(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 Class 7 Maths Chapter 2 Exercise 2.6 Question 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
(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
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. Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 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.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
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. Class 7 Maths Chapter 2 Exercise 2.5 QuesRead more
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.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
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 kmRead more
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
(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 = 2x1/10 = 2 tenths (iv) Place value of 2 in 9.42 = 2x1/100 = 2 hundredth (v) Place value of 2 in 63.352 = 2x1/1000 = 2 thousandth Class 7 Maths Chapter 2 Exercise 2.5 QuestRead more
(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 = 2×1/10 = 2 tenths
(iv) Place value of 2 in 9.42 = 2×1/100 = 2 hundredth
(v) Place value of 2 in 63.352 = 2×1/1000 = 2 thousandth
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
(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 Class 7 Maths Chapter 2 Exercise 2.6 Question 1, 2, 3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
(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
Class 7 Maths Chapter 2 Exercise 2.6 Question 1, 2, 3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
How much less is 28 km than 42.6 km?
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. Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 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.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
See lessShyam 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?
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. Class 7 Maths Chapter 2 Exercise 2.5 QuesRead more
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.
Class 7 Maths Chapter 2 Exercise 2.5 Question 8, 9
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
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?
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 kmRead more
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
Class 7 Maths Chapter 2 Exercise 2.5 Question 5, 6, 7
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/
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
(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 = 2x1/10 = 2 tenths (iv) Place value of 2 in 9.42 = 2x1/100 = 2 hundredth (v) Place value of 2 in 63.352 = 2x1/1000 = 2 thousandth Class 7 Maths Chapter 2 Exercise 2.5 QuestRead more
(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 = 2×1/10 = 2 tenths
(iv) Place value of 2 in 9.42 = 2×1/100 = 2 hundredth
(v) Place value of 2 in 63.352 = 2×1/1000 = 2 thousandth
Class 7 Maths Chapter 2 Exercise 2.5 Question 5, 6, 7
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-2/