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Kerala Syllabus Class 6 Mathematics - Unit 5 Decimal Forms - Questions and Answers | Teaching Manual 

Questions and Answers for Class 6 Mathematics - Unit 5 Decimal Forms - Study Notes | Text Books Solution STD 6 - Maths: Unit 5 Decimal Forms - Questions and Answers | ഗണിതം - ദശാംശരൂപങ്ങൾ - ചോദ്യോത്തരങ്ങൾ 

ആറാം ക്ലാസ്സ്‌  Mathematics - Unit 5 Decimal Forms എന്ന പാഠം ആസ്പദമാക്കി തയ്യാറാക്കിയ ചോദ്യോത്തരങ്ങള്‍. ഈ അധ്യായത്തിന്റെ Teachers Handbook, Teaching Manual എന്നിവ ഡൗൺലോഡ് ചെയ്യാനുള്ള ലിങ്ക് ചോദ്യോത്തരങ്ങളുടെ അവസാനം നൽകിയിട്ടുണ്ട്. പുതിയ അപ്‌ഡേറ്റുകൾക്കായി ഞങ്ങളുടെ Telegram Channel ൽ ജോയിൻ ചെയ്യുക.

Kerala Syllabus STD 6 Maths: Unit 5 Decimal Forms - Textbook Solutions

♦ In the decimal form of a number, the dot separates the whole number
part and the fractional part. Digits to the left of the dot show the
multiples of ones, tens, hundreds and so on; the digits to the right show
the multiples of tenths, hundredths, thousandths and so on.

♦ Textbook Activities (Textbook Page No: 69)
♦ Split the numbers below according to place value:
(i) 4.5 
(ii) 4.57 
(iii) 4.572 
(iv) 45.72 
(v) 457.2

(i) 4.5 
1  ⅒
 4.5    4   5 

(ii) 4.57 
1  ⅒  ¹⁄₁₀₀ 
 4.57    4   5    7 

(iii) 4.572 
1  ⅒  ¹⁄₁₀₀   ¹⁄₁₀₀₀ 
 4.572    4   5    7    2 

(iv) 45.72 
10  1  ¹⁄₁₀   ¹⁄₁₀₀ 
 45.72    4   5    7    2 

(iv) 457.2 
100  10  1   ¹⁄₁₀ 
 457.2    4   5    7    2 

♦ Textbook Activities (Textbook Page No: 71)
♦ Now try to write 4 kilograms and 55 grams as kilograms in decimal form.
Convert the measures below into the measures specified, using fractions and decimal forms.
Answer:
♦ Textbook Activities (Textbook Page No: 74)
(1) The decimal form of some numbers are given below. Write each of them as a fraction with denominator 10, 100 or 1000.
(i) 3.7 (ii) 3.07 (iii) 30.7 (iv) 3.72 (v) 37.2 (vi) 3.072 (vii) 30.72

(i) 3.7 = ³⁷⁄₁₀

(ii) 3.07 = ³⁰⁷⁄₁₀₀

(iii) 30.7  = ³⁰⁷⁄₁₀

(iv) 3.72 = ³⁷²⁄₁₀₀

(v) 37.2 = ³⁷²⁄₁₀

(vi) 3.072 = ³⁰⁷²⁄₁₀₀₀

(vii) 30.72 = ³⁰⁷²⁄₁₀₀

(2)  Write the decimal form of the fractions given below.
(i) ⁵¹⁄₁₀ (ii) ⁵¹³⁄₁₀ (iii) ⁵¹³⁄₁₀₀ (iv) ⁵¹³⁄₁₀₀₀ (v) ⁵¹³⁰⁄₁₀₀₀₀

(i) ⁵¹⁄₁₀  = 5.1

(ii) ⁵¹³⁄₁₀ = 51.3

(iii) ⁵¹³⁄₁₀₀ = 5.13

(iv) ⁵¹³⁄₁₀₀₀ = 0.513

(v) ⁵¹³⁰⁄₁₀₀₀₀ = 5.13

♦ Addition and subtraction

♦ Textbook Activities (Textbook Page No: 79)
1) Anu made an 8.5 metre long festoon, and Sarah made a 7.8 metre long one to decorate their classroom for the school anniversary. What is the total length of the festoon they made?
The length of the festoon made by Anu = 8.5 metres 
  = ⁸⁵⁄₁₀ metres
The length of the festoon made by Sarah = 7.8 metres 
  = ⁷⁸⁄₁₀ metres
Total length of the festoon = ⁸⁵⁄₁₀ + ⁷⁸⁄₁₀ 
   = 85 + 78/10 = ¹⁶³⁄₁₀ = 16.3 meters

(2) Amal needs 2.25 metres of cloth and Sagar, 1.85 metres for school uniform. How many metres of cloth in all?
Length of the cloth needed for Amal = 2.25 metres = ²²⁵⁄₁₀₀ metres
Length of the cloth needed for Sagar = 1.85 metres = ¹⁸⁵⁄₁₀₀ metres
Total length of uniform cloth = ²²⁵⁄₁₀₀ + ¹⁸⁵⁄₁₀₀ = 225 +185/100
   = ⁴¹⁰⁄₁₀₀ = ⁴¹⁄₁₀ = 4.1 metres

(3) A tin weighs 2.85 kilograms, and it is filled with 12.5 kilograms of rice. What is the total weight?
Weight of the tin = 2.85 kilograms = ²⁸⁵⁄₁₀₀ kilograms
Weight of the rice = 12.5 kilograms = ¹²⁵⁄₁₀ kilograms = ¹²⁵⁰⁄₁₀₀ kilograms
  ∴ Total weight = ²⁸⁵⁄₁₀₀ + ¹²⁵⁰⁄₁₀₀ = 285 + 1250/100
  = ¹⁵³⁵⁄₁₀₀ = 15.35 kilograms

(4) Bakul walks 2.25 kilometres in the morning and 1.5 kilometres in the evening every day. What is the total distance she walks each day?
Distance walked in the morning = 2.25 kilometres = ²²⁵⁄₁₀₀ kilometres
Distance walked in the evening = 1.5 kilometres = ¹⁵⁄₁₀ kilometres
= ¹⁵⁰⁄₁₀₀ kilometres
∴ Total distance walked every day = ²²⁵⁄₁₀₀ + ¹⁵⁰⁄₁₀₀ = 225 + 150/100
= ³⁷⁵⁄₁₀₀ = 3.75 kilometres

(5) Two small bottles contain 0.850 litres and 0.375 litres of honey. If both the bottles are emptied into a large bottle, how much honey does it contain?
Amount of honey in the first small bottle = 0.850 litres = ⁸⁵⁰⁄₁₀₀₀ litres
Amount of honey in the second small bottle = 0.375 litres = ³⁷⁵⁄₁₀₀₀ litres
Amount of honey in the large bottle = ⁸⁵⁰⁄₁₀₀₀ + ³⁷⁵⁄₁₀₀₀ = 850 + 375/1000
= ¹²²⁵⁄₁₀₀₀ = 1.225 litres

♦ Textbook Activities (Textbook Page No: 82)
(1) From a rod 14.7 metres long, a piece 7.75 metres long is cut off. What is the length of the remaining piece?
Total length of the rod = 14.7 metres
Total length of the rod cut off = 7.75 metres
Length of the remaining piece = 14.7 − 7.75 = ¹⁴⁷⁄₁₀ − ⁷⁷⁵⁄₁₀₀
= ¹⁴⁷⁰⁄₁₀₀ − ⁷⁷⁵⁄₁₀₀ = 1470 − 775/100
= ⁶⁹⁵⁄₁₀₀ = 6.95 metres

(2) There was 38.7 kilograms of rice in a sack, and 12.350 kilograms of this is used up. How much rice remains in the sack? 
Total amount of rice in the sack = 38.7 kilograms
Total amount of rice used = 12.350 kilograms
Total amount of rice left = 38.7 − 12.350 = ³⁸⁷⁄₁₀ − ¹²³⁵⁰⁄₁₀₀₀
= ³⁸⁷⁄₁₀ − ¹²³⁵⁄₁₀₀ = ³⁸⁷⁰⁄₁₀₀ − ¹²³⁵⁄₁₀₀ 
= 3870 − 1235/100 = ²⁶³⁵⁄₁₀₀
= 26.35 kilograms

(3) The perimeter of a rectangle is 24 centimetres, and the length of one side is 6.4 centimetres. What is the length of the other side?
Perimeter of the rectangle = 24 centimetres
2 (length + width) = 24
length + width = ²⁴⁄₂
length + width = 12 cm
length = 6.4 cm
width = 12 − 6.4 = 12 − ⁶⁴⁄₁₀
= ¹²⁰⁄₁₀ − ⁶⁴⁄₁₀ = 120 − 64/10
= ⁵⁶⁄₁₀ = 5.6 cm

(4) There was 2.50 litres of oil in a bottle, and 0.475 litres of this was used for cooking. How much oil is left in the bottle?
Total quantity of oil in the bottle = 2.50 litres
Quantity of oil used for cooking = 0.475 litre
Quantity of oil remaining = 2.50 − 0.475 = ²⁵⁰⁄₁₀₀ − ⁴⁷⁵⁄₁₀₀₀
= ²⁵⁰⁰⁄₁₀₀₀ − ⁴⁷⁵⁄₁₀₀₀ = 2500 − 475/1000
= ²⁰²⁵⁄₁₀₀₀ = 2.025 litres

(5) What number we must add to 14.32 to get 16.43?
Number to be added = 16.43 − 14.32
= ¹⁶⁴³⁄₁₀₀ − ¹⁴³²⁄₁₀₀ = 1643 − 1432/100
= ²¹¹⁄₁₀₀ = 2.11
OR
16.43 − 14.32 = 2.11
∴ Number to be added = 2.11