KnowledgeBoat Logo
|
OPEN IN APP

Chapter 2

Fractions

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

Exercise 2(B)

Question 1(i)

Find the reciprocal of :

1317\dfrac{13}{17}

Answer

We have:

1317\dfrac{13}{17}

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

1317\dfrac{13}{17} = 1713\dfrac{17}{13}

= 1413[Converting improper to mixed fraction]1\dfrac{4}{13} \hspace{2cm}\text{[Converting improper to mixed fraction]}

∴ The reciprocal of 1317\dfrac{13}{17} is 14131\dfrac{4}{13}

Question 1(ii)

Find the reciprocal of :

3313\dfrac{3}{313}

Answer

We have:

3313\dfrac{3}{313}

Interchanging its numerator and denominator we get,

3313\dfrac{3}{313} = 3133\dfrac{313}{3}

= 10413[Converting improper to mixed fraction]104\dfrac{1}{3} \hspace{2cm}\text{[Converting improper to mixed fraction]}

∴ The reciprocal of 3313\dfrac{3}{313} is 10413104\dfrac{1}{3}

Question 1(iii)

Find the reciprocal of :

217

Answer

We have:

217

This can be written as 2171\dfrac{217}{1}

Interchanging its numerator and denominator we get,

2171\dfrac{217}{1} = 1217\dfrac{1}{217}

∴ The reciprocal of 2171\dfrac{217}{1} is 1217\dfrac{1}{217}

Question 1(iv)

Find the reciprocal of :

11024\dfrac{1}{1024}

Answer

We have:

11024\dfrac{1}{1024}

Interchanging its numerator and denominator we get,

11024\dfrac{1}{1024} = 10241\dfrac{1024}{1} = 1024

∴ The reciprocal of 11024\dfrac{1}{1024} is 1024

Question 1(v)

Find the reciprocal of :

3133\dfrac{1}{3}

Answer

We have:

313=103[Converting mixed to improper fraction]3\dfrac{1}{3} = \dfrac{10}{3} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}

Interchanging its numerator and denominator we get,

103\dfrac{10}{3} = 310\dfrac{3}{10}

∴ The reciprocal of 103\dfrac{10}{3} is 310\dfrac{3}{10}

Question 1(vi)

Find the reciprocal of :

1111111\dfrac{1}{11}

Answer

We have:

1111111\dfrac{1}{11} = 12211[Converting mixed to improper fraction]\dfrac{122}{11} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}

Interchanging its numerator and denominator we get,

12211=11122\dfrac{122}{11} = \dfrac{11}{122}

∴ The reciprocal of 12211\dfrac{122}{11} is 11122\dfrac{11}{122}

Question 1(vii)

Find the reciprocal of :

1121\dfrac{1}{2}

Answer

We have:

112=32[Converting mixed to improper fraction]1\dfrac{1}{2} = \dfrac{3}{2} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}

Interchanging its numerator and denominator we get,

32\dfrac{3}{2} = 23\dfrac{2}{3}

∴ The reciprocal of 32\dfrac{3}{2} is 23\dfrac{2}{3}

Question 1(viii)

Find the reciprocal of :

12518125\dfrac{1}{8}

Answer

We have:

12518=10018[Converting mixed to improper fraction]125\dfrac{1}{8} = \dfrac{1001}{8} \hspace{1.5cm}\text{[Converting mixed to improper fraction]}

Interchanging its numerator and denominator we get,

10018\dfrac{1001}{8} = 81001\dfrac{8}{1001}

∴ The reciprocal of 10018\dfrac{1001}{8} is 81001\dfrac{8}{1001}

Question 2(i)

Divide :

12\dfrac{1}{2} ÷ 13\dfrac{1}{3}

Answer

We have:

=12÷13=12×31[Reciprocal of 13 is 31]=1×32×1=32=112\begin{array}{ll} \phantom{=} \dfrac{1}{2} ÷ \dfrac{1}{3} \\\\ = \dfrac{1}{2} \times \dfrac{3}{1} & [\text{Reciprocal of } \dfrac{1}{3} \text{ is } \dfrac{3}{1}] \\\\ = \dfrac{1 \times 3}{2 \times 1} \\\\ = \dfrac{3}{2} \\\\ = 1\dfrac{1}{2} \end{array}

∴ The answer is 1121\dfrac{1}{2}

Question 2(ii)

Divide :

314\dfrac{3}{14} ÷ 1142\dfrac{11}{42}

Answer

We have:

=314÷1142=314×4211[Reciprocal of 1142 is 4211]=3×4214×11=3×31×11[Simplifying 42 and 14 ⇒ Divide by 14]=911\begin{array}{ll} \phantom{=} \dfrac{3}{14} ÷ \dfrac{11}{42} \\\\ = \dfrac{3}{14} \times \dfrac{42}{11} & [\text{Reciprocal of } \dfrac{11}{42} \text{ is } \dfrac{42}{11}] \\\\ = \dfrac{3 \times 42}{14 \times 11} \\\\ = \dfrac{3 \times 3}{1 \times 11} & \text{[Simplifying 42 and 14 ⇒ Divide by 14]} \\\\ = \dfrac{9}{11} \\\\ \end{array}

∴ The answer is 911\dfrac{9}{11}

Question 2(iii)

Divide :

1 ÷ 6236\dfrac{2}{3}

Answer

We have:

=1÷623=1÷203[Converting mixed to improper fraction]=1×320[Reciprocal of 203 is 320]=11×320=1×31×20=320\begin{array}{ll} \phantom{=} 1 ÷ 6\dfrac{2}{3} \\\\ = 1 ÷ \dfrac{20}{3} & \text{[Converting mixed to improper fraction]} \\\\ = 1 \times \dfrac{3}{20} & [\text{Reciprocal of } \dfrac{20}{3} \text{ is } \dfrac{3}{20}] \\\\ = \dfrac{1}{1} \times \dfrac{3}{20} \\\\ = \dfrac{1 \times 3}{1 \times 20} \\\\ = \dfrac{3}{20} \\\\ \end{array}

∴ The answer is 320\dfrac{3}{20}

Question 2(iv)

Divide :

131313\dfrac{1}{3} ÷ 10

Answer

We have:

=1313÷10=403÷10[Converting mixed to improper fraction]=403×110[Reciprocal of 10 is 110]=43×11[Simplifying 40 and 10 ⇒ Divide by 10]=4×13×1=43=113\begin{array}{ll} \phantom{=} 13\dfrac{1}{3} ÷ 10 \\\\ = \dfrac{40}{3} ÷ 10 & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{40}{3} \times \dfrac{1}{10} & [\text{Reciprocal of } 10 \text{ is } \dfrac{1}{10}] \\\\ = \dfrac{4}{3} \times \dfrac{1}{1} & \text{[Simplifying 40 and 10 ⇒ Divide by 10]} \\\\ = \dfrac{4 \times 1}{3 \times 1} \\\\ = \dfrac{4}{3} \\\\ = 1\dfrac{1}{3} \end{array}

∴ The answer is 1131\dfrac{1}{3}

Question 2(v)

Divide :

1211212\dfrac{1}{12} ÷ 536\dfrac{5}{36}

Answer

We have:

=12112÷536=14512÷536[Converting mixed to improper fraction]=14512×365[Reciprocal of 536 is 365]=2912×361[Simplifying 145 and 5 ⇒ Divide by 5]=291×31[Simplifying 36 and 12 ⇒ Divide by 12]=29×31×1=871=87\begin{array}{ll} \phantom{=} 12\dfrac{1}{12} ÷ \dfrac{5}{36} \\\\ = \dfrac{145}{12} ÷ \dfrac{5}{36} & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{145}{12} \times \dfrac{36}{5} & [\text{Reciprocal of } \dfrac{5}{36} \text{ is } \dfrac{36}{5}] \\\\ = \dfrac{29}{12} \times \dfrac{36}{1} & \text{[Simplifying 145 and 5 ⇒ Divide by 5]} \\\\ = \dfrac{29}{1} \times \dfrac{3}{1} & \text{[Simplifying 36 and 12 ⇒ Divide by 12]} \\\\ = \dfrac{29 \times 3}{1 \times 1} \\\\ = \dfrac{87}{1} \\\\ = 87 \end{array}

∴ The answer is 87

Question 2(vi)

Divide :

1346\dfrac{13}{46} ÷ 21112\dfrac{1}{11}

Answer

We have:

=1346÷2111=1346÷2311[Converting mixed to improper fraction]=1346×1123[Reciprocal of 2311 is 1123]=13×1146×23=1431058\begin{array}{ll} \phantom{=} \dfrac{13}{46} ÷ 2\dfrac{1}{11} \\\\ = \dfrac{13}{46} ÷ \dfrac{23}{11} & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{13}{46} \times \dfrac{11}{23} & [\text{Reciprocal of } \dfrac{23}{11} \text{ is } \dfrac{11}{23}] \\\\ = \dfrac{13 \times 11}{46 \times 23} \\\\ = \dfrac{143}{1058} \end{array}

∴ The answer is 1431058\dfrac{143}{1058}

Question 2(vii)

Divide :

2031\dfrac{20}{31} ÷ 54

Answer

We have:

=2031÷54=2031×154[Reciprocal of 54 is 154]=1031×127[Simplifying 20 and 54 ⇒ Divide by 2]=10×131×27=10837\begin{array}{ll} \phantom{=} \dfrac{20}{31} ÷ 54 \\\\ = \dfrac{20}{31} \times \dfrac{1}{54} & [\text{Reciprocal of } 54 \text{ is } \dfrac{1}{54}] \\\\ = \dfrac{10}{31} \times \dfrac{1}{27} & \text{[Simplifying 20 and 54 ⇒ Divide by 2]} \\\\ = \dfrac{10 \times 1}{31 \times 27} \\\\ = \dfrac{10}{837} \end{array}

∴ The answer is 10837\dfrac{10}{837}

Question 2(viii)

Divide :

9459\dfrac{4}{5} ÷ 323253\dfrac{23}{25}

Answer

We have:

=945÷32325=495÷9825[Converting mixed to improper fraction]=495×2598[Reciprocal of 9825 is 2598]=15×252[Simplifying 49 and 98 ⇒ Divide by 49]=11×52[Simplifying 25 and 5 ⇒ Divide by 5]=1×51×2=52=212\begin{array}{ll} \phantom{=} 9\dfrac{4}{5} ÷ 3\dfrac{23}{25} \\\\ = \dfrac{49}{5} ÷ \dfrac{98}{25} & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{49}{5} \times \dfrac{25}{98} & [\text{Reciprocal of } \dfrac{98}{25} \text{ is } \dfrac{25}{98}] \\\\ = \dfrac{1}{5} \times \dfrac{25}{2} & \text{[Simplifying 49 and 98 ⇒ Divide by 49]} \\\\ = \dfrac{1}{1} \times \dfrac{5}{2} & \text{[Simplifying 25 and 5 ⇒ Divide by 5]} \\\\ = \dfrac{1 \times 5}{1 \times 2} \\\\ = \dfrac{5}{2} \\\\ = 2\dfrac{1}{2} \end{array}

∴ The answer is 2122\dfrac{1}{2}

Question 2(ix)

Divide :

321232\dfrac{1}{2} ÷ 8348\dfrac{3}{4}

Answer

We have:

=3212÷834=652÷354[Converting mixed to improper fraction]=652×435[Reciprocal of 354 is 435]=132×47[Simplifying 65 and 35 ⇒ Divide by 5]=131×27[Simplifying 4 and 2 ⇒ Divide by 2]=13×21×7=267=357\begin{array}{ll} \phantom{=} 32\dfrac{1}{2} ÷ 8\dfrac{3}{4} \\\\ = \dfrac{65}{2} ÷ \dfrac{35}{4} & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{65}{2} \times \dfrac{4}{35} & [\text{Reciprocal of } \dfrac{35}{4} \text{ is } \dfrac{4}{35}] \\\\ = \dfrac{13}{2} \times \dfrac{4}{7} & \text{[Simplifying 65 and 35 ⇒ Divide by 5]} \\\\ =\dfrac{13}{1} \times \dfrac{2}{7} & \text{[Simplifying 4 and 2 ⇒ Divide by 2]} \\\\ = \dfrac{13 \times 2}{1 \times 7} \\\\ = \dfrac{26}{7} \\\\ = 3\dfrac{5}{7} \end{array}

∴ The answer is 3573\dfrac{5}{7}

Question 2(x)

Divide :

81218\dfrac{1}{21} ÷ 1671\dfrac{6}{7}

Answer

We have:

=8121÷167=16921÷137[Converting mixed to improper fraction]=16921×713[Reciprocal of 137 is713]=1321×71[Simplifying 169 and 13 ⇒ Divide by 13]=133×11[Simplifying 7 and 21 ⇒ Divide by 7]=13×13×1=133=413\begin{array}{ll} \phantom{=} 8\dfrac{1}{21} ÷ 1\dfrac{6}{7} \\\\ = \dfrac{169}{21} ÷ \dfrac{13}{7} & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{169}{21} \times \dfrac{7}{13} & [\text{Reciprocal of } \dfrac{13}{7} \text{ is} \dfrac{7}{13}] \\\\ = \dfrac{13}{21} \times \dfrac{7}{1} & \text{[Simplifying 169 and 13 ⇒ Divide by 13]} \\\\ = \dfrac{13}{3} \times \dfrac{1}{1} & \text{[Simplifying 7 and 21 ⇒ Divide by 7]} \\\\ = \dfrac{13 \times 1}{3 \times 1} \\\\ = \dfrac{13}{3} \\\\ = 4\dfrac{1}{3} \end{array}

∴ The answer is 4134\dfrac{1}{3}

Question 2(xi)

Divide :

63166\dfrac{3}{16} ÷ 3173\dfrac{1}{7}

Answer

We have:

=6316÷317=9916÷227[Converting mixed to improper fraction]=9916×722[Reciprocal of 227 is 722]=916×72[Simplifying 99 and 22 ⇒ Divide by 11]=9×716×2=6332=13132\begin{array}{ll} \phantom{=} 6\dfrac{3}{16} ÷ 3\dfrac{1}{7} \\\\ = \dfrac{99}{16} ÷ \dfrac{22}{7} & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{99}{16} \times \dfrac{7}{22} & [\text{Reciprocal of } \dfrac{22}{7} \text{ is } \dfrac{7}{22}] \\\\ = \dfrac{9}{16} \times \dfrac{7}{2} & \text{[Simplifying 99 and 22 ⇒ Divide by 11]} \\\\ = \dfrac{9 \times 7}{16 \times 2} \\\\ = \dfrac{63}{32} \\\\ = 1\dfrac{31}{32} \end{array}

∴ The answer is 131321\dfrac{31}{32}

Question 2(xii)

Divide :

311153\dfrac{11}{15} ÷ 193519\dfrac{3}{5}

Answer

We have:

=31115÷1935=5615÷985[Converting mixed to improper fraction]=5615×598[Reciprocal of 985 is 598]=563×198[Simplifying 5 and 15 ⇒ Divide by 5]=43×17[Simplifying 56 and 98 ⇒ Divide by 14]=4×13×7=421\begin{array}{ll} \phantom{=} 3\dfrac{11}{15} ÷ 19\dfrac{3}{5} \\\\ = \dfrac{56}{15} ÷ \dfrac{98}{5} & \text{[Converting mixed to improper fraction]} \\\\ = \dfrac{56}{15} \times \dfrac{5}{98} & [\text{Reciprocal of } \dfrac{98}{5} \text{ is } \dfrac{5}{98}] \\\\ = \dfrac{56}{3} \times \dfrac{1}{98} & \text{[Simplifying 5 and 15 ⇒ Divide by 5]} \\\\ = \dfrac{4}{3} \times \dfrac{1}{7} & \text{[Simplifying 56 and 98 ⇒ Divide by 14]} \\\\ = \dfrac{4 \times 1}{3 \times 7} \\\\ = \dfrac{4}{21} \end{array}

∴ The answer is 421\dfrac{4}{21}

Question 3

By what fraction should 935\dfrac{9}{35} be multiplied to get 715\dfrac{7}{15} ?

Answer

Given:

The fraction by which 935\dfrac{9}{35} must be multiplied to get 715\dfrac{7}{15} = ?

Let the required fraction be x

Now we have,

935×x=715\dfrac{9}{35} \times x = \dfrac{7}{15}

x = 715÷935\dfrac{7}{15} ÷ \dfrac{9}{35}

=715×359[Reciprocal of 935 is 359]=73×79[Simplifying 35 and 15 ⇒ Divide by 5]=7×73×9=4927=12227\begin{array}{ll} = \dfrac{7}{15} \times \dfrac{35}{9} & [\text{Reciprocal of } \dfrac{9}{35} \text{ is } \dfrac{35}{9}] \\\\ = \dfrac{7}{3} \times \dfrac{7}{9} & \text{[Simplifying 35 and 15 ⇒ Divide by 5]} \\\\ = \dfrac{7 \times 7}{3 \times 9} \\\\ = \dfrac{49}{27} \\\\ = 1\dfrac{22}{27} \end{array}

∴ The answer is 122271\dfrac{22}{27}

Question 4

If the cost of 6146\dfrac{1}{4} m of cloth is ₹1421781421\dfrac{7}{8}, then find the cost of 1 metre of cloth.

Answer

Given:

Total cost = ₹142178=₹1137581421\dfrac{7}{8} = ₹\dfrac{11375}{8}

Total length = 614 m=254 m6\dfrac{1}{4}\text{ m} = \dfrac{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 = ₹ 113758÷254\dfrac{11375}{8} ÷ \dfrac{25}{4} m

=₹(113758×425)[Reciprocal of 254 is 425]=₹(4558×41)[Simplifying 11375 and 25 ⇒ Divide by 25]=₹(4552×11)[Simplifying 4 and 8 ⇒ Divide by 4]=₹455×12×1=₹4552=₹22712\begin{array}{ll} = ₹\Big(\dfrac{11375}{8} \times \dfrac{4}{25}\Big) & [\text{Reciprocal of } \dfrac{25}{4} \text{ is } \dfrac{4}{25}] \\\\ = ₹\Big(\dfrac{455}{8} \times \dfrac{4}{1}\Big) & \text{[Simplifying 11375 and 25 ⇒ Divide by 25]} \\\\ = ₹\Big(\dfrac{455}{2} \times \dfrac{1}{1}\Big) & \text{[Simplifying 4 and 8 ⇒ Divide by 4]} \\\\ = ₹\dfrac{455 \times 1}{2 \times 1} \\\\ = ₹\dfrac{455}{2} \\\\ = ₹227\dfrac{1}{2} \end{array}

∴ The cost of 1 metre of cloth = ₹22712227\dfrac{1}{2}

Question 5

How many pieces of length 3573\dfrac{5}{7}m each can be cut from a wall paper 52 m long?

Answer

Given:

Piece length = 3573\dfrac{5}{7} m = 267\dfrac{26}{7} m [Converting mixed to improper fraction]\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 ÷ 267\dfrac{26}{7} m

=(521×726)[Reciprocal of 267 is 726]=(21×71)[Simplifying 52 and 26 ⇒ Divide by 26]=(2×71×1)=141=14\begin{array}{ll} = \Big(\dfrac{52}{1} \times \dfrac{7}{26}\Big) & [\text{Reciprocal of } \dfrac{26}{7} \text{ is } \dfrac{7}{26}] \\\\ = \Big(\dfrac{2}{1} \times \dfrac{7}{1}\Big) & \text{[Simplifying 52 and 26 ⇒ Divide by 26]} \\\\ = \Big(\dfrac{2 \times 7}{1 \times 1}\Big) \\\\ = \dfrac{14}{1} \\\\ = 14 \end{array}

∴ Number of pieces = 14

Question 6

A log of wood 6276\dfrac{2}{7}m in length, was cut into 11 pieces. What is the length of each piece?

Answer

Given :

Total length of a log of wood = 6276\dfrac{2}{7} m = 447\dfrac{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 = 447\dfrac{44}{7} m ÷ 11

=(447×111) m[Reciprocal of 11 is 111]=(47×11) m[Simplifying 44 and 11 ⇒ Divide by 11]=47 m\begin{array}{ll} = \Big(\dfrac{44}{7} \times \dfrac{1}{11}\Big)\text{ m} & [\text{Reciprocal of } 11 \text{ is } \dfrac{1}{11}] \\\\ = \Big(\dfrac{4}{7} \times \dfrac{1}{1}\Big)\text{ m} & \text{[Simplifying 44 and 11 ⇒ Divide by 11]} \\\\ = \dfrac{4}{7}\text{ m} \\\\ \end{array}

∴ The length of each piece = 47\dfrac{4}{7} m

Question 7

A car covers 64114641\dfrac{1}{4}km in 431843\dfrac{1}{8} litres of fuel. How much distance can this car cover in 1 litre of fuel ?

Answer

Given:

Total distance = 64114 km=25654641\dfrac{1}{4}\text{ km} = \dfrac{2565}{4} km

Total fuel = 4318 litres=3458 litres43\dfrac{1}{8}\text{ litres} = \dfrac{345}{8}\text{ litres}

Distance per litre = ?

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

Substituting the values in above, we get:

Distance per litre = 25654 km÷3458 litres\dfrac{2565}{4}\text{ km} ÷ \dfrac{345}{8}\text{ litres}

=(25654×8345) km[Reciprocal of 3458 is 8345]=(25651×2345) km[Simplifying 8 and 4 ⇒ Divide by 4]=(1711×223) km[Simplifying 2565 and 345 ⇒ Divide by 15]=171×21×23 km=34223 km=142023 km\begin{array}{ll} = \Big(\dfrac{2565}{4} \times \dfrac{8}{345}\Big)\text{ km} & [\text{Reciprocal of } \dfrac{345}{8} \text{ is } \dfrac{8}{345}] \\\\ = \Big(\dfrac{2565}{1} \times \dfrac{2}{345}\Big)\text{ km} & \text{[Simplifying 8 and 4 ⇒ Divide by 4]} \\\\ = \Big(\dfrac{171}{1} \times \dfrac{2}{23}\Big)\text{ km} & \text{[Simplifying 2565 and 345 ⇒ Divide by 15]} \\\\ = \dfrac{171 \times 2}{1 \times 23} \text{ km} \\\\ = \dfrac{342}{23} \text{ km} \\\\ = 14\dfrac{20}{23} \text{ km} \end{array}

∴ The distance per litre of petrol = 14202314\dfrac{20}{23} km

Question 8

The area of a rectangular plot of land is 81745817\dfrac{4}{5}sq. m. If its breadth is 213421\dfrac{3}{4}m, find its length.

Answer

Given:

Area of rectangular plot of land = 81745 m2=40895 m2817\dfrac{4}{5}\text{ m}^2 = \dfrac{4089}{5}\text{ m}^2

Breadth = 2134 m=874 m21\dfrac{3}{4}\text{ m} = \dfrac{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 = 40895 m2÷874 m\dfrac{4089}{5}\text{ m}^2 ÷ \dfrac{87}{4} \text{ m}

=(40895×487) m[Reciprocal of 874 is 487]=(475×41) m[Simplifying 2565 and 345 ⇒ Divide by 15]=47×45×1 m=1885 m=3735 m\begin{array}{ll} = \Big(\dfrac{4089}{5} \times \dfrac{4}{87}\Big)\text{ m} & [\text{Reciprocal of } \dfrac{87}{4} \text{ is } \dfrac{4}{87}] \\\\ = \Big(\dfrac{47}{5} \times \dfrac{4}{1}\Big)\text{ m} & \text{[Simplifying 2565 and 345 ⇒ Divide by 15]} \\\\ = \dfrac{47 \times 4}{5 \times 1} \text{ m} \\\\ = \dfrac{188}{5} \text{ m} \\\\ = 37\dfrac{3}{5}\text{ m} \end{array}

∴ The breadth of rectangular plot of land = 373537\dfrac{3}{5} m

Question 9

The area of a sheet of paper is 623710623\dfrac{7}{10} sq. cm. If its length is 2971029\dfrac{7}{10}cm, find its width.

Answer

Given:

Area of a sheet of paper = 623710 cm2=623710 cm2623\dfrac{7}{10}\text{ cm}^2 = \dfrac{6237}{10}\text{ cm}^2

Length = 29710 cm=29710 cm29\dfrac{7}{10}\text{ cm} = \dfrac{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 = 623710 cm2\dfrac{6237}{10}\text{ cm}^2 ÷ 29710 cm\dfrac{297}{10}\text{ cm}

=(623710×10297) cm[Reciprocal of 29710 is 10297]=(62371×1297) cm[Simplifying 10 and 10 ⇒ Divide by 1]=6237×11×297 cm=6237297 cm=21 cm\begin{array}{ll} = \Big(\dfrac{6237}{10} \times \dfrac{10}{297}\Big)\text{ cm} & [\text{Reciprocal of } \dfrac{297}{10} \text{ is } \dfrac{10}{297}] \\\\ = \Big(\dfrac{6237}{1} \times \dfrac{1}{297}\Big)\text{ cm} & \text{[Simplifying 10 and 10 ⇒ Divide by 1]} \\\\ = \dfrac{6237 \times 1}{1 \times 297} \text{ cm} \\\\ = \dfrac{6237}{297}\text{ cm} \\\\ = 21 \text{ cm} \end{array}

∴ The width of a sheet of paper = 21 cm

Question 10

A bundle of 500 sheets of paper has a net weight of 23102\dfrac{3}{10}kg. Find the weight of 1 sheet of paper.

Answer

Given:

Total weigth = 2310 kg=2310 kg2\dfrac{3}{10} \text{ kg} = \dfrac{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 = 2310\dfrac{23}{10} kg ÷ 500

=(2310×1500) kg[Reciprocal of 500 is 1500]=23×110×500 kg=235000 kg\begin{array}{ll} = \Big(\dfrac{23}{10} \times \dfrac{1}{500}\Big)\text{ kg} & [\text{Reciprocal of } 500 \text{ is } \dfrac{1}{500}] \\\\ = \dfrac{23 \times 1}{10 \times 500} \text{ kg} \\\\ = \dfrac{23}{5000} \text{ kg} \end{array}

∴ The weigth of 1 sheet of paper = 235000\dfrac{23}{5000} kg

Question 11

A car travels 28312283\dfrac{1}{2} km in 4234\dfrac{2}{3} hours. How far does this car go in 1 hour, travelling at the same speed?

Answer

Given:

Total distance = 28312 km=5672 km283\dfrac{1}{2}\text{ km} = \dfrac{567}{2}\text{ km}

Total time = 423 hours=143 hours4\dfrac{2}{3}\text{ hours} = \dfrac{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 = 5672 km÷143 hours\dfrac{567}{2}\text{ km} ÷ \dfrac{14}{3}\text{ hours}

=(5672×314) km[Reciprocal of 143 is 314]=(812×32) km[Simplifying 567 and 14 ⇒ Divide by 7]=81×32×2 km=2434 km=6034 km\begin{array}{ll} = \Big(\dfrac{567}{2} \times \dfrac{3}{14}\Big) \text{ km} & [\text{Reciprocal of } \dfrac{14}{3} \text{ is } \dfrac{3}{14}] \\\\ = \Big(\dfrac{81}{2} \times \dfrac{3}{2}\Big) \text{ km} & \text{[Simplifying 567 and 14 ⇒ Divide by 7]} \\\\ = \dfrac{81 \times 3}{2 \times 2}\text{ km} \\\\ = \dfrac{243}{4}\text{ km} \\\\ = 60\dfrac{3}{4}\text{ km} \end{array}

∴ The distance covered in 1 hour = 603460\dfrac{3}{4} km

Question 12

The product of two fractions is 87258\dfrac{7}{25}. If one of them is 31153\dfrac{1}{15} find the other.

Answer

Given:

Let the two fractions be p and q.

p = 3115=46153\dfrac{1}{15} = \dfrac{46}{15}

Let the other fraction be q.

p x q = 8725=207258\dfrac{7}{25} = \dfrac{207}{25}

q = 20725\dfrac{207}{25} ÷ p [Solving for q]\hspace{2cm}\text{[Solving for q]}

Substituting the value of p, we get:

q = 20725÷4615\dfrac{207}{25} ÷ \dfrac{46}{15}

=20725×1546[Reciprocal of 4615 is 1546]=2075×346[Simplifying 15 and 25 ⇒ Divide by 5]=95×32[Simplifying 207 and 46 ⇒ Divide by 23]=9×35×2=2710=2710\begin{array}{ll} = \dfrac{207}{25} \times \dfrac{15}{46} & [\text{Reciprocal of } \dfrac{46}{15} \text{ is } \dfrac{15}{46}] \\\\ = \dfrac{207}{5} \times \dfrac{3}{46} & \text{[Simplifying 15 and 25 ⇒ Divide by 5]} \\\\ = \dfrac{9}{5} \times \dfrac{3}{2} & \text{[Simplifying 207 and 46 ⇒ Divide by 23]} \\\\ = \dfrac{9 \times 3}{5 \times 2} \\\\ = \dfrac{27}{10} \\\\ = 2\dfrac{7}{10} \end{array}

∴ The other fraction = 27102\dfrac{7}{10}

Exercise 2(C)

Question 1

Simplify :

123+561\dfrac{2}{3} + \dfrac{5}{6} of 2425\dfrac{24}{25}

Answer

We have:

123+561\dfrac{2}{3} + \dfrac{5}{6} of 2425\dfrac{24}{25}

= 53\dfrac{5}{3} + 56\dfrac{5}{6} of 2425\dfrac{24}{25} [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

=53+56×2425=53+16×245=53+2430=53+45[Of simplified]=25+1215=3715[Addition simplified]=2715[Converting improper to mixed fraction]\begin{array}{ll} = \dfrac{5}{3} + \dfrac{5}{6} \times \dfrac{24}{25} \\\\ = \dfrac{5}{3} + \dfrac{1}{6} \times \dfrac{24}{5} \\\\ = \dfrac{5}{3} + \dfrac{24}{30} \\\\ = \dfrac{5}{3} + \dfrac{4}{5} & \text{[Of simplified]} \\\\ = \dfrac{25 + 12}{15} = \dfrac{37}{15} & \text{[Addition simplified]} \\\\ = 2\dfrac{7}{15} & \text{[Converting improper to mixed fraction]} \end{array}

∴ The answer is 27152\dfrac{7}{15}

Question 2

Simplify :

13\dfrac{1}{3} of 423÷213×1124\dfrac{2}{3} ÷ 2\dfrac{1}{3}\times 1\dfrac{1}{2}

Answer

We have:

13\dfrac{1}{3} of 4234\dfrac{2}{3} ÷ 213×1122\dfrac{1}{3}\times 1\dfrac{1}{2}

= 13\dfrac{1}{3} of 143\dfrac{14}{3} ÷ 73×32\dfrac{7}{3}\times \dfrac{3}{2} [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

=13×143÷73×32=1×143×3÷73×32=149÷73×32[Of simplified]=149×37×32[Reciprocal of 73 is s37]=143×17×32=23×11×32=2×13×1×32=23×32[Division simplified]=2×33×2=66=11=1[Multiplication simplified]\begin{array}{ll} = \dfrac{1}{3} \times \dfrac{14}{3} ÷ \dfrac{7}{3}\times \dfrac{3}{2} \\\\ = \dfrac{1 \times 14}{3 \times 3} ÷ \dfrac{7}{3}\times \dfrac{3}{2} \\\\ = \dfrac{14}{9} ÷ \dfrac{7}{3}\times \dfrac{3}{2} & \text{[Of simplified]} \\\\ = \dfrac{14}{9} \times \dfrac{3}{7}\times \dfrac{3}{2} & [\text{Reciprocal of } \dfrac{7}{3} \text{ is }s \dfrac{3}{7}] \\\\ = \dfrac{14}{3} \times \dfrac{1}{7} \times \dfrac{3}{2} \\\\ = \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{3}{2} \\\\ = \dfrac{2 \times1}{3 \times 1} \times \dfrac{3}{2} = \dfrac{2}{3} \times \dfrac{3}{2} & \text{[Division simplified]} \\\\ = \dfrac{2 \times 3}{3 \times 2} = \dfrac{6}{6}\\\\ = \dfrac{1}{1} = 1 & \text{[Multiplication simplified]} \end{array}

∴ The answer is 1

Question 3

Simplify :

214+116−123÷2232\dfrac{1}{4} + 1\dfrac{1}{6} - 1\dfrac{2}{3} ÷ 2\dfrac{2}{3} of 3343\dfrac{3}{4}

Answer

We have:

214+116−123÷2232\dfrac{1}{4} + 1\dfrac{1}{6} - 1\dfrac{2}{3} ÷ 2\dfrac{2}{3} of 3343\dfrac{3}{4}

= 94\dfrac{9}{4} + 76\dfrac{7}{6} - 53\dfrac{5}{3} ÷ 83\dfrac{8}{3} of 154\dfrac{15}{4} [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

=94+76−53÷83×154=94+76−53÷8×153×4=94+76−53÷12012=94+76−53÷101[Of simplified]=94+76−53×110[Reciprocal of 101 is 110]=94+76−5×13×10=94+76−530=94+76−16=27+1412−16=4112−16[Addition simplified]=41−212=3912=134[Subtraction simplified]=314[Converting improper to mixed fraction]\begin{array}{ll} = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{8}{3}\times \dfrac{15}{4} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{8 \times 15}{3 \times 4} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{120}{12} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} ÷ \dfrac{10}{1} & \text{[Of simplified]} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{3} \times \dfrac{1}{10} & [\text{Reciprocal of } \dfrac{10}{1} \text{ is } \dfrac{1}{10}] \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5 \times 1}{3 \times 10} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{5}{30} \\\\ = \dfrac{9}{4} + \dfrac{7}{6} - \dfrac{1}{6} \\\\ = \dfrac{27 + 14}{12} - \dfrac{1}{6} \\\\ = \dfrac{41}{12} - \dfrac{1}{6} & \text{[Addition simplified]} \\\\ = \dfrac{41 - 2}{12} \\\\ = \dfrac{39}{12} \\\\ = \dfrac{13}{4} & \text{[Subtraction simplified]} \\\\ = 3\dfrac{1}{4} & \text{[Converting improper to mixed fraction]} \\\\ \end{array}

∴ The answer is 3143\dfrac{1}{4}

Question 4

Simplify :

112×234÷1471\dfrac{1}{2}\times2\dfrac{3}{4} ÷ 1\dfrac{4}{7} of 2582\dfrac{5}{8}

Answer

We have:

112×234÷1471\dfrac{1}{2}\times2\dfrac{3}{4} ÷ 1\dfrac{4}{7} of 2582\dfrac{5}{8}

= 32×114\dfrac{3}{2}\times \dfrac{11}{4} ÷ 117\dfrac{11}{7} of 218\dfrac{21}{8} [Converting mixed to improper fraction]

According to BODMAS rule, we solve "of" first

=32×114÷117×218=32×114÷111×38=32×114÷11×31×8=32×114÷338[Of simplified]=32×114×833[Reciprocal of 338 is 833]=32×14×83=32×11×23=32×1×21×3=32×23[Division simplified]=66=1[Multiplication simplified]\begin{array}{ll} = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{11}{7} \times \dfrac{21}{8} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{11}{1} \times \dfrac{3}{8} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{11 \times 3}{1 \times 8} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} ÷ \dfrac{33}{8} & \text{[Of simplified]} \\\\ = \dfrac{3}{2}\times \dfrac{11}{4} \times \dfrac{8}{33} & [\text{Reciprocal of } \dfrac{33}{8} \text{ is } \dfrac{8}{33}] \\\\ = \dfrac{3}{2}\times \dfrac{1}{4} \times \dfrac{8}{3} \\\\ = \dfrac{3}{2}\times \dfrac{1}{1} \times \dfrac{2}{3}\\\\ = \dfrac{3}{2}\times \dfrac{1 \times 2}{1 \times 3} \\\\ = \dfrac{3}{2}\times \dfrac{2}{3} & \text{[Division simplified]} \\\\ = \dfrac{6}{6} = 1 & \text{[Multiplication simplified]} \\\\ \end{array}

∴ The answer is 1

Question 5

Simplify :

(234+156)\Big(2\dfrac{3}{4} + 1\dfrac{5}{6}\Big) ÷ 2152\dfrac{1}{5} of 3133\dfrac{1}{3}

Answer

We have:

(234+156)\Big(2\dfrac{3}{4} + 1\dfrac{5}{6}\Big) ÷ 2152\dfrac{1}{5} of 3133\dfrac{1}{3}

= (114+116)\Big(\dfrac{11}{4} + \dfrac{11}{6}\Big) ÷ 115\dfrac{11}{5} of 103\dfrac{10}{3} [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

=(33+2212)÷115 of 103=5512÷115 of 103[Brackets simplified]=5512÷115×103=5512÷11015=5512÷223[Of simplified]=5512×322[Reciprocal of 223 is 322]=512×32[Dividing 55 and 22 by 11]=1524=58[Division simplified]\begin{array}{ll} = \Big(\dfrac{33 + 22}{12}\Big) ÷ \dfrac{11}{5} \text{ of } \dfrac{10}{3} \\\\ = \dfrac{55}{12} ÷ \dfrac{11}{5} \text{ of } \dfrac{10}{3} & \text{[Brackets simplified]} \\\\ = \dfrac{55}{12} ÷ \dfrac{11}{5} \times \dfrac{10}{3} \\\\ = \dfrac{55}{12} ÷ \dfrac{110}{15} \\\\ = \dfrac{55}{12} ÷ \dfrac{22}{3} & \text{[Of simplified]} \\\\ = \dfrac{55}{12} \times \dfrac{3}{22} & [\text{Reciprocal of } \dfrac{22}{3} \text{ is } \dfrac{3}{22}] \\\\ = \dfrac{5}{12} \times \dfrac{3}{2} & \text{[Dividing 55 and 22 by 11]} \\\\ = \dfrac{15}{24} = \dfrac{5}{8} & \text{[Division simplified]} \end{array}

∴ The answer is 58\dfrac{5}{8}

Question 6

Simplify :

715\dfrac{7}{15} of (23+712)\Big(\dfrac{2}{3} + \dfrac{7}{12}\Big) ÷ (56−35)\Big(\dfrac{5}{6}-\dfrac{3}{5}\Big)

Answer

We have:

715\dfrac{7}{15} of (23+712)\Big(\dfrac{2}{3} + \dfrac{7}{12}\Big) ÷ (56−35)\Big(\dfrac{5}{6}-\dfrac{3}{5}\Big)

According to BODMAS rule, we simplify brackets first

=715 of (8+712)÷(56−35)=715 of (1512)÷(56−35)=715 of 512÷(56−35)[First bracket simplified]=715 of 1512÷(25−1830)=715 of 1512÷730[Second bracket simplified]=715×1512÷730=712÷730[Of simplified]=712×307[Reciprocal of 730 is 307]=112×301[Dividing 7 and 7 by 7]=12×51[Dividing 30 and 12 by 6]=52[Division simplified]=212\begin{array}{ll} = \dfrac{7}{15}\text{ of }\Big(\dfrac{8 + 7}{12}\Big) ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) \\\\ = \dfrac{7}{15}\text{ of } \Big(\dfrac{15}{12}\Big) ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) \\\\ = \dfrac{7}{15}\text{ of } \dfrac{5}{12} ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) & \text{[First bracket simplified]} \\\\ = \dfrac{7}{15}\text{ of } \dfrac{15}{12} ÷ \Big(\dfrac{25 - 18}{30}\Big) \\\\ = \dfrac{7}{15}\text{ of } \dfrac{15}{12} ÷ \dfrac{7}{30} & \text{[Second bracket simplified]} \\\\ = \dfrac{7}{15} \times \dfrac{15}{12} ÷ \dfrac{7}{30} \\\\ = \dfrac{7}{12} ÷ \dfrac{7}{30} & \text{[Of simplified]} \\\\ = \dfrac{7}{12} \times \dfrac{30}{7} & [\text{Reciprocal of } \dfrac{7}{30} \text{ is } \dfrac{30}{7}] \\\\ = \dfrac{1}{12} \times \dfrac{30}{1} & \text{[Dividing 7 and 7 by 7]} \\\\ = \dfrac{1}{2} \times \dfrac{5}{1} & \text{[Dividing 30 and 12 by 6]} \\\\ = \dfrac{5}{2} & \text{[Division simplified]} \\\\ = 2\dfrac{1}{2} \end{array}

∴ The answer is 2122\dfrac{1}{2}

Question 7

Simplify :

(22÷512)\Big(22÷5\dfrac{1}{2}\Big) ÷ 2152\dfrac{1}{5} of 313+15113\dfrac{1}{3} + 1\dfrac{5}{11}

Answer

We have:

(22÷512)\Big(22÷5\dfrac{1}{2}\Big) ÷ 2152\dfrac{1}{5} of 313+15113\dfrac{1}{3} + 1\dfrac{5}{11}

= (22÷112)\Big(22÷\dfrac{11}{2}\Big) ÷ 115\dfrac{11}{5} of 103\dfrac{10}{3} + 1611\dfrac{16}{11} [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

=(22×211)÷115 of 103+1611[Reciprocal of 112 is 211]=(4411)÷115 of 103+1611[Divide 44 and 11 by 11]=41÷115 of 103+1611[Bracket simplified]=41÷115×103+1611=41÷11015+1611[Divide 110 and 15 by 5]=41÷223+1611[Of simplified]=41×322+1611[Reciprocal of 223 is 322]=1222+1611[Divide 12 and 22 by 2]=611+1611[Division simplified]=6+1611=2211[Divide 22 and 11 by 11]=2[Division simplified]\begin{array}{ll} = \Big(22 \times \dfrac{2}{11}\Big) ÷ \dfrac{11}{5}\text{ of }\dfrac{10}{3} + \dfrac{16}{11} & [\text{Reciprocal of } \dfrac{11}{2} \text{ is } \dfrac{2}{11}] \\\\ = \Big(\dfrac{44}{11}\Big) ÷ \dfrac{11}{5}\text{ of }\dfrac{10}{3} + \dfrac{16}{11} & \text{[Divide 44 and 11 by 11]} \\\\ = \dfrac{4}{1} ÷ \dfrac{11}{5}\text{ of }\dfrac{10}{3} + \dfrac{16}{11} & \text{[Bracket simplified]} \\\\ = \dfrac{4}{1} ÷ \dfrac{11}{5} \times \dfrac{10}{3} + \dfrac{16}{11} \\\\ = \dfrac{4}{1} ÷ \dfrac{110}{15} + \dfrac{16}{11} & \text{[Divide 110 and 15 by 5]} \\\\ = \dfrac{4}{1} ÷ \dfrac{22}{3} + \dfrac{16}{11} & \text{[Of simplified]} \\\\ = \dfrac{4}{1} \times \dfrac{3}{22} + \dfrac{16}{11} & [\text{Reciprocal of } \dfrac{22}{3} \text{ is } \dfrac{3}{22}] \\\\ = \dfrac{12}{22} + \dfrac{16}{11} & \text{[Divide 12 and 22 by 2]} \\\\ = \dfrac{6}{11} + \dfrac{16}{11} & \text{[Division simplified]} \\\\ = \dfrac{6 + 16}{11} = \dfrac{22}{11} & \text{[Divide 22 and 11 by 11]} \\\\ = 2 & \text{[Division simplified]} \end{array}

∴ The answer is 2

Question 8

Simplify :

6136\dfrac{1}{3} ÷ (215+312)\Big(2\dfrac{1}{5} + 3\dfrac{1}{2}\Big) of 3133\dfrac{1}{3}

Answer

We have:

6136\dfrac{1}{3} ÷ (215+312)\Big(2\dfrac{1}{5} + 3\dfrac{1}{2}\Big) of 3133\dfrac{1}{3}

= 193\dfrac{19}{3} ÷ (115+72)\Big(\dfrac{11}{5} + \dfrac{7}{2}\Big) of 103\dfrac{10}{3} [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

=193÷(22+3510) of 103=193÷5710 of 103[Bracket simplified]=193÷5710×103=193÷57030=193÷191[Of simplified]=193×119[Reciprocal of 191 is 119]=13[Division simplified]\begin{array}{ll} = \dfrac{19}{3} ÷ \Big(\dfrac{22 + 35}{10}\Big)\text{ of }\dfrac{10}{3} \\\\ = \dfrac{19}{3} ÷ \dfrac{57}{10}\text{ of }\dfrac{10}{3} & \text{[Bracket simplified]} \\\\ = \dfrac{19}{3} ÷ \dfrac{57}{10} \times \dfrac{10}{3} \\\\ = \dfrac{19}{3} ÷ \dfrac{570}{30} \\\\ = \dfrac{19}{3} ÷ \dfrac{19}{1} & \text{[Of simplified]} \\\\ = \dfrac{19}{3} \times \dfrac{1}{19} & [\text{Reciprocal of } \dfrac{19}{1} \text{ is } \dfrac{1}{19}] \\\\ = \dfrac{1}{3} & \text{[Division simplified]} \end{array}

∴ The answer is 13\dfrac{1}{3}

Exercise 2(D) - Multiple Choice Questions

Question 1

Which of the following fractions can be expressed as a mixed fraction?

  1. 35\dfrac{3}{5}

  2. 1200\dfrac{1}{200}

  3. 1110\dfrac{11}{10}

  4. 3240\dfrac{32}{40}

Answer

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.

Question 2

Which of the following is an improper fraction?

  1. 713\dfrac{7}{13}

  2. 32\dfrac{3}{2}

  3. 4200\dfrac{4}{200}

  4. 11121\dfrac{11}{121}

Answer

An improper fraction has numerator greater than denominator.

Hence, option 2 is the correct option.

Question 3

Three-fourths of 3 is

  1. a unit fraction
  2. a proper fraction
  3. an improper fraction
  4. not a fraction

Answer

Given:

Three-fourths of 3

= 34\dfrac{3}{4} of 3

= 34×\dfrac{3}{4} \times 3 = 94\dfrac{9}{4}

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

Hence, option 3 is the correct option.

Question 4

If the cost of a pen is ₹162316\dfrac{2}{3}, then, the cost of 60 pens will be

  1. ₹960
  2. ₹480
  3. ₹500
  4. ₹1000

Answer

Given:

Cost of 1 pen = ₹162316\dfrac{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 = ₹162316\dfrac{2}{3} x 60

= ₹503\dfrac{50}{3} x 60 = ₹30003\dfrac{3000}{3} = ₹1000

Hence, option 4 is the correct option.

Question 5

If a sheet of paper having an area of 421442\dfrac{1}{4} cm2 is cut into 13 strips of equal area, then the area of each strip will be

  1. 4144\dfrac{1}{4}cm2

  2. 3143\dfrac{1}{4}cm2

  3. 41524\dfrac{1}{52}cm2

  4. 71137\dfrac{1}{13}cm2

Answer

Given:

Area of sheet = 421442\dfrac{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 = 421442\dfrac{1}{4} cm2 ÷ 13

=1694 cm2÷13=(1694×113) cm2=(134) cm2=314 cm2\begin{array}{ll} = \dfrac{169}{4} \text{ cm}^2 ÷ 13 \\\\ = \Big(\dfrac{169}{4} \times \dfrac{1}{13}\Big) \text{ cm}^2 \\\\ = \Big(\dfrac{13}{4}\Big) \text{ cm}^2 \\\\ = 3\dfrac{1}{4} \text{ cm}^2 \end{array}

Hence, option 2 is the correct option.

Exercise 2(D) - Mental Maths

Question 1

Fill in the blanks :

(i) If ab\dfrac{a}{b} is a fraction, then a and b are ............... .

(ii) In an improper fraction, the numerator is ............... the denominator.

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

(iv) ab\dfrac{a}{b} ÷ cd\dfrac{c}{d} equal to the product of ab\dfrac{a}{b} and ............... .

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

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

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

Answer

(i) If ab\dfrac{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) ab\dfrac{a}{b} ÷ cd\dfrac{c}{d} equal to the product of ab\dfrac{a}{b} and the reciprocal of cd\dfrac{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 .

Question 2

Write true (T) or false (F):

(i) In an improper fraction, the numerator is always greater than the denominator.

(ii) The product of two proper fractions can be an improper fraction.

(iii) Any improper fraction is always greater than any proper fraction.

(iv) Every unit fraction is equal to 1.

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

(vi) The product of two proper fractions is always greater than each of the two proper fractions.

(vii) There exists a fraction whose multiplicative inverse is equal to the fraction itself.

(viii) The product of a proper fraction and and an improper fraction is always less than the improper fraction.

Answer

(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 15,19\dfrac{1}{5}, \dfrac{1}{9}. Only the fraction 11\dfrac{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 11\dfrac{1}{1}. Its multiplicative inverse (reciprocal) is also 11\dfrac{1}{1}, which is equal to the original fraction. (Similarly, −1-1 or −11\dfrac{-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.

Exercise 2(D) - Case Study Based Questions

Question 1

When Prabhu Dayal died, he left all his money for his grand-children, 3 of whom were boys and 5 were girls. In his will he insisted that each grand-child must get equal share of the total amount of ₹56,00,000.

(1) What was the share of each child ?

  1. ₹8,00,000

  2. ₹11,20,000

  3. ₹6,40,000

  4. ₹7,00,000

(2) What fraction of the money did the girls receive ?

  1. 53\dfrac{5}{3}

  2. 35\dfrac{3}{5}

  3. 58\dfrac{5}{8}

  4. 38\dfrac{3}{8}

(3) How much did the boys receive in total ?

  1. ₹21,00,000

  2. ₹24,00,000

  3. ₹35,00,000

  4. ₹40,00,000

(4) If one of the girls did not take her share and the money is divided among the remaining grand-children, the fraction of the money received by the boys is :

  1. 35\dfrac{3}{5}

  2. 25\dfrac{2}{5}

  3. 47\dfrac{4}{7}

  4. 37\dfrac{3}{7}

Answer

(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

= ₹56000008\dfrac{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

= 58\dfrac{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 = 37\dfrac{3}{7}

Hence, option 4 is the correct option.

Question 2

Amar is an electrician. He bought 7127\dfrac{1}{2} bundles of an electric cable where each bundle had 20245202\dfrac{4}{5} m of cable.

(1) Find the total length of the cable purchased by Amar.

  1. 1427 m
  2. 1521 m
  3. 1605 m
  4. 1717 m

(2) If the cost of cable is ₹6236\dfrac{2}{3} per metre, find the amount paid by Amar.

  1. ₹8830
  2. ₹9520
  3. ₹10140
  4. ₹11280

(3) Amar used 2122\dfrac{1}{2} bundles of cable for electric connections in the top floor of the building. What length of cable was used for the top floor ?

  1. 476 m
  2. 507 m
  3. 625 m
  4. 712 m

(4) Amar cut a length of 134513\dfrac{4}{5}m from a bundle and divided the remaining cable of this bundle into pieces of 21 m, length each. How many pieces of 21 m did he get from this bundle ?

  1. 9
  2. 10
  3. 11
  4. 12

Answer

(1) Given:

Total bundles = 712=1527 \dfrac{1}{2} = \dfrac{15}{2}

Length per bundle = 20245=10145202 \dfrac{4}{5} = \dfrac{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 = 152\dfrac{15}{2} x 10145\dfrac{1014}{5} m

= 32\dfrac{3}{2} x 10141\dfrac{1014}{1} m

= 31\dfrac{3}{1} x 5071\dfrac{507}{1} m

= 1521 m

Hence, option 2 is the correct option.

(2) The cost of cable per meter = ₹623=₹2036\dfrac{2}{3} = ₹\dfrac{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 = ₹203\dfrac{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 = 212=522\dfrac{1}{2} = \dfrac{5}{2}

Length per bundle = 10145\dfrac{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 = 52\dfrac{5}{2} x 10145\dfrac{1014}{5} m = 507 m

Hence, option 2 is the correct option.

(4) Given:

Length of one bundle = 10145\dfrac{1014}{5} m

Cut length = 134513\dfrac{4}{5} m = 695\dfrac{69}{5} m

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

Substituting the values in above, we get:

Remaining length = (10145−695)\Big(\dfrac{1014}{5} - \dfrac{69}{5}\Big) m

= 1014−695\dfrac{1014 - 69}{5} m = 9455\dfrac{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 = 18921\dfrac{189}{21}

= 9

Hence, option 1 is the correct option.

Exercise 2(D) - Assertions and Reasons

Question 1

Assertion: Reciprocal of an improper fraction is a proper fraction.

Reason: Reciprocal is also known as multiplicative inverse.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true but Reason (R) is false.
  4. Assertion (A) is false but Reason (R) is true.

Answer

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., 53\dfrac{5}{3}).

Its reciprocal is obtained by interchanging numerator and denominator:

53→35\dfrac{5}{3} \rightarrow \dfrac{3}{5}

Since 35\dfrac{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.

Question 2

Assertion: The product of two proper fractions is less than each of the fractions.

Reason: For any two fractions ab\dfrac{a}{b} and cb\dfrac{c}{b}, we have, ab×cb=acb\dfrac{a}{b} \times \dfrac{c}{b} = \dfrac{ac}{b}.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true but Reason (R) is false.
  4. Assertion (A) is false but Reason (R) is true.

Answer

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:

12×13=16\dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}

16\dfrac{1}{6} is smaller than both 12\dfrac{1}{2} and 13\dfrac{1}{3}

So, the Assertion is true.

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

ab×cb=acbd\dfrac{a}{b} \times \dfrac{c}{b} = \dfrac{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.

PrevNext