NCERT Solutions Class 6th Maths Chapter – 7 Fractions All Examples

NCERT Solutions Class 6th Maths Chapter – 7 Fractions All Examples
Last Doubt

NCERT Solutions Class 6th Maths Chapter – 7 Fractions

TextbookNCERT
Class6th
SubjectMathematics
Example7th
Example NameFractions
CategoryClass 6th Mathematics
MediumEnglish
SourceLast Doubt

NCERT Solutions Class 6th Maths Chapter – 7 Fractions All Examples help the students who aspire to obtain a good academic score in exams. Experts at Last Doubt designed these solutions to boost the confidence of students by assisting them in understanding the concepts covered in this chapter. Ncert Solutions for Class 6 contains the methods to solve problems present in the book quickly and easily. These materials are prepared based on Class 6th NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration.

NCERT Solutions Class 6th Maths Chapter – 7 Fractions

Chapter – 7

Fractions

Example

Example 1. Express the following as mixed fractions-

(a) 17/4
(b) 11/3
(c) 27/5
(d) 7/3

Solution –

(a) 17/4

NCERT Solutions Class 6th Maths Chapter – 6 Integers All Examples

i.e. 4 whole and 1/4 more, or 4 1/4

(b) 11/3

NCERT Solutions Class 6th Maths Chapter – 6 Integers All Examples

i.e. 3 whole and 2/3 more, or 3 2/3

[Alternatively, 11/3 = 9 + 2/3 = 9/3 + 2/3 = 3 + 2/3 = 3 2/3]

Try (c) and (d) using both the methods for yourself.
Thus, we can express an improper fraction as a mixed fraction by dividing the numerator by denominator to obtain the quotient and the remainder. Then the mixed fraction will be written as Quotient Remainder/Divisor.

Example 2. Express the following mixed fractions as improper fractions-

(a) 2 3/4
(b) 7 1/9
(c) 5 3/7

Solution –

(a) 2 3/4 = 2 + 3/4 = 2 × 4/4 + 3/4 = 11/4

(b) 7 1/9 = (7 × 9) + 1/9 = 64/9

(c) 5 3/7 = (5 × 7) + 3/7 = 38/7

Thus, we can express a mixed fraction as an improper fraction as
(Whole × Denominator) + Numerator/Denominator.

Example 3. Find the equivalent fraction of 2/5 with numerator 6.

Solution –

We know 2 × 3 = 6. This means we need to multiply both the numerator and the denominator by 3 to get the equivalent fraction.

Hence, 2/5 = 2 × 3/5 × 3 = 6/15 ; 6/15 is the required equivalent fraction.

Example 4. Find the equivalent fraction of 15/35 with denominator 7.

Solution –

We have 15/35 = ▢/7

We observe the denominator and find 35 ÷ 5 = 7. We, therefore, divide both the numerator and the denominator of 15/35 by 5.

Thus, 15/35 = 15 ÷ 5/35 ÷ 5 = 3/7.

Example 5. Find the equivalent fraction of 2/9 with denominator 63.

Solution –

We have 2/9 = ▢/63
For this, we should have, 9 × ▢ = 2 × 63.
But 63 = 7 × 9, so 9 × ▢ = 2 × 7 × 9 = 14 × 9 = 9 × 14
or 9 × ▢ = 9 × 14
By comparison, ▢ = 14.
Therefore, 2/9 = 14/63.

Example 6. Compare 4/5 and 5/6.

Solution –

The fractions are unlike fractions. Their numerators are different too. Let us write their equivalent fractions.

4/5 = 8/10 = 12/15 = 16/20 = 20/25 = 24/30 = 28/35 = ………..
and 5/6 = 10/12 = 15/18 = 20/24 = 25/30 = 30/36 = ………..

The equivalent fractions with the same denominator are-
4/5 = 24/30 and 5/6 = 25/30
Since, 25/30 > 24/30 so, 5/6 > 4/5

Note that the common denominator of the equivalent fractions is 30 which is 5 × 6. It is a common multiple of both 5 and 6.

So, when we compare two unlike fractions, we first get their equivalent fractions with a denominator which is a common multiple of the denominators of both the fractions.

Example 7. Compare 5/6 and 13/15.

Solution –

The fractions are unlike. We should first get their equivalent fractions with a denominator which is a common multiple of 6 and 15.

Now, 5 × 5/6 × 5 = 25/30, 13 × 2/15 × 2 = 26/30
Since 26/30 > 25/30 we have 13/15 > 5/6.

Example 8. Subtract 3/4 from 5/6.

Solution –

We need to find equivalent fractions of 3/4 and 5/6, which have the same denominator. This denominator is given by the LCM of 4 and 6. The required LCM is 12.

Therefore, 5/6 – 3/4 = 5 × 2/6 × 2 – 3 × 3/4 × 3
= 10/12 – 9/12 = 1/12.

Example 9. Add 2/5 to 1/3.

Solution –

The LCM of 5 and 3 is 15.
Therefore, 2/5 + 1/3 = 2 × 3/5 × 3 + 1 × 5/3 × 5
= 6/15 + 5/15 = 11/15.

Example 10. Simplify 3/5 = 7/20

Solution –

The LCM of 5 and 20 is 20.
Therefore, 3/5 – 7/20 = 3 × 4/5 × 4 – 7/20 = 12/20 – 7/20
= 12 – 7/20 = 5/20 = 1/4.

Example 11. Add 2 ⅘  and 3 ⅚

Solution – 

2 ⅘ + 3 ⅚
= 2 + 4/5 + 3 + 5/6 = 5 + 4/5 + 5/6

Now 4/5 + 5/6 = 4 × 6/5 × 6 + 5 × 5/6 × 5
(Since LCM of 5 and 6 = 30)

24/30 + 25/30 = 49/30 = 30 + 19/30 = 1 + 19/30
Thus, 5 + 4/5 + 5/6 = 5 + 1 + 19/30 = 6 + 19/30 = 6 19⁄30

And, therefore, 2 ⅘ + 3 ⅚ = 6 19⁄30. 

Example 12. Find 4 ⅖ – 2 ⅕

Solution –

The whole numbers 4 and 2 and the fractional numbers
2/5 and 1/5 can be subtracted separately.
(Note that 4 > 2 and 2/5 > 1/5)
So, 4 ⅖ – 2 1⅕ = (4 – 2) + (2/5 – 1/5) = 2 + 1/5 = 2 ⅕.

Example 13. Simplify- 8 1⁄4 = 2 5⁄6

Solution –

Here 8 > 2 but 1/4 < 5/6. We proceed as follows-
8 1⁄4 = (8 × 4) + 1/4 = 33/4 and 2 5⁄6 = 2 × 6 + 5/6 = 17/6

Now, 33/4 – 17/6 = 33 × 3/12 – 17 × 2/12 (Since LCM of 4 and 6 = 12)
= 99 – 34/12 = 65/12 = 5 5⁄12.

NCERT Solution Class 6th Maths All Chapters With Answer

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