\text{m}^2 \\\\ = 374\dfrac{6}{11}\text{m}^2 & \text{[Converting improper to mixed fraction]} \end{array}

∴ The area of a rectangular plot of land is 374611374\dfrac{6}{11} m2

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