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

(b) – 10

(c) + 8

(d) – 1

(e) – 6

**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.**

**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.**

**(c) Which is the coolest place?**

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

**Solutions:**

**(a)**

Siachin – 10^{0} C

Shimla – 2^{0} C

Ahmedabad + 30^{0} C

Delhi + 20^{0} C

Srinagar – 5^{0} C

(b)

(c) Siachin is the coolest place

(d) Ahmedabad and Delhi are the places where the temperatures are above 10^{0}C

**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)**

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

(b)

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

(c)

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

(d)

-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)**

Hence, 8

(b)** **

Hence, 0

(c)

Hence, -4

**(**d)

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)**

Hence, 3

(b)

Hence, -6

(c)

Hence, -8

(d)

Hence, 5

(e)

Hence, -6

(f)

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:**

**(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:

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.**

**(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:**

**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 =

= 3 / 2

Fraction of cake Najma got =

= 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

=

**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.**

**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

(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

**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 =

= 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