NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals: Students must go through Fractions and Decimals Class 7 Maths Chapter 2 Solutions to understand the topic more deeply. The faculty of Physics Wallah have developed these Fractions and Decimals Class 7 Maths Chapter 2 Solutions to help the students with their preparation for the CBSE board exam. Fractions and Decimals Class 7 Maths Chapter 2 Solutions includes multiplication and division of fractions and multiplication and division of decimals. Students can check the Fractions and Decimals Class 7 Maths Chapter 2 Solutions in this article below.
Check out: Class 7th Books
Fractions and Decimals Class 7 Maths Chapter 2 Solutions
Students can check the Fractions and Decimals Class 7 Maths Chapter 2 Solutions to prepare for the CBSE board
1. Solve:
(i) 2 – (3/5)
Solution:-
For subtraction of two unlike fractions, first change them to like fractions.
LCM of 1, 5 = 5
Now, let us change each of the given fractions into an equivalent fraction having 5 as the denominator.
= [(2/1) × (5/5)] = (10/5)
= [(3/5) × (1/1)] = (3/5)
Now,
= (10/5) – (3/5)
= [(10 – 3)/5]
= (7/5)
(ii) 4 + (7/8)
Solution:-
For addition of two unlike fractions, first change them to like fractions.
LCM of 1, 8 = 8
Now, let us change each of the given fractions into an equivalent fraction having 8 as the denominator.
= [(4/1) × (8/8)] = (32/8)
= [(7/8) × (1/1)] = (7/8)
Now,
= (32/8) + (7/8)
= [(32 + 7)/8]
= (39/8)
(iii) (3/5) + (2/7)
Solution:-
For addition of two unlike fractions, first change them to like fractions.
LCM of 5, 7 = 35
Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.
= [(3/5) × (7/7)] = (21/35)
= [(2/7) × (5/5)] = (10/35)
Now,
= (21/35) + (10/35)
= [(21 + 10)/35]
= (31/35)
(iv) (9/11) – (4/15)
Solution:-
For subtraction of two unlike fractions, first change them to like fractions.
LCM of 11, 15 = 165
Now, let us change each of the given fractions into an equivalent fraction having 165 as the denominator.
= [(9/11) × (15/15)] = (135/165)
= [(4/15) × (11/11)] = (44/165)
Now,
= (135/165) – (44/165)
= [(135 – 44)/165]
= (91/165)
(v) (7/10) + (2/5) + (3/2)
Solution:-
For addition of two unlike fractions, first change them to like fractions.
LCM of 10, 5, 2 = 10
Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.
= [(7/10) × (1/1)] = (7/10)
= [(2/5) × (2/2)] = (4/10)
= [(3/2) × (5/5)] = (15/10)
Now,
= (7/10) + (4/10) + (15/10)
= [(7 + 4 + 15)/10]
= (26/10)
= (13/5)
(vi) 22/3+31/2
Solution:-
First convert mixed fraction into improper fraction,
=22/3= 8/3
= 3 ½ = 7/2
For addition of two unlike fractions, first change them to like fractions.
LCM of 3, 2 = 6
Now, let us change each of the given fractions into an equivalent fraction having 6 as the denominator.
= [(8/3) × (2/2)] = (16/6)
= [(7/2) × (3/3)] = (21/6)
Now,
= (16/6) + (21/6)
= [(16 + 21)/6]
= (37/6)
(vii) 812−358
Solution:-
First convert mixed fraction into improper fraction,
= 8 ½ = 17/2
=35/7= 29/8
For subtraction of two unlike fractions, first change them to like fractions.
LCM of 2, 8 = 8
Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.
= [(17/2) × (4/4)] = (68/8)
= [(29/8) × (1/1)] = (29/8)
Now,
= (68/8) – (29/8)
= [(68 – 29)/8]
= (39/8)
Read More: NCERT Solutions for Class 7 Maths Chapter 1 Integers
2. Arrange the following in descending order:
(i) 2/9, 2/3, 8/21
Solution:-
LCM of 9, 3, 21 = 63
Now, let us change each of the given fractions into an equivalent fraction having 63 as the denominator.
[(2/9) × (7/7)] = (14/63) [(2/3) × (21/21)] = (42/63) [(8/21) × (3/3)] = (24/63)
Clearly,
(42/63) > (24/63) > (14/63)
Hence,
(2/3) > (8/21) > (2/9)
Hence, the given fractions in descending order are (2/3), (8/21), (2/9)
CBSE Admit Card 2024
(ii) 1/5, 3/7, 7/10
Solution:-
LCM of 5, 7, 10 = 70
Now, let us change each of the given fractions into an equivalent fraction having 70 as the denominator.
[(1/5) × (14/14)] = (14/70) [(3/7) × (10/10)] = (30/70) [(7/10) × (7/7)] = (49/70)
Clearly,
(49/70) > (30/70) > (14/70)
Hence,
(7/10) > (3/7) > (1/5)
Hence, the given fractions in descending order are (7/10), (3/7), (1/5)
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?
4/11 9/11 2/11
3/11 5/11 7/11
8/11 1/11 6/11
Solution:-
Sum along the first row = (4/11) + (9/11) + (2/11) = (15/11)
Sum along the second row = (3/11) + (5/11) + (7/11) = (15/11)
Sum along the third row = (8/11) + (1/11) + (6/11) = (15/11)
Sum along the first column = (4/11) + (3/11) + (8/11) = (15/11)
Sum along the second column = (9/11) + (5/11) + (1/11) = (15/11)
Sum along the third column = (2/11) + (7/11) + (6/11) = (15/11)
Sum along the first diagonal = (4/11) + (5/11) + (6/11) = (15/11)
Sum along the second diagonal = (2/11) + (5/11) + (8/11) = (15/11)
Yes. The sum of the numbers in each row, in each column and along the diagonals is the same, so it is a magic square.
4. A rectangular sheet of paper is 12 ½ cm long and 102/3 cm wide. Find its perimeter.
Solution:-
From the question, it is given that,
Length = 12 ½ cm = 25/2 cm
Breadth =
102/3cm = 32/3 cm
We know that,
Perimeter of the rectangle = 2 × (length + breadth)
= 2 × [(25/2) + (32/3)]
= 2 × {[(25 × 3) + (32 × 2)]/6}
= 2 × [(75 + 64)/6]
= 2 × [139/6]
= 139/3 cm
5. Find the perimeters of (i) triangle ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
Solution:-
From the fig,
AB = (5/2) cm
AE =33/5= 18/5 cm
BE =23/4= 11/4 cm
ED = 7/6 cm
(i) We know that,
Perimeter of the triangle = Sum of all sides
Then,
Perimeter of triangle ABE = AB + BE + EA
= (5/2) + (11/4) + (18/5)
The LCM of 2, 4, 5 = 20
Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator.
= {[(5/2) × (10/10)] + [(11/4) × (5/5)] + [(18/5) × (4/4)]}
= (50/20) + (55/20) + (72/20)
= (50 + 55 + 72)/20
= 177/20
=817/20
(ii) Now, we have to find the perimeter of the rectangle,
We know that,
Perimeter of the rectangle = 2 × (length + breadth)
Then,
Perimeter of rectangle BCDE = 2 × (BE + ED)
= 2 × [(11/4) + (7/6)]
The LCM of 4, 6 = 12
Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator
= 2 × {[(11/4) × (3/3)] + [(7/6) × (2/2)]}
= 2 × [(33/12) + (14/12)]
= 2 × [(33 + 14)/12]
= 2 × (47/12)
= 47/6
Finally, we have to find which one is having a greater perimeter.
Perimeter of triangle ABE = (177/20)
Perimeter of rectangle BCDE = (47/6)
The two perimeters are in the form of unlike fractions.
Changing perimeters into like fractions we have,
(177/20) = (177/20) × (3/3) = 531/60
(43/6) = (43/6) × (10/10) = 430/60
Clearly, (531/60) > (430/60)
Hence, (177/20) > (43/6)
∴ Perimeter of Triangle ABE > Perimeter of Rectangle (BCDE)
Check out: PW School Books
6. Salil wants to put a picture in a frame. The picture is 73/5 cm wide. To fit in the frame the picture cannot be more than 73/10 cm wide. How much should the picture be trimmed?
Solution:-
From the question, it is given that,
Picture having a width of = 73/5 = 38/5 cm
Frame having a width of = 73/10= 73/10 cm
∴ The picture should be trimmed by = [(38/5) – (73/10)]
The LCM of 5, 10 = 10
Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator.
= [(38/5) × (2/2)] – [(73/10) × (1/1)]
= (76/10) – (73/10)
= (76 – 73)/10
= 3/10 cm
Thus, the picture should be trimmed by (3/10) cm
7. Ritu ate (3/5) part of an apple and the remaining apple was eaten by her brother Somu. What part of the apple did Somu eat? Who had the larger share? By how much?
Solution:-
From the question, it is given that,
Part of the apple eaten by Ritu is = (3/5)
Part of the apple eaten by Somu is = 1 – Part of the apple eaten by Ritu
= 1 – (3/5)
The LCM of 1, 5 = 5
Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator.
= [(1/1) × (5/5)] – [(3/5) × (1/1)]
= (5/5) – (3/5)
= (5 – 3)/5
= 2/5
∴ Part of the apple eaten by Somu is (2/5)
So, (3/5) > (2/5) hence, Ritu ate larger size of the apple.
Now, the difference between the 32 shares = (3/5) – (2/5)
= (3 – 2)/5
= 1/5
Thus, Ritu’s share is larger than the share of Somu by (1/5)
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?
Solution:-
From the question, it is given that,
Time taken by the Michael to colour the picture is = (7/12)
Time taken by the Vaibhav to colour the picture is = (3/4)
The LCM of 12, 4 = 12
Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.
(7/12) = (7/12) × (1/1) = 7/12
(3/4) = (3/4) × (3/3) = 9/12
Clearly, (7/12) < (9/12)
Hence, (7/12) < (3/4)
Thus, Vaibhav worked for longer time.
So, Vaibhav worked longer time by = (3/4) – (7/12)
= (9/12) – (7/12)
= (9 – 7)/12
= (2/12)
= (1/6) of an hour.
Exercise 2.2 Page: 36
1. Which of the drawings (a) to (d) show:
(i) 2 × (1/5) (ii) 2 × ½ (iii) 3 × (2/3) (iv) 3 × ¼
Solution:-
(i) 2 × (1/5) represents the addition of 2 figures, each represents 1 shaded part out of the given 5 equal parts.
∴ 2 × (1/5) is represented by fig (d).
(ii) 2 × ½ represents the addition of 2 figures, each represents 1 shaded part out of the given 2 equal parts.
∴ 2 × ½ is represented by fig (b).
(iii) 3 × (2/3) represents the addition of 3 figures, each represents 2 shaded parts out of the given 3 equal parts.
∴ 3 × (2/3) is represented by fig (a).
(iii) 3 × ¼ represents the addition of 3 figures, each represents 1 shaded part out of the given 4 equal parts.
∴ 3 × ¼ is represented by fig (c).
2. Some pictures (a) to (c) are given below. Tell which of them show:
(i) 3 × (1/5) = (3/5) (ii) 2 × (1/3) = (2/3) (iii) 3 × (3/4) = 2 ¼
Solution:-
(i) 3 × (1/5) represents the addition of 3 figures, each represents 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts.
∴ 3 × (1/5) = (3/5) is represented by fig (c).
(ii) 2 × (1/3) represents the addition of 2 figures, each represents 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts.
∴ 2 × (1/3) = (2/3) is represented by fig (a).
(iii) 3 × (3/4) represents the addition of 3 figures, each represents 3 shaded parts out of the given 4 equal parts and 2 ¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts.
∴ 3 × (3/4) = 2 ¼ is represented by fig (b).
3. Multiply and reduce to lowest form and convert into a mixed fraction:
(i) 7 × (3/5)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (7/1) × (3/5)
= (7 × 3)/ (1 × 5)
= (21/5)
(ii) 4 × (1/3)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4/1) × (1/3)
= (4 × 1)/ (1 × 3)
= (4/3)
(iii) 2 × (6/7)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/1) × (6/7)
= (2 × 6)/ (1 × 7)
= (12/7)
(iv) 5 × (2/9)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/1) × (2/9)
= (5 × 2)/ (1 × 9)
= (10/9)
(v) (2/3) × 4
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2/3) × (4/1)
= (2 × 4)/ (3 × 1)
= (8/3)
(vi) (5/2) × 6
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (5/2) × (6/1)
= (5 × 6)/ (2 × 1)
= (30/2)
= 15
(vii) 11 × (4/7)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (11/1) × (4/7)
= (11 × 4)/ (1 × 7)
= (44/7)
(viii) 20 × (4/5)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (20/1) × (4/5)
= (20 × 4)/ (1 × 5)
= (80/5)
= 16
(ix) 13 × (1/3)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (13/1) × (1/3)
= (13 × 1)/ (1 × 3)
= (13/3)
(x) 15 × (3/5)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (15/1) × (3/5)
= (15 × 3)/ (1 × 5)
= (45/5)
= 9
= (3/20)
(c) (4/3)
We have,
= ¼ × (4/3)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= ¼ × (4/3)
= (1 × 4)/ (4 × 3)
= (4/12)
= 1/3
(ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10
Solution:-
(a) 2/9
We have,
= (1/7) × (2/9)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (1/7) × (2/9)
= (1 × 2)/ (7 × 9)
= (2/63)
(b) 6/5
We have,
= (1/7) × (6/5)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (1/7) × (6/5)
= (1 × 6)/ (7 × 5)
= (6/35)
(c) 3/10
We have,
= (1/7) × (3/10)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (1/7) × (3/10)
= (1 × 3)/ (7 × 10)
= (3/70)
2. Multiply and reduce to lowest form (if possible):
(i) (2/3) × 22/3
Solution:-
First convert the given mixed fraction into improper fraction.
=22/3= 8/3
Now,
= (2/3) × (8/3)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2 × 8)/ (3 × 3)
= (16/9)
(ii) (2/7) × (7/9)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2 × 7)/ (7 × 9)
= (2 × 1)/ (1 × 9)
= (2/9)
(iii) (3/8) × (6/4)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (3 × 6)/ (8 × 4)
= (3 × 3)/ (4 × 4)
= (9/16)
(iv) (9/5) × (3/5)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (9 × 3)/ (5 × 5)
= (27/25)
(v) (1/3) × (15/8)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (1 × 15)/ (3 × 8)
= (1 × 5)/ (1 × 8)
= (5/8)
(vi) (11/2) × (3/10)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (11 × 3)/ (2 × 10)
= (33/20)
(vii) (4/5) × (12/7)
Solution:-
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (4 × 12)/ (5 × 7)
= (48/35)
Check out: PW Experiment Kits
Fractions and Decimals Class 7 Maths Chapter 2 Solutions Summary
In Class 7 Maths Chapter 2, we learn about fractions and decimals. Fractions represent parts of a whole. For example, if you divide a pizza into 4 equal slices and eat 1, you have eaten 1/4 of the pizza. A fraction has two parts: the numerator (the top number) and the denominator (the bottom number). The numerator tells us how many parts we have, and the denominator tells us how many parts the whole is divided into.
We also learn about decimals, which are another way to show parts of a whole. Decimals use a point to separate the whole number from the fractional part. For example, 0.5 is the same as 1/2, and 0.75 is the same as 3/4. Decimals can be added, subtracted, multiplied, and divided just like whole numbers.
Converting between fractions and decimals is also important. To convert a fraction to a decimal, divide the numerator by the denominator. For example, 1/4 becomes 0.25 when divided. To convert a decimal to a fraction, write the decimal as a fraction with a power of 10 in the denominator, and then simplify.
Understanding these concepts helps in comparing fractions and decimals, making calculations, and solving real-life problems. Practice with different examples to become more comfortable with fractions and decimals.
Fractions and Decimals Class 7 Maths Chapter 2 Solutions FAQs
Q1. What are fractions and decimals?
Ans. A part of a whole and it is expressed in the form of a numerator and a denominator is called a Fraction and the numerical representations of fractions where the denominator is a power of ten like 10, 100, 1000, etc is Decimal.
Q2. What is the name of chapter 2 of Maths class 7 solutions?
Ans. The name of chapter 2 of Maths class 7 is Fractions and Decimals.
Q3. Which is the easiest chapter in Maths?
Ans. The easy chapters in Maths includes Introduction to Trigonometry, Arithmetic Progressions, Real Numbers, Polynomials, and Probability.
