RD Sharma 2014 Solutions for Class 7 Math Chapter 2 Fractions are provided here with simple step-by-step explanations. These solutions for Fractions are extremely popular among class 7 students for Math Fractions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2014 Book of class 7 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2014 Solutions. All RD Sharma 2014 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 2.12:

Question 1:

Compare the following fractions by using the symbol > or < or =:
(i) 79 and 813
(ii) 119 and 59
(iii) 3741 and 1930
(iv) 1715 and 119105

Answer:


First, we need to find the LCM of denominators in each case. After that, we will equate the denominators in order to compare the two fractions. 

(i) 
     LCM of 9 and 13 is 117.

     Now make both fraction equivalent with denominator as 11779=79×131379=91117813=813×99813=72117we know91>7291117>7211779>813

(ii)

both fraction have same denominator as 9we know11>5119>59
(iii)

LCM of 41 and 30 is 1230Now convert both fraction to their equivalent fractions with denominator as 12303741=3741×30303741=111012301930=1930×41411930=7791230we know1110>77911101230>77912303741>1930

(iv)

LCM of 15 and 105 is 105.

Now convert fraction to its equivalent fractions with denominator as 1051715=1715×771715=119105 

Page No 2.12:

Question 2:

Arrange the following fractions in ascending order:
(i) 38,56,6824,13
(ii) 43,38,612,516

Answer:

(i)
LCM of the denominators 8, 6, 4 and 3 is 24.Now, convert all fractions into their equivalent fractions with denominator 24.38=38×3338=92456=56×4456=202468=68×3368=182424=24×6624=122413=13×8813=824We know:8<9<12<18<20824<924<1224<1824<202413<38<24<68<56
(ii)
LCM of the denominators 8, 12, 16 and 3 is 48.Now, convert all fractions into their equivalent fractions with denominator 48.43=43×161643=644838=38×6638=1848612=612×44612=2448516=516×33516=1548we know15<18<24<641548<1848<2448<6448516<38<612<43

Page No 2.12:

Question 3:

Arrange the following fractions in descending order:
(i) 45,710,1115,1720
(ii) 27,1135,914,1328

Answer:

(i)
LCM of the denominators 5, 10, 15 and 20 is 60.Now, convert all fractions to their equivalent fractions with denominator 60.45=45×1212=4860710=710×66=42601115=1115×44=44601720=1720×33=5160We know:51>48>44>425160>4860>4460>42601720>45>1115>710

(ii)
LCM of the denominators 7, 35, 14 and 28 is 140.Now, convert all fractions to their equivalent fractions with denominator 140.27=27×2020=401401135=1135×44=44140914=914×1010=901401328=1328×55=65140We know:90>65>44>4090140>65140>44140>40140914>1328>1135>27

Page No 2.12:

Question 4:

Write five equivalent fractions of 35.

Answer:

Five equivalent fractions of 35 are:

(i) 35=35×2235=610(ii) 35=35×3335=915(iii) 35=35×4435=1220(iv) 35=35×5535=1525(v) 35=35×6635=1830

Page No 2.12:

Question 5:

Find the sum:
(i) 58+310
(ii) 434+925
(iii) 56+3+34
(iv) 235+4710+2415

Answer:

(i)

58+310LCM of 8,10  is 40.(5×5)+(3×4)40=3740
(ii)

434+925 or (4×4)+34+(9×5)+25194+475LCM of 5,4 is 20.(19×5)+(47×4)20=28320
(iii)

56+3+34 LCM of 6,4 is 24.(5×4)+(3×24)+(3×6)24=11024
(iv)

235+4710+2415 or (2×5)+35+(4×10)+710+(2×15)+415135+4710+3415LCM of 15,10 and 5 is 30.(13×6)+(47×3)+(34×2)30=28730

Page No 2.12:

Question 6:

Find the difference of
(i) 1324 and 716
(ii) 6 and 233
(iii) 2125 and 1820
(iv) 3310 and 2715

Answer:

(i)

1324-716LCM of 24 and 16 is 48.(13×2)-(7×3)48=548

(ii)

233-6LCM of 3 and 1 is 3.(23×1)-(6×3)3=53

(iii)

1820-2125LCM of 20,25 is 100.(18×5)-(21×4)100=6100=350

(iv)

3310-2715(3×10)+310-(2×15)+7153310-3715LCM of 10 and 15 is 30.(33×3)-(37×2)30=253056



Page No 2.13:

Question 7:

Find the difference:
(i) 67-911
(ii) 8-59
(iii) 9-523
(iv) 4310-1215

Answer:

(i)
67-911LCM of 7 and 11 is 77.67-911(6×11)-(9×7)77377

(ii)
8-59LCM of 1 and 9 is 9.81-59(8×9)-(5×1)9679

(iii)
9-52391-(5×3)+23LCM of 1 and 3 is 3.91-173(9×3)-(17×1)3103

(iv)
4310-1215(4×10)+310-(15×1)+215LCM of 10 and 15 is 30.4310-1715(43×3)-(17×2)309530196

Page No 2.13:

Question 8:

Simplify:
(i) 23+16-29
(ii) 12-312
(iii) 756-438+2712

Answer:

(i)

23+16-29LCM of 3.6 and 9 is 18.=23+16-29=(2×6)+(1×3)-(2×2)18=1118
(ii)

12-312=121-(3×2)+12LCM of 2 and 1 is 2.=121-72=(12×2)-(7×1)2=172

(iii)

756-438+2712=(7×6)+56-(4×8)+38+(2×12)+712LCM of 6,8 and 12 is 24.=476-358+3112=(47×4)-(35×3)+(31×2)24=14524

Page No 2.13:

Question 9:

What should be added to 537 to get 12?

Answer:

Let x be the required fraction.

According to the question:

x+537=12x+(5×7)+37=12x=12-387x=(12×7)-(38×1)7467

Page No 2.13:

Question 10:

What should be added to 5415 to get 1235?

Answer:

Let x be the required fraction.

According to the question:

x+5415=1235x+(15×5)+415=(12×5)+35x=635-7915LCM of 5 and 15 is 15.x=(63×3)-(79×1)1511015x=11015223

Page No 2.13:

Question 11:

Suman studies for 523 hours daily. She devotes 245 hours of her time for Science and Mathematics. How much time does she devote for other subjects?

Answer:

Suman studies for  523hours daily. Therefore, we have
523hours =(5×3)+23=173hours
She studies science and mathematics for  245hours. Therefore, we have
245hours=(2×5)+45=145hours
Time devoted to other subjects = Total study time - Time devoted to science and mathematics 

173-145=(17×5)-(14×3)15=4315hours

Page No 2.13:

Question 12:

A piece of wire is of length 1234m. If it is cut into two pieces in such a way that the length of one piece is 514m, what is the length of the other piece?

Answer:

Let the length of second piece be x.

Total length of wire = Length of one piece + Length of second piece

1234=514+x(12×4)+34=(5×4)+14+xx=(12×4)+34-(5×4)+14x=514-214304x=304152

Page No 2.13:

Question 13:

A rectangular sheet of paper is 1212 cm long and 1023 cm wide. Find its perimeter.

Answer:

Perimeter of rectangle = 2(length + width)

2×1212+1023=2×(12×2)+12+(10×3)+23=2×(25×3)+(32×2)6=2×1396=1393 cm

Page No 2.13:

Question 14:

In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic  square?

411 911 211
311 511 711
811 111 611

Answer:

Sum along columns and rows:411+911+211=1511311+511+711=1511811+111+611=1511411+311+811=1511911+511+111=1511211+711+611=1511Sum along diagonals : 411+511+611=1511211+511+811=1511Since, all the sums in the square   are equal along rows,columns and diagonals,it is a magic square.

Page No 2.13:

Question 15:

The cost of Mathematics book is Rs 2534 and that of Science book is Rs 2012. Which costs more and by how much?

Answer:

Cost of mathematics book =Rs 2534=(25×4)+34=Rs1034Cost of Science book =Rs 2012=(20×2)+12 = 412=412×22 =Rs824We know        82<103=>824<1034Thus, Mathematics book costs more.Difference in the cost of Mathematics and Science book =cost of Mathematics book-Cost of Science book=1034-824= 214 = Rs 514So, Mathematics book costs more by Rs 514

Page No 2.13:

Question 16:

(i) Provide the number in box     and also give its simplest from in each of the following:
(i) 23×   =1030
(ii) 35×   =2475.

Answer:

(i) 23×x=1030x=1030×32=12

(ii) 35×x=2475x=2475×53=815



Page No 2.19:

Question 1:

Multiply:
(i) 711 by 35
(ii) 35 by 25
(iii) 3415 by 24
(iv) 318 by × 41014

Answer:

(i)
711×35=7×311×52155

(ii)
35×25=3×255×175515

(iii)
3415×24(3×15)+415×24=49×2415×13925= 7825

(iv)
318×41011(3×8)+18×(4×11)+1011=25×548×1113508867544 = 151544

Page No 2.19:

Question 2:

Find the product:
(i) 47×1425
(ii) 712×2415
(iii) 367×423
(iv) 61114×312

Answer:

(i)
47×1425=4×147×25825

(ii)
712×2415(7×2)+12×(2×15)+415152×3415171
(iii)
367×423(3×7)+67×(4×3)+232797×1423181
(iv)
61114×312(14×6)+1114×(3×2)+1295142×72954



Page No 2.20:

Question 3:

Simplify:
(i) 1225×1528× 3536
(ii) 1027× 3956× 2865
(iii) 2217× 729× 13352

Answer:

(i)
1225×1528×353612×153×355255×284×3633×55×4×314

(ii)
1027×3956×286510×3931×28279×562×65519
(iii)
2217×729×13352(2×17)+217×(7×9)+29×(52×1)+3352364×655×85517×9×524251

Page No 2.20:

Question 4:

Find:
(i) 12 of 429
(ii) 58 of 923
(iii) 23 of 916

Answer:

(i) 12 of 429=12×(429)=12×(4×9)+29=1×382×9=199=219(ii) 58 of 923=58×(923)=58×(9×3)+23=5×298×3=14524=6124(iii) 23 of 916=2×93×16=38

Page No 2.20:

Question 5:

Which is greater? 12 of 67 or 23 of 37

Answer:

12of 67=12×67=61423of 37=23×37=27
convert 27to its equivalent fraction with denominator as 14
27=27×22=414
we know6>4
614>27
12of 67>23of 37

Page No 2.20:

Question 6:

Find:
(i) 711 of Rs 330
(ii) 59 of 108 metres
(iii) 37 of 42 litres
(iv) 112 of an hour
(v) 56 of an year
(vi) 320 of a kg
(vii) 720 of a litre
(viii) 56 of a day
(ix) 27 of a week

Answer:

(i) 711of Rs 330=7×33011Rs 210(ii) 59of 108 metres=59×10860 metres(iii) 37of 42 litres=37×4218 litres(iv) 112of 1 hour=112×1112hour1 hour =60 minutes112hour=112×60=5 minutes(v) 56of 1 year=56×156year1 year =12 months56year=56×12=10 months(vi) 320of 1 kg=320×1320kg1 kg =1000 g320kg=320×1000=150 g(vii) 720of 1 litre=720×1720litre1 l =1000 ml720litre=720×1000=350 ml(viii) 56of 1 day=56×156day1 day =24 hours56day=56×24=20 hours(ix) 27of 1 week=27×127week1 week =7 days27week=27×7=2 days

Page No 2.20:

Question 7:

Shikha plants 4 saplings in a row in her garden. The distance between two adjacent saplings is 34m. Find the distance between the first and the last sapling.

Answer:

Distance between the first and second saplings = 34m
Distance between the first and third saplings = 2×34m=32m
Distance between the first and fourth saplings 3×34m=94m

Page No 2.20:

Question 8:

Ravish reads 13 part of a book in 1 hour. How much part of the book will he read in 215 houurs?

Answer:

215hours=(2×5)+15=115hoursIn 1 hour Ravish reads 13 of the bookPart of book Ravish will read in 115hours = Part read in 1 hour ×115Part of book Ravish will read in 115hours =13×115=1115

Page No 2.20:

Question 9:

Lipika reads a book for 134 hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Answer:

In one day, Lipika reads for 134hours.
134hours=(1×4)+34=74hours
Total hours required = 74×6=424hours = 212hours = 1012hours

Page No 2.20:

Question 10:

Find the area of a rectangular park which is 4123 m long and 1835 m broad.

Answer:

Length of rectangular park = 4123m =(41×3)+23mLength of rectangular park = 1253mWidth of rectangular park = 1835m =(18×5)+35mWidth of rectangular park = 935mArea of rectangular park = Length of rectangular park × Width of rectangular parkArea of rectangular park = 1253×935=775m2

Page No 2.20:

Question 11:

If milk is available at Rs 1734 per litre, find the cost of 725 litres of milk.

Answer:

Cost of 1 litre milk = Rs. 1734
1734=(17×4)+34=714
Cost of 1 litre milk =Rs. 714
725litres=(7×5)+25=375litresCost of 375litres milk = Cost of 1 litre milk × 375Cost of 375litres milk=714× 375=Rs. 262720= Rs. 131720

Page No 2.20:

Question 12:

Sharda can walk 813 km in one hour. How much distance will she cover in 225 hours?

Answer:

813km=(8×3)+13813km=253km

Distance covered by ​Sharda in 1 hour =  253km
225hours =(2×5)+25=125hours
Distance covered by Sharda in 125hours = Distance covered in 1 hour ×125
Distance covered by Sharda in 125hours = 253×125=20 km

Page No 2.20:

Question 13:

A sugar bag contains 30 kg of sugar. After consuming 23 of it, how much sugar is left in the bag?

Answer:

Total amount of sugar in the bag = 30 kg23 of 30 kg =23×30 = 20 kgAmount of sugar consumed = 20 kgSugar left in the bag = Total amount of sugar in the bag - Amount of sugar consumed Sugar left in the bag = 30-20 =10 kg

Page No 2.20:

Question 14:

Each side of a square is 623 m long. Find its area.

Answer:

623m = (6×3)+23m623m = 203mArea of square=Side × SideArea of square=203 ×203 Area of square=4009m2 =4449m2

Page No 2.20:

Question 15:

There are 45 students in a class and 35 of them are boys. How many girls are there in the class?

Answer:

35 of 45 =35×45 =27Therefore, 27 students are boys. Number of girls=Total number of students in the class - Number of boysNumber of girls=45-27=18Therefore, 18 students are girls.



Page No 2.24:

Question 1:

Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers:
(i) 37
(ii) 58
(iii) 97
(iv) 65
(v) 127
(vi) 18

Answer:

Reciprocal of a non-zero fraction ab is ba
(i)
Reciprocal of 37 is 73It's improper fraction because numerator is greater than denominator

(ii)
Reciprocal of 58 is 85It's improper fraction because numerator is greater than denominator

(iii)
Reciprocal of 97 is 79It's proper fraction because numerator is less than denominator

(iv)
Reciprocal of 65 is 56It's proper fraction because numerator is less than denominator

(v)
Reciprocal of 127 is 712It's proper fraction because numerator is less than denominator

(vi)
Reciprocal of 18 is 81=8It is a whole number



Page No 2.25:

Question 2:

Divide:
(i) 38 by 59
(ii) 314 by 23
(iii) 78 by 412
(iv) 614 by 235

Answer:

(i)
38÷59=38×953×98×52740

(ii)
314÷23=(3×4)+14×32134×32398=478

(iii)
78÷412=78÷(4×2)+12784×2974×19736
(iv)
614÷235=(6×4)+14÷(2×5)+35254÷135=254×51312552=22152

Page No 2.25:

Question 3:

Divide:
(i) 38 by 4
(ii) 916 by 6
(iii) 9 by 316
(iv) 10 by 1003

Answer:

(i)
38÷4=38×14=332

(ii)
916÷6=916×16=9316×62=332

(iii)
9÷316=91×16393×163=48

(iv)
10÷1003=101×310010×310010=310

Page No 2.25:

Question 4:

Simplify:
(i) 310÷103
(ii) 435÷45
(iii) 547÷1310
(iv) 4÷225

Answer:

(i)
310÷103=310×3103×310×109100
(ii)
435÷45=(4×5)+35÷45435÷45=235÷45235×54234=534

(iii)
547÷1310=(5×7)+47÷(1×10)+310547÷1310=397÷13103937×1013307=427

(iv)
4÷225=41÷(2×5)+2541×512353=123

Page No 2.25:

Question 5:

A wire of length 1212m is cut into 10 pieces of equal length. Find the length of each piece.

Answer:

1212m=(12×2)+12m=252m

Length of one piece = Length of wire10=252×110=2520
Length of one piece = 255204=54m

Page No 2.25:

Question 6:

The length of a rectangular plot of area 6513m2 is 1214m. What is the width of the plot?

Answer:

Area of rectangle = Length of rectangle × Width of rectangle
6513=1214×Width of the rectangle(65×3)+13=(12×4)+14×Width of the rectangle1963=494×Width of the rectangleWidth of the rectangle=19643×449=163m

Page No 2.25:

Question 7:

By what number should 629 be multiplied to get 449?

Answer:

Let the required number be x.

According to the question:

629×x=449(6×9)+29×x=(4×9)+49569×x=409x=409×956x=4056=57 

Page No 2.25:

Question 8:

The product of two numbers is 2556. If one of the numbers is 623, find the other.

Answer:

Let the required number be x.

According to the question:

623×x=2556(6×3)+23×x=(25×6)+56203×x=1556x=320×1556=3×15531204×62=318=378

Page No 2.25:

Question 9:

The cost of 614 kg of apples is Rs 400. At what rate per kg are the apples being sold?

Answer:

614kg=(6×4)+14kg254kg
Cost of  254kg of apples = Rs. 400
Cost of 1 kg of apples = 400÷254=400×254 = Rs. 64

Page No 2.25:

Question 10:

By selling oranges at the rate of Rs 514 per orange, a fruit-seller gets Rs 630. How many dozens of oranges does he sell?

Answer:

Cost of 1 orange = Rs 514=(5×4)+14 =Rs214
Number of oranges sold = 630÷214
63030×421=120 
 12 oranges = 1 dozen

 120 oranges = 12012=10 dozen

Page No 2.25:

Question 11:

In mid-day meal scheme 310 litre of milk is given to each student of a primary school. If 30 litres of milk is distributed every day in the school, how many students are there in the school?

Answer:

Number of students in the school = Total amount of milk distributed per dayAmount of milk given to one student
=30÷310=3010×103=100 

Page No 2.25:

Question 12:

In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs 5034, how many tikets were sold?

Answer:

Number of tickets sold = Total amount of money collectedPrice of one ticket
Price of one ticket:

5034(50×4)+34Rs 2034

Number of tickets sold = 6496÷2034
649632×4203128 



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