Rbse Solutions for Class 6 maths Chapter 6: Integers

Exercise 6.1 page no: 120 ncert Solutions for Class 6 maths Chapter 6: Integers

1. Write opposites of the following:

(a) Increase in weight

(b) 30 km north

(c) 80 m east

(d) Loss of Rs 700

(e) 100 m above sea level

Solutions:

(a) The opposite of increase in weight is decrease in weight

(b) The opposite of 30 km north is 30 km south

(c) The opposite of 80 m east is 80 m west

(d) The opposite of loss of Rs 700 is gain of Rs 700

(e) The opposite of 100 m above sea level is 100 m below sea level

2. Represent the following numbers as integers with appropriate signs.

(a) An aeroplane is flying at a height two thousand metre above the ground.

(b) A submarine is moving at a depth, eight hundred metre below the sea level.ncert Solutions for Class 6 maths Chapter 6: Integers

(c) A deposit of rupees two hundred.

(d) Withdrawal of rupees seven hundred.

Solutions:

(a) + 2000 m

(b) – 800 m

(c) + Rs 200

(d) – Rs 700

3. Represent the following numbers on a number line:

(a) + 5

(b) – 10

(c) + 8

(d) – 1

(e) – 6

Solutions:

(a) + 5

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 1

(b) – 10

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 2

(c) + 8

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 3

(d) – 1

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 4

(e) – 6

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 5

4. Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points:ncert Solutions for Class 6 maths Chapter 6: Integers

(a) If point D is + 8, then which point is – 8?

(b) Is point G a negative integer or a positive integer?

(c) Write integers for points B and E.

(d) Which point marked on this number line has the least value?

(e) Arrange all the points in decreasing order of value.

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 6

Solutions:

(a) If point D is +8, then point F is -8

(b) Point G is a negative integer

(c) Point B is 4 and point E is – 10

(d) The least value on this number line is point E as it represents – 10

(e) The points in decreasing order of value are D, C, B, A, O, H, G, F, E

5. Following is the list of temperatures of five places in India on a particular day of the year.

Place Temperature

Siachin 10°C below 0°C ……………..

Shimla 2°C below 0°C ……………..

Ahmedabad 30°C above 0°C ……………..

Delhi 20°C above 0°C ……………..

Srinagar 5°C below 0°C ……………..

(a) Write the temperatures of these places in the form of integers in the blank column.

(b) Following is the number line representing the temperature in degree Celsius.

Plot the name of the city against its temperature.

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 7

(c) Which is the coolest place?

(d) Write the names of the places where temperatures are above 10°C.

Solutions:

(a)

Siachin – 100 C

Shimla – 20 C

Ahmedabad + 300 C

Delhi + 200 C

Srinagar – 50 C

(b)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 8

(c) Siachin is the coolest place

(d) Ahmedabad and Delhi are the places where the temperatures are above 100C

6. In each of the following pairs, which number is to the right of the other on the number line?

(a) 2, 9

(b) – 3, – 8

(c) 0, – 1

(d) – 11, 10

(e) – 6, 6

(f) 1, – 100

Solutions:

(a) 9 lies to the right on the number line (9 > 2)

(b) – 3 lies to the right on the number line (- 3 > – 8)

(c) 0 lies to the right on the number line (0 > -1)

(d) 10 lies to the right on the number line (10 > -11)

(e) 6 lies to the right on the number line (6 > -6)

(f) 1 lies to the right on the number line (1 > -100)

7. Write all the integers between the given pairs (write them in the increasing order.)

(a) 0 and – 7

(b) – 4 and 4

(c) – 8 and – 15

(d) – 30 and – 23

Solutions:

(a) -6, -5, -4, -3, -2, -1 are the integers between 0 and -7

(b) -3, -2, -1, 0, 1, 2, 3 are the integers between -4 and 4

(c) -14, -13, -12, -11, -10, -9 are the integers between -8 and -15

(d) -29, -28, -27, -26, -25, -24 are the integers between -30 and -23

8. (a) Write four negative integers greater than – 20.

(b) Write four integers less than – 10.

Solutions:

(a) -19, -18, -17, -16 are the integers greater then -20

(b) -11, -12, -13, -14 are the integers less than -10

9. For the following statements, write True (T) or False (F). If the statement is false, correct the statement.

(a) – 8 is to the right of – 10 on a number line.

(b) – 100 is to the right of – 50 on a number line.

(c) Smallest negative integer is – 1.

(d) – 26 is greater than – 25.

Solutions:

(a) True as (-8 > -10)

(b) False. (-50 is greater than -100). Hence, -100 lies to the left of -50 on the number line

(c) False. -1 is the greater negative integer.

(d) False. -26 is smaller than -25

10. Draw a number line and answer the following:

(a) Which number will we reach if we move 4 numbers to the right of – 2.

(b) Which number will we reach if we move 5 numbers to the left of 1.

(c) If we are at – 8 on the number line, in which direction should we move to reach – 13?

(d) If we are at – 6 on the number line, in which direction should we move to reach – 1?

Solutions:

(a)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 9

If we move 4 numbers to the right of -2, we will reach at 2

(b)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 10

If we move 5 numbers to the left of 1, we will reach at -4

(c)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 11

-13 lies to the left of -8 on the number line. Hence, we should move towards left direction

(d)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.1 - 12

-1 lies to the right of -6 on the number line. So, we should move towards right direction.


Exercise 6.2 Page no: 128

1. Using the number line write the integer which is:

(a) 3 more than 5

(b) 5 more than –5

(c) 6 less than 2

(d) 3 less than –2

Solutions:

(a)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 1

Hence, 8

(b) 

Hence, 0

(c)
ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 3

Hence, -4

(d)
ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 4

Hence, -5

2. Use number line and add the following integers:

(a) 9 + (–6)

(b) 5 + (–11)

(c) (–1) + (–7)

(d) (–5) + 10

(e) (–1) + (–2) + (–3)

(f) (–2) + 8 + (–4)

Solutions:

(a)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 5

Hence, 3

(b)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 6

Hence, -6

(c)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 7

Hence, -8

(d)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 8

Hence, 5

(e)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 9

Hence, -6

(f)

ncert Solutions for Class 6 Maths Chapter 6 Exercise 6.2 - 10

Hence, 2

3. Add without using number line:

(a) 11 + (–7)

(b) (–13) + (+18)

(c) (–10) + (+19)

(d) (–250) + (+150)

(e) (–380) + (–270)

(f) (–217) + (–100)

Solutions:

(a) 11 + (-7) = 4

(b) (-13) + (+18) = 5

(c) (-10) + (+19) = 9

(d) (-250) + (+150) = -100

(e) (-380) + (-270) = -650

(f) (-217) + (-100) = -317

4. Find the sum of:

(a) 137 and – 354

(b) – 52 and 52

(c) – 312, 39 and 192

(d) – 50, – 200 and 300

Solutions:

(a) 137 and -354

(137) + (-354) = (137) + (-137) + (-217)

= 0 + (-217) [(137) + (-137) = 0]

= (-217)

= -217

(b) -52 and 52

(-52) + (+52) = 0 [(-a) + (+a) = 0]

(c) -312, 39 and 192

(-312) + (+39) + (+192) = (-231) + (-81) + (+39) + (+192)

= (-231) + (-81) + (+231)

= (-231) + (+231) + (-81)

= 0 + (-81) [(-a) + (+a) = 0]

= -81

(d) -50, -200 and 300

(-50) + (-200) + (+300) = (-50) + (-200) + (+200) + (+100)

= (-50) + 0 + (+100) [(-a) + (+a) = 0]

= (-50) + (+100)

= (-50) + (+50) + (+50)

= 0 + (+50) [(-a) + (+a) = 0]

= 50

5. Find the sum:

(a) (–7) + (–9) + 4 + 16

(b) (37) + (–2) + (–65) + (–8)

Solutions:

(a) (-7) + (-9) + 4 + 16

= (-7) + (-9) + 4 + (+7) + (+9)

= (-7) + (+7) + (-9) + (+9) + 4

= 0 + 0 + 4 [(-a) + (+a) = 0]

= 4

(b) (37) + (-2) + (-65) + (-8)

= (+37) + (-75)

= (+37) + (-37) + (-38)

= 0 + (-38) [(-a) + (+a) = 0]

= -38


Exercise 6.3 page no: 131

1. Find

(a) 35 – (20)

(b) 72 – (90)

(c) (-15) – (-18)

(d) (-20) – (13)

(e) 23 – (-12)

(f) (-32) – (-40)

Solutions:

(a) 35 – (20)

35 – 20

15

(b) 72 – (90)

= 72 – 90

= -18

(c) (-15) – (-18)

= -15 + 18

= 3

(d) (-20) – (13)

= -20 – 13

= -33

(e) 23 – (-12)

= 23 + 12

= 35

(f) (-32) – (-40)

= -32 + 40

= 8

2. Fill in the blanks with >, < or = sign.

(a) (–3) + (–6) ______ (–3) – (–6)

(b) (–21) – (–10) _____ (–31) + (–11)

(c) 45 – (– 11) ______ 57 + (– 4)

(d) (–25) – (–42) _____ (– 42) – (–25)

Solutions:

(a) (-3) + (-6) = -9

(-3) – (-6) = -3 + 6 = 3

-9 < 3

Therefore (-3) + (-6) < (-3) – (-6)

(b) -21 – (-10) = -21 + 10 = -11

-31 + (-11) = -42

-11 > -42

Therefore (-21) – (-10) > (-31) + (-11)

(c) 45 – (-11) = 45 + 11 = 56

57 + (-4) = 57 – 4 = 53

56 > 53

Therefore 45 – (-11) > 57 + (-4)

(d) (-25) – (-42) = -25 + 42 = 17

-42 – (-25) = -42 + 25 = -17

17 > -17

Therefore (-25) – (-42) > (-42) – (-25)

3. Fill in the blanks.

(a) (–8) + _____ = 0

(b) 13 + _____ = 0

(c) 12 + (–12) = ____

(d) (–4) + ____ = –12

(e) ____ – 15 = – 10

Solutions:

(a) (-8) + 8 = 0

(b) 13 + (-13) = 0

(c) 12 + (-12) = 0

(d) (-4) + (-8) = -12

(e) 5 – 15 = -10

4. Find

(a) (–7) – 8 – (–25)

(b) (–13) + 32 – 8 – 1

(c) (–7) + (–8) + (–90)

(d) 50 – (–40) – (–2)

Solutions:

(a) (-7) – 8 – (-25) = -7 -8 + 25

= -15 + 25

= 10

(b) (-13) + 32 – 8 – 1 = -13 + 32 – 8 – 1

= 32 – 22

= 10

(c) (-7) + (-8) + (-90) = -7 – 8 – 90

= – 105

(d) 50 – (-40) – (-2) = 50 + 40 + 2

= 92

Exercise 7.4 page no: 152

1. Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘’ between the fractions:

ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.4 - 1

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

5 / 6 ☐ 2 / 6, 3 / 6 ☐ 0, 1 / 6 ☐ 6 / 6, 8 / 6 ☐ 5 / 6

Solutions:

(a) First circle shows 3 shaded parts out of 8 equal parts. Hence, the fraction is 3 / 8

Second circle shows 6 shaded parts out of 8 equal parts. Hence, the fraction is 6 / 8

Third circle shows 4 shaded parts out of 8 equal parts. Hence, the fraction is 4 / 8

Fourth circle shows 1 shaded parts out of 8 equal parts. Hence, the fraction is 1 / 8

The arranged fractions are:

1 / 8 < 3 / 8 < 4 / 8 < 6 / 8

(b) First square shows 8 shaded parts out of 9 equal parts. Hence, the fraction is 8 / 9

Second square shows 4 shaded parts out of 9 equal parts. Hence, the fraction is 4 / 9

Third square shows 3 shaded parts out of 9 equal parts. Hence, the fraction is 3 / 9

Fourth square shows 6 shaded parts out of 9 equal parts. Hence, the fraction is 6 / 9

The arranged fractions are:

3 / 9 < 4 / 9 < 6 / 9 < 8 / 9

(c) Each unit length should be divided into 6 equal parts to represent the fractions 2 / 6, 4 / 6, 8 / 6 and

6 / 6 on number line. These fractions can be represented as follows:

ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.4 - 2

5 / 6 > 2 / 6

3 / 6 > 0

1 / 6 < 6 / 6

8 / 6 > 5 / 6

2. Compare the fractions and put an appropriate sign.

(a) 3 / 6 ☐ 5 / 6

(b) 1 / 7 ☐ 1 / 4

(c) 4 / 5 ☐ 5 / 5

(d) 3 / 5 ☐ 3 / 7

Solutions:

(a) Here both fractions have same denominators. So, the fraction with greater numerator is the highest factor

∴ 3 / 6 < 5 / 6

(b) Multiply by 4

1 / 7 = (1 × 4) / (7 × 4)

= 4 / 28

Multiply by 7

1 / 4 = (1 × 7) / (4 × 7)

= 7 / 28

Here 4 < 7

∴ 1 / 7 < 1 / 4

(c) Here both fractions have same denominators. So, the fraction with greater numerator is the highest factor

∴ 4 / 5 < 5 / 5

(d) Here both numerators are same. So, the fraction having less denominator will be the highest factor

∴ 3 / 7 < 3 / 5

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

Solutions:

(a) 5 / 8 < 6 / 8

Here, the denominators are same. So, the fraction having greater numerator is the highest factor

(ii) 5 / 8 > 2 / 8

Here, the denominators are same. So, the fraction having greater numerator is the highest factor

(iii) 6 / 13 > 6 / 18

Here, the numerators are same. So, the fraction having lesser denominator will be the highest factor

(iv) 5 / 25 > 3 / 25

Here, the denominators are same. So, the fraction having greater numerator is the highest factor

(v) 9 / 50 < 9 / 45

Here, the numerators are same. So, the fraction having lesser denominator will be the highest factor

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

ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.4 - 3

(a) 1 / 6 ☐ 1 / 3

(b) 3 / 4 ☐ 2 / 6

(c) 2 / 3 ☐ 2 / 4

(d) 6 / 6 ☐ 3 / 3

(e) 5 / 6 ☐ 5 / 5

Solutions:

(a) Here, the numerators are same. So, the fraction having lesser denominator is the greater

∴ 1 / 6 < 1 / 3

(b) 3 / 4 = (3 × 3) / (4 × 3)

= 9 / 12

2 / 6 = (2 × 2) / (6 × 2)

= 4 / 12

Between 4 / 12, 9 / 12

Both fractions have same denominators. So, the fraction having greater numerator will be the greater

∴ 9 / 12 > 4 / 12

3 / 4 > 2 / 6

(c) Here, the numerators are same. So, the fraction having lesser denominator is the greater

∴ 2 / 3 > 2 / 4

(d) We get 6 / 6 = 1 and 3 / 3 = 1

So, 6 / 6 = 3 / 3

(e) Here, the numerators are same. So, the fraction having lesser denominator is the greater

∴ 5 / 6 < 5 / 5

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

(a) 1 / 2 ☐ 1 / 5

(b) 2 / 4 ☐ 3 / 6

(c) 3 / 5 ☐ 2 / 3

(d) 3 / 4 ☐ 2 / 8

(e) 3 / 5 ☐ 6 / 5

(f) 7 / 9 ☐ 3 / 9

(g) 1 / 4 ☐ 2 / 8

(h) 6 / 10 ☐ 4 / 5

(i) 3 / 4 ☐ 7 / 8

(j) 6 / 10 ☐ 3 / 5

(k) 5 / 7 ☐ 15 / 21

Solutions:

(a) Here, the numerators are same. So, the fraction having lesser denominator is the greater

∴ 1 / 2 > 1 / 5

(b) 2 / 4 = 1 / 2 and 3 / 6 = 1 / 2

∴ 2 / 4 = 3 / 6

(c) 3 / 5 = (3 × 3) / (5 × 3)

= 9 / 15

2 / 3 = (2 × 5) / 3 × 5)

= 10 / 15

Here, between 9 / 15 and 10 / 15 both have same denominators. Hence, the fraction having greater numerator will be the greater.

∴ 3 / 5 < 2 / 3

(d) Here, 2 / 8 = 1 / 4

As, 3 / 4 and 1 / 4 have same denominators. Hence, the fraction having greater numerator will be the greater

∴ 3 / 4 > 2 / 8

(e) Here, the denominators are same. So, the fraction having greater numerator will be the greater

∴ 3 / 5 < 6 / 5

(f) Here, the denominators are same. So, the fraction having greater numerator will be the greater

∴ 7 / 9 > 3 / 9

(g) We know 2 / 8 = 1 / 4

Hence, 1 / 4 = 2 / 8

(h) 6 / 10 = (3 × 2) / (5 × 2)

= 3 / 5

Between 3 / 5 and 4 / 5

Both have same denominators. So, the fraction having greater numerator will be greater

∴ 6 / 10 < 4 / 5

(i) 3 / 4 = (3 × 2) / (4 × 2)

= 6 / 8

Between 6 / 8 and 7 / 8

Both have same denominators. So, the fraction having greater numerator will be greater

∴ 3 / 4 < 7 / 8

(j) 6 / 10 = (3 × 2) / (5 × 2)

= 3 / 5

∴ 6 / 10 = 3 / 5

(k) 5 / 7 = (5 × 3) / (7 × 3)

= 15 / 21

∴ 5 / 7 = 15 / 21

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.

(a) 2 / 12 (b) 3 / 15 (c) 8 / 50 (d) 16 / 100 (e) 10 / 60 (f) 15 / 75

(g) 12 / 60 (h) 16 / 96 (i) 12 / 75 (j) 12 / 72 (k) 3 / 18 (l) 4 / 25

Solutions:

(a) 2 / 12 = (1 × 2) / (6 × 2)

= 1 / 6

(b) 3 / 15 = (1 × 3) / (5 × 3)

= 1 / 5

(c) 8 / 50 = (4 × 2) / (25 × 2)

= 4 / 25

(d) 16 / 100 = (4 × 4) / (25 × 4)

= 4 / 25

(e) 10 / 60 = (1 × 10) / (6 × 10)

= 1 / 6

(f) 15 / 75 = (1 × 15) / (5 × 15)

= 1 / 5

(g) 12 / 60 = (1 × 12) / (5 × 12)

= 1 / 5

(h) 16 / 96

= (1 × 16) / (6 × 16)

= 1 / 6

(i) 12 / 75 = (4 × 3) / (25 × 3)

= 4 / 25

(j) 12 / 72 = (1 × 12) / 6 × 12)

= 1 / 6

(k) 3 / 18 = (1 × 3) / (6 × 3)

= 1 / 6

(l) 4 / 25

Totally there are 3 groups of equivalent fractions.

1 / 6 = (a), (e), (h), (j), (k)

1 / 5 = (b), (f), (g)

4 / 25 = (c), (d), (i), (l)

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

(a) Is 5 / 9 equal to 4 / 5

(b) Is 9 / 16 equal to 5 / 9

(c) Is 4 /5 equal to 16 / 20

(d) Is 1 / 15 equal to 4 / 30

Solutions:

(a) 5 / 9, 4 / 5

Convert these fractions into like fractions

5 / 9 = (5 / 9) × (5 / 5)

= 25 / 45

4 / 5 = (4 / 5) × (9 / 9)

= 36 / 45

∴ 25 / 45 ≠ 36 / 45

Hence, 5 / 9 is not equal to 4 / 5

(b) 9 / 16, 5 / 9

Convert into like fractions

9 / 16 = (9 / 16) × (9 / 9)

= 81 / 144

5 / 9 = (5 / 9) × (16 / 16)

= 80 / 144

∴ 81 / 144 ≠ 80 / 144

Hence, 9 / 16 is not equal to 5 / 9

(c) 4 / 5, 16 / 20

16 / 20 = (4 × 4) / (5 × 4)

= 4 / 5

∴ 4 / 5 = 16 / 20

Hence, 4 / 5 is equal to 16 / 20

(d) 1 / 15, 4 / 30

4 / 30 = (2 × 2) / (15 × 2)

= 2 / 15

∴ 1 / 15 ≠ 4 / 30

Hence, 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?

Solutions:

Total number of pages a book has = 100 pages

Lalita read = 2 / 5 × 100 = 40 pages

Ila read = 25 pages

∴ Ila read less than Lalita.

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

Solutions:

Rafiq exercised = 3 / 6 of an hour

Rohit exercised = 3 / 4 of a hour

3 / 6, 3 / 4

Convert these into like fractions

3 / 6 = (3 × 2) / (6 × 2)

= 6 / 12

3 / 4 = (3 × 3) / (4 × 3)

= 9 / 12

Clearly, 9 / 12 > 6 / 12

∴ 3 / 4 > 3 / 6

Therefore Rohit exercised for a longer time than Rafiq.

10. In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting with 60% or more marks?

Solutions:

Total number of students in Class A = 25

Students passed in first class in Class A = 20

Hence, fraction = 20 / 25

= 4 / 5

Total number of students in Class B = 30

Students passed in first class in Class B = 24

Hence, fraction = 24 / 30

= 4 / 5

∴ An equal fraction of students passed in first class in both the classes


Exercise 7.5 page no: 157

1. Write these fractions appropriately as additions or subtractions:

ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.5 - 1

Solutions:

(a) Total number of parts each rectangle has = 5

No. of shaded parts in first rectangle = 1 i.e 1 / 5

No. of shaded parts in second rectangle = 2 i.e 2 / 5

No. of shaded parts in third rectangle = 3 i.e 3 / 5

Clearly, fraction represented by third rectangle = Sum of the fractions represented by first and second rectangle

Hence, 1 / 5 + 2 / 5 = 3 / 5

(b) Total number of parts each circle has = 5

We may observe that first, second and third circles represent 5, 3 and 2 shaded parts out of 5 equal parts respectively. Clearly, fraction represented by third circle is the difference between the fractions represented by first and second circles.

Hence, 5 / 5 – 3 / 5 = 2 / 5

(c) Here we may observe that first, second and third rectangles represents 2, 3 and 5 shaded parts out of 6 equal parts respectively. Clearly, fraction represented by third rectangle is the sum of fractions represented by first and second rectangles.

Hence, 2 / 6 + 3 / 6 = 5 / 6

2. Solve:

(a) 1 / 18 + 1 / 18

(b) 8 / 15 + 3 / 15

(c) 7 / 7 – 5 / 7

(d) 1 / 22 + 21 / 22

(e) 12 / 15 – 7 / 15

(f) 5 / 8 + 3 / 8

(g) 1 – 2 / 3 (1 = 3 / 3)

(h) 1 / 4 + 0 / 4

(i) 3 – 12 / 5

Solutions:

(a) 1 / 18 + 1 / 18

= (1 + 1) / 18

= 2 / 18

= 1 / 9

(b) 8 / 15 + 3 / 15

= (8 + 3) / 15

= 11 / 15

(c) 7 / 7 – 5 / 7

= (7 – 5) / 7

= 2 / 7

(d) 1 / 22 + 21 / 22

= (1 + 21) / 22

= 22 / 22

= 1

(e) 12 /15 – 7 / 15

= (12 – 7) / 15

= 5 / 15

= 1 / 3

(f) 5 / 8 + 3 / 8

= (5 + 3) / 8

= 8 / 8

= 1

(g) 1 – 2 / 3

= 3 / 3 – 2 / 3

= (3 – 2) / 3

= 1 / 3

(h) 1 / 4 + 0

= 1/ 4

(i) 3 – 12 / 5

= 15 / 5 – 12/ 5

= (15 – 12) / 5

= 3 / 5

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?

Solutions:

Wall space painted by Shubham in a room = 2 / 3

Wall space painted by Madhavi in a room = 1 / 3

Total space painted by both = (2 / 3 + 1 / 3)

= (2 + 1) / 3

= 3 / 3

= 1

∴ Shubham and Madhavi together painted 1 complete wall in a room.

4. Fill in the missing fractions.

(a) 7 / 10 – ▯ = 3 / 10

(b) ▯ – 3 / 21 = 5 / 21

(c) ▯ – 3 / 6 = 3 / 6

(d) ▯ + 5 / 27 = 12 / 27

Solutions:

(a) Given 7 / 10 – ▯ = 3 / 10

▯ = 7 / 10 – 3 / 10

▯ = (7 – 3) / 10

▯ = 4 / 10

▯ = 2 / 5

(b) Given ▯ – 3 / 21 = 5 / 21

▯ = 5 / 21 + 3 / 21

▯ = (5 + 3) / 21

▯ = 8 / 21

(c) Given ▯ – 3 / 6 = 3 / 6

▯ = 3 / 6 + 3 / 6

▯ = (3 + 3) / 6

▯ = 6 / 6

▯ = 1

(d) Given ▯ + 5 / 27 = 12 / 27

▯ = 12 / 27 – 5 /27

▯ = (12 – 5) / 27

▯ = 7 /27

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

Solutions:

Fraction of oranges given to Javed = 5 / 7

Fraction of oranges left in the basket = 1 – 5 / 7

= 7 / 7 – 5 / 7

= (7 – 5) / 7

= 2 / 7


Exercise 7.6 page no: 160

1. Solve

(a) 2 / 3 + 1 / 7

(b) 3 / 10 + 7 / 15

(c) 4 / 9 + 2 / 7

(d) 5 / 7 + 1 / 3

(e) 2 / 5 + 1 / 6

(f) 4 / 5 + 2 / 3

(g) 3 / 4 – 1 / 3

(h) 5 / 6 – 1 / 3

(i) 2 / 3 + 3 / 4 + 1 / 2

(j) 1/ 2 + 1 / 3 + 1 / 6

(k) 

(l) 

(m) 16 / 5 – 7 / 5

(n) 4 / 3 – 1 / 2

Solutions:

(a) 2 / 3 + 1/ 7

Taking LCM[(2 × 7) + (1 × 3)] / 21

= (14 + 3) / 21

= 17 / 21

(b) 3 / 10 + 7 / 15

Taking LCM 30

= [(3 × 3) + (7 × 2)] / 30

= (9 + 14) / 30

= 23 / 30

(c) 4 / 9 + 2/ 7

Taking LCM 63

= [(4 × 7) + (2 × 9)] / 63

= (28 + 18) / 63

= 46 / 63

(d) 5 / 7 + 1 / 3

Taking LCM 21

= [(5 × 3) + (1 × 7)] / 21

= (15 + 7) / 21

= 22 / 21

(e) 2 / 5 + 1 / 6

Taking LCM 30

= [(2 × 6) + (1 × 5)] / 30

= (12 + 5) / 30

= 17 / 30

(f) 4 / 5 + 2 / 3

Taking LCM 15

= [(4 × 3) + (2 × 5)] / 15

= (12 + 10) / 15

= 22 / 15

(g) 3 / 4 – 1 / 3

Taking LCM 12

= [(3 × 3) – (1 × 4)] / 12

= (9 – 4) / 12

= 5 / 12

(h) 5 / 6 – 1 / 3

Taking LCM 6

= [(5 × 1) – (1 × 2)] / 6

= (5 – 2) / 6

= 3 / 6

= 1 / 2

(i) 2 / 3 + 3 / 4 + 1 / 2

Taking LCM 12

= [(2 × 4) + (3 × 3) + (1 × 6)] / 12

= (8 + 9 + 6) / 12

= 23 / 12

(j) 1 / 2 + 1 / 3 + 1 / 6

Taking LCM 6

= [(1 × 3) + (1 × 2) + (1 × 1)] / 6

= (3 + 2 + 1) / 6

= 6 / 6

= 1

(k) 

= [(3 × 1) + 1] / 3 + [(3 × 3) + 2] / 3

= (3 + 1) / 3 + (9 + 2) / 3

= 4/ 3 + 11 / 3

= (4 + 11) / 3

= 15 / 3

= 5

(l) 

= [(3 × 4) + 2] / 3 + [(3 × 4) + 1] / 4

= 14 / 3 + 13 / 4

= [(14 × 4) + (13 × 3)] / 12

= (56 + 39) / 12

= 95 / 12

(m) 16 / 5 – 7 / 5

= (16 – 7) / 5

= 9 / 5

(n) 4 /3 – 1 / 2

Taking LCM 6

= [(4 × 2) – (1 × 3)] / 6

= (8 – 3) /6

= 5 / 6

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

Solutions:

Ribbon length bought by Sarita = 2 / 5 metre

Ribbon length bought by Lalita = 3 / 4 metre

Total length of the ribbon bought by both of them = 2 / 5 + 3 / 4

Taking LCM 20

= [(2 × 4) + (3 × 5)] / 20

= (8 + 15) / 20

= 23 / 20 metre

∴ Total length of the ribbon bought by both Sarita and Lalita is 23 / 20 metre

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

Solutions:

Fraction of cake Naina got =
ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.6 - 7= 3 / 2

Fraction of cake Najma got =
ncert Solutions for Class 6 Maths Chapter 7 Exercsie 7.6 -8= 4 / 3

Total amount of cake given to both of them = 3 / 2 + 4 / 3

= [(3 × 3) + (4 × 2)] / 6

= (9 + 8) / 6

= 17 / 6

=ncert Solutions for Class 6 Maths Chapter 7 Exercsie 7.6 - 9

4. Fill in the boxes:

(a) ▯ – 5 / 8 = 1 / 4

(b) ▯ – 1 / 5 = 1 / 2

(c) 1 / 2 – ▯ = 1 / 6

Solutions:

(a) ▯ – 5 / 8 = 1 / 4

▯ = 1 / 4 + 5 / 8

▯ = [(1 × 2 + 5)] / 8

▯ = 7 / 8

(b) ▯ – 1 / 5 = 1 / 2

▯ = 1 / 2 + 1 / 5

▯ = [(1 × 5) + (1 × 2)] / 10

▯ = (5 + 2) / 10

▯ = 7 / 10

(c) 1 / 2 – ▯ = 1 / 6

▯ = 1 / 2 – 1 / 6

▯ = [(1 × 3) – (1 × 1)] / 6

▯ = (3 – 1) / 6

▯ = 2 / 6

▯ 1 / 3

5. Complete the addition and subtraction box.

ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.6 - 10

Solutions:

(a) 2 / 3 + 4 / 3

= (2 + 4) / 3

= 6 / 3

= 2

1 / 3 + 2 / 3

= (1 + 2) / 3

= 3 / 3

= 1

2 / 3 – 1 / 3

= (2 – 1) / 3

= 1 / 3

4 / 3 – 2 / 3

= (4 – 2) / 3

= 2 / 3

1 / 3 + 2 / 3

= (1 + 2) / 3

= 3 / 3

= 1

Hence, the complete given box is

ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.6 - 11

(b) 1 / 2 + 1 / 3

= [(1 × 3) + (1 × 2)] / 6

= (3 + 2) / 6

= 5 / 6

1 / 3 + 1 / 4

= [(1 × 4) + (1 × 3)] / 12

= (4 + 3) / 12

= 7 / 12

1 / 2 – 1 / 3

= [(1 × 3) – (1 × 2)] / 6

= (3 – 2) / 6

= 1 / 6

1 / 3 – 1 / 4

= [(1 × 4) – (1 ×3)] / 12

= (4 – 3) / 12

= 1 / 12

1 / 6 + 1 / 12

= [(1 × 2) + 1] / 12

= (2 + 1) / 12

= 3 / 12

= 1 / 4

Hence, the complete given box is

ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.6 - 12

6. A piece of wire 7 / 8 metre long broke into two pieces. One piece was 1 / 4 metre long. How long is the other piece?

Solutions:

Total length of wire = 7 / 8 metre

Length of one piece of wire = 1 / 4 metre

Length of other piece of wire = Length of the original wire and this one piece of wire

= 7 / 8 – 1 / 4

= [(7 × 1) – (1 × 2)] / 8

= (7 – 2) / 8

= 5 / 8

∴ Length of the other piece of wire = 5 / 8 metre

7. Nandini’s house is 9 / 10 km from her school. She walked some distance and then took a bus for 1 / 2 km to reach the school. How far did she walk?

Solutions:

Distance of the school from house = 9 / 10 km

Distance she travelled by bus = 1 / 2 km

Distance walked by Nandini = Total distance of the school – Distance she travelled by bus

= 9 / 10 – 1 / 2

= [(9 × 1) – (1 × 5)] / 10

= (9 – 5) / 10

= 4 / 10

= 2 / 5 km

∴ Distance walked by Nandini is 2 / 5 km

8. Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is 5 / 6 th full and Samuel’s shelf is 2/ 5 th full. Whose bookshelf is more full? By what fraction?

Solutions:

Fraction of Asha’s bookshelf = 5 / 6

Fraction of Samuel’s bookshelf = 2 / 5

Convert these fractions into like fractions

5 / 6 = 5 / 6 × 5 / 5

= (5 × 5) / (6 × 5)

= 25 / 30

2 / 5 = 2 / 5 × 6 / 6

= (2 × 6) / (5 × 6)

= 12 / 30

25 / 30 > 12 / 30

5 / 6 > 2 / 5

∴ Asha’s bookshelf is more full than Samuel’s bookshelf

Difference = 5 / 6 – 2 / 5

= 25 / 30 – 12 / 30

= 13 / 30

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

Solutions:

Time taken by Jaidev to walk across the school ground =
ncert Solutions for Class 6 Maths Chapter 7 Exercise 7.6 - 14= 11 / 5 minutes

Time taken by Rahul to walk across the school ground = 7 / 4 minutes

Convert these fractions into like fractions

11 / 5 = 11 / 5 × 4 / 4

= (11 × 4) / (5 × 4)

= 44 / 20

7 / 4 = 7 / 4 × 5 / 5

= (7 × 5) / (4 × 5)

= 35 / 20

Clearly, 44 / 20 > 35 / 20

11 / 5 > 7 / 4

∴ Rahul takes less time than Jaidev to walk across the school ground

Difference = 11 / 5 – 7 / 4

= 44 / 20 – 35 / 20

= 9 / 20

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