NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals
Here, we have provided NCERT Solutions for Class 7 Maths Chapter 2. Students can view these NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals before exams for better understanding of the chapter.
Neha Tanna27 Feb, 2024
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NCERT Solutions for Class 7 Maths Chapter 2
NCERT Solutions for Class 7 Maths Chapter 2: Here are the NCERT solutions for Class 7 Maths Chapter 2 Fractions and Decimals. To maximize their exam scores, students can practice NCERT Solutions for Class 7 Maths Chapter 2 questions. In previous classes, students have learned how to add and subtract decimals as well as fractions. Students will study multiplication and division of decimals as well as fractions in this lesson.
These NCERT Solutions for Class 7 Maths Chapter 2, which include fractions and decimals, have been developed by our subject experts to help students in preparing for their exams. Students can download NCERT Solutions for Class 7 Maths Chapter 2 PDF.NCERT Solutions for Class 7 Maths Chapter 2 PDF
Students can always find the most important NCERT Solutions for Class 7 Maths Chapter 2 right here. Through the link offered, they can obtain a NCERT Solutions for Class 7 Maths Chapter 2 PDF version of each question and associated solution.NCERT Solutions for Class 7 Maths Chapter 2 PDF
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals
Below we have provided NCERT Solutions for Class 7 Maths Chapter 2 for students to help them understand the NCERT Solutions for Class 7 Maths Chapter 2 chapter better and to score good marks in their examination.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)
Solution:-
First convert mixed fraction into improper fraction, =
= 8/3
= 3 ½ = 7/2
For addition of two unlike fractions, first change them to like fractions.
LCM of 3, 2 = 6
Now, let us change each of the given fractions into an equivalent fraction having 6 as the denominator.
= [(8/3) × (2/2)] = (16/6)
= [(7/2) × (3/3)] = (21/6)
Now,
= (16/6) + (21/6)
= [(16 + 21)/6]
= (37/6)
=
(vii)
812−358
Solution:-
First convert mixed fraction into improper fraction, = 8 ½ = 17/2 =
= 29/8
For subtraction of two unlike fractions, first change them to like fractions.
LCM of 2, 8 = 8
Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator.
= [(17/2) × (4/4)] = (68/8)
= [(29/8) × (1/1)] = (29/8)
Now,
= (68/8) – (29/8)
= [(68 – 29)/8]
= (39/8)
=
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)(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
cm wide. Find its perimeter.
Solution:-
From the question, it is given that, Length = 12 ½ cm = 25/2 cm Breadth =
cm = 32/3 cm
We know that,
Perimeter of the rectangle = 2 × (length + breadth)
= 2 × [(25/2) + (32/3)]
= 2 × {[(25 × 3) + (32 × 2)]/6}
= 2 × [(75 + 64)/6]
= 2 × [139/6]
= 139/3 cm
Hence, the perimeter of the sheet of paper is
cm
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 =
= 18/5 cm
BE =
= 11/4 cm
ED = 7/6 cm
(i) We know that,
Perimeter of the triangle = Sum of all sides
Then,
Perimeter of triangle ABE = AB + BE + EA
= (5/2) + (11/4) + (18/5)
The LCM of 2, 4, 5 = 20
Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator.
= {[(5/2) × (10/10)] + [(11/4) × (5/5)] + [(18/5) × (4/4)]}
= (50/20) + (55/20) + (72/20)
= (50 + 55 + 72)/20
= 177/20
=
cm
(ii) Now, we have to find the perimeter of the rectangle,
We know that,
Perimeter of the rectangle = 2 × (length + breadth)
Then,
Perimeter of rectangle BCDE = 2 × (BE + ED)
= 2 × [(11/4) + (7/6)]
The LCM of 4, 6 = 12
Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator
= 2 × {[(11/4) × (3/3)] + [(7/6) × (2/2)]}
= 2 × [(33/12) + (14/12)]
= 2 × [(33 + 14)/12]
= 2 × (47/12)
= 47/6
=
Finally, we have to find which one is having a greater perimeter.
Perimeter of triangle ABE = (177/20)
Perimeter of rectangle BCDE = (47/6)
The two perimeters are in the form of unlike fractions.
Changing perimeters into like fractions we have,
(177/20) = (177/20) × (3/3) = 531/60
(43/6) = (43/6) × (10/10) = 430/60
Clearly, (531/60) > (430/60)
Hence, (177/20) > (43/6)
∴ Perimeter of Triangle ABE > Perimeter of Rectangle (BCDE)
| CBSE Syllabus Class 7 | |
| CBSE Class 7 English Syllabus | CBSE Class 7 Math Syllabus |
| CBSE Class 7 Social Science Syllabus | CBSE Class 7 Science Syllabus |
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?
Solution:-
From the question, it is given that, Picture having a width of =
= 38/5 cm
Frame having a width of =
= 73/10 cm
∴ The picture should be trimmed by = [(38/5) – (73/10)]
The LCM of 5, 10 = 10
Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator.
= [(38/5) × (2/2)] – [(73/10) × (1/1)]
= (76/10) – (73/10)
= (76 – 73)/10
= 3/10 cm
Thus, the picture should be trimmed by (3/10) cm
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.CBSE Board Exam Centre List 2024
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) =
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(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) ×
Solution:-
First convert the given mixed fraction into improper fraction. =
= 8/3
Now,
= (2/3) × (8/3)
By the rule Multiplication of fraction,
Product of fraction = (product of numerator)/ (product of denominator)
Then,
= (2 × 8)/ (3 × 3)
= (16/9)
=
(ii) (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) =
NCERT Solutions for Class 7 Maths Chapter 2 FAQs
What is the name of the chapter 2 of maths class 7?
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals.
Why is math class 7 important?
Class 7 Math and its topics are crucial for a student's academic journey as they lay the foundation for future mathematical learning.
Is Class 7 maths tough?
The difficulty level of Mathematics in CBSE Class 7 can be considered moderate.
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