7 x 3/5 = 7x3/5 = 21/5 = (4)1/5 Exercise 2.2 Question 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Time taken by Michael to colour the picture = 7/12 Time taken by Vaibhav to colour the picture = 3/4 Converting both fractions in like fractions, 7/12 and 3x3/4x3 = 9/12 Here, 7/12<9/12 ⇒ 7/12<3/4 Thus, Vaibhav worked longer time. Vaibhav worked longer time by 3/4 - 7/4 = 9-7/12 = 2/12 = 1/6 hRead more
Time taken by Michael to colour the picture = 7/12
Time taken by Vaibhav to colour the picture = 3/4
Converting both fractions in like fractions, 7/12 and 3×3/4×3 = 9/12
Here, 7/12<9/12 ⇒ 7/12<3/4
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by 3/4 – 7/4 = 9-7/12 = 2/12 = 1/6 hour.
Thus, Vaibhav took 1/6 hour more than Michael.
The part of an apple eaten by Ritu = 3/5 The part of an apple eaten by Somu = 1-3/5 = 5-3/5 =2/5 Comparing the parts of apple eaten by both Ritu and Somu 3/5>2/5 Larger share will be more by 3/5 - 2/5 = 1/5 part. Thus, Ritu’s part is 1/5 more than Somu’s part. Exercise 2.1 Question 5, 6, 7, 8 forRead more
The part of an apple eaten by Ritu = 3/5
The part of an apple eaten by Somu = 1-3/5 = 5-3/5 =2/5
Comparing the parts of apple eaten by both Ritu and Somu 3/5>2/5
Larger share will be more by 3/5 – 2/5 = 1/5 part.
Thus, Ritu’s part is 1/5 more than Somu’s part.
Given: The width of the picture = (7)3/5 and the width of picture frame = (7)3/10 Therefore, the picture should be trimmed = (7)3/5-(7)3/10=38/5-73/10 = 76-73/10 = 3/10 cm Thus, the picture should be trimmed by 3/10 cm. Exercise 2.1 Question 5, 6, 7, 8 for more answers vist to: https://www.tiwariacaRead more
Given: The width of the picture = (7)3/5
and the width of picture frame = (7)3/10
Therefore, the picture should be trimmed = (7)3/5-(7)3/10=38/5-73/10
= 76-73/10 = 3/10 cm
Thus, the picture should be trimmed by 3/10 cm.
(i) ∆ABE, AB = 5/2 cm, BE= (2)3/4 cm, AE= (3)3/5 cm The perimeter of ∆ABE = AB + BE + AE = 5/2+(2)3/4+(3)3/5 = 5/2+11/4+18/5 = 50+55+72/20 = 177/20 = (8)17/20 cnm Thus, the perimeter of ∆ABE is (8)17/20 cnm. (ii) In rectangle BCDE, BE = (2)3/4 cm, ED =7/6 cm Perimeter of rectangle = 2 (length + breaRead more
(i) ∆ABE, AB = 5/2 cm, BE= (2)3/4 cm, AE= (3)3/5 cm
The perimeter of ∆ABE = AB + BE + AE
= 5/2+(2)3/4+(3)3/5 = 5/2+11/4+18/5
= 50+55+72/20 = 177/20 = (8)17/20 cnm
Thus, the perimeter of ∆ABE is (8)17/20 cnm.
(ii) In rectangle BCDE, BE = (2)3/4 cm, ED =7/6 cm
Perimeter of rectangle = 2 (length + breadth)
=2((2)3/4+7/6) = 2(11/4+7/6)
=2(33+14/12) = 47/6= (7)5/6 cm
Thus, the perimeter of rectangle BCDE is (7)5/6 cm.
Comparing the perimeter of triangle and that of rectangle,
(8)17/20cm > (7)5/6cm
Therefore, the perimeter of triangle ABE is greater than that of rectangle BCDE.
Multiply and reduce to lowest form and convert into a mixed fraction: 7 x 3/5
7 x 3/5 = 7x3/5 = 21/5 = (4)1/5 Exercise 2.2 Question 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
7 x 3/5 = 7×3/5 = 21/5 = (4)1/5
Exercise 2.2 Question 1, 2, 3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
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?
Time taken by Michael to colour the picture = 7/12 Time taken by Vaibhav to colour the picture = 3/4 Converting both fractions in like fractions, 7/12 and 3x3/4x3 = 9/12 Here, 7/12<9/12 ⇒ 7/12<3/4 Thus, Vaibhav worked longer time. Vaibhav worked longer time by 3/4 - 7/4 = 9-7/12 = 2/12 = 1/6 hRead more
Time taken by Michael to colour the picture = 7/12
Time taken by Vaibhav to colour the picture = 3/4
Converting both fractions in like fractions, 7/12 and 3×3/4×3 = 9/12
Here, 7/12<9/12 ⇒ 7/12<3/4
Thus, Vaibhav worked longer time.
Vaibhav worked longer time by 3/4 – 7/4 = 9-7/12 = 2/12 = 1/6 hour.
Thus, Vaibhav took 1/6 hour more than Michael.
Exercise 2.1 Question 5, 6, 7, 8
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
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?
The part of an apple eaten by Ritu = 3/5 The part of an apple eaten by Somu = 1-3/5 = 5-3/5 =2/5 Comparing the parts of apple eaten by both Ritu and Somu 3/5>2/5 Larger share will be more by 3/5 - 2/5 = 1/5 part. Thus, Ritu’s part is 1/5 more than Somu’s part. Exercise 2.1 Question 5, 6, 7, 8 forRead more
The part of an apple eaten by Ritu = 3/5
The part of an apple eaten by Somu = 1-3/5 = 5-3/5 =2/5
Comparing the parts of apple eaten by both Ritu and Somu 3/5>2/5
Larger share will be more by 3/5 – 2/5 = 1/5 part.
Thus, Ritu’s part is 1/5 more than Somu’s part.
Exercise 2.1 Question 5, 6, 7, 8
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Salil wants to put a picture in a frame. The picture is (7)3/5 cm wide. To fit in the frame the picture cannot be more than (7)3/10 cm cm wide. How much should the picture be trimmed?
Given: The width of the picture = (7)3/5 and the width of picture frame = (7)3/10 Therefore, the picture should be trimmed = (7)3/5-(7)3/10=38/5-73/10 = 76-73/10 = 3/10 cm Thus, the picture should be trimmed by 3/10 cm. Exercise 2.1 Question 5, 6, 7, 8 for more answers vist to: https://www.tiwariacaRead more
Given: The width of the picture = (7)3/5
and the width of picture frame = (7)3/10
Therefore, the picture should be trimmed = (7)3/5-(7)3/10=38/5-73/10
= 76-73/10 = 3/10 cm
Thus, the picture should be trimmed by 3/10 cm.
Exercise 2.1 Question 5, 6, 7, 8
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
Find the perimeter of (i) ∆ABE, (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
(i) ∆ABE, AB = 5/2 cm, BE= (2)3/4 cm, AE= (3)3/5 cm The perimeter of ∆ABE = AB + BE + AE = 5/2+(2)3/4+(3)3/5 = 5/2+11/4+18/5 = 50+55+72/20 = 177/20 = (8)17/20 cnm Thus, the perimeter of ∆ABE is (8)17/20 cnm. (ii) In rectangle BCDE, BE = (2)3/4 cm, ED =7/6 cm Perimeter of rectangle = 2 (length + breaRead more
(i) ∆ABE, AB = 5/2 cm, BE= (2)3/4 cm, AE= (3)3/5 cm
The perimeter of ∆ABE = AB + BE + AE
= 5/2+(2)3/4+(3)3/5 = 5/2+11/4+18/5
= 50+55+72/20 = 177/20 = (8)17/20 cnm
Thus, the perimeter of ∆ABE is (8)17/20 cnm.
(ii) In rectangle BCDE, BE = (2)3/4 cm, ED =7/6 cm
Perimeter of rectangle = 2 (length + breadth)
=2((2)3/4+7/6) = 2(11/4+7/6)
=2(33+14/12) = 47/6= (7)5/6 cm
Thus, the perimeter of rectangle BCDE is (7)5/6 cm.
Comparing the perimeter of triangle and that of rectangle,
(8)17/20cm > (7)5/6cm
Therefore, the perimeter of triangle ABE is greater than that of rectangle BCDE.
Exercise 2.1 Question 5, 6, 7, 8
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/