Fractions And Decimals
Class 7th Mathematics CBSE Solution
- Solve: (i) 2- 3/5 (ii) 4+ 7/8 (iii) 3/5 + 2/7 (iv) 9/11 - 4/15 (v) 7/10 + 2/5 +…
- Arrange the following in descending order: (i) 2/9 , 2/3 , 8/21 (ii) 1/5 , 3/7 ,…
- In a magic square, the sum of the numbers in each row, in each column and along…
- A rectangular sheet of paper is 12 1/2 cm long and 10 2/3 cm wide. Find its…
- Find the perimeters of: (i) delta ABC (ii) The rectangle BCDE in this figure.…
- Salil wants to put a picture in a frame. The picture is 7 3/5 cm wide. To fit in…
- Ritu ate 3/5 part of an apple and the remaining apple was eaten by her brother…
- Michael finished coloring a picture in 7/12 hour. Vaibhav finished colouring the…
- Which of the drawings (a) to (d) show: (i) 2 x 1/5 (ii) 2 x 1/2 (iii) 3 x 2/3…
- Some pictures (a) to (c) are given below. Tell which of them show: (i) 3 x 1/5 =…
- Multiply and reduce to lowest form and convert into a mixed fraction: (i) 7 x…
- Shade: (i) 1/2 of the circles in box (a) (ii) 2/3 of the triangles in box (b)…
- (a) 1/2 of (i) 24 (ii) 46 (b) 2/3 of (i) 18 (ii) 27 (c) 3/4 of (i) 16 (ii) 36…
- Multiply and express as a mixed fraction: (a) 3 x 5 1/5 (b) 5 x 6 3/4 (c) 7 x 2…
- (a) 1/2 of (i) 2 3/4 (ii) 4 2/9 (b) 5/8 of (i) 3 5/6 (ii) 9 2/3 Find:…
- Vidya and Pratap went for a picnic. Their mother gave them a water bag that…
- Find: (i) 1/4 of (a) 1/4 (b) 3/5 (c) 4/5 (ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10…
- Multiply and reduce to lowest form (if possible) : (i) 2/3 x 2 2/3 (ii) 2/7 x…
- Multiply the following fractions: (i) 2/5 x 5 1/4 (ii) 6 2/5 x 7/9 (iii) 3/2 x 5…
- 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…
- Saili plants 4 saplings, in a row, in her garden. The distance between two…
- Lipika reads a book for 1 3/4 hours every day. She reads the entire book in 6…
- A car runs 16 km using 1 litre of petrol. How much distance will it cover using…
- (i) Provide the number in the box square , such that 2/3 x square = 10/30 .…
- (i) Provide the number in the box square , such that 3/5 x square = 24/75 .…
- (i) 12 + 3/4 (ii) 14 + 5/6 (iii) 8 + 7/3 (iv) 4 + 8/3 (v) 3+2 1/3 (vi) 5+3 4/7…
- find the reciprocal of each of the following fractions. Classify the reciprocals…
- (i) 7/3 + 2 (ii) 4/9 + 5 (iii) 6/13 + 7 (iv) 4 1/3 + 3 (v) 3 1/2 + 4 (vi) 4 3/7…
- Find: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
- Which is greater? (i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7 (iv) 1.37 or…
- Express are rupees using decimals: (i) 7 paise (ii) 7 rupees 7 paise (iii) 77…
- Express 5 cm in meter and kilometer
- Express 35 mm in cm, m and km
- Express in kg: (i) 200 g (ii) 3470 kg (iii) 4 kg 8 g
- Write the following decimal numbers in the expanded form: (i) 20.03 (ii) 2.03…
- Write the place value of 2 in the following decimal numbers: (i) 2.56 (ii) 21.37…
- Dinesh went from place A to place B and from there to place C. a is 7.5 km from…
- Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g…
- How much less is 28 km than 42.6 km?
- (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…
- Find the area of a rectangle whose length is 5.7 cm and breadth is 3 cm…
- (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…
- A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much…
- (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…
- (i) 0.4 2 (ii) 0.35 5 (iii) 2.48 4 (iv) 65.4 4 (v) 651.2 4 (vi) 14.49 7 (vii)…
- (i) 4.8 10 (ii) 52.5 10 (iii) 0.7 10 (iv) 33.1 10 (v) 272.23 10 (vi) 0.56 10…
- (i) 2.7 100 (ii) 0.3 100 (iii) 0.78 100 (iv) 432.6 100 (v) 23.6 100 (vi) 98.53…
- (i) 7.9 1000 (ii) 26.3 1000 (iii) 38.53 1000 (iv) 128.9 1000 (v) 0.5 1000 Find:…
- (i) 7 35 (ii) 36 0.2 (iii) 3.25 0.5 (iv) 30.94 0.7 (v) 0.5 0.25 (vi) 7.75 0.25…
- A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much…
Exercise 2.1
Question 1.Solve:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
Answer:
(i)
We have,
2 – 
= 
–
= 
–
= 
= 
(ii)
We have,
4 + 
= 
+
= 
+
= 
= 
= 4
(iii)
We have,
+ 
= 
+ 
= 
+ 
= 
= 
(iv)
We have,
– 
= 
– 
= 
– 
= 
= 
(v)
We have,
+ 
+ 
= + 
+ 
= 
= 
= 
= 2
(vi)
We have,
= 
+ 
= 
+ 
= 
+ 
= 
= 
= 6
(vii)
We have,
= 
– 
= 
–
= 
–
= 
=
Question 2.
Arrange the following in descending order:
(i) 
(ii) 
Answer:
(i) We have,
, 
, 
First of all, we have to change them into like fractions:
For doing that we will make the denominators of the fractions to be equal. This will be done by making them the L.C.M of (9, 3, 21)
L.C.M of (9, 3, 21) = 63
Therefore,
= 
= 
Also,
= 
= 
And,
= 
= 
Since,
All have the same denominator and:
42 > 24 > 14
Therefore,
> >
(ii)We have,
, 
,
First of all, we have to change them into like fractions:
This will be done by making the denominators equal. For this make the denominators equal to the L.C.M of (5, 7, 10)
L.C.M(5, 7, 10) = 70
Therefore,
= 
= 
,
= 
= 
And,
= 
= 
Since,
All have the same denominator and:
49 > 30 > 14
Therefore,
> >
Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?
Answer:
From the above question, we have:
Sum along the 1st row = 
+ + 
= 
Sum along the 2nd row = 
+ 
+ 
=
Sum along the 3rd row = 
+ 
+ 
=
Sum along the 1st column = + + =
Sum along the 2nd column = + + =
Sum along the 3rd column = + + =
Sum along the 1st diagonal = + + =
Sum along the 2nd diagonal = + + =
Therefore, the sum of all the rows, columns and diagonals of this square are equal
Hence,
It is a magic square.
Question 4.
A rectangular sheet of paper is 
cm long and 
cm wide. Find its perimeter.
Answer:
It is given in the question that,
Length of the rectangular sheet = 
cm = 
cm
Also,
Breadth of rectangular sheet = 
cm = 
cm
We know that,
Perimeter of rectangle = 2 × (Length + Breadth)
= 2 × ( + )
= 2 × [
]
= 2 × (
)
= 2 × 
= 
cm
Question 5.
Find the perimeters of:
(i) 
ABE
(ii) The rectangle BCDE in this figure.
Whose perimeter is greater?
Answer:
(i) First of all we have to find out the perimeter of triangle ABE:
We know that,
Perimeter of triangle = Sum of all sides
Therefore,
Perimeter of triangle ABE = AB + BE + EA
cm
(ii) Secondly, we have find out the perimeter of rectangle BCDE:
We know that,
Perimeter of rectangle = 2 × (Length + Breadth)
Therefore,
Perimeter of rectangle BCDE = 2 × (
+ 
)
cm
Now, we have to change the perimeter of both triangle and rectangle into like fractions:
Perimeter of triangle = 
= 
= 
Also,
Perimeter of rectangle = 47/6 = 
Since,
531 > 470
Therefore,
> 
Hence,
Perimeter of triangle is greater than the perimeter of rectangle
Question 6.
Salil wants to put a picture in a frame. The picture is 
cm wide. To fit in the frame, the picture cannot be more than 
cm wide. How much should the picture be trimmed?
Answer:
It is given in the question that,
Width of picture = 7 = 
cm
Also,
Required width = 7
= 
cm
Therefore,
The picture should be trimmed by = (
)
= (
– )
= 
= 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:
It is given in the question that,
Part of apple eaten by Ritu = 3/5
Also,
Part of apple eaten by Somu = 1 – Part of apple eaten by Ritu
= 1 –
= 
=
Therefore,
Somu ate part of the apple
Since,
> , therefore, Ritu ate the larger size of apple
Now, the difference between the 2 shares = –
= 
=
Hence,
Share of Ritu is larger than the share of Somu by
Question 8.
Michael finished coloring 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 a picture = 
hour
Also,
Time taken by Vaibhav to colour a picture = 
hour
First of all, we have to convert them into like fractions, we get:
= 
= 
Since,
>
Therefore,
Vaibhav worked for longer period of time
Hence, the difference = –
= 
= 
=
Therefore,
Vaibhav worked for a longer period of time by a fraction of .
Exercise 2.2
Question 1.Which of the drawings (a) to (d) show:
(i) 
(ii) 
(iii) 
(iv) 
(a) 
(b) 
(c) 
(d) 
Answer:
(i) In the first part,
2 × clearly represents the addition of 2 figures out of which each part represents 1 shaded part out of the given 5 equal parts
Therefore,
2 × is represented by option (d)
(ii) In the second part,
2 × 
clearly represents the addition of 2 figures out of which each part represents 1 shaded part out of the given 2 equal parts
Therefore,
2 × is represented by option (b)
(iii) In the third part,
3 × clearly represents the addition of 3 figures out of which each part represents 2 shaded parts out of the given 3 equal parts
Therefore,
3 × is represented by option (a)
(iv) In the fourth part,
3 × 
clearly represents the addition of 3 figures out of which each part represents 1 shaded part out of the given 4 equal parts
Therefore,
3 × is represented by option (c)
Question 2.
Some pictures (a) to (c) are given below. Tell which of them show:
(i) 
(ii) 
(iii) 
(a) 
(b) 
(c) 
Answer:
(i) In the first part,
3 × clearly represents the addition of 3 figures out of which each part represents 1 shaded part out of the given 5 equal parts
Also,
represents 3 shaded parts out of 5 equal parts
Therefore,
3 × = is represented by option (c)
(ii) In the second part,
2 × 
clearly represents the addition of 2 figures out of which each part represents 1 shaded part out of the given 3 equal parts
Also,
represents 2 shaded parts out of 3 equal parts
Therefore,
2 × = is represented by option (a)
(iii) In the first part,
3 × clearly represents the addition of 3 figures out of which each part represents 3 shaded parts out of the given 4 equal parts
Also,
2 represents 2 fully shaded parts and 1 figure having 1 part as shaded out of 4 equal parts
Therefore,
3 × = 2 is represented by option (b)
Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction:
(i) 
(ii) 
(iii) 
(iv)
(v) 
(vi)
(vii) 
(viii) 
(ix) 
(x) 
Answer:
(i) We have,
7 ×
= 
= 4
(ii) We have,
4 ×
= 
= 1
(iii) We have,
2 × 
= 
= 1 
(iv) We have,
5 ×
= 
= 1 
(v) We have,
× 4
=
= 2
(vi) We have,
× 6
= 15
(vii) We have,
11 × 
= 
= 6
(viii) We have,
20 × 
= 16
(ix) We have,
13 ×
= 
= 4
(x) We have,
15 ×
= 3 × 3
= 9
Question 4.
Shade:
(i) 1/2 of the circles in box (a)
(ii) 2/3 of the triangles in box (b)
(iii) 3/5 of the squares in box (c).
(a) 
(b) 
(c) 
Answer:
(a) Here,
As per the question,
We can observe that,
There are in all twelve circles in the box.
And,
We have to shade 1/2 of these circles.
Now,
Since,
12 × = 6
Therefore,
We have to shade 6 circles
(b) Here,
As per the question,
We can observe that,
There are in all nine triangles in the box.
And,
We have to shade 2/3 of these triangles.
Now,
Since,
9× = 6
Therefore,
We have to shade 6 triangles
(c) Here,
As per the question,
We can observe that,
There are in all fifteen squares in the box.
And,
We have to shade 3/5 of these squares.
Now,
Since,
15× () = 9
Therefore,
We have to shade 9 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) We have,
(i) of 24
= × 24
= 12
Also
(ii) of 46
= × 46
= 
(b) We have,
(i) of 18
= × 18
= 2 × 6
= 12
Also,
(ii) of 27
= × 27
= 2 × 9
= 18
(c) We have,
(i) of 16
= × 16
= 3 × 4
= 12
Also,
(ii) × 36
= × 36
= 3 × 9
= 27
(d) We have,
(i) of 20
= × 20
= 4 × 4
= 16
Also,
(ii) of 35
= × 35
= 4 × 7
= 28
Question 6.
Multiply and express as a mixed fraction:
(a) 
(b) 
(c)
(d) 
(e) 
(f) 
Answer:
(a) We have,
3 × 5
= 3 × 
= 
= 
= 15
(b) We have,
5 × 6
= 5 × 
= 
= 
= 33
(c) We have,
7 × 2
= 7 × 
= 
= 
= 15
(d) We have,
4 × 6
= 4 × 
= 
= 
= 25
(e) We have,
3 × 6
= 
× 6
= 
= 
= 
= 19
(f) We have,
3 × 8
= 
× 8
= 
= 
= 27
Question 7.
Find:
(a) 1/2 of (i) 
(ii) 
(b) 5/8 of (i) 
(ii) 
Answer:
(a) We have,
(i) of 2
= × 2
= ×
= 
= 1
Also,
(ii) of 4
= × 4
= × 
= 
= 2
(b) We have,
(i) 
of 3
= × 3
= × 
= 
= 2
Also,
(ii) of 9
= × 9
= × 
= 
= 6
Question 8.
Vidya and Pratap went for a picnic. Their mother gave them a water bag that contained 5 liters 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:
(i) It is given in the question that,
Total amount of water = 5 litres
Also,
Amount of water consumed by Vidya = of the total water
Therefore,
Amount of water that Vidya drink = × 5
= 2 litres
(ii) It is given in the question that,
The total amount of water = 5 litres
Also,
Amount of water consumed by Vidya = 
of the total water
Therefore,
Amount of water consumed by Pratap = 1 –
= 
of the total water
Exercise 2.3
Question 1.Find:
(i) 1/4 of (a) 1/4 (b) 3/5 (c) 4/5
(ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10
Answer:
(i) We have,
(a) of
= ×
= 
(b) of
= ×
= 
(c) Also,
of 
= × 
=5
(ii) We have,
(a)
of
= ×
= 
(b) of 
= ×
= 
(c) Also,
of
= ×
= 
Question 2.
Multiply and reduce to lowest form (if possible) :
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
Answer:
(i) We have,
× 2
= ×
= 
= 1
(ii) We have,
×
(iii) We have,
× 
(iv) We have,
×
= 
= 1
(v) We have,
× 
=
(vi) We have,
×
= 
= 1
(vii) We have,
×
= 
= 1
Question 3.
Multiply the following fractions:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
Answer:
(i) We have,
× 5
× 
We have an improper fraction and now it can be written in terms of the mixed fraction is as follows:
= 2
(ii) We have,
6 ×
×
We have an improper fraction and now it can be written in terms of mixed fraction is as follows:
= 4
(iii) × 5
× 
8
In this question, we have a whole number
(iv) × 2
× 
We have an improper fraction and now it can be written in terms of mixed fraction is as follows:
= 2
(v) 3 ×
×
We have an improper fraction and now it can be written in terms of the mixed fraction is as follows:
= 1
(vi) 2 × 3
× 3
We have an improper fraction and now it can be written in terms of the mixed fraction is as follows:
= 7
(vii) 3 ×
×
We have an improper fraction and now it can be written in terms of mixed fraction is as follows:
= 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) We have,
of
×
Also,
of
×
Now converting the above fractions into like fraction, we get:
= 
= 
Also,
= 
= 
Since,
>
Therefore,
>
Hence,
is greater than
(ii) We have,
of
×
Also,
of
×
Since,
>
Hence,
is greater than
Question 5.
Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 3/4m. Find the distance between the first and the last sapling.
Answer:
It is given in the question that,
Length of 1 gap =
m
Also, from the figure it can be observed that:
Gaps between first and last saplings = 3
Therefore,
Distance between First and last sapling = 3 ×
=
= 
m
Question 6.
Lipika reads a book for 
hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Answer:
It is given in the question that,
Number of hours Lipika reads the book =
= 
hours
Also,
Total days in which she completes the book = 6
Therefore,
Total number of hours required by her to complete the book = 
× 6
= 
= 10 1/2 = 10.5 hours
Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 
litres of petrol?
Answer:
It is given in the question that,
Distance travelled by a car in 1 litre of petrol = 16 km
Also,
Total quantity of petrol = 2 litre
= litres
Therefore,
Distance travelled by the car in litres of petrol = × 16
= 44 km
Hence,
The car will cover a distance of 44 kms in 2 litres of petrol
Question 8.
(i) Provide the number in the box 
, such that 
.
(ii) The simplest form of the number obtained in Is __________.
Answer:
(i) We have,
× 
= 
Hence,
The number in the box will be
(ii) We have,
The simplest form of :
=
Question 9.
(i) Provide the number in the box 
, such that 
.
(ii) The simplest form of the number obtained in is ________.
Answer:
(i) We have,
× 
= 
Hence,
The number in the box will be
(ii) From above,
is itself in a simplest form and it cannot be further simplified
Exercise 2.4
Question 1.Find:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Answer:
(i) We have,
12 ÷
= 12 ×
= 16
(ii) We have,
14 
= 14 ×
= 
(iii) We have,
8 
= 8 ×
= 
(iv) We have,
4 
= 4 ×
=
(v) We have,
3 
2
= 3
= 3 ×
= 
(vi) We have,
5 3
= 5 
= 5 × 
=
Question 2.
find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
Answer:
(i) We have,
We know that,
A proper fraction is that fraction in which denominator of the fraction is greater than the numerator of the fraction.
Also,
Improper fraction is that in which numerator is greater than its denominator
And,
Whole numbers are collection of all positive integers including 0.
Now,
Reciprocal of = 
Hence,
It is a improper fraction
(ii) We have,
Reciprocal of = 
Hence,
It is a improper fraction
(iii) We have,
Reciprocal of =
Hence,
It is a proper fraction
(iv) We have,
Reciprocal of =
Hence,
It is a proper fraction
(v) We have,
Reciprocal of =
Hence,
It is a proper fraction
(vi) We have,
Reciprocal of = 
Hence,
It is a whole number
(vii) We have,
Reciprocal of = 
Hence,
It is a whole number
Question 3.
Find:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Answer:
(i) We have,
2
= ×
=
(ii) We have,
5
= 
×
= 
(iii) We have,
7
= 
×
= 
(iv) We have,
4
3
= 
3
= ×
= 
(v) We have,
3
4
= 
4
= ×
=
(vi) We have,
4
7
= 
7
= 
×
= 
Question 4.
Find:
(i) 
(ii) 
(iii) 
(iv)
(v) 
(vi)
(vii)
(viii) 
Answer:
(i) We have,
= × 2
=
(ii) We have,
= ×
=
(iii) We have,
= ×
=
(iv) We have,
= 
= × 
= 
(v) We have,
= 
= ×
= 
(vi) We have,
= 
= ×
=
(vii) We have,
= 
= 
×
= 
(viii) We have,
= 
= 
×
= 
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
Answer:
(i) We have,
0.5 or 0.05
Firstly, we have to convert these decimals into equivalent fractions:
So,
0.5 =
= 
= 
And,
0.05 = 
Now, from the above fractions it can be observed that both have same denominator
Since,
50 > 5
Therefore,
0.5 > 0.05
(ii) We have,
0.7 or 0.5
Firstly, we have to convert these decimals into equivalent fractions:
So,
0.7 =
And,
0.5 =
Now, from the above fractions it can be observed that both have same denominator
Since,
7 > 5
Therefore,
0.7 > 0.5
(iii) We have,
7 or 0.7
Firstly, we have to convert these decimals into equivalent fractions:
So,
7 = 
= 
= 
And,
0.7 =
Now, from the above fractions it can be observed that both have same denominator
Since,
70 > 7
Therefore,
7 > 0.7
(iv) We have,
1.37 or 1.49
Firstly, we have to convert these decimals into equivalent fractions:
So,
1.37 = 
And,
1.49 = 
Now, from the above fractions it can be observed that both have same denominator
Since,
137 < 149
Therefore,
1.37 < 1.49
(v) We have,
2.03 or 2.30
Firstly, we have to convert these decimals into equivalent fractions:
So,
2.03 = 
And,
2.30 = 
Now, from the above fractions it can be observed that both have same denominator
Since,
203 < 230
Therefore,
2.03 < 2.30
(vi) We have,
0.8 or 0.88
Firstly, we have to convert these decimals into equivalent fractions:
So,
0.8 = 
= 
= 
And,
0.88 = 
Now, from the above fractions it can be observed that both have same denominator
Since,
80 < 88
Therefore,
0.8 < 0.88
Question 2.
Express are rupees using decimals:
(i) 7 paise
(ii) 7 rupees 7 paise
(iii) 77 rupees 77 paise
(iv) 50 paise
(v) 235 paise
Answer:
(i) We have,
7 paise
We know that,
There are 100 paise in 1 rupee
So, for converting paise into rupees we have to divide paise by 100
Therefore,
7 paise = Rs 
= Rs 0.07
(ii) We have,
7 Rs 7 paise
We know that,
There are 100 paise in 1 rupee
So, for converting paise into rupees we have to divide paise by 100
Therefore,
7 Rs 7 paise = Rs 7 + Rs
= Rs 7.07
(iii) We have,
77 Rs 77 paise
We know that,
There are 100 paise in 1 rupee
So, for converting paise into rupees we have to divide paise by 100
Therefore,
77 Rs 77 paise = Rs 77 + Rs
= Rs 77.77
(iv) We have,
We have,
50 paise
We know that,
There are 100 paise in 1 rupee
So, for converting paise into rupees we have to divide paise by 100
Therefore,
50 paise = Rs 
= Rs 0.50
(v) We have,
235 paise
We know that,
There are 100 paise in 1 rupee
So, for converting paise into rupees we have to divide paise by 100
Therefore,
235 paise = Rs 
= Rs 2.35
Question 3.
Express 5 cm in meter and kilometer
Answer:
We have,
5 cm
We know that,
1 m = 100 cm
Also,
1 km = 100000 cm
Therefore,
5 cm = m
= 0.05 m
Also,
5 cm = 
km
= 0.00005 km
Question 4.
Express 35 mm in cm, m and km
Answer:
We have,
35 mm
We know that,
1 cm = 10 mm
1 m = 1000 mm
Also,
1 km = 1000000 mm
Therefore,
35 mm = 
cm
= 3.5 cm
Also,
35 mm = 
m
= 0.035 m
And,
35 mm = 
km
= 0.000035 km
Question 5.
Express in kg:
(i) 200 g (ii) 3470 kg (iii) 4 kg 8 g
Answer:
(i) We have,
200 g
We know that,
1 kg = 1000 g
Therefore,
200 g = 
kg
= 0.2 kg
(ii) We have,
3470 g
We know that,
1 kg = 1000 g
Therefore,
3470 g = 
kg
= 3.470 kg
(iii) We have,
4 kg 8 g
We know that,
1 kg = 1000 g
Therefore,
4 kg + 8 g = 4 kg + 
kg
= 4.008 kg
Question 6.
Write the following decimal numbers in the expanded form:
(i) 20.03 (ii) 2.03 (iii) 200.03 (iv) 2.034
Answer:
(i) We have,
20.03
We have to expand the above given decimal number:
20.03 = 2 × 10 + 0 × 1 + 0 × + 3 × 
(ii) We have,
2.03
We have to expand the above given decimal number:
2.03 = 2 × 1 + 0 × + 3 ×
(iii) We have,
200.03
We have to expand the above given decimal number:
200.03 = 2 × 100 + 0 × 10 + 0 × 1 + 0 × + 3 ×
(iv) We have,
2.034
We have to expand the above given decimal number:
2.034 = 2 × 1 + 0 × + 3 × + 4 × 
Question 7.
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) We have,
2.56
We have to write the place value of 2 in the given decimal number
Therefore,
Place value of 2 in 2.56 = Ones
(ii) We have,
21.37
We have to write the place value of 2 in the given decimal number
Therefore,
Place value of 2 in 21.37 = Tens
(iii) We have,
10.25
We have to write the place value of 2 in the given decimal number
Therefore,
Place value of 2 in 10.25 = Tenths
(iv) We have,
9.42
We have to write the place value of 2 in the given decimal number
Therefore,
Place value of 2 in 9.42 = Hundredths
(v) We have,
63.352
We have to write the place value of 2 in the given decimal number
Therefore,
Place value of 2 in 63.352 = Thousandths
Question 8.
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?
Answer:
It is given in the question that,
Distance travelled by Dinesh = AB + BC
= (7.5 + 12.7) km
7.5
+ 12.7
20.2
Therefore,
Distance travelled by Dinesh = 20.2 km
Also,
Distance travelled by Ayub = AD + DC
= (9.3 + 11.8) km
9.3
+ 11.8
21.1
Therefore,
Distance travelled by Ayub = 21.1 km
Hence,
It is clear that Ayub has travelled more distance
Now,
Difference = (21.1 – 20.2) km
21.1
- 20.2
0.9
Therefore,
Difference of their distance = 0.9
Question 9.
Shyama 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:
It is given in the question that,
Total fruits bought by Shyama = 5 kg300 g + 3 kg 250 g
= 8 kg 550 g
= (8 + 
) kg
= 8.550 kg
Also,
Total fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g
= 8 kg 950 g
= (8 + 
) kg
= 8.950 kg
Therefore,
Sarala bought more fruits as compared to Shyama
Question 10.
How much less is 28 km than 42.6 km?
Answer:
For the above question, we have
Therefore,
From the above results, it is clear that 28 km is 14.6 km less 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
Answer:
(i) We have,
0.2 × 6
Therefore,
0.2 × 6 = 
× 6
= 
= 1.2
(ii) We have,
8 × 4.6
Therefore,
8 × 4.6 = 8 × 
= 
= 36.8
(iii) We have,
2.71 × 5
Therefore,
2.71 × 5 = 
× 5
= 
= 13.55
(iv) We have,
20.1 × 4
Therefore,
20.1 × 4 = 
× 4
= 
= 80.4
(v) We have,
0.05 × 7
Therefore,
0.05 × 7 = × 7
= 
= 0.35
(vi) We have,
211.02 × 4
Therefore,
211.02 × 4 = 
× 4
= 
= 844.08
(vii) We have,
2 × 0.86
Therefore,
2 × 0.86 = 2 × 
= 
= 1.72
Question 2.
Find the area of a rectangle whose length is 5.7 cm and breadth is 3 cm
Answer:
It is given in the question that,
Length of rectangle = 5.7 cm
Also,
Breadth of rectangle = 3 cm
We know that,
Area of rectangle = Length × Breadth
Therefore,
Area = Length × Breadth
= 5.7 × 3
= 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
Answer:
(i) We have
1.3 × 10
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side.
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.
Therefore,
The product of 1.3 and 10 can be calculated as follows:
1.3 × 10 = 13
(ii) We have
36.8 × 10
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 36.8 and 10 can be calculated as follows:
36.8 × 10 = 368
(iii) We have
153.7 × 10
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 153.7 and 10 can be calculated as follows:
153.7. × 10 = 1537
(iv) We have
168.07 × 10
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 168.07 and 10 can be calculated as follows:
168.07 × 10 = 1680.7
(v) We have
31.1 × 100
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 31.1 and 100 can be calculated as follows:
31.1 × 100 = 3110
(vi) We have
156.1 × 100
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 156.1 and 100 can be calculated as follows:
156.1 × 100 = 15610
(vii) We have
3.62 × 100
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 3.62 and 100 can be calculated as follows:
3.62 × 100 = 362
(viii) We have
43.07 × 100
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 43.07 and 100 can be calculated as follows:
43.07 × 100 = 4307
(ix) We have
0.5 × 10
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 0.5 and 10 can be calculated as follows:
0.5 × 10 = 5
(x) We have
0.08 × 10
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 0.08 and 10 can be calculated as follows:
0.08 × 10 = 0.8
(xi) We have
0.9 × 100
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 0.9 and 100 can be calculated as follows:
0.9 × 100 = 90
(xii) We have
0.03 × 1000
We know that,
Whenever any decimal number is multiplied by 10, 100 or 1000 then the decimal number which is at the product is shifted towards the right side
There would be a shift according to the number of zeroes. Like if a number is multilpied by 10, the decimal will shift to 1 place right.Therefore,
The product of 0.03 and 1000 can be calculated as follows:
0.03 × 1000 = 30
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:
It is given in the question that,
Distance covered by two wheeler in 1 litre of petrol = 55.3 km
Therefore,
Distance covered by two wheeler in 10 litres of petrol = 55.3 × 10
= 553 km
Hence,
Two - wheeler will cover a distance of 553 km 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
Answer:
(i) We have,
2.5 × 0.3
= 
×
= 
= 0.75
(ii) We have,
0.1 × 51.7
= × 
= 
= 5.17
(iii) We have,
0.2 × 316.8
= × 
= 
= 63.36
(iv) We have,
1.3 × 3.1
= 
× 
= 
= 4.03
(v) We have,
0.5 × 0.05
= ×
= 
= 0.025
(vi) We have,
11.2 × 0.15
= 
× 
= 
= 1.680
= 1.68
(vii) We have,
1.07 × 0.02
= 
× 
= 
= 0.0214
(viii) We have,
10.05 × 1.05
= 
× 
= 
= 10.5525
(ix) We have,
101.01 × 0.01
= 
×
= 
= 1.0101
(x) We have,
100.01 × 1.1
= 
× 
= 
= 110.011
Exercise 2.7
Question 1.Find:
(i) 0.4 
2 (ii) 0.35 
5
(iii) 2.48 4 (iv) 65.4 4
(v) 651.2 4 (vi) 14.49 7
(vii) 3.96 4 (viii) 0.80 5
Answer:
(i) We have,
0.4 2
= 
2
= 
× 
= 
= 0.2
(ii) We have,
0.35 5
= 
5
= 
× 
=
= 0.07
(iii) We have,
2.48 4
= 
4
= 
× 
= 
= 0.62
(iv) We have,
65.4 6
= 
6
= 
× 
= 
= 10.9
(v) We have,
651.2 4
= 
4
= 
× 
= 
= 162.8
(vi) We have,
14.49 7
= 
7
= 
× 
= 
= 2.07
(vii) We have,
3.96 4
= 
4
= 
×
= 
= 0.99
(viii) We have,
0.80 5
= 
5
= 
×
= 
= 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) We have,
4.8 10
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
4.8 10
= 
= 0.48
(ii) We have,
52.5 10
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
52.5 10
= 
= 5.25
(iii) We have,
0.7 10
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
0.7 10
= 
= 0.07
(iv) We have,
33.1 10
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
33.1 10
= 
= 3.31
(v) We have,
272.23 10
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
272.23 10
= 
= 27.223
(vi) We have,
0.56 10
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
0.56 10
= 
= 0.056
(vii) We have,
3.97 10
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
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
Answer:
(i) We have,
2.7 100
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
2.7 100
= 
= 0.027
(ii) We have,
0.3 100
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
0.3 100
= 
= 0.003
(iii) We have,
0.78 100
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
0.78 100
= 
= 0.0078
(iv) We have,
432.6 100
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
432.6 100
= 
= 4.326
(v) We have,
23.6 100
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
23.6 100
= 
= 0.236
(vi) We have,
98.53 100
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
98.53 100
= 
= 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) We have,
7.9 1000
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
7.9 1000
= 
= 0.0079
(ii) We have,
26.3 1000
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
26.3 1000
= 
= 0.0263
(iii) We have,
38.53 1000
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
38.53 1000
= 
= 0.03853
(iv) We have,
128.9 1000
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
128.9 1000
= 
= 0.1289
(v) We have,
0.5 1000
We know that,
Whenever any decimal number is divided by 10, 100 or 1000 then the decimal point of that number will be shifted towards the left side as many as there are zeros in the number
Therefore,
0.5 1000
= 
= 0.0005
Question 5.
Find:
(i) 7 35 (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) We have,
7 35
= 7 
= 7 ×
= 
= 2
(ii) We have,
36 0.2
= 36 
= 36 × 
= 18 × 10
= 180
(iii) We have,
3.25 0.5
= 
= 
×
= 
= 6.5
(iv) We have,
30.94 0.7
= 
= 
× 
= 
= 44.2
(v) We have,
0.5 0.25
= 
= × 
= 
= 2
(vi) We have,
7.75 0.25
= 
= 
×
= 
= 31
(vii) We have,
76.5 0.15
= 
= 
× 
= 
= 510
(viii) We have,
37.8 1.4
= 
= 
× 
= 27
(ix) We have,
2.73 1.3
= 
= 
× 
= 
= 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 of petrol?
Answer:
It is given in the question that,
Distance covered by a vehicle in 2,4 litres of petrol = 43.2 km
Therefore,
Distance travelled by a vehicle in 1 litre of petrol = 43.2 2.4
= 
= 
× 
= 
× 
= 
= 18
Hence,
The vehicle will cover a distance of 18 km in 1 litre of petrol