NCERT Solutions for Class 7 Maths Chapter 2 - Fractions and Decimals

From the question, it is given that,

Time taken by the Michael to colour the picture is = (7/12)

Time taken by the Vaibhav to colour the picture is = (3/4)

The LCM of 12, 4 = 12

Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.

(7/12) = (7/12) × (1/1) = 7/12

(3/4) = (3/4) × (3/3) = 9/12

Clearly, (7/12) < (9/12)

Hence, (7/12) < (3/4)

Thus, Vaibhav worked for longer time.

So, Vaibhav worked longer time by = (3/4) – (7/12)

= (9/12) – (7/12)

= (9 – 7)/12

= (2/12)

= (1/6) of an hour.

 

Sum along the first row = (4/11) + (9/11) + (2/11) = (15/11)

Sum along the second row = (3/11) + (5/11) + (7/11) = (15/11)

Sum along the third row = (8/11) + (1/11) + (6/11) = (15/11)

Sum along the first column = (4/11) + (3/11) + (8/11) = (15/11)

Sum along the second column = (9/11) + (5/11) + (1/11) = (15/11)

Sum along the third column = (2/11) + (7/11) + (6/11) = (15/11)

Sum along the first diagonal = (4/11) + (5/11) + (6/11) = (15/11)

Sum along the second diagonal = (2/11) + (5/11) + (8/11) = (15/11)

Yes. The sum of the numbers in each row, in each column and along the diagonals is the same, so it is a magic square.

 

From the question, it is given that,

Length = 12 ½ cm = 25/2 cm

Breadth =
cm = 32/3 cm

We know that,

Perimeter of the rectangle = 2 × (length + breadth)

= 2 × [(25/2) + (32/3)]

= 2 × {[(25 × 3) + (32 × 2)]/6}

= 2 × [(75 + 64)/6]

= 2 × [139/6]

= 139/3 cm

Hence, the perimeter of the sheet of paper is
cm

 

From the fig,

AB = (5/2) cm

AE == 18/5 cm

BE == 11/4 cm

ED = 7/6 cm

(i) We know that,

Perimeter of the triangle = Sum of all sides

Then,

Perimeter of triangle ABE = AB + BE + EA

= (5/2) + (11/4) + (18/5)

The LCM of 2, 4, 5 = 20

Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator.

= {[(5/2) × (10/10)] + [(11/4) × (5/5)] + [(18/5) × (4/4)]}

= (50/20) + (55/20) + (72/20)

= (50 + 55 + 72)/20

= 177/20

=cm

(ii) Now, we have to find the perimeter of the rectangle,

We know that,

Perimeter of the rectangle = 2 × (length + breadth)

Then,

Perimeter of rectangle BCDE = 2 × (BE + ED)

= 2 × [(11/4) + (7/6)]

The LCM of 4, 6 = 12

Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator

= 2 × {[(11/4) × (3/3)] + [(7/6) × (2/2)]}

= 2 × [(33/12) + (14/12)]

= 2 × [(33 + 14)/12]

= 2 × (47/12)

= 47/6

=

Finally, we have to find which one is having a greater perimeter.

Perimeter of triangle ABE = (177/20)

Perimeter of rectangle BCDE = (47/6)

The two perimeters are in the form of unlike fractions.

Changing perimeters into like fractions we have,

(177/20) = (177/20) × (3/3) = 531/60

(43/6) = (43/6) × (10/10) = 430/60

Clearly, (531/60) > (430/60)

Hence, (177/20) > (43/6)

∴ Perimeter of Triangle ABE > Perimeter of Rectangle (BCDE)

 

From the question, it is given that,

Picture having a width of == 38/5 cm

Frame having a width of == 73/10 cm

∴ The picture should be trimmed by = [(38/5) – (73/10)]

The LCM of 5, 10 = 10

Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator.

= [(38/5) × (2/2)] – [(73/10) × (1/1)]

= (76/10) – (73/10)

= (76 – 73)/10

= 3/10 cm

Thus, the picture should be trimmed by (3/10) cm

 

From the question, it is given that,

Part of the apple eaten by Ritu is = (3/5)

Part of the apple eaten by Somu is = 1 – Part of the apple eaten by Ritu

= 1 – (3/5)

The LCM of 1, 5 = 5

Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator.

= [(1/1) × (5/5)] – [(3/5) × (1/1)]

= (5/5) – (3/5)

= (5 – 3)/5

= 2/5

∴ Part of the apple eaten by Somu is (2/5)

So, (3/5) > (2/5) hence, Ritu ate larger size of the apple.

Now, the difference between the 32 shares = (3/5) – (2/5)

= (3 – 2)/5

= 1/5

Thus, Ritu’s share is larger than the share of Somu by (1/5)

 

(i) For subtraction of two unlike fractions, first change them to like fractions.

LCM of 1, 5 = 5

Now, let us change each of the given fractions into an equivalent fraction having 5 as the denominator.

= [(2/1) × (5/5)] = (10/5)

= [(3/5) × (1/1)] = (3/5)

Now,

= (10/5) – (3/5)

= [(10 – 3)/5]

= (7/5)

(ii) For addition of two unlike fractions, first change them to like fractions.

LCM of 1, 8 = 8

Now, let us change each of the given fractions into an equivalent fraction having 8 as the denominator.

= [(4/1) × (8/8)] = (32/8)

= [(7/8) × (1/1)] = (7/8)

Now,

= (32/8) + (7/8)

= [(32 + 7)/8]

= (39/8)

=

(iii) For addition of two unlike fractions, first change them to like fractions.

LCM of 5, 7 = 35

Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.

= [(3/5) × (7/7)] = (21/35)

= [(2/7) × (5/5)] = (10/35)

Now,

= (21/35) + (10/35)

= [(21 + 10)/35]

= (31/35)

(iv) For subtraction of two unlike fractions, first change them to like fractions.

LCM of 11, 15 = 165

Now, let us change each of the given fractions into an equivalent fraction having 165 as the denominator.

= [(9/11) × (15/15)] = (135/165)

= [(4/15) × (11/11)] = (44/165)

Now,

= (135/165) – (44/165)

= [(135 – 44)/165]

= (91/165)

(v) For addition of two unlike fractions, first change them to like fractions.

LCM of 10, 5, 2 = 10

Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.

= [(7/10) × (1/1)] = (7/10)

= [(2/5) × (2/2)] = (4/10)

= [(3/2) × (5/5)] = (15/10)

Now,

= (7/10) + (4/10) + (15/10)

= [(7 + 4 + 15)/10]

= (26/10)

= (13/5)

=

(vi) First convert mixed fraction into improper fraction,

== 8/3

= 3 ½ = 7/2

For addition of two unlike fractions, first change them to like fractions.

LCM of 3, 2 = 6

Now, let us change each of the given fractions into an equivalent fraction having 6 as the denominator.

= [(8/3) × (2/2)] = (16/6)

= [(7/2) × (3/3)] = (21/6)

Now,

= (16/6) + (21/6)

= [(16 + 21)/6]

= (37/6)

=

(vii) First convert mixed fraction into improper fraction,

= 8 ½ = 17/2

== 29/8

For subtraction of two unlike fractions, first change them to like fractions.

LCM of 2, 8 = 8

Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.

= [(17/2) × (4/4)] = (68/8)

= [(29/8) × (1/1)] = (29/8)

Now,

= (68/8) – (29/8)

= [(68 – 29)/8]

= (39/8)

=

 

 

(i) LCM of 9, 3, 21 = 63

Now, let us change each of the given fractions into an equivalent fraction having 63 as the denominator.

[(2/9) × (7/7)] = (14/63) [(2/3) × (21/21)] = (42/63) [(8/21) × (3/3)] = (24/63)

Clearly,

(42/63) > (24/63) > (14/63)

Hence,

(2/3) > (8/21) > (2/9)

Hence, the given fractions in descending order are (2/3), (8/21), (2/9)

(ii) LCM of 5, 7, 10 = 70

Now, let us change each of the given fractions into an equivalent fraction having 70 as the denominator.

[(1/5) × (14/14)] = (14/70) [(3/7) × (10/10)] = (30/70) [(7/10) × (7/7)] = (49/70)

Clearly,

(49/70) > (30/70) > (14/70)

Hence,

(7/10) > (3/7) > (1/5)

Hence, the given fractions in descending order are (7/10), (3/7), (1/5)

 

 

(a) First, convert the given mixed fraction into an improper fraction.

== 26/5

Now,

= 3 × (26/5)

= 78/5

=

(b) First, convert the given mixed fraction into an improper fraction.

= 6 ¾ = 27/4

Now,

= 5 × (27/4)

= 135/4

= 33 ¾

(c) First, convert the given mixed fraction into an improper fraction.

= 2 ¼ = 9/4

Now,

= 7 × (9/4)

= 63/4

= 15 ¾

(d) First, convert the given mixed fraction into an improper fraction.

== 19/3

Now,

= 4 × (19/3)

= 76/3

=

(e) First, convert the given mixed fraction into an improper fraction.

= 3 ¼ = 13/4

Now,

= (13/4) × 6

= (13/2) × 3

= 39/2

= 19 ½

(f) First, convert the given mixed fraction into an improper fraction.

== 17/5

Now,

= (17/5) × 8

= 136/5

=

 

(i) 2 × (1/5) represents the addition of 2 figures, each representing 1 shaded part out of the given 5 equal parts.

∴ 2 × (1/5) is represented by fig (d).

(ii) 2 × ½ represents the addition of 2 figures, each representing 1 shaded part out of the given 2 equal parts.

∴ 2 × ½ is represented by fig (b).

(iii) 3 × (2/3) represents the addition of 3 figures, each representing 2 shaded parts out of the given 3 equal parts.

∴ 3 × (2/3) is represented by fig (a).

(iii) 3 × ¼ represents the addition of 3 figures, each representing 1 shaded part out of the given 4 equal parts.

∴ 3 × ¼ is represented by fig (c).

 

(i) 3 × (1/5) represents the addition of 3 figures, each representing 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts.

∴ 3 × (1/5) = (3/5) is represented by fig (c).

(ii) 2 × (1/3) represents the addition of 2 figures, each representing 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts.

∴ 2 × (1/3) = (2/3) is represented by fig (a).

(iii) 3 × (3/4) represents the addition of 3 figures, each representing 3 shaded parts out of the given 4 equal parts, and 2 ¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts.

∴ 3 × (3/4) = 2 ¼ is represented by fig (b).

 

(i) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (7/1) × (3/5)

= (7 × 3)/ (1 × 5)

= (21/5)

=

(ii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4/1) × (1/3)

= (4 × 1)/ (1 × 3)

= (4/3)

=

(iii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/1) × (6/7)

= (2 × 6)/ (1 × 7)

= (12/7)

=

(iv) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/1) × (2/9)

= (5 × 2)/ (1 × 9)

= (10/9)

=

(v) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/3) × (4/1)

= (2 × 4)/ (3 × 1)

= (8/3)

=

(vi) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/2) × (6/1)

= (5 × 6)/ (2 × 1)

= (30/2)

= 15

(vii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (11/1) × (4/7)

= (11 × 4)/ (1 × 7)

= (44/7)

=

(viii) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (20/1) × (4/5)

= (20 × 4)/ (1 × 5)

= (80/5)

= 16

(ix) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (13/1) × (1/3)

= (13 × 1)/ (1 × 3)

= (13/3)

=

(x) By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (15/1) × (3/5)

= (15 × 3)/ (1 × 5)

= (45/5)

= 9

 

(i) From the question,

We may observe that there are 12 circles in the given box. So, we have to shade ½ of the circles in the box.

∴ 12 × ½ = 12/2

= 6

So we have to shade any 6 circles in the box.

(ii) From the question,

We may observe that there are 9 triangles in the given box. So, we have to shade 2/3 of the triangles in the box.

∴ 9 × (2/3) = 18/3

= 6

So we have to shade any 6 triangles in the box.

(iii) From the question,

We may observe that there are 15 squares in the given box. So, we have to shade 3/5 of the squares in the box.

∴ 15 × (3/5) = 45/5

= 9

So we have to shade any 9 squares in the box.

 

(i) From the question, it is given that,

Amount of water in the water bottle = 5 litres

Amount of water consumed by Vidya = 2/5 of 5 litres

= (2/5) × 5

= 2 litres

So, the total amount of water drank by Vidya is 2 litres

(ii) From the question, it is given that,

Amount of water in the water bottle = 5 litres

Then,

Amount of water consumed by Pratap = (1 – water consumed by Vidya)

= (1 – (2/5))

= (5-2)/5

= 3/5

∴ The total amount of water consumed by Pratap = 3/5 of 5 litres

= (3/5) × 5

= 3 litres

So, the total amount of water drank by Pratap is 3 litres

 

(a) (i) 24

We have,

= ½ × 24

= 24/2

= 12

(ii) 46

We have,

= ½ × 46

= 46/2

= 23

(b) (i) 18

We have,

= 2/3 × 18

= 2 × 6

= 12

(ii) 27

We have,

= 2/3 × 27

= 2 × 9

= 18

(c) (i) 16

We have,

= ¾ × 16

= 3 × 4

= 12

(ii) 36

We have

= ¾ × 36

= 3 × 9

= 27

(d) (i) 20

We have,

= 4/5 × 20

= 4 × 4

= 16

(ii) 35

We have,

= 4/5 × 35

= 4 × 7

= 28

 

(a) (i) 2 ¾

First, convert the given mixed fraction into an improper fraction.

= 2 ¾ = 11/4

Now,

= ½ × 11/4

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ½ × (11/4)

= (1 × 11)/ (2 × 4)

= (11/8)

=

(ii)   

First, convert the given mixed fraction into an improper fraction.

== 38/9

Now,

= ½ × (38/9)

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ½ × (38/9)

= (1 × 38)/ (2 × 9)

= (38/18)

= 19/9

=

(b)  (i)   

First, convert the given mixed fraction into an improper fraction.

== 23/6

Now,

= (5/8) × (23/6)

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) × (23/6)

= (5 × 23)/ (8 × 6)

= (115/48)

=

(ii)

First, convert the given mixed fraction into an improper fraction.

== 29/3

Now,

= (5/8) × (29/3)

By the rule Multiplication of fractions,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) × (29/3)

= (5 × 29)/ (8 × 3)

= (145/24)

=

 

(i) We have,

= (2/7) × (3/4) and (3/5) × (5/8)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/7) × (3/4)

= (2 × 3)/ (7 × 4)

= (1 × 3)/ (7 × 2)

= (3/14) … [i]

And,

= (3/5) × (5/8)

= (3 × 5)/ (5 × 8)

= (3 × 1)/ (1 × 8)

= (3/8) … [ii]

Now, convert [i] and [ii] into like fractions,

LCM of 14 and 8 is 56

Now, let us change each of the given fractions into an equivalent fraction having 56 as the denominator.

[(3/14) × (4/4)] = (12/56) [(3/8) × (7/7)] = (21/56)

Clearly,

(12/56) < (21/56)

Hence,

(3/14) < (3/8)

(ii) We have,

= (1/2) × (6/7) and (2/3) × (3/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/2) × (6/7)

= (1 × 6)/ (2 × 7)

= (1 × 3)/ (1 × 7)

= (3/7) … [i]

And,

= (2/3) × (3/7)

= (2 × 3)/ (3 × 7)

= (2 × 1)/ (1 × 7)

= (2/7) … [ii]

By comparing [i] and [ii],

Clearly,

(3/7) > (2/7)

 

From the question, it is given that,

Lipika reads the book for = 1 ¾ hours every day = 7/4 hours

Number of days she took to read the entire book = 6 days

∴ Total number of hours required by her to complete the book = (7/4) × 6

= (7/2) × 3

= 21/2

= 10 ½ hours

Hence, the total number of hours required by her to complete the book is 10 ½ hours.

 

From the question, it is given that,

The total number of distance travelled by a car in 1 liter of petrol = 16 km

Then,

Total quantity of petrol = 2 ¾ liter = 11/4 liters

Total number of distance travelled by car in 11/4 liters of petrol = (11/4) × 16

= 11 × 4

= 44 km

∴ Total number of distance travelled by car in 11/4 liters of petrol is 44 km.

 

(i) First convert the given mixed fraction into improper fraction.

== 8/3

Now,

= (2/3) × (8/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 8)/ (3 × 3)

= (16/9)

=

(ii) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 7)/ (7 × 9)

= (2 × 1)/ (1 × 9)

= (2/9)

(iii) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 × 6)/ (8 × 4)

= (3 × 3)/ (4 × 4)

= (9/16)

(iv) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (9 × 3)/ (5 × 5)

= (27/25)

=

(v) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1 × 15)/ (3 × 8)

= (1 × 5)/ (1 × 8)

= (5/8)

(vi) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (11 × 3)/ (2 × 10)

= (33/20)

=

(vii) By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (4 × 12)/ (5 × 7)

= (48/35)

=

 

From the question, it is given that,

The distance between two adjacent saplings = ¾ m

Number of saplings planted by Saili in a row = 4

Then, number of gap in saplings = ¾ × 4

= 3

∴ The distance between the first and the last saplings = 3 × ¾

= (9/4) m

= 2 ¼ m

Hence, the distance between the first and the last saplings is 2 ¼ m.

 

(i) (a) ¼

We have,

= ¼ × ¼

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × ¼

= (1 × 1)/ (4 × 4)

= (1/16)

(b) 3/5

We have,

= ¼ × (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × (3/5)

= (1 × 3)/ (4 × 5)

= (3/20)

(c) (4/3)

We have,

= ¼ × (4/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × (4/3)

= (1 × 4)/ (4 × 3)

= (4/12)

= 1/3

(ii) (a) 2/9

We have,

= (1/7) × (2/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) × (2/9)

= (1 × 2)/ (7 × 9)

= (2/63)

(b) 6/5

We have,

= (1/7) × (6/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) × (6/5)

= (1 × 6)/ (7 × 5)

= (6/35)

(c) 3/10

We have,

= (1/7) × (3/10)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/7) × (3/10)

= (1 × 3)/ (7 × 10)

= (3/70)

 

(i) First convert the given mixed fraction into improper fraction.

= 5 ¼ = 21/4

Now,

= (2/5) × (21/4)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 21)/ (5 × 4)

= (1 × 21)/ (5 × 2)

= (21/10)

=

(ii) First convert the given mixed fraction into improper fraction.

== 32/5

Now,

= (32/5) × (7/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (32 × 7)/ (5 × 9)

= (224/45)

=

(iii) First convert the given mixed fraction into improper fraction.

== 16/3

Now,

= (3/2) × (16/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 × 16)/ (2 × 3)

= (1 × 8)/ (1 × 1)

= 8

(iv) First convert the given mixed fraction into improper fraction.

== 17/7

Now,

= (5/6) × (17/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 × 17)/ (6 × 7)

= (85/42)

=

(v) First convert the given mixed fraction into improper fraction.

== 17/5

Now,

= (17/5) × (4/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (17 × 4)/ (5 × 7)

= (68/35)

=

(vi) First convert the given mixed fraction into improper fraction.

== 13/5

Now,

= (13/5) × (3/1)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (13 × 3)/ (5 × 1)

= (39/5)

=

(vii) First convert the given mixed fraction into improper fraction.

== 25/7

Now,

= (25/7) × (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (25 × 3)/ (7 × 5)

= (5 × 3)/ (7 × 1)

= (15/7)

=

 

(a) (i)

Let the required number be x,

Then,

= (2/3) × (x) = (10/30)

By cross multiplication,

= x = (10/30) × (3/2)

= x = (10 × 3) / (30 × 2)

= x = (5 × 1) / (10 × 1)

= x = 5/10

∴ The required number in the box is (5/20)

(ii) The number in the box is 5/10

Then,

The simplest form of 5/10 is ½

(b) (i)

Let the required number be x,

Then,

= (3/5) × (x) = (24/75)

By cross multiplication,

= x = (24/75) × (5/3)

= x = (24 × 5) / (75 × 3)

= x = (8 × 1) / (15 × 1)

= x = 8/15

∴ The required number in the box is (8/15)

(ii) The number in the box is 8/15

Then,

The simplest form of 8/15 is 8/15

 

 

(i) We have

= 12 × reciprocal of ¾

= 12 × (4/3)

= 4 × 4

= 16

(ii) We have

= 14 × reciprocal of (5/6)

= 14 × (6/5)

= 84/5

(iii) We have

= 8 × reciprocal of (7/3)

= 8 × (3/7)

= (24/7)

(iv) We have

= 4 × reciprocal of (8/3)

= 4 × (3/8)

= 1 × (3/2)

= 3/2

(v) While dividing a whole number by a mixed fraction, first convert the mixed fraction into an improper fraction.

We have

== 7/3

Then,

= 3 ÷ (7/3)

= 3 × reciprocal of (7/3)

= 3 × (3/7)

= 9/7

(vi) While dividing a whole number by a mixed fraction, first convert the mixed fraction into an improper fraction.

We have

== 25/7

Then,

= 5 ÷ (25/7)

= 5 × reciprocal of (25/7)

= 5 × (7/25)

= 1 × (7/5)

= 7/5

 

(i) We have

= (2/5) × reciprocal of ½

= (2/5) × (2/1)

= (2 × 2) / (5 × 1)

= 4/5

(ii) We have

= (4/9) × reciprocal of (2/3)

= (4/9) × (3/2)

= (4 × 3) / (9 × 2)

= (2 × 1) / (3 × 1)

= 2/3

(iii) We have

= (3/7) × reciprocal of (8/7)

= (3/7) × (7/8)

= (3 × 7) / (7 × 8)

= (3 × 1) / (1 × 8)

= 3/8

(iv) First, convert the mixed fraction into an improper fraction.

We have,

== 7/3

Then,

= (7/3) × reciprocal of (3/5)

= (7/3) × (5/3)

= (7 × 5) / (3 × 3)

= 35/9

(v) First, convert the mixed fraction into an improper fraction.

We have

= 3 ½ = 7/2

Then,

= (7/2) × reciprocal of (8/3)

= (7/2) × (3/8)

= (7 × 3) / (2 × 8)

= 21/16

(vi) First, convert the mixed fraction into an improper fraction.

We have

= 1 ½ = 3/2

Then,

= (2/5) × reciprocal of (3/2)

= (2/5) × (2/3)

= (2 × 2) / (5 × 3)

= 4/15

(vii) First, convert the mixed fraction into an improper fraction.

We have

== 16/5

== 5/3

Then,

= (16/5) × reciprocal of (5/3)

= (16/5) × (3/5)

= (16 × 3) / (5 × 5)

= 48/25

(viii) First, convert the mixed fraction into an improper fraction.

We have

== 11/5

== 6/5

Then,

= (11/5) × reciprocal of (6/5)

= (11/5) × (5/6)

= (11 × 5) / (5 × 6)

= (11 × 1) / (1 × 6)

= 11/6

(i) Reciprocal of (3/7) is (7/3) [∵ ((3/7) × (7/3)) = 1]

So, it is an improper fraction.

An improper fraction is a fraction in which the numerator is greater than its denominator.

(ii) Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]

So, it is an improper fraction.

An improper fraction is a fraction in which the numerator is greater than its denominator.

(iii) Reciprocal of (9/7) is (7/9) [∵ ((9/7) × (7/9)) = 1]

So, it is a proper fraction.

A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.

(iv) Reciprocal of (6/5) is (5/6) [∵ ((6/5) × (5/6)) = 1]

So, it is a proper fraction.

A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.

(v) Reciprocal of (12/7) is (7/12) [∵ ((12/7) × (7/12)) = 1]

So, it is a proper fraction.

A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.

(vi) Reciprocal of (1/8) is (8/1) or 8 [∵ ((1/8) × (8/1)) = 1]

So, it is a whole number.

Whole numbers are collections of all positive integers, including 0.

(vii) Reciprocal of (1/11) is (11/1) or 11 [∵ ((1/11) × (11/1)) = 1]

So, it is a whole number.

Whole numbers are collections of all positive integers, including 0.

 

(i) We have

= (7/3) × reciprocal of 2

= (7/3) × (1/2)

= (7 × 1) / (3 × 2)

= 7/6

=

(ii) We have

= (4/9) × reciprocal of 5

= (4/9) × (1/5)

= (4 × 1) / (9 × 5)

= 4/45

(iii) We have

= (6/13) × reciprocal of 7

= (6/13) × (1/7)

= (6 × 1) / (13 × 7)

= 6/91

(iv) First, convert the mixed fraction into an improper fraction.

We have,

== 13/3

Then,

= (13/3) × reciprocal of 3

= (13/3) × (1/3)

= (13 × 1) / (3 × 3)

= 13/9

(v) First, convert the mixed fraction into an improper fraction.

We have

= 3 ½ = 7/2

Then,

= (7/2) × reciprocal of 4

= (7/2) × (1/4)

= (7 × 1) / (2 × 4)

= 7/8

(vi) First, convert the mixed fraction into an improper fraction.

We have

== 31/7

Then,

= (31/7) × reciprocal of 7

= (31/7) × (1/7)

= (31 × 1) / (7 × 7)

= 31/49

 

(i) We know that,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 5 cm = (5/100)

= 0.05 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.05 m = (0.05/1000)

= 0. 00005 km

(ii) We know that,

= 1 cm = 10 mm

Then,

= 1 mm = (1/10) cm

= 35 mm = (35/10) cm

= 3.5 cm

And,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 3.5 cm = (3.5/100) m

= (35/1000) m

= 0.035 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.035 m = (0.035/1000)

= 0. 000035 km

 

(i) We have,

20.03 = (2 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(ii) We have,

2.03 = (2 × 1) + (0 × (1/10)) + (3 × (1/100))

(iii) We have,

200.03 = (2 × 100) + (0 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(iv) We have,

2.034 = (2 × 1) + (0 × (1/10)) + (3 × (1/100)) + (4 × (1/1000))

 

From the question, it is given that,

Distance travelled by Dinesh = AB + BC

= 7.5 + 12.7

= 20.2 km

∴Dinesh travelled 20.2 km

Distance travelled by Ayub = AD + DC

= 9.3 + 11.8

= 21.1 km

∴Ayub travelled 21.1km

Clearly, Ayub travelled more distance by = (21.1 – 20.2)

= 0.9 km

∴Ayub travelled 0.9 km more than Dinesh.

 

From the question, it is given that,

Fruits bought by Shyama = 5 kg 300 g

= 5 kg + (300/1000) kg

= 5 kg + 0.3 kg

= 5.3 kg

Fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g

= (4 + (800/1000)) + (4 + (150/1000))

= (4 + 0.8) kg + (4 + .150) kg

= 4.8 kg + 4.150kg

= 8.950 kg

So, Sarala bought more fruits.

 

Now, we have to find the difference of 42.6 km and 28 km

42.6

-28.0

14.6

∴ 14.6 km less is 28 km than 42.6 km.

(i) By comparing whole number, 0 = 0

By comparing the tenths place digit, 5 > 0

∴ 0.5 > 0.05

(ii) By comparing whole number, 0 = 0

By comparing the tenths place digit, 7 > 5

∴ 0.7 > 0.5

(iii) By comparing whole number, 7 > 0

∴ 7 > 0.7

(iv) By comparing whole number, 1 = 1

By comparing the tenths place digit, 3 < 4

∴ 1.37 < 1.49

(v) By comparing whole number, 2 = 2

By comparing the tenths place digit, 0 < 3

∴ 2.03 < 2.30

(vi) By comparing whole number, 0 = 0

By comparing the tenths place digit, 8 = 8

By comparing the hundredths place digit, 0 < 8

∴ 0.8 < 0.88

 

(i) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 paise = ₹ (7/100)

= ₹ 0.07

(ii) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 rupees 7 paise = ₹ 7 + ₹ (7/100)

= ₹ 7 + ₹ 0.07

= ₹ 7.07

(iii) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 77 rupees 77 paise = ₹ 77 + ₹ (77/100)

= ₹ 77 + ₹ 0.77

= ₹ 77.77

(iv) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 50 paise = ₹ (50/100)

= ₹ 0.50

(v) We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 235 paise = ₹ (235/100)

= ₹ 2.35

 

(i) We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 200 g = (200/1000) kg

= (2/10)

= 0.2 kg

(ii) We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 3470 g = (3470/1000) kg

= (3470/100)

= 3.470 kg

(iii) We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 4 kg 8 g = 4 kg + (8/1000) kg

= 4 kg + 0.008

= 4.008 kg

 

(i) From the question, we observe that,

The place value of 2 in 2.56 is ones

(ii) From the question, we observe that,

The place value of 2 in 21.37 is tens

(iii) From the question, we observe that,

The place value of 2 in 10.25 is tenths.

(iv) From the question, we observe that,

The place value of 2 in 9.42 is hundredth.

(v) From the question, we observe that,

The place value of 2 in 63.352 is thousandth.

 

(i) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 1.3 × 10 = 13

(ii) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 36.8 × 10 = 368

(iii) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 153.7 × 10 = 1537

(iv) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 168.07 × 10 = 1680.7

(v) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 31.1 × 100 = 3110

(vi) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 156.1 × 100 = 15610

(vii) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 3.62 × 100 = 362

(viii) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 43.07 × 100 = 4307

(ix) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.5 × 10 = 5

(x) On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.08 × 10 = 0.8

(xi) On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 0.9 × 100 = 90

(xii) On multiplying a decimal by 1000, the decimal point is shifted to the right by three places.

We have,

= 0.03 × 1000 = 30

 

From the question,

Length of the rectangle = 5.7 cm

Breadth of the rectangle = 3 cm

Then,

Area of the rectangle = length × Breadth

= 5.7 × 3

= 17.1 cm2

 

It is given that,

Distance covered by two-wheeler in 1 litre of petrol = 55.3 km

Then,

Distance covered by two-wheeler in 10 litres of petrol = (10 × 55.3)

= 553 km

∴ The two-wheeler will cover a distance of 553 km in 10 litres of petrol.

 

(i) We have,

= (2/10) × 6

= (12/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 1.2

(ii) We have,

= (8) × (46/10)

= (368/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 36.8

(iii) We have,

= (271/100) × 5

= (1355/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 13.55

(iv) We have,

= (201/10) × 4

= (804/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 80.4

(v) We have,

= (5/100) × 7

= (35/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.35

(vi) We have,

= (21102/100) × 4

= (84408/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 844.08

(vii) We have,

= (2) × (86/100)

= (172/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 1.72

 

(i) We have,

= (25/10) × (3/10)

= (75/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.75

(ii) We have,

= (1/10) × (517/10)

= (517/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 5.17

(iii) We have,

= (2/10) × (3168/10)

= (6336/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 63.36

(iv) We have,

= (13/10) × (31/10)

= (403/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 4.03

(v) We have,

= (5/10) × (5/100)

= (25/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 0.025

(vi) We have,

= (112/10) × (15/100)

= (1680/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 1.680

(vii) We have,

= (107/100) × (2/100)

= (214/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 0.0214

(viii) We have,

= (1005/100) × (105/100)

= (105525/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 10.5525

(ix) We have,

= (10101/100) × (1/100)

= (10101/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 1.0101

(x) We have,

= (10001/100) × (11/10)

= (110011/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 110.011

 

From the question, it is given that,

Total distance covered by the vehicle in 2.4 litres of petrol = 43.2 km

Then,

Distance covered in 1 litre of petrol = 43.2 ÷ 2.4

= (432/10) ÷ (24/10)

= (432/10) × (10/24)

= (432 × 10)/ (10 × 24)

= (36 × 1)/ (1 × 2)

= (18 × 1)/ (1 × 1)

= 18 km

∴ The total distance covered in 1 litre of petrol is 18 km.

 

(i) We have,

= (4/10) ÷ 2

Then,

= (4/10) × (1/2)

= (2/10) × (1/1)

= (2/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 0.2

(ii) We have,

= (35/100) ÷ 5

Then,

= (35/100) × (1/5)

= (7/100) × (1/1)

= (7/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.07

(iii) We have,

= (248/100) ÷ 4

Then,

= (248/100) × (1/4)

= (62/100) × (1/1)

= (62/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.62

(iv) We have,

= (654/10) ÷ 6

Then,

= (654/10) × (1/6)

= (109/10) × (1/1)

= (109/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 10.9

(v) We have,

= (6512/10) ÷ 4

Then,

= (6512/10) × (1/4)

= (1628/10) × (1/1)

= (1628/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 162.8

(vi) We have,

= (1449/100) ÷ 7

Then,

= (1449/100) × (1/7)

= (207/100) × (1/1)

= (207/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 2.07

(vii) We have,

= (396/100) ÷ 4

Then,

= (396/100) × (1/4)

= (99/100) × (1/1)

= (99/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.99

(viii) We have,

= (80/100) ÷ 5

Then,

= (80/100) × (1/5)

= (16/100) × (1/1)

= (16/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.16

 

(i) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 4.8 ÷ 10

= (4.8/10)

= 0.48

(ii) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 52.5 ÷ 10

= (52.5/10)

= 5.25

(iii) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.7 ÷ 10

= (0.7/10)

= 0.07

(iv) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 33.1 ÷ 10

= (33.1/10)

= 3.31

(v) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 272.23 ÷ 10

= (272.23/10)

= 27.223

(vi) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.56 ÷ 10

= (0.56/10)

= 0.056

(vii) On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 3.97 ÷ 10

= (3.97/10)

= 0.397

 

(i) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 2.7 ÷ 100

= (2.7/100)

= 0.027

(ii) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.3 ÷ 100

= (0.3/100)

= 0.003

(iii) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.78 ÷ 100

= (0.78/100)

= 0.0078

(iv) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 432.6 ÷ 100

= (432.6/100)

= 4.326

(v) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 23.6 ÷ 100

= (23.6/100)

= 0.236

(vi) On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 98.53 ÷ 100

= (98.53/100)

= 0.9853

 

(i) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 7.9 ÷ 1000

= (7.9/1000)

= 0.0079

(ii) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 26.3 ÷ 1000

= (26.3/1000)

= 0.0263

(iii) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 38.53 ÷ 1000

= (38.53/1000)

= 0.03853

(iv) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 128.9 ÷ 1000

= (128.9/1000)

= 0.1289

(v) On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 0.5 ÷ 1000

= (0.5/1000)

= 0.0005

 

(i) We have,

= 7 ÷ (35/10)

= 7 × (10/35)

= 1 × (10/5)

= 2

(ii) We have,

= 36 ÷ (2/10)

= 36 × (10/2)

= 18 × 10

= 180

(iii) We have,

= (325/100) ÷ (5/10)

= (325/100) × (10/5)

= (325 × 10)/ (100 × 5)

= (65 × 1)/ (10 × 1)

= 65/10

= 6.5

(iv) We have,

= (3094/100) ÷ (7/10)

= (3094/100) × (10/7)

= (3094 × 10)/ (100 × 7)

= (442 × 1)/ (10 × 1)

= 442/10

= 44.2

(v) We have,

= (5/10) ÷ (25/100)

= (5/10) × (100/25)

= (5 × 100)/ (10 × 25)

= (1 × 10)/ (1 × 5)

= 10/5

= 2

(vi) We have,

= (775/100) ÷ (25/100)

= (775/100) × (100/25)

= (775 × 100)/ (100 × 25)

= (155 × 1)/ (1 × 5)

= (31 × 1)/ (1 × 1)

= 31

(vii) We have,

= (765/10) ÷ (15/100)

= (765/10) × (100/15)

= (765 × 100)/ (10 × 15)

= (51 × 10)/ (1 × 1)

= 510

(viii) We have,

= (378/10) ÷ (14/10)

= (378/10) × (10/14)

= (378 × 10)/ (10 × 14)

= (27 × 1)/ (1 × 1)

= 27

(ix) We have,

= (273/100) ÷ (13/10)

= (273/100) × (10/13)

= (273 × 10)/ (100 × 13)

= (21 × 1)/ (10 × 1)

= 21/10

= 2.1