## NCERT solutions for class 6 Maths chapter 7

fraction class 6 represent parts of a whole or a collection of objects. This chapter introduces you to the basic concepts of fractions.

**Key points:**

**Fraction:**A number that represents a portion of a whole. It can be written in two parts: numerator (top number) and denominator (bottom number).**Types of Fractions:****Proper Fraction:**Numerator is smaller than the denominator (e.g., 1/4, 3/7).**Improper Fraction:**Numerator is greater than or equal to the denominator (e.g., 5/4, 7/7). Improper fractions can be converted to mixed numbers (whole numbers + fractions).**Mixed Fraction:**A combination of a whole number and a proper fraction (e.g., 1 ½, 2 ¾).

**Representing Fractions:**Fractions can be represented visually using fraction models (like circles or rectangles divided into parts), number lines, or fraction notation.**Comparing Fractions:**We can compare fractions with the same denominator by looking at the numerators. The larger numerator represents the larger fraction.**Operations on Fractions:****Addition and Subtraction:**Requires fractions to have the same denominator. We add/subtract the numerators and keep the denominator the same.**Multiplication:**We multiply both the numerator and denominator of one fraction by the numerator and denominator of the other fraction.**Division:**We flip (reverse) the second fraction (divisor) and then multiply.

**Additional concepts you might encounter:**

- Simplifying fractions: Reducing fractions to their lowest terms by dividing the numerator and denominator by the greatest common factor (GCD).
- Equivalent fractions: Fractions that represent the same quantity even though they are written differently (e.g., 1/2 is equivalent to 2/4).

NCERT solutions for class 6 Maths chapter 7

**Exercise 7.1**

**Write the fraction representing the shaded portion.**

**Ans : **

(i) Total parts = 4

shaded parts = 2

∴ Fraction = 2/4

(ii) Total parts = 9

shaded parts = 8

∴ Fraction = 8/9

(iii) Total parts = 8

shaded parts = 4

∴ Fraction = 4/8

(iv) Total parts = 4

shaded parts = 1

∴ Fraction = 1/4

(v) Total parts = 7

shaded parts = 3

∴ Fraction = 3/7

(vi) Total parts = 12

shaded parts = 3

∴ Fraction = 3/12

(vii) Total parts = 10

shaded parts = 10

∴ Fraction = 10/10

(viii) Total parts = 9

Number of shaded parts = 4

∴ Fraction = 4/9

(ix) Total parts = 8

shaded parts = 4

∴ Fraction = 4/8

(x) Total parts = 2

shaded part = 1

∴ Fraction = 1/2

**3. Identify the error, if any.**

**Ans : **

(a) shaded part is not half.

∴ This is not 1/2

(b) parts are not equal.

∴ Shaded part is not 1/4

(c) part are not equal.

∴ Shaded part is not 3/4

**4. What fraction of a day is 8 hours?**

**Ans : **

A day has 24 hours.

8 hours is a portion of the 24 hours.

Therefore, the fraction representing 8 hours is: **8/24**.

**5. What fraction of a hour is 40 minutes?**

**Ans : **

An hour has 60 minutes.

40 minutes is a portion of the 60 minutes.

Therefore, the fraction representing 40 minutes is: **40/60**.

**6. Arya, Abhimanyu and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of Jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.**

**(a) How can Arya divide his sandwiches so that each person has an equal share?**

**(b) What part of a sandwich will each boy receive?**

**Ans : **

Since there are 3 people sharing (Arya, Abhimanyu, and Vivek), Arya needs to divide each of his sandwiches (vegetable and jam) into 3 equal parts.

Since each sandwich is divided into 3 parts, each boy will get 1/3 of the vegetable sandwich and 1/3 of the jam sandwich.

**7. Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?**

**Ans : **

Fraction of dresses finished = Number of dresses finished / Total number of dresses

Fraction of dresses finished = 20 dresses / 30 dresses

This simplifies to **2/3**.

**8. Write the natural numbers from 2 to 12. What fraction of them are prime numbers?**

**Ans : **

**List the natural numbers from 2 to 12:**2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12**Identify the prime numbers:**

A prime number has exactly two factors: 1 and itself. By checking each number’s divisibility:

- Prime numbers: 2, 3, 5, 7, 11 (These numbers only have 1 and themselves as factors)

**Calculate the fraction of prime numbers:**

- Total numbers = 11 (from 2 to 12)
- Prime numbers = 5

There are 5 prime numbers out of 11 total numbers. We can express this as a fraction:

**Fraction of prime numbers = Number of prime numbers / Total numbers**

**Fraction of prime numbers = 5 / 11**

Therefore, 5/11 of the natural numbers from 2 to 12 are prime numbers.

**9. Write the natural numbers from 102 to 113. What fraction of them are prime numbers?**

**Ans : **

**Analyze the numbers:**

- Most even numbers are not prime (divisible by 2).
- We can quickly eliminate all even numbers (102, 104, 106, 108, 110, 112) from consideration of being prime.

**Identify the remaining odd numbers:**

- This leaves us with odd numbers: 103, 105, 107, 109, 111, 113.

**Check for primality in the remaining odd numbers:**

- 105 is divisible by 3 (3 x 35).
- 111 is divisible by 3 (3 x 37).

**Prime numbers:**

Therefore, the prime numbers among 102 to 113 are: 103, 107, 109, 113.

**Calculate the fraction:**

- Total numbers: 113 – 102 + 1 = 12 (from 102 to 113)
- Prime numbers: 4 (103, 107, 109, 113)

**Fraction of prime numbers = 4 / 12 = 1 / 3**

Therefore, 1/3 of the natural numbers from 102 to 113 are prime numbers.

**10 . What fraction of these circles have X’s in them?**

**Ans : **Total circles = 8

Number of circles having X’s = 4

Required fraction = 4/8 = 1/2

**11. Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?**

**Ans : **

**(a) Fraction of CDs bought:**

- Kristin bought 3 CDs.
- We don’t know the total number of CDs yet, so let’s represent it as “x”.
- The fraction representing the bought CDs is the number of CDs bought divided by the total number:
**3 CDs / x**.

**(b) Fraction of CDs received as gifts:**

- Kristin received 5 CDs as gifts.
- Again, using “x” for the total number of CDs, the fraction representing the gifted CDs is:
**5 CDs / x**.

**(c) Total fraction of CDs:**

- To find the total fraction of CDs Kristin has (both bought and gifted), we need to add the fractions for bought and gifted CDs:

**Total fraction = Fraction bought + Fraction received as gifts**

**Total fraction = (3 CDs / x) + (5 CDs / x)**

**Simplifying the expression (optional):**

Since both fractions have the same denominator (x), we can simply add the numerators:

**Total fraction = (3 + 5) CDs / x**

**Total fraction = 8 CDs / x**

Therefore, Kristin has 8 CDs / x of her total CDs from buying and receiving gifts.

NCERT solutions for class 6 Maths chapter 7

**Exercise 7.2**

**1. Draw number lines and locate the points on them.**

(a) 1/2 , 1/4 , 3/4 , 4/4 ,

(b) 1/8 , 2/8 , 3/8 , 7/8

(c) 2/5 , 3/5 , 8/5 , ⅘

**Ans : **

**2. Express the following as mixed fractions:**

**(a) 20/3**

**(b) 11/5**

**(c) 17/7**

**(d) 28/5**

**(e) 19/6**

**(f) 35/9**

**Ans : **

(a) **20/3 = 6 2/3**

- Divide 20 by 3 (the denominator of the fraction). We get 6 with a remainder of 2.
- The quotient (6) becomes the whole number part of the mixed fraction.
- The remainder (2) becomes the numerator of the fractional part.
- The denominator of the fractional part stays the same (3), representing the original denominator.

(b) **11/5 = 2 1/5**

Following the same process:

- Divide 11 by 5. We get 2 with a remainder of 1.
- So, the mixed fraction is 2 with 1/5 remaining.

(c) **17/7 = 2 3/7**

- Divide 17 by 7. We get 2 with a remainder of 3.
- The mixed fraction becomes 2 3/7.

(d) **28/5 = 5 3/5**

- Divide 28 by 5. We get 5 with a remainder of 3.
- So, the mixed fraction is 5 3/5.

(e) **19/6 = 3 1/6**

- Divide 19 by 6. We get 3 with a remainder of 1.
- The mixed fraction becomes 3 1/6.

(f) **35/9 = 3 8/9**

- Divide 35 by 9. We get 3 with a remainder of 8.
- So, the mixed fraction is 3 8/9.

**3. Express the following as improper fractions:**

**(a) 7 3/4**

**(b)5 6/7**

**(c) 2 5/6**

**(d) 10 3/5**

**(e) 9 3/7**

**(f) 8 4/9**

**Ans : **

(a) **7 3/4:**

- Convert the whole part (7) to a fraction with the same denominator (4) as the fractional part: 7 * (4/4) = 28/4
- Add the whole number fraction (28/4) and the fractional part (3/4): 28/4 + 3/4 =
**31/4**

(b) **5 6/7:**

- Follow the same approach:
- Whole number part as a fraction: 5 * (7/7) = 35/7
- Add both fractions: 35/7 + 6/7 =
**41/7**

(c) **2 5/6:**

- Convert the whole number: 2 * (6/6) = 12/6
- Add the fractions: 12/6 + 5/6 =
**17/6**

(d) **10 3/5:**

- Convert the whole number: 10 * (5/5) = 50/5
- Add the fractions: 50/5 + 3/5 =
**53/5**

(e) **9 3/7:**

- Convert the whole number: 9 * (7/7) = 63/7
- Add the fractions: 63/7 + 3/7 =
**66/7**

(f) **8 4/9:**

- Convert the whole number: 8 * (9/9) = 72/9
- Add the fractions: 72/9 + 4/9 =
**76/9**

NCERT solutions for class 6 Maths chapter 7

**Exercise 7.3**

**1. Write the fractions. Are all these fractions equivalent?**

**Ans : **

**(a)** (i) Shaded part = 1/2

(ii) Shaded Part = 2/4 = 1/2

(iii) Shaded Part = 3/6 = 1/2

(iv) Shaded Part = 4/8 = 1/2

All Fraction equivalent.

**(b) **(i) Shaded Part = 4/12 = 1/3

(ii) Shaded Part = 3/9 = 1/3

(iii) Shaded Part = 2/6 = 1/3

(iv) Shaded Part = 1/3

(v) Shaded Part = ⅖

Since the fractions are in their simplest form and have different denominators, they are not equivalent.

**2. Write the fractions and pair up the equivalent fractions from each row.**

**Ans: **

(a)** **1/2

(b) 4/6 = 2/3

(c) 3/9 = 1/3

(d) 2/8 = 1/4

(e) 3/4

(i) 6/18 = 1/3

(ii) 4/8 = 1/2

(iii) 12/16 = 3/4

(iv) 8/12 = 2/3

(v) 4/16 = 1/4

**Equivalent fractions pair**

(a) = (ii) = 1/2

(b) = (iv) = 2/3

(c) = (i) = 1/3

(d) = (v) = 1/4

(e) = (iii) = 3/4

**3. Replace **** in each of the following by the correct number:**

**Ans :**

(a) 2 *— = 7*8

——- = 7*8/2 =28

2/7 = 8/28

(b) 5/8 = 10/ —- = 5*—–

=8*10

——– = 8*10/5 = 16

5/8 = 10/16

(c) 3/5 = ——-/20

5*—- = 3*20

——– = 3*20/5 = 12

3/5 = 12/20

(d) 45/60 = 15/——

45*—– = 15*60

——– = 15*60/45 = 20

45/60 = 15/20

(e) 18/24 = ——/4

24/—— = 18*4

——– = 18*4/24 = 3

18/24 = 3/4

**4. Find the equivalent fraction of 3/5**** **** having**

**(a) denominator 20**

**(b) numerator 9**

**(c) denominator 30**

**(d) numerator 27**

**Ans : **

(a) Here We have denominator 20

Consider N be the numerator of the fractions

N/20 = 3/5

5 * N = 20*3

N = 20*3/5 =

Therefore fraction is 12/20

(b) Here /3we have numerator 9

Consider D be the denominator of the fraction

9/D = 3/5

3*D = 9*5

D = 9*5/3 = 15

Therefore fraction is 9/15

(c) Here we have denominator 30

Consider N be the numerator of the fraction

N/30=3/5

5*N = 3*30

N = 3*30/5 = 18

Therefore fraction is 18/30

(d) Here we have numerator 27

Consider D be the denominator of the fraction

27/D = 3/5

3*D = 5*27

D = 5*27/3 = 45

Therefore fraction is 27/45

**5. Find the equivalent fraction of 36/48 with**

**(a) numerator 9**

**(b) denominator 4**

**Ans : **

(a) 9/D = 36/48

D * 36 = 9 * 48

D = 9*48/36 = 12

Equivalent fraction is 9/12

(b) N/4 = 36/48

N * 48 = 4*36

N = 4*36/48 = 3

Equivalent fraction is 3/4

**6. Check whether the given fractions are equivalent:**

**Ans : **

a. 5/9 and 30/54

We have 5 x 54 = 270 and 9 x 30 = 270.

Since 5 x 54 = 9 x 30, it follows that

5/9 and 30/54 are equivalent fractions.

b. 3/10 and 12/50

We have 3 x 50 = 150 and 10 x 12 = 120.

Since 3 x 50 ≠ 10 x 12, it follows that

3/10 and 12/50 are not equivalent fractions.

c. 7/13 and 5/11

We have 7 x 11 = 77 and 5 x 13 = 65.

Since 7 x 11 ≠ 5 x 13, it follows that

7/13 and 5/11 are not equivalent fractions.

7. Reduce the following fractions to simplest form:

**Ans : **

**Fraction in (a):**

- The text in image (a) is difficult to discern, but it appears to be 48/60.
- Reduce by finding the greatest common factor (GCD) of 48 and 60.
- GCD(48, 60) = 12 (both 48 and 60 can be divided by 12).
- Divide both numerator and denominator by 12: 48/60 = (48 / 12) / (60 / 12) = 4/5.

**Therefore, the reduced form of the fraction in (a) is 4/5.**

**Fraction in (b):**

- The text in image (b) is also difficult to decipher, but it looks like 150/60.
- Reduce using the GCD: GCD(150, 60) = 30.
- Divide both numerator and denominator by 30: 150/60 = (150 / 30) / (60 / 30) = 5/2.

**Therefore, the reduced form of the fraction in (b) is 5/2.**

**Fraction in (c):**

- The text in image (c) is unclear, but it seems to be 84/98.
- Reduce using the GCD: GCD(84, 98) = 14.
- Divide both numerator and denominator by 14: 84/98 = (84 / 14) / (98 / 14) = 6/7.

**Therefore, the reduced form of the fraction in (c) is 6/7.**

**Fraction in (d):**

- The text in image (d) is difficult to make out, but it appears to be 12/52.
- Reduce using the GCD: GCD(12, 52) = 4.
- Divide both numerator and denominator by 4: 12/52 = (12 / 4) / (52 / 4) = 3/13.

**Therefore, the reduced form of the fraction in (d) is 3/13.**

**Fraction in (e):**

- GCD(7, 28) = 7 (7 is a prime number and doesn’t share any factors with 28).
- Reduced form: 7/28 is already in its simplest form (the numerator and denominator have no common factors).

**Therefore, the reduced form of the fraction in (e) is 7/28 = 1/4**

**8. ****Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils.**

**Ans : **

**Ramesh:**- Pencils used: 10
- Total pencils: 20
- Fraction used: Pencils used / Total pencils = 10 pencils / 28 pencils
- Simplifying the fraction: We can’t simplify 10 and 28 further. Therefore, the fraction remains 1/2.

**Sheelu:**- Pencils used: 25
- Total pencils: 50
- Fraction used: Pencils used / Total pencils = 25 pencils / 50 pencils
- Simplifying the fraction: Dividing both numerator and denominator by 5 gives us 5/10. We can further simplify to 1/2.

**Jamaal:**- Pencils used: 40
- Total pencils: 80
- Fraction used: Pencils used / Total pencils = 40 pencils / 80 pencils
- Simplifying the fraction: Dividing both numerator and denominator by 40 gives us 1/2.

Yes, each has used up an equal fractions, i.e., 1/2

**9. ****Match the equivalent fractions and write two more for each.**

**Ans : **

250/400 | 5/8 |

180/200 | 9/10 |

660/990 | 2/3 |

180/360 | 1/2 |

220/550 | 2/5 |

NCERT solutions for class 6 Maths chapter 7

**Exercise 7.4**

**1. Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<‘, ‘=’, ‘>’ between the fractions.**

**Ans : **

**(a)** Total number of divisions = 8

(i) Number of shaded parts = 3

∴ Fraction = 3/8

(ii) Number of shaded parts = 6

∴ Fraction = 6/8

(iii) Number of shaded parts = 4

∴ Fraction = 4/8

(iv) Number of shaded parts = 1

∴ Fraction = 1/8

Now the fractions are: 3/8, 6/8, 4/8, and 1/8, all with the same denominator

(b)(i) Total number of divisions = 9

Number of shaded parts = 8

∴ Fraction = 8/9

(ii) Total number of divisions = 9

Number of shaded parts = 4

∴ Fraction = 4/9

(iii) Total number of divisions = 9

Number of shaded parts = 3

∴ Fraction = 3/9

(iv) Total number of divisions = 9

Number of shaded parts = 6

∴ Fraction = 6/9

∴ The fractions are 8/9, 4/9, 3/9, and 6/9, all with the same denominator.

**(c) Show 2/4 , 4/6 , 8/6 and 6/6 on the number line. Put appropriate signs between the fractions given.**

**Ans : **

**2. Compare the fractions and put an appropriate sign.**

**Ans : **

**Solution:**

- Here, denominators of the two fractions are same and 3 < 5.
- Here, numerators of the fractions are same and 7 > 4.
- Here, denominators of the two fractions are same and 4 < 5.
- Here, numerators of the two fractions are same and 5 < 7.

**3. Make five more such pairs and put appropriate signs.**

**Ans : **

**4. Look at the figures and write ’<’, or ’>’ ’=’ between the given pairs of fractions.**

**Make five more such problems and solve them with your friends**

**Ans : **

**5. How quickly can you do this? Fill appropriate sign. ‘<‘, ‘=’, ‘>’.**

**Ans : **

**6. The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.**

**Ans : **

**7. Find answers to the following. Write and indicate how you solved them.**

**Ans : **

**a )**

By cross-multiplying, we get

5 x 5 = 25 and 4 x 9 = 36

Since 25 ≠ 36

5/9 is not equal to ⅘

**b )**

By cross-multiplying, we get

9 x 9 = 81 and 16 x 5 =80

Since 81 ≠ 80

9/16 is not equal to 5/9

**c )**

By cross-multiplying, we get

4 x 20 = 80 and 5 x 16 = 80

Since 80 = 80

4/5 is equal to 16/20

**d )**

By cross-multiplying, we get

1 x 30 = 30 and 4 x 15 = 60

1/15 is not equal to 4/30

**8. Ila read 25 pages of a book containing 100 pages.**

**Lalita read 2/5 of the same book. Who read less?**

**Ans : **

Here’s how we can find out:

**Ila:**

Pages read: 25

Total pages in the book: 100

Fraction of the book Ila read: Pages read / Total pages = 25 / 100= 1/4

**Lalita:**

Fraction of the book read: 2/5

IIa read less book than Lalita

**1/4 < 2/5**

**9. Rafiq exercised for 3/6 of an hour, while Rohit exercised for 3/4 of an hour. Who exercised for a longer time?**

**Ans : **

Rafiq exercised for 3/6of an hour.

Rohit exercised for 3/4 of an hour.

Comparing 3/6 and 3/4

3/6=1/2

3/4=3/4\

We find a common denominator (12):

1/2=6/12

3/4=9/12

Since 6/12<9/12

3/4>3/6

Therefore, Rohit exercised for a longer time.

**10. In a class A of 25 students, 20 passed in first class, in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?**

**Ans : **

Class | Total Students | Students Getting First Class | Fraction of Students Getting First Class |

A | 25 | 20 | 4/5 |

B | 30 | 24 | 4/5 |

NCERT solutions for class 6 Maths chapter 7

**Exercise 7.5**

**1. Write these fractions appropriately as additions or subtractions.**

**Ans : **

**(a) 1/5 + 3/5 = 4/5** (shaded area is the sum of two colored sections)

**(b) 5/5 – 3/5 = 2/5** (shaded area is the difference between the whole pie and a colored section)

**(c) 2/6 + 3/6 = 5/6** (shaded area is the sum of two colored sections)

**2. Solve:**

**Ans : **

**3. Shubham painted 2/3 of the wall space in his room. His sister Madhavi helped and painted 1/3 of the wall space. How much did they paint together?**

**Ans : **

They painted together ⅔ + ⅓ of the wall space.

To find the total area they painted, we can add the fractions:

⅔ + ⅓ = (2 x 1) / (3 x 1) + (1 x 1) / (3 x 1) = 2/3 + 1/3

Since they have the same denominator (3), we simply add the numerators:

2/3 + 1/3 = 3/3

However, 3/3 represents the entire wall space. Since painting the entire wall once is enough, we can simplify the answer:

3/3 = 1

Therefore, Shubham and Madhavi painted 1 (or the whole) wall space together.

**4. Fill in the missing fractions.**

**Ans : **

**(a) 7/10 – □ = 3/10**

- We can start with the whole: 1 (representing the whole)
- Subtract the fraction on the right: 1 – 3/10 = 7/10
- This remaining fraction (7/10) is what needs to be divided into two parts.
- So, the missing fraction on the left is the difference: 7/10 – □ = 3/10.
- Therefore, the missing fraction in (a) is
**4/10**.

**(b) □ – 3/21 = 5/21**

- Similar to (a), we can reason starting with the whole: 1.
- Subtract the fraction on the right: 1 – 3/21 = 18/21
- The missing fraction needs to be subtracted to result in 5/21.
- So, the missing fraction in (b) is
**18/21 – 3/21 = 15/21**.

**(c) □ – 3/6 = 3/6**

- In this case, both fractions have the same denominator (6).
- We want to find a number that, when we subtract 3/6 from it, leaves 3/6 remaining.
- The answer is simply
**6/6**(the whole). - Subtracting 3/6 from 6/6 gives us 3/6 as expected.

**(d) □ + 5/27 = 12/27**

- Here, we need to find a number that, when added to 5/27, gives us 12/27.
- The missing fraction is
**7/27**. - Adding 5/27 and 7/27 gives us 12/27.

**5. Javed was given 5/7 of a basket of oranges. What fraction of oranges was left in the basket?**

Fraction of basket of oranges = 5/7

The fraction of the basket as a whole is 1.

∴ The fraction of the basket of oranges left is:

1−5/7=7/7−5/7=2/7

Thus, the required fraction is 2/7

NCERT solutions for class 6 Maths chapter 7

**Exercise 7.6 **

**1. Solve**

**Ans : **

**2. Sarita bought 3/5 metre of ribbon and Lalita 3/4 metre of ribbon. What is the total length of the ribbon they bought?**

**Ans : **

Length of ribbon bought by Sarita = 2/5

Length of ribbon bought by Lalita = 3/4

∴ Total length of ribbon bought by Sarita and Lalita:

To add these fractions, find a common denominator:

2/5=8/20

3/4 = 15/20

Now, add the fractions:

8/20+15/20=8+15/20=23/20

Hence, the required length is 23/20.

**3. Naina was given 1 ½ piece of cake and Najma was given 1 ⅓ piece of cake. Find the total amount of cake was given to both of them.**

**Ans : **

Cake given to Naina = 1 ½

Piece of cake given to Najma = 1 ⅓

Piece of cake given to Naina and Najma

Hence the total amount of piece given to both = 2 ⅚

**4. Fill in the boxes:**

**Ans :**

**a ) **

**b )**

**c )**

**5. Complete the addition-subtraction box.**

**Ans : **

**The box may be completed as follows:**

**6. A piece of wire ⅞ metre long broke into two pieces. One piece was ¼ metre long. How long is the other piece?**

**Ans : **

**Total Length:** The wire was originally ⅞ meter long.

**First Piece:** One piece is ¼ meter long.

Remaining piece = Total length – Length of first piece

Remaining piece = ⅞ meter – ¼ meter

Hence, the length of the other piece = ⅝ metre.

**7. Nandini’s house is 9/10 km from her school. She walked some distance and then took a bus for ½ km to reach the school. How far did she walk?**

**Ans : **

**Total Distance:** Nandini’s house is 9/10 km from school.

**Bus Distance:** She travelled ½ km by bus.

**Walking Distance:** We want to find the distance she walked.

Walking distance = Total distance – Bus distance Walking distance

= 9/10 km – ½ km

The distance travelled by her on foot = ⅖ km.

**8. Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is ⅚ th full and Samuel’s shelf is ⅖th full. Whose bookshelf is more full? By what fraction?**

**Ans : **

**Compare Fractions:** We need to compare ⅚ (Asha’s) and ⅖ (Samuel’s)

Asha: ⅚ * (5/5) = 25/30

Samuel: ⅖ * (6/6) = 12/30

25/30 (Asha) > 12/30 (Samuel)

Difference = Asha – Samuel = 25/30 – 12/30 = 13/30

Asha’s bookshelf is more full by a fraction of 13/30.

**9. Jaidev takes 2 ⅕ minutes to walk across the school ground. Rahul takes 7/4 minutes to do the same. Who takes less time and by what fraction?**

**Ans : **

**Times:** Jaidev takes 21/5 minutes and Rahul takes 7/4 minutes.

Comparing 2 ⅕ minutes and 7/4 minutes

So, the time take to cover the same distance by Rahul is less than that of Jaidev.

Hence, Rahul takes 9/20 minutes less to across the school ground.

NCERT solutions for class 6 Maths chapter 7

**FAQs**

**What topics are covered in NCERT solutions for Class 6 Maths Chapter 7?**

The NCERT solutions for Class 6 Maths Chapter 7 cover various aspects of fractions. This includes understanding fractions, types of fractions, comparison of fractions, and operations on fractions.

**How can I understand fractions better with NCERT solutions for Class 6 Maths Chapter 7?**

The NCERT solutions for Class 6 Maths Chapter 7 provide detailed explanations and step-by-step solutions to help you grasp the concepts of fractions. Practice problems and examples are included to enhance your understanding.

**Are there any exercises specifically focused on fraction class 6 in NCERT solutions for Chapter 7?**

Yes, Chapter 7 of the Class 6 Maths NCERT book includes several exercises focused on fractions. These exercises cover different types of fractions and operations on them to reinforce learning.

**Where can I find NCERT solutions for Class 6 Maths Chapter 7?**

NCERT solutions for Class 6 Maths Chapter 7 can be found in various educational books and online platforms. They are designed to help students understand the concepts of fractions in detail.

**How do NCERT solutions for Class 6 Maths Chapter 7 help in exam preparation?**

NCERT solutions for Class 6 Maths Chapter 7 provide thorough practice and clear explanations of fractions, which are crucial for exams. Solving these problems can improve your problem-solving skills and help you perform better in exams.

**What are the key features of fractions covered in NCERT solutions for Class 6 Maths Chapter 7?**

The key features of fractions covered include types of fractions (proper, improper, and mixed), equivalent fractions, comparison of fractions, and arithmetic operations involving fractions.

**Can I use NCERT solutions for Class 6 Maths Chapter 7 to learn fractions independently?**

Yes, the solutions are designed to be self-explanatory, allowing students to learn and understand fractions independently. The step-by-step approach makes it easier to follow and grasp the concepts.