Kerala Syllabus Class 6 Mathematics - Unit 2 One Fraction Many Forms - Questions and Answers | Teaching Manual 

Questions and Answers for Class 6 Mathematics - Unit 2 One Fraction Many Forms - Study Notes | Text Books Solution STD 6 - Maths: Unit 2 One Fraction Many Forms - Questions and Answers | ഗണിതം - ഒരു ഭിന്നം പല രൂപം - ചോദ്യോത്തരങ്ങൾ 

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Kerala Syllabus STD 6 Maths: Unit 2 One Fraction Many Forms - Textbook Solutions

♦ Textbook Activities (Textbook Page No: 24, 25, 26)
(1) Find the fractions specified below :
(i) The form of ½ with denominator 24 
(ii) The form of ½ with numerator 24 
(iii) The form of ⅓ with denominator 24 
(iv) The form of ⅓ with numerator 24 
(v) The form of ¼ with numerator 100 
Answer:
(i) The form of ½ with denominator 24 
To get ½ of the whole 24 equal parts, we must take ²⁴⁄₂ = 12 of these
The form of ½ with denominator 24 = ¹²⁄₂₄

(ii) The form of ½ with numerator 24 
To get ½ of the whole 2 × 24 = 48 equal parts, we must take ⁴⁸⁄₂ = 24 of these
∴ The form of ½ with numerator 24 = ²⁴⁄₄₈

(iii) The form of ⅓ with denominator 24 
To get ⅓ of the whole 24 equal parts, we must take ²⁴⁄₃ = 8 of these
∴ The form of ⅓ with denominator 24 = ⁸⁄₂₄

(iv) The form of ⅓ with numerator 24 
To get ⅓ of the whole 3 × 24 = 72 equal parts, we must take ⁷²⁄₃ = 24 of these
∴ The form of ⅓ with numerator 24 = ²⁴⁄₇₂

(v) The form of ¼ with numerator 100 
To get ¼ of the whole 4 × 100 = 400 equal parts, we must take ⁴⁰⁰⁄₄ = 100 of these
∴ The form of ¼ with numerator 100 = ¹⁰⁰⁄₄₀₀

(2) Write three different forms of  ¼.
Answer:
¼ = ¹ ˣ ²⁄₄ ₓ ₂ = ²⁄₈
¼ = ¹ ˣ ³⁄₄ ₓ ₃ = ³⁄₁₂
¼ = ¹ ˣ ⁴⁄₄ ₓ ₄ = ⁴⁄₁₆

(3) For each of  the pair of  fractions below, find three forms with the same denominator: 
(i) ½, ⅓ 
(ii) ½, ¼
(iii) ⅓, ¼
Answer:
(i) ½, ⅓ 
1) ½ = ¹ ˣ ³⁄₂ ₓ ₃ = ³⁄₆
    ⅓ = ¹ ˣ ²⁄₃ ₓ ₂ = ²⁄₆
2) ½ = ¹ ˣ ⁶⁄₂ ₓ ₆ = ⁶⁄₁₂
    ⅓ = ¹ ˣ ⁴⁄₃ ₓ ₄ = ⁴⁄₁₂
3) ½ = ¹ ˣ ⁹⁄₂ ₓ ₉ = ⁹⁄₁₈
    ⅓ = ¹ ˣ ⁶⁄₃ ₓ ₆ = ⁶⁄₁₈

(ii) ½, ¼
1) ½ = ¹ ˣ ⁴⁄₂ ₓ ₄ = ⁴⁄₈
    ¼ = ¹ ˣ ²⁄₄ ₓ ₂ = ²⁄₈
2) ½ = ¹ ˣ ⁸⁄₂ ₓ ₈ = ⁸⁄₁₆
    ¼ = ¹ ˣ ⁴⁄₄ ₓ ₄ = ⁴⁄₁₆
3) ½ = ¹ ˣ ¹²⁄₂ ₓ ₁₂ = ¹²⁄₂₄
    ¼ = ¹ ˣ ⁶⁄₄ ₓ ₆ = ⁶⁄₂₄

(iii) ⅓, ¼
1) ⅓ = ¹ ˣ ⁴⁄₃ ₓ ₄ = ⁴⁄₁₂
    ¼ = ¹ ˣ ³⁄₄ ₓ ₃ = ³⁄₁₂
2) ⅓ = ¹ ˣ ⁸⁄₃ ₓ ₈ = ⁸⁄₂₄
    ¼ = ¹ ˣ ⁶⁄₄ ₓ ₆ = ⁶⁄₂₄
3) ⅓ = ¹ ˣ ¹²⁄₃ ₓ ₁₂ = ¹²⁄₃₆
    ¼ = ¹ ˣ ⁹⁄₄ ₓ ₉ = ⁹⁄₃₆

(4) Does ⅓ have another form with denominator 10, 100 or 1000? Give reasons.
Answer:
 ⅓ = ¹ ˣ ³⁄₃ ₓ ₃ = ³⁄₉
 ⅓ = ¹ ˣ ³³⁄₃ ₓ ₃₃ = ³³⁄₉₉
 ⅓ = ¹ ˣ ³³³⁄₃ ₓ ₃₃₃ = ³³³⁄₉₉₉
Since 10, 100, 1000 are not multiples of 3, ⅓ can't be written as a fraction with those denominators.

♦ Now, can't you fill up the following table by multiplying the numerator and denominator by a number?
♦ Textbook Activities (Textbook Page No: 28)
(1) Write each of  the following fractions in lowest terms: 
(i) ³²⁄₆₄
(ii) ²⁷⁄₈₁
(iii) ³⁰⁄₄₅
(iv) ¹²⁄₂₁
(v) ⁴⁵⁄₅₄
Answer:
(i) ³²⁄₆₄
32 = 2 x 2 x 2 x 2 x 2
64 = 2 x 2 x 2 x 2 x 2 x 2
Here, 32 is 2 multiplied 5 times and 64 is 2 multiplied 6 times.
If we remove the common factors of 32 and 64, we get ³²⁄₆₄ = ½.

(ii) ²⁷⁄₈₁
27 = 3 x 3 x 3
81 = 3 x 3 x 3 x 3
Here, 27 is 3 multiplied 3 times and 81 is 3 multiplied 4 times.
If we remove the common factors of 27 and 81, we get ²⁷⁄₈₁ = ⅓.

(iii) ³⁰⁄₄₅
30 = 2 x 3 x 5
45 = 3 x 3 x 5
If we remove the common factors of 30 and 45, we get ³⁰⁄₄₅ = ⅔.

(iv) ¹²⁄₂₁
12 = 2 x 2 x 3
21 = 3 x 7
If we remove the common factors of 12 and 21, we get ¹²⁄₂₁ = ² ˣ ²⁄₇ = ⁴⁄₇.

(v) ⁴⁵⁄₅₄
45 = 3 x 3 x 5
54 = 2 x 3 x 3 x 3
If we remove the common factors of 45 and 54, we get ⁴⁵⁄₅₄ = ⁵⁄₂ ₓ ₃ = ⅚.

♦ Textbook Activities (Textbook Page No: 32)
Fraction as division 

(1) 20 litres of water is used to fill 8 identical bottles. How much litres of water is there in each bottle?
Answer:
20 = 4 x 5
8 = 4 x 2
Let's divide the 8 bottles into 4 sets of 2 bottles each. If each set gets 5 litres of water, and the 2 bottles in each set are filled with 5 litres, then each bottle will have ⁵⁄₂ litres.
ie, ²⁰⁄₈ = ⁵⁄₂
Now divide 5 by 2 and write as multiple and remainder.
5 = (2 x 2) + 1
Dividing the reminder 1 also by 2, we can write 1 ÷ 2 = ½
So, ²⁰⁄₈ = ⁵⁄₂ = 2½
Thus, each bottle contains 2½ litres of water.

(2) A rope of length 140 centimetres is cut into 16 equal pieces. What is the length of each piece? 
Answer:
140 = 4 x 35
16 = 4 x 4
¹⁴⁰⁄₁₆ = (⁴ ˣ ⁵ ˣ ⁷⁄₄ ₓ ₄) = ³⁵⁄₄
First, divide 35 by 4 and write as multiple and remainder.
35 = (4 x 8) + 3
Dividing the remainder 3 also by 4, we can write 3 ÷ 4 = ¾
So ¹⁴⁰⁄₁₆ = ³⁵⁄₄ = 8¾
Length of each piece = 8¾ cm

(3) If 215 kilograms of rice is divided equally among 15 people, how much kilogram of rice would each get?
Answer:
215 = 5 x 43
15 = 5 x 3
²¹⁵⁄₁₅ = ⁵ ˣ ⁴³⁄₅ ₓ ₃ = ⁴³⁄₃
Divide 43 by 3 and write as multiple and remainder.
43 = (3 x 14) + 1
Dividing the remainder 1 also by 3, we can write 1 ÷ 3 = ⅓.
So, ²¹⁵⁄₁₅ = ⁴³⁄₃ = 14⅓
Thus, each gets 14⅓ kg