Types of Fractions-Introduction

1. Types of Fractions-Introduction INTRODUCTION 27 Sep 2021 The study of fractions includes proper, improper and mixed fractions as well as their addition and subtraction. We also studied comparison of fractions, equivalent fractions, representation of fractions on the number line and ordering of fractions. Our study of decimals included, their comparison, their representation on the number line and their addition and subtraction. Now we shall next level multiplication and division of fractions as well as of decimals. Numerator and Denominator \frac{2}{6}-\frac{\text { Numerator }}{\text { Denominator }} Numerator: The upper part of the fraction is called Numerator. It tells the number of parts we have. Denominator The lower part of the fraction is called Denominator. It tells the total parts in a whole.

2. Representation of Fractions on Number Line For Example: Mark 12,13,14,16 and 18 on the different lines. Types Of Fractions Proper Fractions: If the numerator is less than the denominator then it is called proper fraction. If we represent a proper fraction on the number line than it will always lie between 0 and 1. Example: 14,47,19 etc. Improper or Mixed Fractions:

3. When the numerator is greater than the denominator then it is called Improper fraction. Example: 54 The above fraction is made by adding one whole part and one-fourth part. 1+14=114=54 The fraction made by the combination of whole and a part is called Mixed fraction. Like fractions: Fractions which have same denominators are known as Like fractions. Example: 513,313,113 etc. Unlike fractions: Fractions which have different denominators are known as unlike fractions. Example: 23,59,1317 etc. Fraction as an operator 'of' 'of' represent multiplication So, 12 of 4=12×4=2 Multiple Of A Fraction By A Whole Number Multiplication of a Fraction by a Whole Number:

4. To multiply a whole number with a proper or an improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator same. 2×43=2×43=83 To multiply a mixed fraction to whole numbers, first convert the mixed fraction to an improper fraction and then multiple. 3×257=3×197=577=817 Multiplication of two fractions = Product of Numerators Production of Denominators Example: =35∗17=3∗15∗7=335 Value of Product of fractions: The value of the product of two proper fractions is smaller than each of the two fractions. The value of the product of two improper fractions is more than each of the two fractions. Reciprocal Of A Number Reciprocal of any number n is written as 1n Reciprocal of a fraction is obtained by interchanging the numerator and denominator. Reciprocal of a fraction is obtained by interchanging the numerator and Reciprocal of 2/5 is 5/2 Example = Reciprocal of 35 is 53 Although zero divided by any number means zero itself, we cannot find reciprocals as a number divided by 0 is undefined. Example = Reciprocal of 08 is 80 (undefined)

5. Division Of A Fraction Division of a whole number by a fraction: To divide a whole number by any fraction, multiply that whole number by the reciprocal of that fraction. 523=5×32=152 Division of a fraction by a whole number: We multiply the fraction with the reciprocal of the whole number. 8114=811×14=211 Division of a fraction by another fraction: To divide two fractions, multiply the first fraction by the reciprocal of the second fraction. 2346=23×64=1212=1 Decimals Whenever we use dots to write singe or many numbers then that dot is the decimal point. Example: 0.5, 1.125, 5.0 How to show Decimals on Number Line? We have to divide the gap of each number into 10 equal parts as the decimal shows the tenth part of the number. Example

6. Q. Show 0.3, 0.5 and 0.8 on the number line. Solution All the three numbers are greater than 0 and less than 1. So we have to make a number line with 0 and 1 and divide the gap into 10 equal parts. Then mark as shown below. Place Value Chart It show the place value of each digit in the decimal number. It makes it easy to write numbers in decimal form. A fraction can be written as a decimal: 53 is a part of 100 , so the fraction will be 53100 In the decimal form we will write it as 0.53 Multiplication And Division Of Decimals

7. Multiplication of decimal numbers with whole numbers: Multiply both the numbers as whole Numbers. Now Count the number of digits to the right of the decimal point in the one decimal. Put the decimal point in the product by counting the digits from its rightmost place. Example = 1.3 × 5 = 6.5 Multiplication of decimals with powers of 10: Example = 53.12 × 10 = 531.20 Multiplication of decimals with decimals: Example: 1.53 × 2.63 = 4.0239 Division of decimal numbers with whole numbers: We first divide them as whole numbers. Then place the decimal point in the quotient as in the decimal number. Example: 8.44=2.1 Division of decimals with powers of 10: The digit in the division remains as same as in the decimal number. Just the decimal point is shifted towards left by the same places as there are zeroes over one. 523.1100=5.231 Division of decimals with decimals:

8. First shift the decimal points to the right by equal number of places in both, to convert the divisor to a natural number and then divide. Example: 1.441.2=14.412=1.2 NCERT SOLUTIONS EXERCISE 2.1 1. Solve: (i) 2−v5 (ii) 4+78 (iii) 35+27 (iv) 911−415 (v) 710+25+32 (vi) 223+312 (vii) 812−358 Solution: (i) 2−35=21−35=2×5−3×11×5 =10−35=75=125 Hence, 2−35=125 (ii) 4+78=41+78=4×8+1×71×8 =32+78=398=478 Hence, 4+78=478 (iii) 35+27=3×7+2×55×7=21+1035=3135

9. Hence, 35+27=3135 (iv) 911−415=9×15−4×1111×15=135−44165=91165 Hence, 911−415=91165 (v) 710+25+32=7+4+1510 [LCM of 10,5 and 2=10 ] =2610=26÷10÷2=135=235 Hence, 710+25+32=235 \[\begin{array}{l} (v i) 2 \frac{2}{3}+3 \frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{8 \times 2+3 \times 7}{3 \times 2} \\ =\frac{16+21}{6}=\frac{37}{6}=6 \frac{1}{6} \end{array}\] Hence, 223+312=616 (vii)812−358=172−298 [LCM of 2 and 8=8] \[\begin{array}{l} =\frac{17 \times 4-29 \times 1}{8} \\ =\frac{68-29}{8}=\frac{39}{8}=4 \frac{7}{8} \end{array}\] Hence, 812−358=478 2. Arrange the following in descending order: (i) 29,23,821 (ii) 15,37,710 Solution:

10. (i) 29,23,821 LCM of 9,3 and 21=3×3×7=63 Making the denominator same, we have 29×77=1463,23×2121=4263 \[\text { and } \frac{8}{21} \times \frac{3}{3}=\frac{24}{63} \mid \\ \text { Since } 42>24>14 \\ \text { Thus } \frac{42}{63}>\frac{24}{63}>\frac{14}{63} \\ \text { Hence, } \frac{2}{3}>\frac{8}{21}>\frac{2}{9}\] (ii) 15,37,710 LCM of 5,7 and 10 \begin{tabular}{c|c} 2 & 5,7,10 \\ \hline 5 & 5,7,5 \\ \hline 7 & 1,7,1 \\ \hline & 1,1,1 \end{tabular} LCM of 5,7 and 10 =2×5×7

11. =70 Making the denominator same, we have \[\frac{1}{5} \times \frac{14}{14}=\frac{14}{70} & {[\because 70 \div 5=14]} \\ \frac{3}{7} \times \frac{10}{10}=\frac{30}{70} & {[\because 70 \div 7=10]} \\ \frac{7}{10} \times \frac{7}{7}=\frac{49}{70} & {[\because 70 \div 10=7]} \\ \text { Since } 49>30>14 & \\ \text { Thus } \frac{49}{70}>\frac{30}{70}>\frac{14}{70} \\ \text { Hence, } \frac{7}{10}>\frac{3}{7}>\frac{1}{5}\] 3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square? \begin{tabular}{|c|c|c|} \hline411 & 911 & 211 \\ \hline311 & 511 & 711 \\ \hline811 & 111 & 611 \\ \hline \end{tabular} Solution: Along first row, 411+911+211=1511 Along second row, 311+511+711=1511 Along third row, 811+111+611=1511 Along first column, 411+311+811=1511 Along second column, 911+511+111=1511 Along third column, 211+711+611=1511

12. Along first diagonal, 411+511+611=1511 Along second diagonal, 211+511+811=1511 Since the sum of all the fraction row wise, column wise and the diagonal 1511 Hence, it is a magic square. 4. A rectangular sheet of paper is 1212 cm long and 1023cm wide. Find its perimeter. Solution: Length of sheet =1212 cm=252 cm Breadth of the sheet =1023=323 cm Perimeter =2× [length + breadth ] =2×[252+323]cm Undefined control sequence \= [LCM of 2 and 3=6] \[= 2 \times \frac{139}{6_{3}}=\frac{139}{3} \\=46 \frac{1}{3} \mathrm{~cm} \\ \text { Hence, the required perimeter }=46 \frac{1}{3} \mathrm{~cm}.\] NOTE:

13. How to divide -> 5. Find the perimeters of (i) Triangle ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater? Solution: (i) Perimeter of Triangle ABE =AB+BE+AE=52 cm+234 cm+335 cm=(52+114+185)cm=(5×10+11×5+18×420)cm [ LCM of 2,4,5=20] Unknown environment 'tabular' =(50+55+7220)cm =17720 cm=81720 cm Hence, the perimeter of △ABE=81720 cm. (ii) Perimeter of rectangle BCDE =2×[ Length + Breadth ] =2×[234+76]cm=2×[114+76]cm=2×[11×3+7×212]cm [LCM of 4 and 6=12] =2×[33+1412]cm=2×4712=476 cm=756 cm Hence, the required perimeter

14. 756 cm Since 81720>756 Thus perimeter of △ABE is greater than the perimeter of the rectangle BCDE. 6. Salil wants to put a picture in a frame. The picture is 755 cm wide. To fit in the frame the picture cannot be more than 7310 cm wide. How much should the picture be trimmed? Solution: The width of the picture =735 cm=385 cm The required width of the frame =7310 cm=7310 cm ∴ The width of the picture to be trimmed of \[=\frac{38}{5} \mathrm{~cm}-\frac{73}{10} \mathrm{~cm}=\left(\frac{38}{5}- \frac{73}{10}\right) \mathrm{cm} \\ =\left(\frac{2 \times 38-73 \times 1}{10}\right) \mathrm{cm}\] [LCM of 5 and 10=10] =(76−7310)cm=310 cm Hence, the required width to be trimmed =310 cm. 7. Ritu ate (3/5) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?

15. Solution: Let the whole part of the apple be 1. Part of the apple eaten by Ritu =35 ∴ Part of the apple eaten by her brother Somu \[=1-\frac{3}{5}=\frac{1}{1}-\frac{3}{5}=\frac{1 \times 5-3 \times 1}{5} \\ =\frac{5-3}{5}=\frac{2}{5} \\ \text { Since } \frac{3}{5}>\frac{2}{5}\] Thus, the share of Ritu was larger. Difference between the two parts =35−25=15 part. 8. Michael finished colouring a picture in (7/12) hour. Vaibhav finished colouring the same picture in (3/4) hour. Who worked longer? By what fraction was it longer? Solution: Time taken by Michael =712 hour Time taken by Vaibhav =34 hour Comparing 712 and 34, we have 7×112 and 3×312⇒712 and 912 [LCM of 12 and 4=12] Since 912>712⇒34 hour >712 hour Hence, time taken by Vaibhav was longer. Difference =34−712=3×34×3−7×112=912−712=212=16 hour longer.

16. Exercise 3 1. Find: (i) 14 of (a) 14 (b) 3/5 (c) 4/3 (ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10 Solution: (i) (a) 14 of 14=14×14=1×14×4=116 (b) 14 of 35=14×35=1×34×5=320 (c) 14 of 43=14×43=13 (ii) (a) 17 of 29=17×29=1×27×9=263 (b) 17 of 65=17×65=1×67×5=635 (c) 17 of 310=17×310=1×37×10=370 2. Multiply and reduce to lowest form (if possible): (i) 23×223 (ii) 27×79 (iii) 38×64 (iv)95×35 (v) 13×158 (vi) 112×310 (vii) 45×127 Solution: (i) 23×223=23×83=2×83×3 =169=179 (ii) 27×79=2×77×9=1463=14÷763÷7=29

17. (iii) 38×64=3×68×4=1832=18÷232÷2=916 (iv) 95×35=9×35×5=2725=1225 (v) 13×158=1×153×8=1524=15÷324÷3=58 (vi) 112×310=11×32×10=3320=11320 (vii) 45×127=4×125×7=4835=11335 3. Multiply the following fractions: (i) 25×514 (ii) 625×79 (iii) 32×513 (iv) 56×237 (v) 325×47 (vi) 235×3 (vii) 347×35 Solution: (i) 25×514=25×21A2=1×215×2 =2110=2110

18. (ii) 625×79=325×79=32×75×9 =22445=44445 (iii) 32×513=32×163=8 (iv)56×237=56×177=8542=2142 (v) 325×47=175×47=6835=13335 (vi) 235×3=135×3=395=745 (vii) 347×35=257×35=5×37 =157=217 4. Which is greater: (i) 27 of 34 or 35 of 58 (ii) 12 of 67 or 23 of 37 Solution: (i) 27 of 34=27×34=1×37×2=314 35 of 58=35×58=38 Since in 314 and 38, their numerators are same and 14>8 ∴314<38 or 38>314

19. Hence, 35 of 58>27 of 34 (ii) 12 of 67 or 23 of 37 12 of 67=12×67=1×62×7=614=37 23 of 37=23×37=27 Here, denominators are same. ∴27<37 or 37>27 Hence, 12 of 67>23 of 37 5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling. Solution: Number of saplings =4 Distance between two adjacent saplings =34 m =34 m+34 m+34 m =3×34 m =94 m=214 m 6. Lipika reads a book for 1 ¾ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book? Solution: In 1 day Lipika needs 134 hours In 6 days Lipika will need 6×134 hours =6×74 hours =3×72 hours =212 hours =1012 hours Hence the required hours =1012 7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 ¾ litres of petrol. Solution: In 1 litre of petrol, the car covers 16 km distance in 234 litres o petrol, the car will cover 2 34×16 km distance =234×16 km =114×16 km =11×4 km=44 km Hence, the required distance =44 km. 8. Fill in the blanks

20. (a) (i) provide the number in the box [], such that (2/3)×[]=(10/30). (ii) The simplest form of the number obtained in [] is....... (b) (i) Provide the number in the box [], such that 3/5×[]=(24/75). (ii) The simplest form of the number obtained in [] is Solution: (a) (i)23×□=1030⇒23×510=1030 Hence, the required number in □is510 (ii) The simplest form of the number obtained in□is510=12 (b) (i)35×□=2475⇒35×815=2475 Hence, the required number in the box □ is 2475 Simplest form of /248/525=825 (ii) The simplest form of the number obtained in □ is 825 Exercise 2 EXERCISE 2.2 1. Which of the drawings (a) to (d) show: (i) 2×15 (ii) 2×12 (iii) 3×23 (iv) 3×14 (a)

21. (b) (c) (d) Solution: (i) 2×15 represents drawing (d) (ii) 2×12 represents drawing (b) (iii) 3×23 represents drawing (a) (iv) 3×14 represents drawing (c) 2. Some pictures (a) to (c) are given below. Tell which of them show: (i) 3×15=35 (ii) 2×13=23 (iii) 3×34=214 (a) (b)

22. (c) Solution: (i) 3×15=35 represents figure (c) (ii) 2×13=23 represents figure (a) (iii) 3×34=214 represents figure (b) 3. Multiply and reduce to lowest from and convert into a mixed fraction: (i) 7×35 (ii) 4×13 (iii) 2×67 (iv) 5×29 (v) 23×4 (vi) 52×6 (vii) 11×47 (viii) 20×45 (ix) 13×13 (x) 15×35 Solution: (i) 7×35=215=415 (ii) 4×13=43=113 (iii) 2×67=127=157

23. (iv) 5×29=109=119 (v) 23×4=83=223 ) (vi)52×6=15 ( (vii) 11×47=447=627 (viii) 20×45=16 (i x) 13×13=133=413

24. (x) 15×35=9 4. Shade: (i) 12 of the circles in box (a) (ii) 23 of the circles in box (b) (iii) 35 of the circles in box (c) (a) (b) (c) Solution: (i) 12 of the circles

25. =12×12=6 = 6 triangles are to be shaded (ii) 23 of the triangles =23×9 = 6 triangles are to be shaded (iii) 35 of the squares =35×15=9 =9 squares are to be shaded

26. 5. Find: (a) 12 of (i) 24 (ii) 46 (b) 23 of (i) 18 (ii) 27 (c) 34 of (i) 16 (ii) 36 (d) 45 of (i) 20 (ii) 35 Solution: (a) (i)12 of 24=12×24=12 (ii) 12 of 46=12×46=23 (b) (i)23 of 18=23×18=12 (ii) 23 of 27=23×27=18 (c) (i)34 of 16=34×16=12 (i i) 34 of 36=34×36=27 (d) (i) 45 of 20=34×36=16 (ii) 45 of 35=45×37=28

27. 6. Multiply and express as a mixed fraction: (a) 3×515 (b) 5×634 (c) 7×214 (d) 4×613 (e) 314×6 (f) 325×8 Solution: (a) 3×515=3×265=785 =1535 (b) 5×634=5×274 =1354 =3334 (c) 7×214=7×94

28. =634 =1534 (d) 4×613=4×193 =763 =2513 (e) 314×6=134×6 =13×32 =392

29. =1912 (f)325×8=175×8 =1365 3275 7. Find: Options (a) 12 of (i) 234 (ii) 429 (b) 58 of (i) 356 (ii) 923 Solution:

30. (a) (i) 12 of 234=12×114=118=138 (ii) 12 of 429=12×389=199=219 (b) (i) 58×356=58×236=11548=21948 (ii) 58×923=58×293=14524=6124 8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water. (i) How much water did Vidya drink? (ii) What fraction of the total quantity of water did Pratap drink? Solution: (i) Water consumed by Vidya =25 of 5 litres 25×5 litres =2 litres Water consumed by Pratap =5 litres −2 litres =3 litres (ii) Fraction of water consumed by Pratap =35 litres Exercise 4

31. EXERCISE 2.4 1. Find (i) 12÷34 (ii) 14÷56 (iii) 8÷73 (iv) 4÷83 (v) 3÷213 (vi) 5÷347 Solution: (i) 12÷34=12×43=4×4=16 (ii) 14÷56=14×65 (iii) 8÷73=8×37=247=337 (iv) 4÷83=4×382=32=112

32. (v) 3÷213=3÷73=3×37 =97=127 (vi) 5÷347=5÷257= 5×725 =75=125 2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers. (i) 37 (ii) 58 (iii) 97 (iv)65 (v)127 (vi)18

33. (vii) 111 Solution: (i) Reciprocal of 37=73, which is improper fraction. (ii) Reciprocal of 58=85, which is improper fraction. (iii) Reciprocal of 97=79, which is proper fraction. (iv) Reciprocal of 65=56, which is proper fraction. (v) Reciprocal of 127=712, which is proper fraction. (vi) Reciprocal of 18=8, which is whole number. (vii) Reciprocal of 111=11, which is whole number. 3. Find: (i) 73÷2 (ii) 49÷5 (iii) 613÷7 (iii) 613÷7 (iv) 413÷3 (v) 312÷4 (vi) 437÷7 Solution: (i) 73÷2=73×12=76=116 (ii) 49÷5=49×15=445

34. (iii) 613÷7=613×17=691 (iv) 413÷3=133÷3=133×13 =139=149 (v) 312÷4=72÷4=72×14=78 ( vi )437÷7=317÷7=317×17=3149 4. Find: (i) 25÷12 (ii) 49÷23 (iii) 37÷87 (iv) 213÷35 (v) 312÷83 (vi) 25÷112 (vii) 315÷123 (viii) 215÷115 Solution: (i) 25÷12=25×21=45 (ii) 49÷23=49×32=23 (iii) 37÷87=37×78=38

35. (iv) 113÷35=73÷35=73×53 =359=389 (v) 312÷83=72÷83=72×38 =2116=1516 (vi) 25÷112=25÷32=25×23=415 (vii) 315÷123=165÷53=165×35 =4825=12325 (viii) 215÷115=115÷65=115×56 =116=156

36. Exercise 5 EXERCISE 2.5 1. Which is greater? (i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7 Solution: (i) 0.5 or 0.05 Comparing the tenths place, we get 5>0 ∴0.5>0.05 (ii) 0.7 or 0.5 Comparing the tenths place, we get 7>5 ∴0.7>0.5 (iii) 7 or 0.7 Comparing the one's place, we get 7>0 ∴7>0.7 (iv) 1.37 or 1.49 Comparing the tenths place, we get 3>4 ∴1.37<1.49 or 1.49>1.37 (v) 2.03 or 2.30 Comparing the tenths place, we get 0>3 ∴2.03<2.30 or 2.30>2.03

37. (vi) 0.8 or 0.88⇒0.80 or 0.88∣ Since tenths place is same. Comparing the hundredth place, we get 0>8 ∴0.80<0.88 or 0.88>0.80 2. Express as rupees as decimals: (i) 7 paise (ii) 7 rupees 77 paise (iii) 77 rupees 77 paise (iv) 50 paise (v) 235 paise Solution: (i) Since 1 rupee =100 paise and 1 paise =1100 rupees 7 paise =7100 rupees =0.07 rupees (ii) 7 rupees 7 paise =7 rupees +7100 rupees =7.07 rupees iii) 77 rupees 77 paise =77 rupees +77100 rupees =77.77 rupees iv) 50 paise =50100 rupees =0.50 rupees (v) 235 paise =235100 rupees =2.35 rupees 3. Do as directed below: (i) Express 5 cm in meter and kilometer

38. ii) Express 35 mm in cm,m and km. Solution: (i) 1 metre =100 cm 1 kilometre =1000 metre =100×1000 cm=100000 cm ∴5 cm=5100 metre =0.05 metre 5 cm=5100000 km=0.00005 km Hence, 5 cm=0.05 m and 0.00005 km (ii) 1 cm=10 mm and 1 km=100000 cm ∴35 mm=3510 cm=3.5 cm 35 mm=351000 m=0.035 m 35 mm=351000000 km=0.000035 km 4. Express in kg (i) 200 g (ii) 3470 g (iii) 4 kg 8 g Solution: (i) 200 g=20010000=0.02 kg ∵1 kg=1000 g (ii) 3470 g=347010000=3.47 kg ∵1 kg=1000 g (iii) 4 kg8 g=4 kg+810000 kg ∵1 kg=1000 g

39. =4 kg+0.008 kg=4.008 kg 5. Write the following decimal numbers in the expanded form: (i) 20.03 (ii) 2.03 (iii) 200.03 (iv) 2.034 Solution: (i) 20.03=2×10+0×1+0×1100+3×1100 (i) 2.03=2×1+0×110+3×1100 (iii) 200.03=2×100+0×10+0×1+0×110+3×1100 (iv) 2.034=2×1+0×110+3×110+4×11000 6. Write the place value of 2 in the following decimal numbers: (i) 2.56 (ii) 21.37 (iii) 10.25 (iv) 9.42 (v) 63.352 Solution: (i) Place value of 2 in 2.56=2×1=2 is, ones (ii) Place value of 2 in 21.37=2×10=20 i.e. tens (iii) Place value of 2 in 10.25=210=0.2 i.e. tenths

40. (iv) Place value of 2 in 9.42=2100=0.02 i.e. hundredths (v) Place value of 2 in 63.352=21000=0.002 i.e. thousandths. 7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much? Solution: Distance travelled by Dinesh from A to C =AB+BC =7.5 km+12.7 km =20.2 km Distance travelled by Ayub from A to C =AD+DC =9.3 km+11.8 km =21.1 km Since 21.1 km>20.2 km. Hence, Ayub travelled more distance. 8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4kg800 g oranges and 4 kg150 g bananas. Who bought more fruits?

41. Solution: Fruits bought by Shyama =5 kg300 g apples +3 kg250 g mangoes =5.300 kg apples +3.250 kg mangoes =8.550 kg of fruits Fruits bought by Sarala =4 kg800 g oranges +4 kg150 g bananas =4.800 kg oranges +4.150 kg bananas =8.950 kg of fruits Since 8.950 kg>8.550 kg Hence, Sarala bought more fruits. 9. How much less is 28 km than 42.6 km ? Solution: Since 28 km<42.6 km 42.6 km−28.0 km 14.6 km Hence, 28 km is less than 42.6 km by 14.6 km.