What Is 3 5 of 9 3 8

Exercise 2.1

Question 1:

Solve:

(i) 2 – 3/5            (ii) 4 + 7/8            (iii) 3/5 + 2/7          (iv) 9/11 – 4/15            (v) 7/10 + 2/5 + 3/2

(vi) 2  + 3   (vii) 8  - 3

Answer:

(i) 2 – 3/5 = 2/1 – 3/5

                   = (2 * 5 – 3 * 1)/5                         [LCM (1, 5) = 5]

                   = (10 - 3)/5

                   = 7/5

                   = 1

(ii) 4 + 7/8 = 4/1 + 7/8

                    = (4 * 8 + 7)/8                               [LCM (1, 8) = 8]

                    = (32 + 7)/8

                    = 39/8

                    = 4

(iii) 3/5 + 2/7 = (3 * 7 + 2 * 5)/35                  [LCM (7, 5) = 35]

                         = (21 + 10)/35

                         = 31/35

(iv) 9/11 – 4/15 = (9 * 15 – 4 * 11)/165       [LCM (11, 15) = 165]

                             = (135 - 44)/165

                             = 91/165

(v) 7/10 + 2/5 + 3/2 = (7 * 1 + 2 * 2 + 3 * 5)/10

                                 = (7 + 4 + 15)/10

                                 = 26/10                        [26 and 10 are divided by 2]

                                 = 13/5

                                 = 2

(vi) 2  + 3  = 8/3 + 7/2

                      = (8 * 2 + 7 * 3)/6                  [LCM (2, 3) = 6]

                      = (16 + 21)/6

                      = 37/6

                      = 6

(vii) 8  - 3  = 17/2 – 29/8

                       = (17 * 4 – 29 * 1)/8            [LCM (2, 8) = 8]

                       = (68 - 19)/8

                      = 39/8

                      = 4

Question 2:

Arrange the following in descending order:

(i) 2/9, 2/3, 8/21                      (ii) 1/5, 3/7, 7/10

Answer:

(i) 2/9, 2/3, 8/21 = (2 * 7, 2 * 21, 8 * 3)/63              [LCM (9, 3, 21) = 63]

                              = (14, 42, 24)/63

                              = 14/63, 42/63, 24/63

Now, 42/63, 24/63, 14/63                                           [Arrange in descending order]

So, 2/3, 8/21, 2/9

(ii) 1/5, 3/7, 7/10 = (1 * 14, 3 * 10, 7 * 7)/70

                                = (14, 30, 49)/70

                                = 14/70, 30/70, 49/70

Now, 49/70, 30/70, 14/70                                            [Arrange in descending order]

So, 7/10, 3/7, 1/5

Question 3:

In a "magic square", the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

Class_7_Maths_Fraction_&_Decimals_Representing_Fractions

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

Answer:

Sum of first row:  4/11 + 9/11 + 2/11 = 15/11    (Given)

Sum of second row:  3/11 + 5/11 + 7/11 = (3 + 5 + 7)/11 = 15/11

Sum of third row:  8/11 + 1/11 + 6/11 = (8 + 1 + 6)/11 = 15/11

Sum of first column:  4/11 + 3/11 + 8/11 = (4 + 3 + 8)/11 = 15/11

Sum of second column:  9/11 + 5/11 + 1/11 = (9 + 5 + 1)/11 = 15/11

Sum of third column:  2/11 + 7/11 + 6/11 = (2 + 7 + 6)/11 = 15/11

Sum of first diagonal (left to right):  4/11 + 5/11 + 6/11 = (4 + 5 + 6)/11 = 15/11

Sum of second diagonal (left to right):  2/11 + 5/11 + /11 = (2 + 5 + 8)/11 = 15/11

Since the sum of fractions in each row, in each column and along the diagonals is same, therefore it is a magic square.

Question 4:

A rectangular sheet of paper is 12  cm long and 10  cm wide. Find its perimeter.

Answer:

Given: The sheet of paper is in rectangular form.

Length of sheet = 12  cm = 25/2 cm

and Breadth of sheet = 10  cm = 32/3

Now, Perimeter of rectangle = 2 (length + breadth)

       = 2(25/2 + 32/2)

       = 2[(25 * 3 + 32 * 3)/6]

       = 2[(75 + 96)/6]

       = 2(139/6)

       = 139/3

       = 46

Thus, the perimeter of the rectangular sheet is 46  cm.

Question 5:

Find the perimeter of (i) Δ ABE, (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Class_7_Maths_Fraction_&_Decimals_Combinationof_Triangle_&_Rectangle

 Answer:

(i) In Δ ABE, AB = 5/2 cm, BE = 2  cm, AE = 3  cm

The perimeter of Δ ABE = AB + BE + AE

         = 5/2 + 2  + 3

         = 5/2 + 11/4 + 18/5

         = (5 * 10 + 11 * 5 + 18 *4)/20

      = (50 + 55 + 72)/20

      = 177/20

     = 8  cm

Thus, the perimeter of Δ ABE is 8  cm.

(ii) In rectangle BCDE, BE = 2  cm, ED = 7/6

Perimeter of rectangle = 2(Length + Breath)

                                          = 2(2  + 7/6)

                                          = 2(11/4 + 7/6)

                                          = 2[(11 * 3 + 7 * 2)/12]

                                          = 2[(33 + 14)/12]

                                          = 2(47/12)

                                          = 47/6

                                          = 7

Thus, the perimeter of rectangle BCDE is 7  cm.

Since, 8  cm > 7  cm

Therefore, the perimeter of Δ ABE is greater than that of rectangle BCDE.

Question 6:

Salil wants to put a picture in a frame. The picture is 7 cm wide. To fit in the frame the picture cannot be more than 7 cm wide.

How much should the picture be trimmed?

Answer:

Given, the width of the picture = 7 cm = 38/5 cm

and the width of the picture frame = 7 cm = 73/10

Therefore, the picture should be trimmed = 38/5 – 73/10

                                                                                = (38 * 2 – 73 * 1)/10

        = (76 – 73)/10

        = 3/10

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

Question 7:

Ritu ate 3/5 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat?

Who had the larger share? By how much?

Answer:

The part of an apple eaten by Ritu = 3/5

The part of an apple eaten by Somu = 1 – 3/5 = (5 - 3)/5 = 2/5

Comparing the parts of an apple eaten by both Ritu and Somu: 3/5 > 2/5

Larger share will be more by 3/5 – 2/5 = 1/5

Thus Ritu's part is 1/5 more than Somu's part.

Question 8:

Michael finished colouring a picture in 7/12 hour. Vaibhav finished colouring the same picture in 3/4 hour.

Who worked longer? By what fraction was it longer?

Answer:

Time taken by Michael to colour the picture = 7/12 hour

Time taken by Vaibhav to colour the picture = 3/4 hour

Converting both fractions in like fractions, 7/12 and (3 *3)/(4 * 3) = 9/12

Here, 7/12 < 9/12

=> 7/12 < 3/4

Thus, Vaibhav worked longer time.

Vaibhav worked longer time by 3/4 - 7/12 = 9/12 – 7/12

                                                                            = (9 - 7)/12

                                                                            = 2/12 = 1/6 hours

Thus, Vaibhav took 1/6 hour more than Michael.

Exercise 2.2

Question 1:

Which of the drawings (a) to (d) show

(i) 2 * 1/5Class_7_Maths_Fraction_&_Decimals_Different_Images6

(ii) 2 * 1/2Class_7_Maths_Fraction_&_Decimals_Different_Images5

(iii) 3 * 2/3Class_7_Maths_Fraction_&_Decimals_Different_Images4

(iv) 3 * 1/4Class_7_Maths_Fraction_&_Decimals_Different_Images3

Answer:

(i) à  (d)            Since 2 * 1/5 = 1/5 + 1/5

(ii) à (b)            Since 2 * 1/2 = 1/2 + 1/2

(iii) à (a)           Since 3 * 2/3 = 2/3 + 2/3 + 2/3

(iv) à (c)           Since3 * 1/4 = 1/4 + 1/4 + 1/4

Question 2:

Some pictures (a) to (c) are given below. Tell which of them show:

(i) 3 * 1/5 = 3/5

Class_7_Maths_Fraction_&_Decimals_Different_Images2

(ii) 2 * 1/3 = 2/3

Class_7_Maths_Fraction_&_Decimals_Different_Images1

(iii) 3 * 3/4 = 2

Class_7_Maths_Fraction_&_Decimals_Different_Images

Answer:

(i)  à (c)          Since 3 * 1/5 = 1/5 + 1/5 + 1/5

(ii) à (a)          Since2 * 1/3 = 1/3 + 1/3

(iii) à (b)        Since 3 * 3/4 = 3/4 + 3/4 + 3/4

Question 3:

Multiply and reduce to lowest form and convert into a mixed fraction:

(i) 7 * 3/5             (ii) 4 * 1/3          (iii) 2 * 6/7           (iv) 5 * 2/9          (v) 2/3 * 4

(vi) 5/2 * 6          (vii) 11 * 4/7      (viii)  20 * 4/5      (ix) 13* 1/3         (x) 15 * 3/5

Answer:

(i) 7 * 3/5 = (7 * 3)/5 = 21/5 = 4

(ii) 4 * 1/3 = (4 * 1)/3 = 4/3 = 1

(iii) 2 * 6/7 = (2 * 6)/7 = 12/7 = 1

(iv) 5 * 2/9 = (5 *2)/9 = 10/9 = 1

(v) 2/3 * 4 = (2 * 4)/3 = 8/3 = 2

(vi) 5/2 * 6 = (5 * 6)/2 = 30/2 = 15

(vii) 11 * 4/7 = (11 * 4)/7 = 44/7 = 6

(viii)  20 * 4/5 = (20 *4)/5 = 80/5 = 16

(ix) 13* 1/3 = (13 * 1)/3 = 13/3 = 4

(x) 15 * 3/5 = (15 * 3)/5 = 45/5 = 9

Question 4:

Shade:

(i) 1/2 of the circles in box

Class_7_Maths_Fraction_&_Decimals_Circles1

(ii) 2/3 of the triangles in box

Class_7_Maths_Fraction_&_Decimals_Triangles1

(ii) 3/5 of the squares in box

Class_7_Maths_Fraction_&_Decimals_Squares1

Answer:

(i) 1/2 of 12 circles

= (1/2) * 12

= 12/2 = 6 circles

Class_7_Maths_Fraction_&_Decimals_Circles

(ii) 2/3 of 9 triangles

= (2/3) * 9

= (2 * 9)/3

= 18/3

= 6 triangles

Class_7_Maths_Fraction_&_Decimals_Triangles

(ii) 3/5 of 15 squares

= (3/5) * 15

= (3 * 15)/5

= 45/5

= 9 squares

Class_7_Maths_Fraction_&_Decimals_Squares

Question 5:

Find:

(a) 1/2 of (i) 24 (ii) 46                       (b) 2/3 of (i) 18 (ii) 27

(c) 3/4 of (i) 16 (ii) 36                       (d) 4/5 of (i) 20 (ii) 35

Answer:

(a) (i) 1/2 of 24 = 1/2 * 12 = 12/2 = 6

(ii) 1/2 of 46 = 1/2 * 46 = 46/2 = 23

(b) (i) 2/3 of 18 = 2/3 * 18 = (2 * 18)/3 = 36/3 = 12

(ii) 2/3 of 27 = 2/3 * 27 = (2 * 27)/3 = 54/3 = 18

(c) (i) 3/4 of 16 = 3/4 * 16 = (3 *16)/4 = 48/4 = 12

(ii) 3/4 of 36 = 3/4 * 36 = 3 * 9 = 27

(d) (i) 4/5 of 20 = 4/5 * 20 = 4 * 4 = 16

(ii) 4/5 of 35 = 4/5 * 35 = 4 * 7 = 28

Question 6:

Multiply and express as a mixed fraction:

(a) 3 * 5   (b) 5 * 6    (c) 7 * 2    (d) 4 * 6   (e) 3  * 6   (f) 3  * 8

Answer:

(a) 3 * 5  = 3 * 26/5 = (3 * 26)/5 = 78/5 = 15

(b) 5 * 6  = 5 * 27/4 = (5 * 27)/4 = 135/6 = 33

(c) 7 * 2  = 7 * 9/4 = (7 * 9)/4 = 63/4 = 15

(d) 4 * 6  = 4 * 19/4 = (4 * 19)/3 = 76/3 = 25

(e) 3  * 6 = 13/4 * 6 = (13 * 6)/4 = 78/4 = 39/2 = 19

(f) 3  * 8 = 17/5 * 8 = (17 * 8)/5 = 136/5 = 27

Question 7:

Find:

(a) 1/2 of (i) 2   (ii) 4   (b) 5/8 of (i) 3   (ii) 9

Answer:

(a)

        (i) 1/2 of 2  = 1/2 * 11/4 = (1 * 11)/(2 * 4) = 11/8 = 1

        (ii) 1/2 of 4  = 1/2 of 38/9 = (1 * 38)/(2 * 9) = 38/18 = 19/9 = 2

(b)

         (i) 5/8 of 3   = 5/8 * 23/6 = (5 * 23)/(8 * 6) = 115/48 = 2

         (ii) 5/8 of 9  = 5/8 * 29/3 = (5 * 29)/(8 * 3) = 145/24 = 6

Question 8:

Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water.

Vidya consumed 2/5 of the water.

Pratap consumed the remaining water.

(i) How much water did Vidya drink?

(ii) What fraction of the total quantity of water did Pratap drink?

Answer:

Given: Total quantity of water in bottle = 5 litres

(i) Vidya consumed = 2/5 of 5 litres = 2/5 * 5 = 2 * 1= 2 litres

Thus, Vidya drank 2 litres water from the bottle.

(ii) Pratap consumed = (1 – 2/5) part of bottle

                                      = (5 - 2)/5

                                      = 3/5 part of bottle

Pratap consumed 3/5 of 5 litres water = 3/5 * 5 = 3 * 1 = 3 litres

Thus, Pratap drank 3/5 part of the total quantity of water.

Exercise 2.3

Question 1:

Find:

(i) 1/4 of                          (a) 1/4                      (b) 3/5                        (c) 4/3

(ii) 1/7 of                         (a) 2/9                      (b) 6/5                        (c) 3/10

Answer:

(i)

      (a) 1/4 of 1/4 = 1/4 * 1/4 = (1 * 1)/(4 * 4) = 1/16

      (b) 1/4 of 3/5 = 1/4 * 3/5 = (1 * 3)/(4 * 5) = 3/20

      (c) 1/4 of 4/3 = 1/4 * 4/3 = (1 * 4)/(4 * 3) = 4/12 = 1/3

(ii)

     (a) 1/7 of 2/9 = 1/7 * 2/9 = (1 * 2)/(7 * 9) = 2/63

     (b) 1/7 of 6/5 = 1/7 * 6/5 = (1 * 6)/(7 * 5) = 6/35

     (c) 1/7 of 3/10 = 1/7 * 3/10 = (1 * 3)/(7 * 10) = 3/70

Question 2:

Multiply and reduce to lowest form (if possible):

(i) 2/3 * 2              (ii) 2/7 * 7/9              (iii) 3/8 * 6/4           (iv) 9/5 * 3/5

(v) 1/3 * 15/8        (vi) 11/2 * 3/10         (vii) 4/5 * 12/7

Answer:

(i) 2/3 * 2  = 2/3 * 8/3 = (2 * 8)/(3 * 3) = 16/9

(ii) 2/7 * 7/9 = (2 * 7)/(2 * 9) = 2/9

(iii) 3/8 * 6/4 = (3 * 6)/(8 * 4) = 18/32 = 9/16

(iv) 9/5 * 3/5 = (9 * 3)/(5 * 5) = 27/25 = 1

(v) 1/3 * 15/8 = (1 * 15)/(3 * 8) = 15/24

(vi) 11/2 * 3/10 = (11 * 3)/(2 * 10) = 33/20 = 1

(vii) 4/5 * 12/7 = (4 * 12)/(5 * 7) = 48/35 = 1

Question 3:

Multiply the following fractions:

(i) 2/5 * 5              (ii) 6   * 7/9           (iii) 3/2 * 5                 (iv) 5/6 * 2

(v) 3   * 4/7          (vi) 2   * 3               (vii) 3   * 3/5

Answer:

(i) 2/5 * 5  = 2/5 * 21/4 = (2 * 21)/(5 * 4) = 21/10 = 2

(ii) 6   * 7/9 = 32/5 * 7/9 = (32 * 7)/(5 * 9) = 224/45 = 4

(iii) 3/2 * 5  = 3/2 * 16/3 = (3 * 16)/(2 * 3) = 48/6 = 8

(iv) 5/6 * 2  = 5/6 * 17/7 = (5 * 17)/(6 * 7) = 85/42 = 2

(v) 3   * 4/7 = 17/5 * 4/7 = (17 * 4)/(5 * 7) = 68/35 = 1

(vi) 2   * 3 = 13/5 * 3/1 = (13 * 3)/(5 * 1) = 39/5 = 7

(vii) 3   * 3/5 = 25/7 * 3/5 = (25 * 3)/(7 * 5) = 75/35 = 15/7 = 2

Question 4:

Which is greater:

(i) 2/7 of 3/4 or 3/5 of 5/8               (ii) 1/2 of 6/7 or 2/3 of 3/7

Answer:

(i) 2/7 of 3/4 or 3/5 of 5/8

= 2/7 * 3/4 or 3/5 * 5/8

= (2 * 3)/(7 * 4) or (3 * 5)/(5 * 8)

= 6/28 or 15/40

= 3/14 or 3/8

Since 3/14 < 3/8

So, 3/5 of 5/8 is greater.

(ii) 1/2 of 6/7 or 2/3 of 3/7

= 1/2 * 6/7 or 2/3 * 3/7

= (1 * 6)/(2 * 7) or (2 * 3)/(3 * 7)

= 6/14 or 6/21

= 3/7 or 2/7

Since 3/7 > 2/7

Hence, 1/2 of 6/7 is greater.

Question 5:

Saili plants 4 saplings in a row in her garden. The distance between two adjacent saplings is 3/4 m.

Find the distance between the first and the last sapling.

Answer:

The distance between two adjacent saplings = 3/4 m

Saili planted 4 saplings in a row, then number of gap in saplings = 3

Class_7_Maths_Fraction_&_Decimals_Saplings_Planted

Therefore,

The distance between the first and the last saplings = 3 * 3/4 = 9/4 = 2  m

Thus the distance between the first and the last saplings is 2  m.

Question 6:

Lipika reads a book for 1  hour every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Answer:

Time taken by Lipika to read a book = 1  hours

She reads entire book in 6 days.

Now, total hours taken by her to read the entire book = 6 * 1  hours

                                                                                                  = 6 * 7/4

                                                                                                   = 42/4

                                                                                                   = 21/2

                                                                                                   = 10  hours

Thus, 10  hours were required by her to read the book.

Question 7:

A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2  litres of petrol?

Answer:

In 1 litre of pertrol, car covers the distance = 16 km

In 2  litres of petrol, car covers the distance = 2  of 16 km

                                                                                = 2  * 16

                                                                                 = 11/4 * 16

                                                                                 = 11 * 4

                                                                                  = 44 km

Thus, the car will cover 44 km distance.

Question 8:

(a) (i) Provide the number in the boxClass_7_Fractions_&_Decimals_Box    , such that 2/3 *Class_7_Fractions_&_Decimals_Box    = 10/30

     (ii) The simplest form of the number obtained inClass_7_Fractions_&_Decimals_Box     is __________.

(b) (i) Provide the number in the boxClass_7_Fractions_&_Decimals_Box   , such that 3/5 *Class_7_Fractions_&_Decimals_Box      = 24/25

     (ii) The simplest form of the number obtained inClass_7_Fractions_&_Decimals_Box    is __________.

Answer:

(a)

(i)2/3 *Class_7_Maths_Fraction_&_Decimals_Box1           = 10/30                (ii) The simplest form of 5/10 = 1/2

(b)

(i)  3/5 *Class_7_Maths_Fraction_&_Decimals_Box2          = 24/75              (ii) The simplest form of 8/15 is 8/15

Exercise 2.4

Question 1:

Find:

(i) 12 ÷ 3/4        (ii) 14 ÷ 5/6       (iii) 8 ÷ 7/3      (iv) 4 ÷ 8/3       (v) 3 ÷ 2          (vi) 5 ÷ 3

Answer:

(i) 12 ÷ 3/4 =12 * 4/3 = 4 * 4 = 16

(ii) 14 ÷ 5/6 = 14 * 6/5 = 84/5 = 16

(iii) 8 ÷ 7/3 = 8 * 3/7 = 24/7 = 3

(iv) 4 ÷ 8/3 = 4 * 3/8 = 3/2 = 1

(v) 3 ÷ 2  = 3 ÷ 7/3 = 3 * 3/7 = 9/7 = 1

(vi) 5 ÷ 3  = 5 ÷ 25/7 = 5 * 7/25 = 7/5 = 1

Question 2:

Find the reciprocal of each of the following fractions. Classify the reciprocals as proper

fraction, improper fractions and whole numbers.

(i) 3/7          (ii) 5/8          (iii) 9/7          (iv) 6/5         (v) 12/7        (vi) 1/8          (vii) 1/11

Answer:

(i) Reciprocal of 3/7 = 7/3   -----------à Improper fraction

(ii) Reciprocal of 5/8 = 8/5 ------------à Improper fraction

(iii) Reciprocal of 9/7 = 7/9 -----------à Proper fraction

(iv) Reciprocal of 6/5 = 5/6 -----------à Proper fraction

(v) Reciprocal of 12/7 = 7/12---------à Proper fraction

(vi) Reciprocal of 1/8 = 8 --------------à Proper fraction

(v) Reciprocal of 1/11 = 11 -----------à Proper fraction

Question 3:

Find:

(i) 7/3 ÷ 2        (ii) 4/9 ÷ 5       (iii) 6/13 ÷ 7      (iv) 4  ÷ 3       (v) 3   ÷ 4        (vi) 4  ÷ 7

Answer:

(i) 7/3 ÷2 =7/3 * 1/2 = (7 * 1)/(3 * 2) = 7/6 = 1

(ii) 4/9 ÷ 5 = 4/9 * 1/5 = (4 * 1)/(9 * 5) = 4/45

(iii) 6/13 ÷ 7 = 6/13 * 1/7 = (6 * 1)/(13 * 7) = 6/91

(iv) 4  ÷ 3 = 13/3 ÷ 3 = 13/3 * 1/3 = (13 * 1)/(3 * 3) = 13/9 = 1

(v) 3   ÷ 4 = 7/2 ÷ 4 = 7/2 * 1/4 = (7 * 1)/(2 * 4) = 7/8

(vi) 4  ÷ 7 = 31/7 ÷ 7 = 31/7 * 1/7 = (31 * 1)/(7 * 7) = 31/49

Question 4:

Find:

(i) 2/5 ÷ 1/2                (ii) 4/9 ÷ 2/3              (iii) 3/7 ÷ 8/7                  (iv) 2  ÷ 3/5

(v) 3   ÷ 8/3               (vi) 2/5 ÷ 1               (vii) 3   ÷ 1                   (viii) 2   ÷ 1

Answer:

(i) 2/5 ÷ 1/2 = 2/5 * 2/1 = (2 * 2)/(4 * 1) = 4/5

(ii) 4/9 ÷ 2/3 = 4/9 * 3/2 = (4 * 3)/(9 * 2) = 12/18 = 2/3

(iii) 3/7 ÷ 8/7 = 3/7 * 7/8 = (3 * 7)/(7 * 8) = 21/56 = 3/8

(iv) 2  ÷ 3/5 = 7/3 ÷ 3/5 = 7/3 * 5/3 = (7 * 5)/(3 * 3) = 35/9 = 3

(v) 3   ÷ 8/3 = 7/2 ÷ 8/3 = 7/2 * 3/8 = (7 * 3)/(2 * 8) = 21/16 = 1

(vi) 2/5 ÷ 1  = 2/5 ÷ 3/2 = 2/5 * 2/3 = (2 * 2)/(5 * 3) = 4/15

(vii) 3   ÷ 1   = 16/5 ÷ 5/3 = 16/5 * 3/5 = (16 * 3)/(5 * 5) = 48/25 = 1

(viii) 2   ÷ 1  = 11/5 ÷ 6/5 = 11/5 * 5/6 = (11 * 5)/(5 * 6) = 55/30 = 11/6 = 1

Exercise 2.5

Question 1:

Which is greater:

(i) 0.5 or 0.05           (ii) 0.7 or 0.07           (iii) 7 or 0.7         (iv) 1.37 or 1.49

(v) 2.03 or 2.30       (vi) 0.8 or 0.88

Answer:

(i) 0.5 = 5/10

and 0.05 = 5/100

Now, 5/10 = (5 * 10)/(10 * 10) = 50/100

Since denominator are same and 50 > 5

So, 0.5 is greater than 0.05

(ii) 0.7 = 7/10

and 0.5 = 5/10

Since denominator are same and 7 > 5

So, 0.7 is greater than 0.5

(iii) 7 = 5/1

and 0.7 = 7/10

Now, 7/1 = (7 * 10)/(1 * 10) = 70/10

Since denominator are same and 70 > 7

So, 7 is greater than 0.7

(iv) 1.37 = 137/100

and 1.49 = 149/100

Since denominator are same and 149 > 137

So, 1.49 is greater than 1.37

(v) 2.03 = 203/100

and 2.30 = 230/100

Since denominator are same and 230 > 203

So,2.30 is greater than 2.03

(vi) 0.8 = 8/10

and 0.88 = 88/100

Now, 8/10 = (8 * 10)/(10 * 10) = 80/100

Since denominator are same and 88 > 80

So, 0.88 is greater than 0.8

Question 2:

Express as rupees using decimals:

(i) 7 paise                      (ii) 7 rupees 7 paise                (iii) 77 rupees 77 paise

(iv) 50 paise                 (v) 235 paise

Answer:

Since 100 paise = Rs 1

So, 1 paisa = Rs 1/100

(i) 7 paise = 7 * 1/100 = 7/100 = Rs 0.07

(ii) 7 paise = 7 * 1/100 = 7/100 = Rs 0.07

Now, 7 rupees 7 paise = Rs 7 + Rs 0.07 = Rs 7.07

(iii) 77 paise = 77 * 1/100 = 77/100 = Rs 0.77

Now, 7 rupees 77 paise = Rs 7 + Rs 0.77 = Rs 7.77

(iv) 50 paise = 50 * 1/100 = 50/100 = Rs 0.50

(v) 235 paise = 235 * 1/100 = 235/100 = Rs 2.35

Question 3:

(i) Express 5 cm in meter and kilometer.

(ii) Express 35 mm in cm, m and km.

Answer:

(i) Since 100 cm = 1 m

So, 1 cm = 1/100 m

Hence, 5 cm = 5 * 1/100 = 5/100 = 0.05 m

Again, 1000 m = 1 km

So, 1 m = 1/1000 km

Hence, 0.05 m = 0.05 * 1/1000 = 0.05/1000 = 0.00005 m

(ii) Since 10 mm = 1 cm

So, 1 mm = 1/10 cm

Hence, 35 mm = 35 * 1/10 = 35/10 = 3.5 cm

Again, 100 cm = 1 m

So, 1 cm = 1/100 m

Hence, 3.5 cm = 3.5 * 1/100 = 3.5/100 = 0.035 m

Again, 1000 m = 1 km

So, 1 m = 1/1000 km

Hence, 0.035 m = 0.035 * 1/1000 = 0.035/1000 = 0.000035 m

Question 4:

Express in kg.:

(i) 200 g                     (ii) 3470 g                      (iii) 4 kg 8 g

Answer:

Since 1000 g = 1 kg

So, 1 g = 1/1000 kg

(i) 200 g = 200 * 1/1000 = 200/1000 = 2/10 = 0.2 kg

(ii) 3470 g = 3470 * 1/1000 = 3470/1000 = 3.470 kg

 (iii) 8 g = 8 * 1/1000 = 8/1000 = 0.008 kg

Now, 4 kg 8 g = 4 + 0.008 = 4.008 kg

Question 5:

Write the following decimal numbers in the expanded form:

(i) 20.03               (ii) 2.03                  (iii) 200.03               (iv) 2.034

Answer:

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

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

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

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

Question 6:

Write the place value of 2 in the following decimal numbers:

(i) 2.56            (ii) 21.37            (iii) 10.25             (iv) 9.42            (v) 63.352

Answer:

(i) Place value of 2 in 2.56 = 2 * 1 = 2 ones

(ii) Place value of 2 in 21.37 = 2 * 10 = 2 tens

(iii) Place value of 2 in 10.25 = 2 * 1/10 = 2 tenths

(iv) Place value of 2 in 9.42 =2 * 1/100 = 2 hundredth

(v) Place value of 2 in 63.352 = 2 * 1/1000 = 2 thousandth

Question 7:

Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C.

Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D.

Who travelled more and by how much?

Class_7_Maths_Fraction_&_Decimals_Distance_Travelled1

Answer:

Distance travelled by Dinesh when he went from place A to place B = 7.5 km and from

place B to C = 12.7 km.

Class_7_Maths_Fraction_&_Decimals_Distance_Travelled

Total distance covered by Dinesh = AB + BC

                                                            = 7.5 + 12.7 = 20.2 km

Total distance covered by Ayub = AD + DC

                                                         = 9.3 + 11.8 = 21.1 km

On comparing the total distance of Ayub and Dinesh, 21.1 km > 20.2 km

Therefore, Ayub covered more distance by 21.1 – 20.2 = 0.9 km

                                                                                                   = 0.9 * 1000    [1 km = 1000 m]

                                                                                                   = 900 m

Question 8:

Shyam bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?

Answer:

Total weight of fruits bought by Shyam = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g

Total weight of fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g = 8 kg 950 g

On comparing the quantity of fruits, 8 kg 550 g < 8 kg 950 g

Therefore, Sarala bought more fruits.

Question 9:

How much less is 28 km than 42.6 km?

Answer:

We have to find the difference of 42.6 km and 28 km.

Difference = 42.6 – 28.0 = 14.6 km

Therefore, 14.6 km less is 28 km than 42.6 km.

Exercise 2.6

Question 1:

Find:

(i) 0.2 * 6       (ii) 8 * 4.6       (iii) 2.71 * 5         (iv) 20.1 * 4       (v) 0.05 * 7        (vi) 211.02 * 4

(vii) 2 * 0.86

Answer:

(i) 0.2 * 6 = 1.2              (ii) 8 * 4.6 = 36.8                (iii) 2.71 * 5 = 13.55         (iv) 20.1 * 4 = 80.4

(v) 0.05 * 7 = 0.35        (vi) 211.02 * 4 = 844.08    (vii) 2 * 0.86 = 1.72

Question 2:

Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.

Answer:

Given, Length of rectangle = 5.7 cm

and breadth of rectangle = 3 cm

Area of rectangle = Length * Breadth = 5.7 * 3 = 17.1 cm2

Thus, the area of rectangle is 17.1 cm2.

Question 3:

Find:

(i) 1.3 * 10              (ii) 36.8 * 10          (iii) 153.7 * 10             (iv) 168.07 * 10        (v) 31.1 * 100

(vi) 156.1 * 100    (vii) 3.62 * 100      (viii) 43.07 * 100          (ix) 0.5 * 10               (x) 0.08 * 10

(xi) 0.9 * 100         (xii) 0.03 * 1000

Answer:

(i) 1.3 * 10 = 13.0      (ii) 36.8 * 10 = 368.0     (iii) 153.7 * 10 = 1537.0     (iv) 168.07 * 10 = 1680.7

(v) 31.1 * 100 = 3110.0       (vi) 156.1 * 100 = 15610.0           (vii) 3.62 * 100 = 362.0

(viii) 43.07 * 100 = 4307.0                (ix) 0.5 * 10 = 5.0             (x) 0.08 * 10 = 0.80

(xi) 0.9 * 100 = 90.0                           (xii) 0.03 * 1000 = 30.0

Question 4:

A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?

Answer:

In one litre, a two-wheeler covers a distance = 55.3 km

So, In 10 litres, a two- wheeler covers a distance = 55.3 * 10 = 553.0 km

Thus, 553 km distance will be covered by it in 10 litres of petrol.

Question 5:

Find:

(i) 2.5 * 0.3     (ii) 0.1 * 51.7     (iii) 0.2 * 316.8    (iv) 1.3 * 3.1     (v) 0.5 * 0.05      (vi) 11.2 * 0.15

(vii) 1.07 * 0.02     (viii) 10.05 * 1.05     (ix) 101.01 * 0.01      (x) 100.01 * 1.1

Answer:

(i) 2.5 * 0.3 = 0.75      (ii) 0.1 * 51.7 = 5.17       (iii) 0.2 * 316.8 = 63.36      (iv) 1.3 * 3.1 = 4.03

(v) 0.5 * 0.05 = 0.025          (vi) 11.2 * 0.15 = 1.680                (vii) 1.07 * 0.02 = 0.0214

(viii) 10.05 * 1.05 = 10.5525     (ix) 101.01 * 0.01 = 1.0101       (x) 100.01 * 1.1 = 110.11

Exercise 2.7

Question 1:

Find:

(i) 0.4 ÷ 2        (ii) 0.35 ÷ 5         (iii) 2.48 ÷ 4      (iv) 65.4 ÷ 6        (v) 651.2 ÷ 4     (v) 14.49 ÷ 7

(vii) 3.96 ÷ 4 (viii) 0.80 ÷ 5

Answer:

(i) 0.4 ÷ 2 = 4/10 ÷ 2 = 4/10 * 1/2 = (4 * 1)/(10 * 2) = 4/20 = 1/5 = 0.4

(ii) 0.35 ÷ 5 = 35/100 ÷ 5 = 35/100 * 1/5 = (35 * 1)/(100 * 5) = 35/500 = 7/100 = 0.07

(iii) 2.48 ÷ 4 = 248/100 ÷ 4 = 248/100 * 1/2 = (248 * 1)/(100 * 2) = 248/200 = 124/100 = 1.24

(iv) 65.4 ÷ 6 = 654/100 ÷ 6 = 654/100 * 1/6 = (654 * 1)/(100 * 6) = 654/600 = 109/100 = 1.09

(v) 651.2 ÷ 4 = 6512/10 ÷ 4 = 6512/10 * 1/4 = (6512 * 1)/(10 * 4) = 6512/40 = 1628/10 = 16.28

(v) 14.49 ÷ 7 = 1449/100 ÷ 7 = 1449/100 * 1/7 = (1449 * 1)/(100 * 7) = 1449/700

                       = 207/100 = 2.07

(vii) 3.96 ÷ 4 = 396/100 ÷ 4 = 396/100 * 1/4 = (396 * 1)/(100 * 4) = 396/400 = 99/100 = 0.99

(viii) 0.80 ÷ 5 = 80/100 ÷ 5 = 80/100 * 1/5 = (80 * 1)/(100 * 5) = 80/500 = 16/100 = 0.16

Question 2:

Find:

(i) 4.8 ÷ 10             (ii) 52.5 ÷ 10         (iii) 0.7 ÷ 10         (iv) 33.1 ÷ 10         (v) 272.23 ÷ 10

(vi) 0.56 ÷ 10        (vii) 3.97 ÷ 10

Answer:

(i) 4.8 ÷ 10 = 48/10 ÷ 10 = 48/10 * 1/10 = (48 * 1)/(10 * 10) = 48/100 = 0.48

(ii) 52.5 ÷ 10 = 525/10 ÷ 10 = 525/10 * 1/10 = (525 * 1)/( 10 * 10) = 525/100 = 5.25

(iii) 0.7 ÷ 10 = 7/10 ÷ 10 = 7/10 * 1/10 = (7 * 1)/(10 * 10) = 7/100 = 0.07

(iv) 33.1 ÷ 10 = 331/10 ÷ 10 = 331/10 * 1/10 = (331 * 1)/(10 * 10) = 331/100 = 3.31

(v) 272.23 ÷ 10 = 27223/100 ÷ 10 = 27223/100 * 1/10 = (27223 * 1)/(100 * 10)

                            = 27223/1000 =27.223

(vi) 0.56 ÷ 10 = 56/100 ÷ 10 = 56/100 * 1/10 = (56 * 1)/(10 * 10) = 56/100 = 0.56

(vii) 3.97 ÷ 10 = 397/100 ÷ 10 = 397/100 * 1/10 = (397 * 1)/(100 * 10) = 397/1000 = 0.397

Question 3:

Find:

(i) 2.7 ÷ 100        (ii) 0.3 ÷ 100             (iii) 0.78 ÷ 100          (iv) 432.6 ÷ 100         (v) 23.6 ÷ 100

(vi) 98.53 ÷ 100

Answer:

(i) 2.7 ÷ 100 = 27/10 ÷ 100 = 27/10 * 1/100 = (27 * 1)/(10 * 100) = 27/1000 = 0.027

(ii) 0.3 ÷ 100 = 3/10 ÷ 100 = 3/10 * 1/100 = (3 * 1)/(10 * 100) = 3/1000 = 0.003

(iii) 0.78 ÷ 100 = 78/100 ÷ 100 = 78/100 * 1/100 = (78 * 1)/(100 * 100) = 78/10000 = 0.0078

(iv) 432.6 ÷ 100 = 4326/10 ÷ 100 = 4326/10 * 1/100 = (4326 * 1)/(10 * 100)

                             = 4326/1000 = 4.326

(v) 23.6 ÷ 100 = 236/10 ÷ 100 = 236/10 * 1/100 = (236 * 1)/(10 * 100) = 236/1000 = 0.236

(vi) 98.53 ÷ 100 = 9853/100 ÷ 100 = 9853/100 * 1/100 = (9853 * 1)/(100 * 100)

                             = 9853/10000 = 0.9853

Question 4:

Find:

(i) 7.9 ÷ 1000          (ii) 26.3 ÷ 1000           (iii) 38.53 ÷ 1000            (iv) 128.9 ÷ 1000

(v) 0.5 ÷ 1000

Answer:

(i) 7.9 ÷ 1000 = 79/10 ÷ 1000 = 79/10 * 1/1000 = (79 * 1)/(10 * 1000) = 79/10000 = 0.0079

(ii) 26.3 ÷ 1000 = 263/10 ÷ 1000 = 263/10 * 1/1000 = (263 * 1)/(10 * 1000)

                            = 263/10000 = 0.0263

(iii) 38.53 ÷ 1000 = 3853/100 ÷ 1000 = 3853/100 * 1/1000 = (3853 * 1)/(100 * 1000)

                               = 3853/100000 = 0.03853

(iv) 128.9 ÷ 1000 = 1289/10 ÷ 1000 = 1289/10 * 1/1000 = (1289 * 1)/(10 * 1000)

                               = 79/10000 = 0.01289

(v) 0.5 ÷ 1000 = 5/10 ÷ 1000 = 5/10 * 1/1000 = (5 * 1)/(10 * 1000) = 5/10000 = 0.0005

Question 5:

Find:

(i) 7 ÷ 3.5 (ii) 36 ÷ 0.2 (iii) 3.25 ÷ 0.5 (iv) 30.94 ÷ 0.7 (v) 0.5 ÷ 0.25 (vi) 7.75 ÷ 0.25

(vii) 76.5 ÷ 0.15 (viii) 37.8 ÷ 1.4 (ix) 2.73 ÷ 1.3

Answer:

(i) 7 ÷ 3.5 = 7 ÷ 35/10 = 7/1 * 10/35 = (7 * 10)/(1 * 35) = 70/35 = 2

(ii) 36 ÷ 0.2 = 36 ÷ 2/10 = 36 * 10/2 = 18 * 10 = 180

(iii) 3.25 ÷ 0.5 = 325/100 ÷ 5/10 = 325/100 * 10/5 = (325 * 10)/(100 * 5)

                         = 325/(10 * 5) = 65/10 = 6.5

(iv) 30.94 ÷ 0.7 = 3094/100 ÷ 7/10 = 3094/100 * 10/7 = (3094 * 10)/(100 * 7)

                            = 3094/(10 * 7) = 442/10 = 44.2

(v) 0.5 ÷ 0.25 = 5/10 ÷ 25/100 = 5/10 * 100/25 = (5 * 100)/(10 * 25) = (5 * 10)/25

                         = 50/25 = 2

(vi) 7.75 ÷ 0.25 = 775/100 ÷ 25/100 = 775/100 * 100/25 = (775 * 100)/(100 * 25) = 775/25 = 31

(vii) 76.5 ÷ 0.15 = 765/10 ÷ 15/100 = 765/10 * 100/15 = (765 * 100)/(10 * 15)

                             = (765 * 10)/15 = 7650/15 = 510

(viii) 37.8 ÷ 1.4 = 378/10 ÷ 14/10 = 378/10 * 10/ 14 = (378 * 10)/(10 * 14) = 378/14 = 27

(ix) 2.73 ÷ 1.3 = 273/100 ÷ 13/10 = 273/100 * 10/13 = (273 * 10)/(100 * 13)

                         = 273/(10 * 13) = 21/10 = 2.1

Question 6:

A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre petrol?

Answer:

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

So, in 1 litre of petrol, distance covered by the vehicle = 43.2 ÷ 2.4

                                                                                                 = 432/10 ÷ 24/10

                                                                                                 = 432/10 * 10/24

                                                                                                 = (432 * 10)/(10 * 24)

                                                                                                 = 432/24

                                                                                                 = 18 km

Thus, it covered 18 km distance in one litre of petrol.

What Is 3 5 of 9 3 8

Source: https://www.examfear.com/cbse-ncert-solution/Class-7/Maths/Fractions-and-Decimals/solutions.htm

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