Rs Aggarwal 2017 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 Rs Aggarwal 2017 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 Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 21:

Answer:

We have the following:

(i) 58 and 712
  
By cross multiplication, we get:
5 × 12 = 60 and 7 × 8 = 56
However, 60 > 56
∴  58>712

(ii) 59and1115
By cross multiplication, we get:
 5 × 15 = 75 and 9 × 11 = 99
However, 75 < 99
∴  59<1115

(iii) 1112and1516
By cross multiplication, we get:
11 × 16 = 176 and 12 × 15 = 180
However, 176 < 180
∴  1112<1516

Page No 21:

Answer:

(i) The given fractions are 34,56,79and1112.
    
LCM of 4, 6, 9 and 12 = 36
    
Now, let us change each of the given fractions into an equivalent fraction with 72 as its denominator.
34=3×94×9=2736 

56=5×66×6=3036

79=7×49×4=2836

1112=11×312×3=3336

Clearly, 2736<2836<3036<3336

Hence,34<79<56<1112

∴ The given fractions in ascending order are 34, 79, 56 and 1112.

(ii) The given fractions are: 45, 710, 1115 and 1720.

LCM of 5, 10, 15 and 20 = 60

Now, let us change each of the given fractions into an equivalent fraction with 60 as its denominator.

45=4×125×12=4860

710=7×610×6=4260

1115=11×415×4=4460

1720=17×320×3=5160

Clearly, 4260<4460<4860<5160

Hence,710<1115<45<1720

∴ The given fractions in ascending order are 710, 1115, 45 and 1720.



Page No 22:

Answer:

We have the following:
(i) The given fractions are 34, 78, 712 and 1724.

LCM of 4,8,12 and 24 = 24

Now, let us change each of the given fractions into an equivalent fraction with 24 as its denominator.
34=3×64×6=1824

78=7×38×3=2124

712=7×212×2=1424

1724=17×124×1=1724

Clearly, 2124>1824>1724>1424

Hence, 78>34>1724>712

∴ The given fractions in descending order are 78, 34, 1724 and 712.

(ii) The given fractions are 23, 35, 710 and 815.
LCM of 3,5,10 and 15 = 30
Now, let us change each of the given fractions into an equivalent fraction with 30 as its denominator.
23=2×103×10=2030

35=3×65×6=1830

710=7×310×3=2130

815=8×215×2=1630

Clearly, 2130>2030>1830>1630

Hence, 710>23>35>815

∴ The given fractions in descending order are 710, 23, 35 and 815.

Page No 22:

Answer:

We will compare the given fractions 27and45 in order to know who got the larger part of the apple.
We have,
By cross multiplication, we get:
2 × 5 = 10  and 4 × 7 = 28

However, 10 < 28
27<45
Thus, Sonal got the larger part of the apple.

Now, 45-27=28-1035=1835

∴ Sonal got 1835 part of the apple more than Reenu.

Page No 22:

Answer:

(i) 59+39=89

(ii) 89+712
    
     = 3236 + 2136            [∵ LCM of 9 and 12 = 36]

      = 32+2136          
         
      = 5336=11736

(iii) 56+78

      = 2024 + 2124                   [∵ LCM of 6 and 8 = 24]

       = 20+2124

       =4124=11724

(iv) 712+1116+924

     2848 + 3348 + 1848             [∵ LCM of 12, 16 and 24 = 48]

      = 28+33+1848

      = 7948=13148

(v) 345+2310+1115

     = 195+2310+1615

   = 11430 + 6930 + 3230            [∵ LCM of 5, 10 and 15 = 30]

     = 114+69+3230

     = 21530=7530 = 716

(vi) 834+1025

      = 354+525

     = 17520 + 20820                  [∵ LCM of 4 and 5 = 20]

      = 175+20820

      = 38320=19320

Page No 22:

Answer:

(i) 57-27=5-27=37

(ii) 56-34

     = 1012 - 912                 [∵ LCM of 6 and 4 = 12]

     = 10-912             
    
      = 112

(iii) 315-710

      = 165-710

     =  3210 - 710         [∵ LCM of 5 and 10 = 10]

      =  32-710
          
      = 2510=52=212

(iv) 7-423

       =  71-143

       = 21-143        [∵ LCM of 1 and 3 = 3]
          
       = 73=213

(v) 3310-1715

      = 3310-2215

      = 99-4430              [∵ LCM of 10 and 15 = 30]
      
      = 5530=116=156

(vi) 259-1715

       = 239-2215

       = 115-6645                [∵ LCM of 9 and 15 = 45]
 
       =4945=1445

Page No 22:

Answer:

(i) 23+56-19

   = 12+15-218    [∵ LCM of 3, 6 and 9 = 18]

  = 27-218=2518=1718

(ii) 8-412-214

 = 81-92-94

 = 32-18-94     [∵ LCM of 1, 2 and 4 = 4]

 = 32-274=54=114

(iii) 856-338+1712

    = 536-278+1912

    =212-81+3824     [∵ LCM of 6, 8 and 12 = 24]

    = 250-8124=16924=7124

Page No 22:

Answer:

Total weight of fruits bought by Aneeta = 334 + 412 kg
Now, we have:

334 + 412 = 154 + 92

                =15 + 184     [∵ LCM of 2 and 4 = 4]

                =15 + 184=334=814

Hence, the total weight of the fruits purchased by Aneeta is 814 kg.

Page No 22:

Answer:

We have:

Perimeter of the rectangle ABCD = AB + BC + CD +DA
= 1534 + 1212 + 1534 + 1212 cm
= 634 + 252 + 634 + 252 cm
= 63 + 50 + 63 + 504 cm          [∵ LCM of 2 and 4 = 4]
= 2264 cm=1132 cm=5612 cm

Hence, the perimeter of ABCD is 5612 cm.

Page No 22:

Answer:

Actual width of the picture = 735cm=385cm
Required width of the picture = 7310cm=7310cm
∴ Extra width = 385-7310cm
                       = 76-7310cm       [∵ LCM of 5 and 10 is 10]
                       = 310cm
Hence, the width of the picture should be trimmed by 310 cm.

Page No 22:

Answer:

Required number to be added = 18-735

                                                =181-385

                                                = 90-385             [∵ LCM of 1 and 5 = 5]
                                                = 525=1025

Hence, the required number is 1025.

Page No 22:

Answer:

Required number to be added = 825-7415

                                                = 425-10915

                                                = 126-10915    [∵ LCM of 5 and 15 = 15]
                                               
                                               =1715=1215

Hence, the required number should be 1215.

Page No 22:

Answer:

Required length of other piece of wire = 334-112m

                                                  =154-32m

                                                  =15-64m    [∵ LCM of 4 and 2 = 4]

                                                  = 94m=214m

Hence, the length of the other piece of wire is 214m.

Page No 22:

Answer:

Actual duration of the film = 323-112hours

                                           = 113-32hours

                                           = 22-96hours   [∵ LCM of 3 and 2 = 6]

                                           = 136hours=216hours

Hence, the actual duration of the film was 216hours.

Page No 22:

Answer:

First we have to compare the fractions: 23 and 59.
By cross multiplication, we have:
2 × 9 = 18 and 5 × 3 = 15

However, 18 > 15
23>59

So, 23 is larger than 59.
Now, 23-59

      = 6-59    [∵ LCM of 3 and 9 = 9]
      =19
Hence, 23 is 19 part more than 59.

Page No 22:

Answer:

First, we have to compare the cost of the pen and the pencil.
Cost of the pen = Rs 1635 = Rs 835

Cost of the pencil = Rs 434 = Rs 194
Now, we have to compare fractions 835 and 194.
By cross multiplication, we get:

83 × 4 = 332 and 19 × 5 = 95

However, 332 > 95

835>194

So, the cost of pen is more than that of the pencil.
Now, Rs 835 - 194

      = Rs 332 - 9520     [∵ LCM of 4 and 5 = 20]

      = Rs 23720 = Rs 111720

∴ The pen costs Rs 111720 more than the pencil.



Page No 26:

Answer:

(i) 35×711=3×75×11=2155

(ii) 58×47=5×48×7=5×12×7=514

(iii) 49×1516=4×159×16=1×53×4=512

(iv) 25×15=25×151=2×155×1=2×31×1=6

(v) 815×20=815×201=8×2015×1=8×43×1=323=1023

(vi) 58×1000=58×10001=5×10008×1=5×1251×1=625

(vii) 318×16=258×161=25×168×1=25×21×1=50

(viii) 2415×12=3415×121=34×1215×1=34×45×1=1365=2715

(ix) 367×423=277×143=27×147×3=9×21×1=18

(x) 912×1919=192×2819=19×282×19=1×141×1=14

(xi) 418×21011=338×3211=33×328×11=3×41×1=12

(xii) 556×157=356×127=35×126×7=5×21×1=10

Page No 26:

Answer:

We have the following:

(i) 23×544×3335=2×5×333×44×35=1×1×111×22×7=1×1×11×2×7=114

(ii)1225×1528×3536=1×3×55×4×3=1×1×11×4×1=14

(iii) 1027×2865×3956=10×1×327×5×2=1×1×327×1×1=327=19

(iv) 147×11322×1115

      =117×3522×1615=11×35×167×22×15=1×5×161×2×15=1×1×81×1×3=83=223

(v) 2217×729×13352

     =3617×659×8552=36×65×8517×9×52=4×5×51×1×4=1×5×51×1×1=25

(vi) 3116×737×12539

      =4916×527×6439=7×4×41×1×3=1123=3713

Page No 26:

Answer:

We have the following:

(i) 13 of 24 = 24×13=241×13=24×11×3=8

(ii) 34 of 32 = 32×34=321×34=32×31×4=8×31×1=24

(iii) 59 of 45 = 45×59=451×59=45×51×9=5×51×1=25

(iv) 750 of 1000 = 1000×750=10001×750=20×71×1=140

(v) 320 of 1020 = 1020×320=10201×320=51×31×1=153

(vi) 511 of Rs 220 = Rs 220×511 = Rs (20 × 5 ) = Rs 100

(vii) 49of 54 m = 49×54m = (4 × 6) m = 24 m

(viii) 67 of 35 L = 67×35L = (6 × 5) L = 30 L

(ix) 16 of 1 h = 16 of 60 min = 60×16 min = 10 min

(x) 56 of an year = 56 of 12 months = 12×56 months = (2 × 5) months = 10 months

(xi) 720 of a kg = 720 of 1000 g = 1000×720 g = (50 × 7) gm = 350 g

(xii) 920 of 1 m = 920 of 100 cm = 100×920 cm = (5 × 9) cm = 45 cm

(xiii) 78 of a day = 78 of 24 h = 24×78 h = (3 × 7) = 21 h

(xiv) 37 of a week = 37 of 7 days = 7×37 days = 3 days

(xv) 750 of 1 L = 750 of 1000 ml = 1000×750 ml = (20 × 7) ml = 140 ml

Page No 26:

Answer:

Cost of 1kg of apples = Rs 1825= Rs 925
∴ Cost of 334 kg of apples = Rs 925 × 334
                                           = Rs 925 × 154 = Rs 23 × 31 × 1 = Rs 69

Hence, the cost of 334 kg of apples is Rs 69.

Page No 26:

Answer:

Cost of 1 m of cloth = Rs 4212 = Rs 852
∴ Cost of 535 m of cloth = Rs 852×535
                                        = Rs 852×285=Rs 85×282×5=Rs 17×14=Rs 238
Hence, the cost of 535m of cloth is Rs 238.

Page No 26:

Answer:

Distance covered by the car in 1 h = 6623 km
Distance covered by the car in 9 h  =6623×9 km
                                                            =2003×9 km=200×93×1 km=200×3 km=600 km

Hence, the distance covered by the car in 9 h will be 600 km.

Page No 26:

Answer:

Capacity of 1 tin = 1234 L=514 L
∴ Capacity of 26 such tins =26×514 L
                                           =261×514 L=26×511×4 L=13×511×2 L=6632 L=33112 L

Hence, 26 such tins can hold 33112 L of oil.

Page No 26:

Answer:

Cost of 1 ticket = Rs 3512= Rs 712
∴ Cost of 308 tickets = Rs 712×308=Rs 712×3081=Rs (71×154)=Rs 10934

Hence, 308 tickets were sold for Rs 10,934.

Page No 26:

Answer:

Thickness of 1 board = 323 cm
∴ Thickness of 9 boards = 9×323 cm
                                        =91×113 cm = (3 × 11) cm = 33 cm

Hence, the height of the stack is 33 cm.



Page No 27:

Answer:

Time taken by Rohit to complete one round of the circular park = 445 min = 245min

∴ Time taken to complete 15 rounds =15×245 min
                                                                      = (3 × 24) min
                                                                      = 72 min
                                                                      = 1 h 12 min   [∵ 1 hr = 60 min]

Hence, Rohit will take 1 h 12 min to make 15 complete rounds of the circular park.

Page No 27:

Answer:

Weight of Amit = 35 kg
Weight of Kavita = 35 of Amit's weight
                             = 35 kg x 35 = 35×35kg=(7×3) kg=21 kg
Hence, Kavita's weight is 21 kg.

Page No 27:

Answer:

Number of boys in the class = 57 of the total no. of students
                                           =57 × 42 = 5×427=5×6=30

∴ Number of girls in the class = 42 − 30 = 12

Hence, there are 12 girls in the class.

Page No 27:

Answer:

Sapna's total monthly income = Rs 12000
Monthly expenditure = 78 of Rs 12000
                                   = Rs 78×12000=Rs (7 × 1500) = Rs 10500
∴ Monthly savings = Rs 12000 − Rs 10500
                                = Rs 1500

Hence, Sapna deposits Rs 1500 in the bank every month.

Page No 27:

Answer:

Side of the square field = 423 m
∴ Area of the square = (side)2
                                  = 423 m2
                                  = 143 m2=143 m ×143 m=14×143×3 m2=1969 m2=2179 m2

Hence, the area of the square field is 2179 m2.

Page No 27:

Answer:

Length of the rectangular park = 4123 m=1253 m

Its breadth = 1835 m=935 m

∴ Its area = length × breadth

               = 1253×935 m2
                = (25 × 31) m = 775 m2

Hence, the area of the rectangular park is 775 m2.



Page No 30:

Answer:

(i) Reciprocal of 58 = 85            [ ∵ 58×85=1]

(ii) Reciprocal of  7 =17             [ ∵ 7×17=1]

(iii) Reciprocal of  112 = 12       [ ∵ 112×12=1]

(iv) Reciprocal of 1235 = Reciprocal of 635 =563            [∵ 635×563=1]

Page No 30:

Answer:

(i) 47÷914=47×149              [∵ Reciprocal of 914 = 149]

    = 89

(ii) 710÷35=710×53            [∵ Reciprocal of 35 = 53]

     = 76=116

(iii) 89÷16=89×116              [∵ Reciprocal of 16 = 116]

       = 118

(iv) 9÷13=9×3                      [∵ Reciprocal of 13 = 3]

     = 27


(v) 24÷67=24×76                [∵ Reciprocal of 67 = 76]

    = 4 × 7 = 28


(vi) 335÷45=185÷45

     = 185×54            [∵ Reciprocal of 45 = 54]

      = 184=92=412

(vii) 337÷821=247÷821

       = 247×218          [∵ Reciprocal of 821 = 218]

       = 3  3 = 9


(viii) 547÷1310 =397÷1310

       = 397×1013             [∵ Reciprocal of 1310 = 1013]

       = 307=427


(ix) 1537÷12349 = 1087÷7249

      = 1087×4972          [∵ Reciprocal of 7249 = 4972]

      = 9×71×6=3×71×2=212=1012

Page No 30:

Answer:

(i)  1124÷78

      = 1124×87                           [∵ Reciprocal of 78 = 87]

      = 1121

(ii) 678÷1116 = 558÷1116

       =558×1611         [∵ Reciprocal of 1116 = 1611]

       = 5 × 2 = 10

(iii) 559÷313 = 509÷103

        = 509×310          [∵ Reciprocal of 103 = 310]

        =  53 = 123

(iv) 32÷135 = 32÷85

      = 32×58                [∵ Reciprocal of 85 = 58]

      = 4 × 5 = 20

(v) 45÷145 = 45÷95

      = 45×59               [∵ Reciprocal of 95 = 59]

      = 5 × 5 = 25

(vi) 63÷214 = 63÷94

      = 63×49            [∵ Reciprocal of 94 = 49]

      = 7 × 4 = 28

Page No 30:

Answer:

Length of the rope = 1312 m =272 m
Number of equal pieces = 9

∴ Length of each piece = 272÷9 m
                                      = 272×19 m      [∵ Reciprocal of 9 = 19]
                                      = 32 m =112 m
Hence, the length of each piece of rope is 112 m.

Page No 30:

Answer:

Weight of 18 boxes of nails = 4912 kg = 992 kg
∴ Weight of 1 box = 992÷18 kg
                            = 992×118 kg         [∵ Reciprocal of 18 = 118]
                            = 99×12×18 kg = 11×12×2 kg =114 kg = 234 kg

Hence, the weight of each box is 234 kg.



Page No 31:

Answer:

Cost of 1 orange = Rs 334 = Rs 154
Total cost of the oranges sold by the man = Rs 210

∴ Required number of oranges = 210÷154
                                                = 210×415      [∵ Reciprocal of 154 = 415]
                                                = (14 × 4) = 56

Hence, the man sold 56 oranges.

Page No 31:

Answer:

Cost of 1 kg of mangoes = Rs 1812= Rs372
Total cost of the required mangoes = Rs 15714 = Rs 6294
∴ Weight of the required mangoes = 6294÷372 kg
                                                   =6294×237 kg      [∵ Reciprocal of 372 = 237]
                                                   = 172 kg = 812 kg
Hence, the weight of the mangoes available for Rs 15714 is 812 kg.  

Page No 31:

Answer:

Distance covered by Vikas in 734 h = 2023 km
∴ Distance covered by him in 1 h = 2023÷734 km
                                                   = 623÷314 km
                                                   = 623×431 km
                                                    = 2×43 km =83 km = 223 km

Hence, the distance covered by Vikas in 1 h is 223 km.

Page No 31:

Answer:

Cost of 812 kg of sugar = Rs 14834
∴ Cost of 1 kg of sugar = Rs 14834÷812
                                    = Rs 5954÷172
                                    = Rs 5954×217 = Rs 352 = Rs 1712

Hence, the cost of 1 kg of sugar is Rs 1712.

Page No 31:

Answer:

Cost of 1 notebook = Rs 734  = Rs 314
∴ Number of notebooks purchased for Rs 6934 = 6934÷314
                                                                        =2794÷314
                                                                        =2794×431    [∵ Reciprocal of 314 =413]
                                                                        =27931 = 9
Hence, 9 notebooks can be purchased for Rs 6934.

Page No 31:

Answer:

Cost of 1 ticket = Rs 1012 = Rs 212
Total amount collected by the boy = Rs 28312 = Rs 5672
∴ Number of tickets sold = 5672÷212

                                         = 5672×221     [∵ Reciprocal of 212 = 221]

                                         = 56721=27

Hence, the boy sold 27 tickets of the charity show.

Page No 31:

Answer:

Amount contributed by 1 student = Rs 6112 = Rs 1232
Total amount collected = Rs 67612 = Rs 13532
∴ Number of students in the group = 13532÷1232

                                                      = 13532×2123        [∵ Reciprocal of 1232 = 2123]

                                                      =1353123=11

Hence, there are 11 students in the group.

Page No 31:

Answer:

Quantity of milk given to each student  = 25 L
Total quantity of milk distributed among all the students = 24 L

∴ Number of students = 24÷25

                                    = 24×52       [∵ Reciprocal of 25 = 52]

                                    = (12 × 5) = 60

Hence, there are 60 students in the hostel.

Page No 31:

Answer:

Capacity of the small jug = 34 L
Capacity of the bucket = 2014 L = 814 L
∴ Required number of small jugs = 814 ÷ 34
                                                   = 814×43      [∵ Reciprocal of 34 = 43]
                                                   = 813 = 27

Hence, the small jug has to be filled 27 times to empty the water from the bucket.

Page No 31:

Answer:

Product of the two numbers = 1556 =956

One of the numbers = 613 =193

∴ The other number = 956 ÷ 193

                                 = 956 × 319     [∵ Reciprocal of 193 = 319]

                               = 52 = 212

Hence, the other number is 212.

Page No 31:

Answer:

Product of the two numbers = 42
One of the numbers = 945 = 495
∴ The other number = 42 ÷ 495
                                 =42 × 549           [∵ Reciprocal of 495 = 549]
                                 =6 × 57 = 307 = 427

Hence, the required number is 427.

Page No 31:

Answer:

Required number = 629 ÷ 423
                            = 569 ÷ 143
                            = 569  × 314     [ ∵ Reciprocal of 143 = 314]
                            = 43=113

Hence, we have to divide 629 by 113 to get 423.

Page No 31:

Answer:

(c) 103

103 is a vulgar fraction, because its denominator is other than 10, 100, 1000, etc.

Page No 31:

Answer:

(c) 97
97 is an improper fraction, because its numerator is greater than its denominator.

Page No 31:

Answer:

(a) 105112

A fraction that is reducible can be reduced by dividing both the numerator and denominator by a common factor.

105÷7112÷7=1516

Thus, 105112 is a reducible fraction.



Page No 32:

Answer:

(c) equivalent fractions

Equivalent fractions are those which are the same but look different.

Thus, 23, 46=23, 69=23, 812=23 are equivalent fractions.

Page No 32:

Answer:

(c) 916 > 1324
The two fraction are 916 and 1324.
By cross multiplication, we have:
9 × 24 = 216 and 13 × 16 = 208
However, 216 > 208
916 > 1324

Page No 32:

Answer:

(d) none of these
Reciprocal of 134 = Reciprocal of 74 = 47

Page No 32:

Answer:

(c) 56

310+815=9+1630        [∵ LCM of 10 and 15 = 30]

               = 2530=56

Page No 32:

Answer:

(d) 1112

314-213 = 134-73
                = 39-2812            [∵ LCM of 4 and 3 = 12]
                = 1112

Page No 32:

Answer:

(d) 144

36÷14=36×4   [∵ Reciprocal of 14= 4]
          = 144

Page No 32:

Answer:

(b) 57

Required number = 167÷235

                             = 137÷135

                             = 137×513    [∵ Reciprocal of 135 = 513]

                             = 57

Page No 32:

Answer:

(d) 214

Required number = 112÷23

                             = 32÷23

                             = 32×32      [∵ Reciprocal of 23 = 32]

                             = 94=214

Page No 32:

Answer:

(c) 225

135÷23=85÷23

           = 85×32        [∵ Reciprocal of 23 = 32]

           = 4×35=125=225

Page No 32:

Answer:

(d) 156

215÷115=115÷65

             = 115×56         [∵ Reciprocal of 65 = 56]

              = 116=156

Page No 32:

Answer:

(d) 35

Reciprocal of 123 = Reciprocal of 53 = 35

Page No 32:

Answer:

(b) 35<23<1415

The given fractions are 35, 23 and1415.

LCM of 5, 3 and 15 = 15

Now, we have:

23×55=1015, 35×33=915 and 1415×11=1415

Clearly, 915 <1015<1415

35<23<1415



Page No 33:

Answer:

(c) 44 km
Distance covered by the car on 234 L of petrol =16×234 km

                                                                   = 16×114 km

                                                                   = (4 × 11) km = 44 km

Page No 33:

Answer:

(a) 1012 hours
Time taken by Lalit to read the entire book = 6×134 h

                                                                   = 6×74 h

                                                                     = 212 h = 1012 h



Page No 34:

Answer:

(i) A number of the form ab, where a and b are natural numbers, is called a natural number.
Here, a is the numerator and b is the denominator.

23 is a fraction with 2 as the numerator and 3 as the denominator.

125 is a fraction with 12 as the numerator and 5 as the denominator.

(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples: 25 and 415

(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Examples: 113 and 4135

Page No 34:

Answer:

Required number to be added = 15-635

                                             = 151-335

                                              = 75-335  [∵ LCM of 1 and 5 = 5]

                                              = 425=825

Hence, the required number is 825.

Page No 34:

Answer:

We have,

956-438+2712

= 596-358+3112

= 236-105+6224   [∵ LCM of 6, 8 and 12 = 24]

= 298-10524= 19324=8124

Page No 34:

Answer:

We have:

(i) 1225 of 1 L = 1225 of 1000 ml = 1000×1225 ml = (40 × 12) ml = 480 ml

(ii)58 of 1 kg = 58 of 1000 g = 1000×58 g = (125 ×5) g = 625 g

(iii) 35 of 1 h = 35 of 60 min = 60×35 min = (12 × 3) min = 36 min

Page No 34:

Answer:

Cost of 1 L of milk = Rs 3734 =  Rs 1514
Cost of 625 L of milk = Rs 1514×625
                                 = Rs 1514×325
                                 = Rs 151×81×5 =  Rs 12085 = Rs 24135
Hence, the cost of 625 L of milk is Rs 24135.

Page No 34:

Answer:

Cost of  514 kg of mangoes = Rs 189
Cost of 1 kg of mango = Rs 189÷514
                                   = Rs 189÷214 
                                   = Rs 189×421      [∵ Reciprocal of 214 = 421]
                                   = Rs (9 × 4) = Rs 36

Hence, the mangoes are being sold at Rs 36 per kg.

Page No 34:

Answer:

We have:

(i)134×225×347

     = 74×125×257

     = 7×12×254×5×7=1×3×51×1×1=15


(ii) 559÷313

     = 509÷103

     = 509×310   [∵ Reciprocal of 103 = 310]

     = 5×13×1=53=123

Page No 34:

Answer:

Required number = 629÷423

                            = 569÷143

                            = 569×314    [∵ Reciprocal of 143 = 314]

                            = 43 = 113

Hence, we have to divide 629 by 113 to obtain 423.

Page No 34:

Answer:

Side of the square = 523 m = 173 m

Its area = (side)2 = 173m2 = 173m×173m=2899m2=3219 m2

Hence, the area of the square is 3219m2.

Page No 34:

Answer:

(d) 58

58 is a vulgar fraction, because its denominator is other than 10, 100, 1000, etc.

Page No 34:

Answer:

(c) 4663

A fraction ab is said to be irreducible or in its lowest terms if the HCF of a and b is 1.
46 = 2 × 23 ×1
63 = 3 × 3× 21 ×1

Clearly, the HCF of 46 and 63 is 1.

Hence, 4663 is an irreducible fraction.

Page No 34:

Answer:

(d) none of these

Reciprocal of 135 = Reciprocal of 85 = 58

Page No 34:

Answer:

(c) 225

135÷23= 85÷23

             = 85×32        [∵ Reciprocal of 23 = 32]

             = 4×35=125=225

Page No 34:

Answer:

(b) 35<23<1115

The given fractions are 23, 35 and 1115.

LCM of 5, 3 and 15 = 15

Now, we have:

23×55=1015, 35×33=915 and 1115×11=1115

Clearly, 915<1015<1115

35<23<1115

Page No 34:

Answer:

(c) 710

Required number = 134÷212

                             = 74÷52

                             = 74×25   [∵ Reciprocal of 52 = 25]

                             = 7×12×5=710



Page No 35:

Answer:

(b) 33 km
Distance covered by the car on 323 L of petrol = 9×323 km
                                                                      = 9×113 km
                                                                       = (3 × 11) km = 33 km

Page No 35:

Answer:

(i) The reciprocal of 825 is 542.

Reciprocal of 825 = Reciprocal of 425 = 542

(ii) 1312÷8=11116

     1312÷8=272×18=2716=11116

(iii) 6934÷734=9

6934÷734=2794÷314

 =2794×431=27931 = 9

(iv) 4123÷1835=775

        4123×1835=1253×935
      
       = 1253×935=25×311×1=775

(v) 8498(irreducible form) = 67
  
The HCF of 84 and 98 is 14.
      
84÷1498÷14=67

Page No 35:

Answer:

(i) F

     By cross multiplication, we have:

     9 × 24 = 216 and 13 × 16 = 208

     However, 216 > 208

     ∴ 916>1324

(ii) F

      The LCM of 5, 35 and 14 is 70.

      Now, 25=25×1414=2870; 1635=1635×22=3270 and 914=914×55=4570

      Clearly, 2870<3270<4570

      ∴ 25<1635<914

(iii) T

       The LCM of 15 and 20 = (5 × 3 × 4) = 60

        ∴ 1115-920=44-2760=1760
(iv) T

       1125 of 1 L = 1125 of 1000 ml = 1000×1125 ml = (40 × 11) ml = 440 ml

(v) F

      1634×625= 674×325=67×324×5=67×85=5365=10715



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