Monday, December 18, 2023

Class – 7 CH-2 FRACTIONS AND DECIMALS MATHS NCERT SOLUTIONS

Class – 7 CH-2 FRACTIONS AND DECIMALS

MATHS NCERT SOLUTIONS

Exercise 2.1

Question 1:

Solve:

(iii) 3 2



5 7 (iv) 9 4 

11 15

(v) 7 2 3

 

10 5 2 (vi) 2 1

2 3 3 2

(vii) 1 5

8 3

2 8

(i) 2  (ii) 4 

SOLUTION 1:

(i) 2  = 103  7 = 1

5 5

(ii) 4  = 327  39 = 4

8 8

3 2

(iii)  =

5 7

9 4 13544 91 (iv)  = 

11 15 165 165

7 2 3 26 13 (v)    = 2

10 5 2 10 5

2 1 8 7

(vi) 2 3 =  = = = 6

3 2 3 2

(vii) 81 35 = 17 29 = = 4

2 8 2 8

Question 2:

Arrange the following in descending order:

2 2 8 1 3 7

(i) , , (ii) , ,

9 3 21 5 7 10

SOLUTION 2:

2 2 8 (i) , ,

9 3 21

14 42 24

 , , [Converting into like fractions]

63 63 63

42 24 14

   [Arranging in descending order]

63 63 63

2 8 2

Therefore,  

3 21 9

1 3 7

(ii) , ,

5 7 10

14 30 49

 , , [Converting into like fractions] 70 70 70

49 30 14

   [Arranging in descending order]

70 70 70

7 3 1

Therefore,  

10 7 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?

Along the first row 4  9  2  15

 11 11 11 11

Sum of first row 4 9 2 15

=    [Given]

SOLUTION 3:

11 11 11 11

3 5 7 3 5 7 15 Sum of second row =    

11 11 11 11 11

8 1 6 8 1 6 15 Sum of third row =    

11 11 11 11 11

4 3 8 4 3 8 15 Sum of first column =    

11 11 11 11 11

9 5 1 9 5 1 15 Sum of second column =    

11 11 11 11 11

2 7 6 2 7 6 15 Sum of third column =    

11 11 11 11 11

4 5 6 4 5 6 15

Sum of first diagonal (left to right) =    

11 11 11 11 11

2 5 8 2 5 8 15

Sum of second diagonal (left to right) =    

11 11 11 11 11

Since the sum of fractions in each row, in each column and along the diagonals are 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.

SOLUTION 4:

Given: The sheet of paper is in rectangular form.

Length of sheet = 12 cm and Breadth of sheet = 10 cm

Perimeter of rectangle = 2 (length + breadth)

= 2 12 121032 = 225 322  3 



= 2 25 3 6322  = 2 756 64 



139 1 = 2 =  46 cm.

3 3

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

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

SOLUTION 5:

(i) In ABE, AB = cm, BE = 2 cm, AE = 3 cm

The perimeter of ABE = AB + BE + AE

5 3 3 5 11 18 =  2 3 =  

2 4 5 2 4 5

50 55 72

= = = 8 cm

Thus, the perimeter of ABEis 8 cm.

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

Perimeter of rectangle = 2 (length + breadth)

= 2 2 3 74 6  = 211 74 6 

= 2 331214  = 476 = 7 56 cm



Thus, the perimeter of rectangle BCDE is 7 cm.

Comparing the perimeter of triangle and that of rectangle,

8 cm > 7 cm

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

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?

SOLUTION 6:

Therefore, the picture should be trimmed 3 3 38 73

= 7 7 = 

5 10 5 10

Given: The width of the picture = 7 cm and the width of picture frame = 7 cm

= = cm

Thus, the picture should be trimmed by cm.

Question 7:

Ritu ate 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?

SOLUTION 7:

The part of an apple eaten by Ritu =

The part of an apple eaten by Somu = 1 3 53  2

5 5 5

3 2

Comparing the parts of apple eaten by both Ritu and Somu 

5 5

3 2 1

Larger share will be more by   part.

5 5 5

Thus, Ritu’s part is more than Somu’s part.

Michael finished colouring a picture in hour. Vaibhav finished colouring the same picture in hour. Who worked longer? By what fraction was it longer?

SOLUTION 8:

Time taken by Michael to colour the picture = hour

Time taken by Vaibhav to colour the picture = hour

Converting both fractions in like fractions, and

Here, <  <

Thus, Vaibhav worked longer time.

3 7 97 2 1

Vaibhav worked longer time by     hour.

4 12 12 12 6

Thus, Vaibhav took hour more than Michael.

Exercise 2.2

Question 1:

Which of the drawings a to d show:

(i) 2

(ii) 2

(iii) 3

(iv) 3

(i) – (d)

Since 1 1 1

2   5 5 5

(ii) – (b)

Since 1 1 1 2  

2 2 2

(iii) – (a)

Since 2 2 2 2

3    3 3 3 3

(iv) – (c) Since 1 1 1 1

3   

4 4 4 4

SOLUTION 1:

Question 2:

Some pictures a to c are given below. Tell which of them show:

1 3

(i) 3 

5 5

1 2

(ii) 2 

3 3

3 1

(iii) 3  2

4 4

(i) – (c)

Since 1 1 1 1

3    5 5 5 5

(ii) – (a)

Since 1 1 1

2   3 3 3

(iii) – (b) Since 3 3 3 3 3   

4 4 4 4

SOLUTION 2:

Question 3:

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

(i) 7 (ii) 4 (iii) 2 (iv) 5

(v) 4 (vi) 6 (vii) 11 (viii) 20

(ix) 13 (x) 15

SOLUTION 3:

21 1 (i) 7 = =  4

5 5

4 1

(ii) 4 = = 1

3 3

12 5

(iii) 2 = = 1

7 7

10 1 (iv) 5 = = 1

9 9

(v) 4 = = = 2

(vi) 6 = 5 3 = 15

(vii) 11 = = = 6

(viii) 20 = 4 x 4 = 16

13 1 (ix) 13 =  4

3 3

(x) 15 = 3 x 3 = 9

Question 4:

Shade:

(i) of the circles in box

(ii) of the triangles in box

(iii) of the squares inbox

SOLUTION 4:

(i) of 12 circles

= 12 = 6 circles

(ii) of 9 triangles

= 9 = 2 x 3 = 6 triangles

(iii) of 15 squares

= 15 3 x 3 = 9 squares

Question 5:

Find:

(a) of (i) 24 (ii) 46 (b) of (i) 18 (ii) 27

(a) (i) of 24 = 12 (ii) of 46 = = 23

(b) (i) of 18 = 18 = 2 x 6 = 12 (ii) of 27 = 27 = 2 x 9 = 18

(d) (i) of 20 = 20 = 4 x 4 = 16 (ii) of 35 = 35 = 4 x 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

SOLUTION 6:

(a) 3 5   1 3 26 3 26  78 15 3

5 5 5 5 5

(b) 5 6   3 5 27 5 27 135 333

4 4 4 4 4

(d) 4 6   1 4 19 4 19  76 25 1

3 3 3 3 3

(e) 3 1   6 13 6 13 3   39 19 1

4 4 2 2 2

(f) 3 2   8 17 8 17 8 136  271

5 5 5 5 5

Question 7:

Find:

(a) of (i) 2 (ii) 4 (b) of (i) 3 (ii) 9

SOLUTION 7:

1 3 1 11 11 3

(a) (i) of 2 = 2 =   1

2 4 2 4 8 8

1 2 1 38 19 1

(ii) of 4 = 4 =    2

2 9 2 9 9 9

5 5 5 23 115 19 (b) (i) of 3 =    3  2

8 6 8 6 48 48

5 2 5 29 145 1 (ii) of 9 =    9  6

8 3 8 3 24 24

Question 8:

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

SOLUTION 8:

Given: Total quantity of water in bottle = 5 litres

(i) Vidya consumed = of 5 litres = 5 = 2 litres Thus, Vidya drank 2 litres water from the bottle. (ii) Pratap consumed = 1 52  part of bottle

52 3

=  part of bottle

5 5

Pratap consumed of 5 litres water = 5 = 3 litres

Thus, Pratap drank part of the total quantity of water.

Question 1:

Find:

(i) of

(ii) of

SOLUTION 1:

(i) (a)

(b)

(c)

(ii) (a)

(b)

(c)

Question 2:

2 2

(i) 2

3 3

1 15 (v) 

3 8

SOLUTION 2:

2 2 (i)    2

3 3

2 7 (ii)   7 9

of of of

Multiply and reduce to lowest form (if possible):

2 8 3 3

 Exercise 2.3

(a) (b)

1 1 1 1 1 =   

4 4 4 4 16

1 3 1 3 3 =   

4 4 4 4 16

1 4 1 4 1 =   

4 3 4 3 3

1 2 1 2 2 =   

7 9 7 9 63

1 6 1 6 6 =   

7 5 7 5 35

1 3 1 3 3 =   

7 10 7 10 70

2 7

(ii)  (iii)

7 9

11 3 (vi)  (vii) 2 10

2 8 16 7  1

3 3 9 9

(c)

3 6



8 4

4 12

 5 7 9 3

(iv)  5 5

3 6 3 6 3 3 9 (iii)    

8 4 8 4 8 2 16

9 3 9 3 27 2 (iv)    1

5 5 5 5 25 25

1 15 1 15 1 5 5 (v)    

3 8 3 8 1 8 8

11 3 11 3 33 3

(vi)

2 10 2 10 20 20

(vii)     48 113 4 12 4 12

5 7 5 7 35 35

Question 3:

Multiply the following fractions:

2 1 2 7

(i) 5 (ii) 6 

5 4 5 9

2 4

(v) 3  (vi) 23

5 7

SOLUTION 3:

(i)    5  1 21 2 1 2 21 2 21

5 4 5 4 5 4 5 2

(ii) 6 4

3 1 3 16 48 (iii)     5 8

2 3 2 3 6

5 3 5 17 85 1 (iv)     2 2

6 7 6 7 42 42

2 4 17 4 68 33 (v) 3     1

5 7 7 7 35 35

(vi) 2    3   39 7 4 3 13 3 13 3

5 5 1 5 1 5 5

(vii) 3       15 2 1 4 3 25 3 5 3

7 5 7 5 7 1 7 7

(iii)

(vii)

21 1   2

10 10

3 1

5

2 3

4 3 3  7 5 5 3

(iv) 2

6 7

Question 4:

Which is greater:

(i) of or of (ii) of or of

SOLUTION 4:

(i) of or of

 x or x

 3 3



14 8

 or

Thus, of is greater.

(ii) of or of

 x or x

 or

 >

Thus, of is greater.

Question 5:

Saili plants 4 saplings in a row in her garden. The distance between two adjacent saplings is m. Find the distance between the first and the last sapling.

SOLUTION 5:

The distance between two adjacent saplings = m

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

Therefore,

The distance between the first and the last saplings = 3 = m = 2 m

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

Question 6:

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

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 = 1 6

7 21 1

=   6 10 hours

4 2 2

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?

SOLUTION 7:

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

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

= 16 = 44 km

Thus, the car will cover 44 km distance.

Question 8:

2 10

(a) (i) Provide the number in the box , such that   .

3 30

(ii) The simplest form of the number obtained in is __________.

3 24

(b) (i) Provide the number in the box , such that   .

5 75

(ii) The simplest form of the number obtained in is __________. SOLUTION 8:

2 5 10 5 1

(a) (i)   (ii) The simplest form of is .

3 10 30 10 2

3 8 24 8 8

(b) (i)   (ii) The simplest form of is .

5 15 75 15 15

Exercise 2.4

Question 1:

Find:

(i) 12 (ii) 14  (iii) 8

(iv) 4 (v) 32 (vi) 53

SOLUTION 1:

6 84 4

(i) 12 = 12 = 16 (ii) 14  = 14  16

5 5 5

3 24 33 3 1

(iii) 8 = 8   3 (iv) 4 = 4  1

7 7 78 2 2

7 3 9 225 7 7 2

(v) 32 = 3    3 1 (vi) 53 = 5    5 1

3 7 7 77 25 5 5

Question 2:

Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fraction, improper fractions and whole numbers.

(i) (ii) (iii) (iv)

(v) (vi) (vii)

(i) Reciprocal of 3 7

  Improper fraction

7 3

(ii) Reciprocal of 5 8

  Improper fraction

8 5

(iii) Reciprocal of 9 7

  Proper fraction

7 9

(iv) Reciprocal of 6 5

  Proper fraction

5 6

(v) Reciprocal of 12 7

  Proper fraction

SOLUTION 2:

7 12

(vi) Reciprocal of  8  Whole number

(vi) Reciprocal of  11  Whole number

Question 3:

Find:

(i)  2 (ii) 5 (iii) 7

(iv) 4 3 (v) 3 4 (vi) 4 7

SOLUTION 3:

(i) 7    2 7 1 7 1  7 1 1

3 3 2 3 2 6 6

4 4 1 4 1 4 (ii)    5 

9 9 5 9 5 45

6 6 1 6 1 6

(iii)    7 

13 13 7 13 7 91

1 13 13 1 13 4

(iv) 4       3 3 1

3 3 3 3 9 9

1 7 7 1 7 (v) 3      4 4

2 2 2 4 8

3 31 31 1 31

(vi) 4      7 7

7 7 7 7 49

Question 4:

Find:

2 1 4 2 3 8

(i)  (ii)  (iii) 

5 2 9 3 7 7

1 3 1 8 2 1

(iv) 2  (v) 3  (vi) 1

3 5 2 3 5 2

1 2 1 1

(vii) 3 1 (viii) 2 1

5 3 5 5

SOLUTION 4:

2 1 2 2 2 2 4 (i)     

5 2 5 1 5 1 5

4 2 4 3 2

(ii)    

9 3 9 2 3

3 8 3 7 3

(iii)    

7 7 7 8 8

1 3 7 3 7 5 35 8

(iv) 2       3

3 5 3 5 3 3 9 9

(v) 3 1      8 7 8 7 3 7 3  21 1 5

2 3 2 3 2 8 2 8 16 16

2 1 2 3 2 2 2 2

(vi)      1   4

5 2 5 2 5 3 5 3 15

(vii) 3 1      12 16 5 16 3 16 3   48 1 23

5 3 5 3 5 5 5 5 25 25

1 1 11 6 11 5 11 5

(viii) 2       1 1

5 5 5 5 5 6 6 6

Exercise 2.5

Question 1:

Which is greater:

(i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7

(iv) 1.37 or 1.49 (v) 2.03 or 2.30 (vi) 0.8 or 0.88

SOLUTION 1:

(i) 0.5 > 0.05 (ii) 0.7 > 0.5 (iii) 7 > 0.7

(iv) 1.37 < 1.49 (v) 2.03 < 2.30 (vi) 0.8 < 0.88

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

SOLUTION 2:

100 paise = ₹1

 1 paisa = ₹

(i) 7 paise = ₹ = ₹ 0.07

(ii) 7 rupees 7 paise = ₹ 7 + ₹ = ₹ 7 + ₹ 0.07 = ₹ 7.07

(iii) 77 rupees 77 paise = ₹ 77 + ₹ = ₹ 77 + ₹ 0.77 = ₹ 77.77

(iv) 50 paise = ₹ = ₹ 0.50

(v) 235 paise = ₹ = ₹ 2.35

Question 3:

(i) Express 5 cm in metre and kilometer. (ii) Express 35 mm in cm, m and km. SOLUTION 3:

(i) Express 5 cm in meter and kilometer.

100 cm = 1 meter

 1 cm = meter

 5 cm = = 0.05 meter.

Now,

1000 meters = 1 kilometers

 1 meter = kilometer

 0.05 meter = = 0.00005 kilometer

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

10 mm = 1 cm

 1 mm = cm

 35 mm = = 3.5 cm

Now, 100 cm = 1 meter

 1 cm = meter

 3.5 cm = = 0.035 meter

Again,

1000 meters = 1 kilometers

 1 meter = kilometer

0.035

 0.035 meter = = 0.000035 kilometer

1000

Express in kg.:

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

SOLUTION 4:

Let us consider,

1000 g = 1 kg

 1 g = kg

(i) 200 g = 2001000 1  kg = 0.2 kg

(ii) 3470 g = 34701000 1  kg = 3.470 kg



(iii) 4 kg 8 g = 4 kg + 81000 1  kg = 4 kg + 0.008 kg = 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

1 1

(i) 20.03 = 2 10 0 1 0      3

10 100

1 1

(ii) 2.03 = 2 1    0 3

10 100

1 1

(iii) 200.03 = 2 100        0 10 0 1 0 3

10 100

1 1 1

(iv) 2.034 = 2 1    0 3  4 10 100 1000

Question 6:

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

(i) 2.56 (ii) 21.37

(iv) 9.42 (v) 63.352 (iii) 10.25

SOLUTION 5:

SOLUTION 6:

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

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

(iii) Place value of 2 in 10.25 = 2 = 2 tenths

(iv) Place value of 2 in 9.42 = 2 = 2 hundredth

(v) Place value of 2 in 63.352 = 2 = 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?

SOLUTION 7:

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.

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 = 900 m

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?

SOLUTION 8:

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?

SOLUTION 9:

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 x 6 (ii) 8 x 4.6 (iii) 2.71 x 5

(iv) 20.1 x 4 (v) 0.05 x 7 (vi) 211.02 x 4

(vii) 2 x 0.86

SOLUTION 1:

(i) 0.2 x 6 = 1.2 (ii) 8 x 4.6 = 36.8

(iii) 2.71 x 5 = 13.55 (iv) 20.1 x 4 = 80.4

(v) 0.05 x 7 = 0.35 (vi) 211.02 x 4 = 844.08

(vii) 2 x 0.86 = 1.72

Question 2:

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

SOLUTION 2:

Given: Length of rectangle = 5.7 cm and

Breadth of rectangle = 3 cm

Area of rectangle = Length x Breadth = 5.7 x 3 = 17.1 cm2 Thus, the area of rectangle is 17.1 cm2.

Question 3:

Find:

(i) 1.3 x 10 (ii) 36.8 x 10 (iii) 153.7 x 10

(iv) 168.07 x 10 (v) 31.1 x 100 (vi) 156.1 x 100

(vii) 3.62 x 100 (viii) 43.07 x 100 (ix) 0.5 x 10 (x) 0.08 x 10 (xi) 0.9 x 100 (xii) 0.03 x 1000

SOLUTION 3:

(i) 1.3 x 10 = 13.0 (ii) 36.8 x 10 = 368.0

(iii) 153.7 x 10 = 1537.0 (iv) 168.07 x 10 = 1680.7

(v) 31.1 x 100 = 3110.0 (vi) 156.1 x 100 = 15610.0

(vii) 3.62 x 100 = 362.0 (viii) 43.07 x 100 = 4307.0

(ix) 0.5 x 10 = 5.0 (x) 0.08 x 10 = 0.80

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

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?

SOLUTION 4:

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

 In 10 litres, a two- wheeler covers a distance = 55.3 x 10 = 553.0 km Thus, 553 km distance will be covered by it in 10 litres of petrol.

Question 5:

Find:

(i) 2.5 x 0.3 (ii) 0.1 x 51.7 (iii) 0.2 x 316.8

(iv) 1.3 x 3.1 (v) 0.5 x 0.05 (vi) 11.2 x 0.15

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

(x) 100.01 x 1.1 SOLUTION 5:

(i) 2.5 x 0.3 = 0.75 (ii) 0.1 x 51.7 = 5.17

(iii) 0.2 x 316.8 = 63.36 (iv) 1.3 x 3.1 = 4.03

(v) 0.5 x 0.05 = 0.025 (vi) 11.2 x 0.15 = 1.680

(vii) 1.07 x 0.02 = 0.0214 (viii) 10.05 x 1.05 = 10.5525

(ix) 101.01 x 0.01 = 1.0101 (x) 100.01 x 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

SOLUTION 1:

4 1 2

(i) 0.4  2 =   = 0.2

10 2 10

35 1 7

(ii) 0.35  5 =   = 0.07

100 5 100

248 1 62

(iii) 2.48  4 =   = 0.62

100 4 100

654 1 109 (iv) 65.4  6 =   = 10.9

10 6 10

6512 1 1628 (v) 651.2  4 =   = 162.8

10 4 10

1449 1 207 (vi) 14.49  7 =   = 2.07

100 7 100

396 1 99

(vii) 3.96  4 =   = 0.99

100 4 100

80 1 16

(viii) 0.80  5 =   = 0.16

100 5 100

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 SOLUTION 2:

4.8

(i) 4.8  10 = = 0.48 (ii) 52.5  10 = = 5.25

0.7

(iii) 0.7  10 = = 0.07 (iv) 33.1  10 = = 3.31

0.56

3.97

(vii) 3.97  10 = = 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

(v) 272.23  10 = = 27.223 (vi) 0.56  10 = = 0.056 10

SOLUTION 3:

27 1 27

(i) 2.7  100 =   = 0.027

10 100 1000

3 1 3

(ii) 0.3  100 =   = 0.003

10 100 1000

78 1 78

(iii) 0.78  100 =   = 0.0078

100 100 10000

4326 1 4326 (iv) 432.6  100 =   = 4.326

10 100 1000

236 1 236

(v) 23.6  100 =   = 0.236

10 100 1000

9853 1 9853 (vi) 98.53  100 =   0.9853

100 100 10000

Question 4:

Find:

(i) 7.9  1000 (ii) 26.3  1000 (iii) 38.53  1000

(iv) 128.9  1000 (v) 0.5  1000 SOLUTION 4:

79 1 79

(i) 7.9  1000 =   = 0.0079

10 1000 10000

263 1 263

(ii) 26.3  1000 =   = 0.0263

10 1000 10000

3853 1 3853

(iii) 38.53  1000 =   = 0.03853

100 1000 100000

1289 1 1289 (iv) 128.9  1000 =   = 0.1289

10 1000 10000

5 1 5

(v) 0.5  1000 =   = 0.0005 10 1000 10000

Question 5:

Find:

(i) 7  3.5 (ii) 36  0.2 (iii)

(iv) 30.94  0.7 (v) 0.5  0.25 (vi) (vii) 76.5  0.15 (viii) 37.8  1.4 (ix)

SOLUTION 5:

35 10 10

(i) 7  3.5 = 7   7 = 2

10 35 5

2 10

(ii) 36  0.2 = 36   36 18 x 10 = 180

10 2

325 5 325 10 65 (iii) 3.25  0.5 =     = 6.5 100 10 100 5 10

3094 7 3094 10 442 (iv) 30.94  0.7 =     = 44.2

100 10 100 7 10

5 25 5 100 10

(v) 0.5  0.25 =     = 2

10 100 10 25 5

775 25 775 100 (vi) 7.75  0.25 =    = 31

100 100 100 25

765 15 765 100

(vii) 76.5  0.15 =    = 51 x 10 = 510

10 100 10 15

378 14 378 10 (viii) 37.8  1.4 =    = 27

10 10 10 14

273 13 273 10 21 (ix) 2.73  1.3 =     = 2.1 100 10 100 13 10

3.25  0.5

7.75  0.25

2.73  1.3

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?

SOLUTION 6:

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

 In 1 litre of petrol, distance covered by the vehicle = 43.2  2.4

432 24 432 24

=  = 

10 10 10 10

= 18 km

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

Class – 7 CH-1 INTEGERS MATHS NCERT SOLUTIONS

Class – 7 CH-1 INTEGERS

MATHS NCERT SOLUTIONS

EXERCISE 1.1

Question 1:

Following number line shows the temperature in degree Celsius (°C) at different places on a particular day:

(a) Observe this number line and write the temperature of the places marked on it.

(b) What is the temperature difference between the hottest and the coldest places among the above?

(d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?

SOLUTION 1:

(a) The temperature of the places marked on it is:

Places Temperature

Lahulspiti –8°C

Srinagar –2°C

Shimla 5°C

Ooty 14°C

Bangalore 22°C

(b) The temperature of the hottest place Bangalore = 22°C

The temperature of the coldest place Lahulspiti = –8°C

Difference = 22°C – (–8°C)

= 22°C + 8°C

= 30°C

The temperature of Lahulspiti = –8°C

Difference = –2°C + (–8°C)

= –2°C –8°C

= 6°C

(d) The temperature of Srinagar and Shimla = 5°C + (–2°C)

= 5°C –2°C = 3°C

The temperature at Shimla = 5°C

Therefore, 3°C < 5°C

Thus, temperature of Srinagar and Shimla taken together is less than the temperature at Shimla. °

Now, Temperature of Srinagar = –2°C

Therefore, 3°C > –2°C

No, THE TEMPERATURE OF SRINAGAR AND SHIMLA TOGETHER is not less than the temperature at Srinagar.

Question 2:

In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, -5, -10,15 and 10, what was his total at the end?

SOLUTION 2:

Jack’s scores in five successive rounds are 25, -5, -10,15 and 10.

Total marks got by Jack = 25 +(- 5) + (-10) +15+10

= 25 – 15 + 25 = 35

Thus, 35 marks are got by Jack in a quiz.

Question 3:

At Srinagar temperature was -5° C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?

SOLUTION 3:

On Monday, temperature at Srinagar = –5°C

On Tuesday, temperature dropped = 2°C

Temperature on Tuesday

= –5°C – 2°C

= –7°C

On Wednesday, temperature rose up = 4°C

Temperature on Wednesday = –7°C + 4°C = –3°C

Thus, temperature on Tuesday and Wednesday was –7°C and –3°C respectively.

Question 4:

A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?

SOLUTION 4:

Height of a place above the sea level = 5000 m

Floating a submarine below the sea level = 1200 m

The vertical distance between the plane and the submarine

= 5000 - (- 1200) = 5000+1200=6200 m

Thus, the vertical distance between the plane and the submarine is 6200 m.

Question 5:

Mohan deposits ₹2,000 in his bank account and withdraws ₹1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s accounts after the withdrawal?

SOLUTION 5:

Deposit amount = ₹2,000 and

Withdrawal amount = ₹1,642

Balance = 2,000 – 1,642 = ₹358

Thus, the balance in Mohan’s account after withdrawal is ₹ 358.

Question 6:

Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?

SOLUTION 6:

According to the number line, Rita moves towards east is represented by a positive integer. But she moves in opposite direction means Rita moves west, is represented by negative integer.

Distance from A to B = 20 km

Distance from B to C = 30 km

Distance from A to C = 20 + ( – 30) = –10 km

Thus, Rita is at final position from A to C is –10 km.

Question 7:

In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square.

SOLUTION 7:

(i) Taking rows 5 + (–1) + (–4) = 5 – 5 = 0

(–5) + (-2) + 7 = –7 + 7 = 0

0 + 3 + (–3) = 3 – 3 = 0

Taking columns

5 + (–5) + 0 = 5 – 5 = 0

(–1) + (–2) + 3 = –3 + 3 = 0

(–4) + 7 + (–3) = 7 – 7 = 0

Taking diagonals

5 + (–2) + (–3) = 5 – 5 = 0

(–4) + (–2) + 0 = –6

This box is not a magic square because all the sums are not equal.

(ii) Taking rows

1 + (–10) + 0 = 1 – 10 = –9

(–4) + (–3) + (–2) = –7 – 2 = –9

(–6) + 4 + (–7) = –2 – 7 = –9

Taking columns

1 + (–4) + (–6) = 1 – 10 = –9

(–10) + (–3) + 4 = –13 + 4 = –9

0 + (–2) + (–7) = 0 – 9 = –9

Taking diagonals

1 + (–3) + (–7) = 1 – 10 = –9

0 + (–3) + (–6) = –9

This box is magic square because all the sums are equal.

Question 8:

Verify a— (—b) = a+ b for the following values of a and b :

a=21. b=18

a=118,b=125

a=75,b=84

a=28, b=11

SOLUTION 8:

(i) Given: a = 2 1,b = 18

We have a— (—b) = a+ b

Putting the values in L.H.S. = a —(—b) = 21—(—18) = 21 + 18 = 39

Putting the values in R.H.S. = a+ b —— 21 + 19 = 39

Since, L.H.S. = R.H.S

Hence, verified.

ii)

Given: a = 118,b = 125

We have a— (—b) = a+ b

Putting the values in L.H.S. = a —(—b) = 118 —(—125) = 118 + 125 = 243

Putting the values in R.H.S. = a + b —— 118 + 125 = 243

Since, L.H.S. = R.H.S

Hence, verified.

iii)

Given: a = 75,b = 84

We have a— (—b) = a+ b

Putting the values in L.H.S. = a —(—b) = 75 —(—84) = 75 + 84 = 159

Putting the values in R.H.S. = a + b —— 75 + 84 = 159

Since, L.H.S. = R.H.S

Hence, verified.

iv)

Given: a = 28,b = 11

We have a— (—b) = a+ b

Putting the values in L.H.S. = a—(—b) = 28 —(— I I) = 28 + 11 = 39

Putting the values in R.H.S. = a+ b -- 28 + 11 = 39

Since, L.H.S. = R.H.S

Hence, verified.

Question 9:

Use the sign of >, < or = in the box to make the statements true:

(a)   8  4  8  4

(b)   3 7 1915 8   9

(d) 39  241536  52  36

(e) 231795139915981 SOLUTION 9:

(a)   8  4  8  4

  8 4 8 4

 124

 12 < 4

(b)   3 7 1915 8   9

   3 7 1915 8 9

 4 1915 17

 152

 15 < 2

  18 112352

 729

 7 > 29

(d) 39  241536  52  36

 39 24 1536 52 36

 39397252

 020

 0 < 20

(e) 231795139915981

  231 130399240

 101159

 101 > 159

Question 10:

A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step:

(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?

(ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?

(iii) If the number of steps moved down is represented by negative integers and the number of steps move up by positive integers, represent his moves in part (i) and (ii) by completing the following:

(a)   3 2 ..........8 (b) 4 2 ........8

In (a) the sum 8 represent going down by eight steps. So, what will the sum 8 in (b) represent?

SOLUTION 10:

(i) He jumps 3 steps down and jumps back 2 steps up. Following number ray shows the jumps of monkey:

First jump = 1 + 3 = 4 steps

Second jump = 4 – 2 = 2 steps

Third jump = 2 + 3 = 5 steps

Fourth jump = 5 – 2 = 3 steps

Fifth jump = 3 + 3 = 6 steps

Sixth jump = 6 – 2 = 4 steps

Seventh jump = 4 + 3 = 7 steps

Eighth jump = 7 – 2 = 5 steps

Ninth jump = 5 + 3 = 8 steps

Tenth jump = 8 – 2 = 6 steps

Eleventh jump = 6 + 3 = 9 steps He will reach ninth steps in 11 jumps.

(ii) He jumps four steps and then jumps down 2 steps. Following number ray shows the jumps of monkey:

Thus monkey reach back on the first step in fifth jump.

(iii) (a)                 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 8 (b) 4       2 4 2 4 2 4 2 8

Thus, sum 8 in (b) represents going up by eight steps.

Exercise 1.2

Question 1:

Write down a pair of integers whose:

(a) sum is 7

(b) difference is 10

(a) sum is 0 SOLUTION 1:

(a) One such pair whose sum is 7:    5  2 7

(b) One such pair whose difference is 10:   2 8 10

Question 2:

(a) Write a pair of negative integers whose difference gives 8.

(b) Write a negative integer and a positive integer whose is 5. (c) Write a negative integer and a positive integer whose difference is 3.

SOLUTION 2:

(a)      2  10 2 10 8

(b)   7 2 5

Question 3:

In a quiz, team A scored 40,10,0 and team B scores 10, 0, 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

SOLUTION 3:

Team A scored 40,10,0

Total score of Team A =    40 10 0 30

Team B scored 10,0,40

Total score of Team B = 10  0  40   10 0 40 30

Thus, scores of both teams are same.

Yes, we can add integers in any order due to commutative property.

Question 4:

(i) 

(ii)

(iii) 17

(iv) 

(v) 

SOLUTION 4

(i)

(ii)

(i)

(ii)

(iii)

Fill in the blanks to make the following statements true:

     5  8  8 .......

 53 .......53

+ ……. = 0

13  12.......13  12  7

 4 15  3   4 15.......

      5  8  8  5

  53 0 53

17  17 0

13 12    7 13   12  7

 4 15  3   4 15  3

[Commutative property]

[Zero additive property]

(Additive identity]

[Associative property]

Exercise 1.3

Question 1:

Find the each of the following products:

(a) 3 x (–1) (b) (–1) x 225

(e) (–15) x 0 x (–18)

(f) (–12) x (–11) x (10)

(g) 9 x (–3) x (–6)

(h) (–18) x (–5) x (–4)

(i) (–1) x (–2) x (–3) x 4

(j) (–3) x (–6) x (2) x (–1)

SOLUTION 1:

(a) 3 x (–1) = –3

(b) (–1) x 225 = –225

(d) (–316) x (–1) = 316

(e) (–15) x 0 x (–18) = 0

(f) (–12) x (–11) x (10) = 132 x 10 = 1320

(g) 9 x (–3) x (–6) = 9 x 18 = 162

(h) (–18) x (–5) x (–4) = 90 x (–4) = –360

(i) (–1) x (–2) x (–3) x 4 = (–6 x 4) = –24

(j) (–3) x (–6) x (2) x (–1) = (–18) x (–2) = 36

Question 2:

Verify the following:

(a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)]

(b) (–21) x [(–4) + (–6)] = [(–21) x (-4)] + [(–21) x (–6)]

SOLUTION 2:

(a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)]

 18 x 4 = 126 + (–54)

 72 = 72

 L.H.S. = R.H.S.

Hence verified.

(b) (–21) x [(–4) + (–6)] = [(–21) x (–4)] + [(–21) x (–6)]

 (–21) x (–10) = 84 + 126

 210 = 210

 L.H.S. = R.H.S.

Hence verified.

Question 3:

(i) For any integer a, what is  1 a equal to?

(ii) Determine the integer whose product with 1 is:

(a) –22 (b) 37 (c) 0

SOLUTION 3:

(i)   1 a a, where a is an integer.

(ii) (a)   1  22  22

(b)   1 37 37

Question 4:

Starting from  1 5, write various products showing some patterns to show

   1  1 1.

SOLUTION 4:

  1 5 5   1 4 4

  1 3 3   1 2 2

  1 1 1   1 0 0

   1  1 1

Thus, we can conclude that this pattern shows the product of one negative integer and one positive integer is negative integer whereas the product of two negative integers is a positive integer.

Question 5:

Find the product, using suitable properties:

(a) 26      48  48  36 (b) 8 53   125

(e) 625    35  62565 (f) 7502

(g)   17  29 (h)    57  19 57

SOLUTION 5:

(a) 26    48  48  36

 4826  36

   48  10

 480

(b) 8 53   125

 53  8  125

 53  1000

 53000

 15  25    4  10

 15  1000

 15000

(d) 41102

  41 100 2

 41100    412

 4100  82

 4182

(e) 625    35  62565

 625  35  65

 625  100

 62500

 7 50  7 2

 350 14 336

[Distributive property]

[Commutative property]

[Distributive property]

(d)   17  29

 17  301 [Distributive property]

  17 30   17 1

 510  17

 493

(e)    57  19 57

     57  19 57 1

 57 x 19 + 57 x 1

 57 x (19 + 1) [Distributive property]

 57 x 20 = 1140

Question 6:

A certain freezing process requires that room temperature be lowered from 40oC at the rate of 5oC every hour. What will be the room temperature 10 hours after the process begins?

SOLUTION 6:

Given: Present room temperature = 40oC

Decreasing the temperature every hour = 5oC

Room temperature after 10 hours = 40oC + 10 x (–5oC )

= 40oC – 50oC

= – 10oC

Thus, the room temperature after 10 hours is – 10oC after the process begins.

Question 7:

In a class test containing 10 questions, 5 marks are awarded for every correct SOLUTION and 2 marks are awarded for every incorrect SOLUTION and 0 for questions not attempted.

(i) Mohan gets four correct and six incorrect SOLUTIONs. What is his score?

(ii) Reshma gets five correct SOLUTIONs and five incorrect SOLUTIONs, what is her score?

(iii) Heena gets two correct and five incorrect SOLUTIONs out of seven questions she attempts. What is her score?

SOLUTION 7:

(i) Mohan gets marks for four correct questions = 4 x 5 = 20

He gets marks for six incorrect questions = 6 x (–2) = –12

Therefore, total scores of Mohan = (4 x 5) + [6 x (–2)]

= 20 – 12 = 8

Thus, Mohan gets 8 marks in a class test.

(ii) Reshma gets marks for five correct questions = 5 x 5 = 25

She gets marks for five incorrect questions = 5 x (–2) = –10 Therefore, total score of Resham = 25 + (–10) = 15 Thus, Reshma gets 15 marks in a class test.

(iii) Heena gets marks for two correct questions = 2 x 5 = 10

She gets marks for five incorrect questions = 5 x (–2) = –10

Therefore, total score of Resham = 10 + (–10) = 0 Thus, Reshma gets 0 marks in a class test.

Question 8:

A cement company earns a profit of ₹8 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold.

(a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?

(b) What is the number of white cement bags it must sell to have neither profit nor loss. If the number of grey bags sold is 6,400 bags. SOLUTION 8:

Given: Profit of 1 bag of white cement = ₹ 8

And Loss of 1 bag of grey cement = ₹ 5

(a) Profit on selling 3000 bags of white cement = 3000 x ₹ 8 = ₹ 24,000

Loss of selling 5000 bags of grey cement = 5000 x ₹ 5 = ₹ 25,000

Since Profit < Loss

Therefore, his total loss on selling the grey cement bags = Loss – Profit

= ₹ 25,000 – ₹ 24,000

= ₹ 1,000

Thus, he has lost of `₹1,000 on selling the grey cement bags.

(b) Let the number of bags of white cement be x. According to question, Loss = Profit

 5 x 6,400 = x x 8

 x = 5000 bags

Thus, he must sell 4000 white cement bags to have neither profit nor loss.

Question 9:

Replace the blank with an integer to make it a true statement:

(a)  3 _______  27 (b) 5_______ 35

SOLUTION 9:

(a)    3  9 27

(b) 5   7 35

(d) 11  12132

Exercise 1.4

Question 1:

Evaluate each of the following:

(a)  30 10 (b) 50  5

(e) 13   2 1 (f) 0  12

(g) 31  30  1 (h) 36123

(i)  6 5     2 1 SOLUTION 1:

1  30 1

(a)  30 10 = 30  3

10 10

1 50  1 10

(b) 50  5 = 50 5  5 



(d)  49 49 = 49 1  491

49 49

(e) 13   2 1 = 13   1 13  1113

(f) 0  12 = 0121120 0

 (g) 31  30  1 = 31  30 1    31  31  31311 31311



(h) 36123 = 3612 3 1  1  1236 1    3  3 13 3 31



1

(i)  6 5     2 1 =            6 5  2 1  1  1  1 1 1

Question 2:

Verify that a b c       a b a c for each of the following values of a b, and c.

(a) a12,b 4,c 2 (b) a  10,b 1,c 1

SOLUTION 2:

(a) Given: a b c       a b a c a12,b 4,c 2

Putting the given values in L.H.S. = 12   4 2

= 12   2 1221 212 6

Putting the given values in R.H.S. = 12  4122

121   6 3 6 3

=  4 

Since, L.H.S.  R.H.S.

Hence verified.

(b) Given: a b c       a b a c a10,b 1,c 1

Putting the given values in L.H.S. =   10 1 1

=  10 2 5

Putting the given values in R.H.S. =     10 1  10 1

=  10 10 20

Since, L.H.S.  R.H.S.

Hence verified.

Question 3:

Fill in the blanks:

(a) 369 _______  369 (b)  75 _______   1

(e) _______ 1 87 (f) _______ 48 1

(g) 20 _______ 2 (h) _______4 3

SOLUTION 3:

(a) 369 1 369 (b)    75 75  1 (c) 206  2061 (d) 87   1 87

(e) 87 1 87 (f) 4848 1

(g) 20   10 2 (h) 1243

Question 4:

Write five pairs of integers a b,  such that a b 3. One such pair is 6,2 because

6    2  3 . SOLUTION 4:

(i)   6 2 3 (ii) 9   3 3

(iii) 12   4 3 (iv)   9 3 3

(v)   15 5 3

Question 5:

The temperature at noon was 10oC above zero. If it decreases at the rate of 2oC per hour until mid-night, at what time would the temperature be 8oC below zero? What would be the temperature at mid-night?

SOLUTION 5:

Following number line is representing the temperature:

The temperature decreases 2oC = 1 hour

The temperature decreases 1oC = hour

The temperature decreases 18oC = 18 = 9 hours

Total time = 12 noon + 9 hours = 21 hours = 9 pm

Thus, at 9 pm the temperature would be 8oC below 0oC.

Question 6:

In a class test (+3) marks are given for every correct answer and -2 marks are given for every incorrect answer and no marks for not attempting any question.

(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?

(ii) Mohini scores -5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?

SOLUTION 6:

(i) Marks given for one correct answer = 3

Marks given for 12 correct answers = 3 x 12 = 36

Radhika scored 20 marks.

Therefore, Marks obtained for incorrect answers = 20 – 36 = –16

Now, marks given for one incorrect answer = –2

Therefore, number of incorrect answers = 16/ 2= 8

Thus, Radhika has attempted 8 incorrect questions.

(ii) Marks given for seven correct answers = 3 x 7 = 21

Mohini scores = –5

Marks obtained for incorrect answers = = –5 – 21 = –26

Now, marks given for one incorrect answer = –2

Therefore, number of incorrect answers = 26/ 2= 13

Thus, Mohini has attempted 13 incorrect questions.

Question 7:

An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 above the ground level, how long will it take to reach -350 m?

SOLUTION 7:

Starting position of mine shaft is 10 m above the ground but it moves in opposite direction so it travels the distance (–350) m below the ground.

So total distance covered by mine shaft = 10 m – (–350) m

= 10 + 350

= 360 m

Now, time taken to cover a distance of 6 m by it = 1 minute

So, time taken to cover a distance of 1 m by it = minute

Therefore, time taken to cover a distance of 360 m = 360 /6

= 60 minutes = 1 hour (Since 60 minutes = 1 hour)

Thus, in one hour the mine shaft reaches –350 below the ground.

Monday, December 18, 2023