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Page No 21:

Question 1:

Compare the fractions:

(i) 58and712
(ii) 59and1115
(iii) 1112and1516

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:

Question 2:

Arrange the following fractions in ascending order:

(i) 34,56,79,1112
(ii) 45,710,1115,1720

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:

Question 3:

Arrange the following fractions in descending order:

(i) 34,78,712,1724
(ii) 23,35,710,815

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:

Question 4:

Reenu got 27 part of an apple while Sonal got 45 part of it. Who got the larger part and by how much?

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:

Question 5:

Find the sum:

(i) 59+39
(ii) 89+712
(iii) 56+78
(iv) 712+1116+924
(v) 345+2310+1115
(vi) 834+1025

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:

Question 6:

Find the difference:

(i) 57-27
(ii) 56-34
(iii) 315-710
(iv) 7-423
(v) 3310-1715
(vi) 259-1715

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:

Question 7:

Simplify:

(i) 23+56-19
(ii) 8-412-214
(iii) 856-338+1712

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:

Question 8:

Aneeta bought 334 kg apples and 412 kg guava. What is the total weight of fruits purchased by her?

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:

Question 9:

A rectangular sheet of paper is 1534 cm long and 1212 cm wide. Find its perimeter.

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:

Question 10:

A picture is 735 cm wide. How much should it be trimmed to fit in a frame 7310 cm wide?

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:

Question 11:

What should be added to 735 to get 18?

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:

Question 12:

What should be added to 7415 to get 825?

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:

Question 13:

A piece of wire 334 m long broke into two pieces. One piece is 112 m long. How long is the other piece?

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:

Question 14:

A film show lasted of 323 hours. Out of this time 112 hours was spent on advertisements. What was the actual duration of the film?

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:

Question 15:

Of 23 and 59 , which is greater and by how much?

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:

Question 16:

The cost of a pen is Rs 1635 and that of a pencil is Rs 434. Which costs more and by how much?

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:

Question 1:

Find the product:

(i) 35×711
(ii) 58×47
(iii) 49×1516
(iv) 25×15
(v) 815×20
(vi) 58×1000
(vii) 318×16
(viii) 2415×12
(ix) 367×423
(x) 912×1919
(xi) 418×21011
(xii) 556×157

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:

Question 2:

Simplify:

(i) 23×544×3335
(ii) 1225×1528×3536
(iii) 1027×2865×3956
(iv) 147×11322×1115
(v) 2217×729×13352
(vi) 3116×737×12539

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:

Question 3:

Find:

(i) 13 of 24
(ii) 34 of 32
(iii) 59 of 45
(iv) 750 of 1000
(v) 320 of 1020
(vi) 511 of Rs 220
(vii) 49 of 54 metres
(viii) 67 of 35 litres
(ix) 16 of an hour
(x) 56  of an year
(xi) 720  of a kg
(xii) 920 of a metre
(xiii) 78 of a day
(xiv) 37 of a week
(xv) 750 of a litre

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:

Question 4:

Apples are sold at Rs 1825 per kg. What is the cost of 334 kg of apples?

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:

Question 5:

Cloth is being sold at Rs 4212 per metre. What is the cost of 535 metres of this cloth?

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:

Question 6:

A car covers a certain distance at a uniform speed of 6623 km per hour. How much distance will it cover in 9 hours?

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:

Question 7:

One tin holds 1234 litres of oil. How many litres of oil can 26 such tins hold?

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:

Question 8:

For a particular show in a circus, each ticket costs Rs 3512. If 308 tickets are sold for the  show, how much amount has been collected?

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:

Question 9:

Nine boards are stacked on the top of each other. The thickness of each board is 323 cm. How high is the stack?

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:

Question 10:

Rohit takes 445 minutes to make complete round of a circular park. How much time will he take to make 15 rounds?

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:

Question 11:

Amit weighs 35 kg. His sister Kavita's weight is 35 of Amit's weight. How much does Kavita weigh?

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:

Question 12:

There are 42 students in a class and 57 of the students are boys. How many girls are there in the class?

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:

Question 13:

Sapna earns Rs 12000 per month. She spends 78 of her income and deposits rest of the money in a bank. How much money does she deposit in the bank each month?

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:

Question 14:

Each side of a square field is 423 m. Find its area.

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:

Question 15:

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

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:

Question 1:

Write down the reciprocal of:

(i) 58
(ii) 7
(iii) 112
(iv) 1235

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:

Question 2:

Simplify:

(i)  47÷914
(ii) 710÷35
(iii) 89÷16
(iv) 9÷13
(v) 24÷67
(vi) 335÷45
(vii) 337÷821
(viii) 547÷1310
(ix) 1537÷12349

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:

Question 3:

Divide:

(i) 1124 by 78
(ii) 678 by 1116
(iii) 559 by 313
(iv) 32 by 135
(v) 45 by 145
(vi) 63 by 214

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:

Question 4:

A rope of length 1312 m has been divided into 9 pieces of the same length. What is the length of each piece?

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:

Question 5:

18 boxes of nails weigh equally and their total weight is 4912 kg. How much does each box weigh?

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:

Question 6:

By selling oranges at the rate of Rs 334 per orange, a man gets Rs 210. How many oranges does he sell?

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:

Question 7:

Mangoes are sold at Rs 1812 per kg. What is the weight of mangoes available for Rs 15714 ?

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:

Question 8:

Vikas can cover a distance of 2023 km in 734 hours on foot. How many km per hour does he walk?

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:

Question 9:

Preeti bought 812 kg of sugar for Rs 14834. Find the price of sugar per kg.

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:

Question 10:

If the cost of a notebook is Rs 734, how many notebooks can be purchased for Rs 6934 ?

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:

Question 11:

At a charity show the price of each ticket was Rs 1012. The total amount collected by a boy was Rs 28312. How many tickets were sold by him?

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:

Question 12:

A group of students arranged a picnic. Each student contributed Rs 6112. The total contribution was Rs 67612. How many students are there in the group?

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:

Question 13:

24 litres of milk was distributed equally among all the students of a hostel. If each student got 25 litre of milk, how many students are there in the hostel?

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:

Question 14:

A bucket contains 2014 litres of water. A small jug has a capacity of 34 litre. How many times the jug has to be filled with water from the bucket to get it emptied?

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:

Question 15:

The product of two numbers is 1556. If one of the numbers is 613, find the other.

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:

Question 16:

By what number should 945 be multiplied to get 42?

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:

Question 17:

By what number should 629 be divided to obtain 423?

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:

Question 1:

Mark (✓) against the correct answer
Which of the following is a vulgar fraction?

(a) 310
(b) 1310
(c) 103
(d) none of these

Answer:

(c) 103

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

Page No 31:

Question 2:

Mark (✓) against the correct answer
Which of the following is an improper fraction?

(a) 710
(b) 79
(c) 97
(d) none of these

Answer:

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

Page No 31:

Question 3:

Mark (✓) against the correct answer
Which of the following is a reducible fraction?

(a) 105112
(b) 104121
(c) 7772
(d) 4663

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:

Question 4:

Mark (✓) against the correct answer
23,46,69,812 are
(a) like fractions
(b) irreducible fractions
(c) equivalent fractions
(d) none of these

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:

Question 5:

Mark (✓) against the correct answer
Which of the following statements is true?

(a) 916=1324
(b) 916<1324
(c) 916>1324
(d) none of these

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:

Question 6:

Mark (✓) against the correct answer
Reciprocal of 134 is

(a) 143
(b) 413
(c) 314
(d) none of these

Answer:

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

Page No 32:

Question 7:

Mark (✓) against the correct answer
310+815=?

(a) 1110
(b) 1115
(c) 56
(d) none of these

Answer:

(c) 56

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

               = 2530=56

Page No 32:

Question 8:

Mark (✓) against the correct answer
314-213=?

(a) 1112
(b) 112
(c) 1111
(d) 1112

Answer:

(d) 1112

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

Page No 32:

Question 9:

Mark (✓) against the correct answer
36÷14=?

(a) 9
(b) 19
(c) 1144
(d) 144

Answer:

(d) 144

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

Page No 32:

Question 10:

Mark (✓) against the correct answer
By what number should 235 be multiplied to get 167 ?

(a) 157
(b) 57
(c) 117
(d) 17

Answer:

(b) 57

Required number = 167÷235

                             = 137÷135

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

                             = 57

Page No 32:

Question 11:

Mark (✓) against the correct answer
By what number should 112 be divided to get 23 ?

(a) 223
(b) 123
(c) 49
(d) 214

Answer:

(d) 214

Required number = 112÷23

                             = 32÷23

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

                             = 94=214

Page No 32:

Question 12:

Mark (✓) against the correct answer
135÷23=?

(a) 1115
(b) 1910
(c) 225
(d) none of these

Answer:

(c) 225

135÷23=85÷23

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

           = 4×35=125=225

Page No 32:

Question 13:

Mark (✓) against the correct answer
215÷115=?

(a) 1
(b) 2
(c) 115
(d) 156

Answer:

(d) 156

215÷115=115÷65

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

              = 116=156

Page No 32:

Question 14:

Mark (✓) against the correct answer
The reciprocal of 123 is
(a) 312
(b) 213
(c) 113
(d) 35

Answer:

(d) 35

Reciprocal of 123 = Reciprocal of 53 = 35

Page No 32:

Question 15:

Mark (✓) against the correct answer
Which one of the following is the correct statement?

(a) 23<35<1415
(b) 35<23<1415
(c) 1415<35<23
(d) none of these

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:

Question 16:

Mark (✓) against the correct answer
A car runs 16 km using 1 litre of petrol. How much distance will it cover in 234 litres of petrol?
(a) 24 km
(b) 36 km
(c) 44 km
(d) 3234 km

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:

Question 17:

Mark (✓) against the correct answer
Lalit reads a book for 134 hours evrey day and reads the entire book in 6 days. How many hours does he take to read the entire book?

(a) 1012 hours
(b) 912 hours
(c) 712 hours
(d) 1112 hours

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:

Question 1:

Define:

(i) Fractions
(ii) Vulgar fractions
(iii) Improper fractions

Give two examples of each.

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:

Question 2:

What should be added to 635 to get 15?

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:

Question 3:

Simplify: 956-438+2712.

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:

Question 4:

Find:

(i) 1225 of a litre
(ii) 58 of a kilogram
(iii) 35 of an hour

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:

Question 5:

Milk is sold at Rs 3734 per litre. Find the cost of 625 litres milk.

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:

Question 6:

The cost of 514 kg of mangoes is Rs 189. At what rate per kg are the mangoes being sold?

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:

Question 7:

Simplify:

(i) 134×225×347
(ii) 559÷313

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:

Question 8:

By what number should 629 be divided to obtain 423?

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:

Question 9:

Each side of a square is 523 m long. Find its area.

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:

Question 10:

Mark (✓) against the correct answer
Which of the following is a vulgar fraction?

(a) 710
(b) 19100
(c) 3310
(d) 58

Answer:

(d) 58

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

Page No 34:

Question 11:

Mark (✓) against the correct answer
Which of the following is an irreducible fraction?

(a) 105112
(b) 6677
(c) 4663
(d) 5185

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:

Question 12:

Mark (✓) against the correct answer
Reciprocal of 135 is
(a) 153
(b) 513
(c) 315
(d) none of these

Answer:

(d) none of these

Reciprocal of 135 = Reciprocal of 85 = 58

Page No 34:

Question 13:

Mark (✓) against the correct answer
135÷23=?

(a) 1910
(b) 1115
(c) 225
(d) none of these

Answer:

(c) 225

135÷23= 85÷23

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

             = 4×35=125=225

Page No 34:

Question 14:

Mark (✓) against the correct answer
Which of the following is correct?

(a) 23<35<1115
(b) 35<23<1115
(c) 1115<35<23
(d) 35<1115<23

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:

Question 15:

Mark (✓) against the correct answer
By what number should 134 be divided to get 212?
(a) 37
(b) 125
(c) 710
(d) 137

Answer:

(c) 710

Required number = 134÷212

                             = 74÷52

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

                             = 7×12×5=710



Page No 35:

Question 16:

Mark (✓) against the correct answer
A car runs 9 km using 1 litre of petrol. How much distance will it cover in 323 litres o petrol?
(a) 36 km
(b) 33 km
(c) 2511 km
(d) 22 km

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:

Question 17:

Fill in the blanks.

(i) Reciprocal of 825 is ...... .
(ii) 1312÷8=......
(iii) 6934÷734=......
(iv) 4123×1835=......
(v) 8498(in irreducible form)= ......

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:

Question 18:

Write 'T' for true and 'F' for false

(i) 916<1324.
(ii) Among 25,1635 and 914, the largest is 1635.
(iii) 1115-920=1760.
(iv) 1125 of a litre = 440 mL.
(v) 1634×625=107310.

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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